PHYSICS 526 NOTES: FLUID DYNAMICS
Jean Eilek
Physics Department, New Mexico Tech
Socorro, NM 87801, U.S.A. and
jeilek@aoc.nrao.edu
Our goal in this course is to explore the basic
physics of fluid systems, both with and without the effects of magnetic fields. Fluid physics by itself – hydrodynamics – applies to any neutral fluid (think about water, or molasses, or the earth’s atmosphere), whether or
not there are magnetic fields around. “Hydro” or “HD”
has broad applications – smooth flows, turbulent flows,
shocks, sound waves, instabilities. If the fluid is ionized
(think of liquid sodium, the higher reaches of the earth’s
atmosphere, or just about any astrophysical system you
can bring to mind), it can (and almost certainly will)
carry a current. The interaction of the current with a
magnetic field (external or self-generated) modifies all
of the phenomena above and adds some new ones. This
is described by magnetohydrodynamics (MHD).
Traditionally, HD and MHD are treated separately,
but they have a lot in common, and I have learned a lot
by comparing and contrasting HD and MHD phenomena. I’m hoping to share that with you as we go through
this course.
A word of caution: you should especially note units
and dimensions. These notes are in cgs. That makes
very little difference for “rocks” (analyses that involve
mass, length, time); but it makes a big difference for
electrodynamics and MHD. The E and B fields, as well
as the fundamental charge, have different dimensions in
cgs than in SI; and the coupling constants in Maxwell’s
equations are different. Both systems appear in the literature; while there is some trend towards SI, many important references are in cgs. I am going to follow my
experience and preference and use cgs; I’ll also give the
SI versions of most critical equations.
Contents
1. BASIC CONCEPTS AND TOOLS
A. Kinematics: How to Describe a Flow
stream function
B. Mass Conservation: The Continuity Equation
lagrangian derivative
C. Momentum Conservation: Euler’s Equation
1. control volumes
2. bernoulli’s relation
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D. Work in a Rotating Frame
E. Dimensional Analysis
F. Appendix: When can we use hydrodynamics?
1. hard sphere collisions
2. plasmas: the coulomb cross section
3. collisionless plasmas
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2. VISCOSITY AND LAMINAR FLOW: BASICS
A. One-dimensional Laminar Flows
B. Steady Flow: Cartesian Applications
1. flow between parallel plates
2. flow in an open channel
3. hele shaw flow
C. Steady Flow: Cylindrical Applications
1. pipe flow
2. circulating flow
D. Viscous Stresses, Generally
1. do it physically first
2. then do it formally
E. The Navier-Stokes Equation (in Cartesian)
F. Appendix: Navier-Stokes in other coordinates
1. Cylindrical coordinates
2. Spherical Polar Coordinates
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3. POTENTIAL FLOW
A. Setups and Assumptions
1. velocity potential
2. stream function
3. other coordinates
B. Two-dimensional Planar Problems
1. sources and sinks, flow past half body
2. flow past a cylinder
C. Axisymmetric 3D Problems
point source and stream flow
flow around a sphere
a line source
D. d’Alembert’s “Paradox”
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4. VORTICITY
A. Vortex Kinematics
B. Vortex Dynamics
C. Conservation of Circulation: Kelvin’s Theorem
D. The Magnus Effect
E. Vortex lines and their behavior
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ii
F. Generation of Vorticity
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5. LAMINAR FLOW: MORE APPLICATIONS
A. Geostrophic Flow
atmospheric circulation
rossby waves
the ekman layer
B. Viscous Flow: Time-Dependent Problems
1. similarity methods in a diffusion equation
2. smoothing out a velocity jump
3. flow above an oscillating plate
4. irrotational vortex decay
C. Creeping Flow
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6. BASICS OF COMPRESSIBLE FLOW
A. Some useful thermodynamic quantities
B. Hydrostatics: gaseous atmospheres
1. constant gravity: the exponential atmosphere
2. variable gravity: the isothermal sphere
3. reality: nonisothermal atmospheres
4. adiabatic atmosphere
C. Convective Stability
1. adiabatic atmosphere
2. potential temperature
3. brunt-väisälä frequency
D. Energetics of Compressible Flow
E. Appendix: Viscous Dissipation
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7. SIGNAL PROPAGATION
A. Sound Waves and the Signal Speed
sound waves: a physical approach
sound waves: a formal approach
B. Why is the sound speed important?
1. when can we assume incompressible flow?
2. the importance of causality
C. Weak Waves and Causality
D. Two examples of simple waves
1. shock tube
2. piston problem
3. waves at boundaries
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8. ONE-DIMENSIONAL STEADY FLOW
A. Two-point Connections in Steady Flow
B. One-dimensional channel flow
1. nozzles and diffusers
2. a smooth transition?
3. normal shocks
4. 1D flows with internal shocks
C. Spherical stellar wind flow
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9. SHOCKS IN FLUID FLOW
A. Jump conditions
B. Normal shocks
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strong shock limit, normal shocks
C. Oblique shocks
1. two possible deflections
2. high mach number limit
D. The Weak Shock Limit
prandtl-meyer function
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10. ONE-DIMENSIONAL UNSTEADY FLOW
A. Shock Formation: The Physical Picture
shock thickness
B. The Method of Characteristics
1. piston motion in a channel
2. connection to shock formation
3. traffic shocks
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11. TWO-DIMENSIONAL STEADY FLOW
A. The Nature of Steady, two-dimensional flow.
B. Signal Propagation in Flows
C. How Does Supersonic Flow Turn a Corner?
D. One example: Prandtl-Meyer flow
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12. SIMILARITY SOLUTIONS
A. Blast Waves: the Sedov-Taylor Solution
B. Prandtl-Meyer flow, revisited
C. Gravitational Collapse: the Shu Solution
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13. FOUNDATIONS OF MHD
A. Remember your E&M?
1. conductivity and ohm’s law: i
2. conductivity and ohm’s law: ii
B. Field Evolution: Induction Equation
1. ideal limit: flux freezing
2. resistive limit: flux annihilation
C. Fluid Equations: Lorentz force
D. Fluid Equations: Energetics
1. energetics of the e and b fields
2. energetics of the fluid
E. Appendix I: Conductivity notes
1. collisional conductivity
2. cross-field conductivity
F. Appendix II: Do It in SI
1. maxwell in SI:
2. induction equation
3. force equation, magnetic tension and pressure
4. energy equation
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14. SIMPLE MHD EQUILIBRIA
A. Potential Fields
B. Plasma Confinement
1. theta pinch
2. bennet pinch or z pinch
3. general screw pinch
C. Force-Free Fields
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1. cylindrical geometry
2. spherical geometry
3. non-linear fields
4. boundaries
D. Gravitational Equilibrium I
1. planar geometry: the galaxy
E. Magnetic Bouyancy
1. “convective” instability
2. scale of unstable perturbations: Parker Instability
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15. NOT-SO-SIMPLE EQUILIBRIA
A. Flux functions I: Gravitational Equilibrium Revisited
1. define the flux functions
2. apply them: solar magnetic arches
B. Flux Functions II: Grad-Shafranov Equation
1. define the flux functions
2. apply them: the g-s equation
3. example: simple pinches
C. Helicity and Taylor relaxation
1. helicity
2. Invariance of the helicity
3. the minimum energy state
4. taylor relaxation
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16. MHD EFFECTS IN FLUID FLOWS
A. Magnetic damping and stirring
1. Magnetic damping.
2. Magnetic stirring.
B. Channel flow with MHD: Hartmann flow
1. The basic setup
2. the solution
3. an mhd generator or an mhd pump?
C. Magnetic Coupling of Two Plasma Slabs.
1. pure mhd
2. current-based approach
3. energetics and equivalent circuit
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17. WAVES AND SHOCKS IN MHD
A. MHD Waves
1. basic structure: linear analysis
2. alfven waves
3. magnetosonic waves
4. validity of mhd wave theory
B. MHD Shocks; Jump Conditions
C. Perpendicular (Normal) Shocks
D. Oblique Shocks
1. do it generally
2. find a useful reference frame
3. now solve the system: i
4. now solve the system: ii
5. back to the physics
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18. MAGNETIC RECONNECTION
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A. Advection vs. dissipation of magnetic field
1. resistive limit: field decay on axis
2. advection limit: field growth on axis
B. Steady, 2D Reconnection
1. why do field lines “break”?
2. sweet-parker model
3. energetics
4. timescales and rates
C. Speed up the reconnection?
1. petschek reconnection
2. compressible flow
3. anomalous diffusion
D. Other approaches
1. spontaneous reconnection
2. driven reconnection
3. non-steady reconnection
4. three-dimensional reconnection
E. Appendix
1. diffusion-only solution
2. advection-only solution
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19. FLUID INSTABILITIES
A. Buoyancy and Thermal Convection
B. The Rayleigh Taylor Instability
1. the physics
2. the math
3. the magnetized case
C. The Kelvin Helmholtz Instability
1. the physics
2. the math
3. the magnetized case
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20. IDEAL MHD INSTABILITIES
A. Overview; Energy Methods
1. energy methods
2. the details
B. Apply: Pinch Instabilities
1. theta pinch
2. z pinch
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21. RESISTIVE MHD INSTABILITIES
A. Tearing Mode: the Physics
B. Tearing Mode: the Math
C. Tearing Mode: the Consequences
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22. TURBULENCE, part I
A. The transition to turbulence
B. Turbulent flows: overview
1. characteristics
2. behavior
C. Homogeneous Turbulence
1. overview
2. eddies and the energy cascade
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3. the kolmogorov scaling arguments
4. what if the fluid is magnetized?
D. Appendix: fun facts from Fourier transforms
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23. TURBULENCE, part II
A. Mean Field Equations
1. the continuity equation
2. the mean momentum equation
3. example: 2D channel flow
4. the mean energy equation
5. what about the turbulent terms?
B. Two-dimensional Turbulence
C. Small scales and intermittency
1. intermittency
2. turbulence on small scales
3. the role of vortex filaments
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24. MHD Turbulence and Dynamos
A. Magnetic fields in Isotropic turbulence
B. The Inertial Range in MHD Turbulence
1. the kraichnan model
2. anisotropy; goldreich-sridhar model
3. what now?
C. Cascades and Related Things
1. ideal invariants
2. cascade directions
3. self-organization in mhd
D. MHD Dynamos
1. cowling’s theorem
2. parker’s solar dynamo
E. Kinematic Dynamos
F. Mean-Field Dynamos
G. Astrophysical dynamos in the lab
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1
1. BASIC CONCEPTS AND TOOLS
The streamline is defined by
dx
dy
dz
=
=
vx
vy
vz
In this course, we will study the dynamics of fluids,
that is, hydrodynamics (HD). What, in this context, is
a fluid? For the purposes of this course, there are three
answers. (1) A normal, everyday liquid – such as water or molasses. The equation of state will depend on
the substance; however, in most applications we will
assume the density of the liquid is constant. (2) A diffuse gas – such as the atmosphere. The density of such
a gas is definitely not constant; the relevant equation of
state is the ideal gas law. (3) An ionized plasma – such
as in laboratory plasma experiments, the upper layer’s
of earth’s atmosphere, and nearly all astrophysical situations. The ideal gas law also applies here. In addition, the free charges in plasmas allow current flow;
we must extend our subject matter to magnetohydrodynamics (MHD).
We will generally assume a fluid can be described
in terms of macroscopic quantites. Our three basic variables will be the density ρ, the pressure p, and the velocity v. We will often also use an equation of state,
relating p and ρ. We will occasionally need transport:
most often the viscosity ν, also thermal conductivity κ,
and the electricial resistivity η. These are often derivable from the microphysics, but are usually considered
as given constants for a particular problem.
A few comments are in order before we get into
specifics.
• My units are cgs, not SI. This matters very little for
normal fluid mechanics, but is critical for MHD. I will
try also to introduce SI variants of basic electrodynamic
terms and equations as they come up.
• My notation is driven somewhat by the vagaries of
TEX. vectors are denoted by bold face roman, e.g., A;
and for greek, either by bold (e.g., ω , or by over-arrows
(e.g., ω
~ ). Tensors are more problematic: possibilities
~
are direct subscripts, Tij ; T~ is possible, ab for a product
of two vectors, and occasionally bold face capitals (as
in P).
A. Kinematics: How to Describe a Flow
Visualizing a complex flow field can be important in
understanding a problem. The most common concept
is the streamline. This is a line (one of an infinite set
of lines) that is everywhere tangent to the velocity vector in the flow. For instance, let ds = (dx, dy, dz) be
the element of arc length along a steamline; and let
v = (vx , vy , vz ) be the velocity vector at that point.
(1.1)
If the velocity field is specified, the streamline can be
found directly from (1.1). In addition, (1.1 is equivalent
to ds × v = 0; no fluid crosses the streamline. We can
also visualize a streamtube: a tubular region within the
fluid, bounded by a set of streamlines (think of them
as pink). Because streamlines can’t intersect, the same
streamlines follow the streamtube everywhere along its
length.
Two other concepts are sometimes used. The path
line is the trajectory of a fluid particle of fixed identity,
over a period of time. (Paint a single fluid element red,
say, and take a series of photos which display its location as time passes.) The streakline, through a fixed
spatial point, is defined as the current location of all
fluid particles which have passed through that point at
some previous time. (Picture injecting dye at a fixed
point in a flow, and taking one photo at some later time.
The dye traces out the streakline). All three of these
“lines” are the same in a steady flow; in a time-varying
flow they will differ.
STREAM FUNCTION
Now consider a special case, when the flow is incompressible. That means constant density, and turns out
to be equivalent to ∇ · v = 0 (we’ll prove this immediately below). Now, we know from E&M that a
vector field with zero divergance can be written as the
curl of some other vector: v = ∇ × A, say. This
isn’t necessarily useful ... unless the flow field is twodimensional. In that case, the vector potential A has
only one non-zero component. To be specific, consider
Cartesian coordinates, and assume the flow field only
depends on (x, y). We can then write the (vector) potential as A = (0, 0, ψ), for some function ψ(x, y); and
the velocity field is
v = −∇ × (ψẑ)
(1.2)
This function ψ(x, y) is called the stream function.
We’ll return to this in more detail in chapter 3, where
we’ll find that streamlines in the flow are also lines of
constant ψ: each ψ value labels one particular streamline.
B. Mass Conservation: The Continuity Equation
This will be one of our most basic tools. Consider an arbitrary volume of fluid, V , bounded by a closed surface,
2
A; let the surface haveRan outward normal, n̂. The mass
within this volume is V ρdV , if ρ is the mass density.
The net rate of change of this mass is
Z
d
ρdV ;
dt V
if there are no sources or sinks of matter, this quantity
must equal zero. Now, there are two ways this integral
can change with time. (i) there can be intrinsic variation
of ρ, ∂ρ/∂t 6= 0; or (ii) there can be flow into or out of
the volume, at a rate ρv · n̂ per surface area. The sum
of (i) and (ii) must balance out to zero:
Z
Z
Z
∂ρ
d
ρdV =
dV +
ρv · n̂dA = 0 (1.3)
dt V
V ∂t
A
R
But
the
surface
integral
can
be
written
as
A ρv· n̂dA =
R
V ∇ · (ρv)dV . Since V is arbitrary, we can set the
integrand to zero, and we get the differential form of
this basic equation:
∂ρ
+ ∇ · (ρv) = 0
∂t
(1.4)
This is, of course, the continuity equation, applied to
mass conservation.
LAGRANGIAN DERIVATIVE
Note, the two terms in the above expression describe
the two “intrinsic” ways in which the mass in the elemental volume can change. We collect them as
∂
D
=
+v·∇
Dt
∂t
(1.5)
A
V
V
V
where we have again used vector identities in the last
step. This balance, (1.7) and (1.8), can then be written
in differential form,
∂
(ρv) + ∇ · (ρvv) = −∇p + ρg
(1.9)
∂t
This is our basic force equation, known as Euler’s equation. Note that the vv term is a tensor.1
It is conventional to simplify this, by expanding the
derivatives on the left hand side and using (1.4); we get
∂v
+ ρ(v · ∇)v = −∇p + ρg
(1.10)
∂t
Another version of Euler’s equation holds if there is no
body force. To get to this, write the pressure in terms of
the pressure tensor (for an isotropic gas pressure):


p 0 0




(1.11)
P = 0 p 0 = pδij


0 0 p
ρ
and note that the divergance of a tensor is found (in
Cartesian) by
and with this, the continuity equation becomes
Dρ
+ ρ∇ · v = 0
Dt
In the second expression, we have used Gauss’s law for
tensors (noting that ρvv is a second-rank tensor).
Now, the net rate of change of momentum in the volume must be equal to the net force exerted on the volume. We consider external forces which act throughout
the volume (“body” forces, such as gravity, electomagnetism, bouyancy, radiation pressure if the fluid is optically thin; we let g be the net force per mass), and also
the force exerted on the surface by the fluid outside V .
The net force on the volume V is, then,
Z
Z
Z
Z
∇pdV (1.8)
ρgdV −
pn̂dA =
ρgdV −
(1.6)
C. Momentum Conservation: Euler’s Equation
Consider again our surface A, enclosing
R volume V .
The momentum within this surface is V ρvdV . The
net rate of change of this quantity again must reflect intrinsic (∂/∂t 6= 0) variation and advection (flow across
the surface). Thus, we write the net rate of change of
momentum as
Z
Z
∂
(ρv)dV + (ρv)v · n̂dA
V ∂t
A
Z
Z
(1.7)
∂
∇ · (ρvv)dV
=
(ρv)dV +
V ∂t
V
(∇ · P)i =
∂Pij
∂xj
(1.12)
with the Einstein summation convention assumed.
With this notation, the Euler equation in the case of
no body force can be written,
∂
(ρv) + ∇ · (ρvv + P) = 0
∂t
1
In general, a “vector product” tensor ab expands out as
2
3
a 1 b 1 a1 b 2 a 1 b 3
ab = 4a2 b1 a2 b2 a2 b3 5
a 3 b 1 a3 b 2 a 3 b 3
(1.13)
3
This, and (1.4), are conservative forms of the basic
equations.
1.
CONTROL VOLUMES
We argued above that equating (1.7) and (1.8) will give
an integral form of the momentum balance equation.
This is true if the volume V and surface A are fixed (not
themselves moving). It can also be useful to consider
a moving volume (such as in a rocket problem). To
specify, let V ∗ (t) be the instantaneous control volume,
and A∗ (t) be its area. Let b be the local velocity of the
boundary. Our integral equations become
Z
Z
d
ρdV +
ρ(v − b) · n̂dA = 0
dt V ∗ (t)
A∗ (t)
and
d
dt
Z
ρvdV +
V ∗ (t)
Z
ρv(v − b) · n̂dA
Z
Z
pn̂dA
ρgdV −
=
A∗ (t)
V ∗ (t)
A∗ (t)
Exercise for the student: can you derive or justify these
relations?
2.
BERNOULLI ’ S RELATION
It might be comforting to prove that we can extract
Bernoulli’s relationship from what we have so far.
Start with Euler’s equation, in the form (1.10). But
now, note two useful facts. The first is that if the fluid is
barotropic – that is if pressure is a function of density
only and vice-versa (as in an adiabatic gas), we have
Z
1
dp
∇p = ∇
(1.14)
ρ
ρ
(this can be verified using the chain rule; take p =
F (ρ), F being some function, and go from there).
Thus, this term is a perfect differential. The second useful fact is that
µ
¶
1 2
v · ∇v = −v × ω + ∇
v
(1.15)
2
(this is easiest to verify by expanding out in Cartesian
coordinates). Thus, this term is also a perfect differential. The first term on the right hand side is written in
terms of ω = ∇ × v, the local vorticity (more in chapter 4 on this). Specify the force to gravity, which can be
expressed in terms of a potential: g = ∇Φg . If we then
consider steady flow, we can rewrite (1.10) as
·
¸
Z
1 2
dp
∇ v +
+ Φg = v × ω
(1.16)
2
ρ
But now: the right hand side of (1.16) is normal to both
the local flow field (that is normal to streamlines) and
to the local vorticity ω
~ . Thus, we have one form of
Bernoulli’s relation: in inviscid, steady flow, the term
in brackets has zero gradient in the direction of the local
velocity field. Thus, we have one version of Bernoulli’s
law:
Z
dp
1 2
v +
+ Φg = constant along streamline
2
ρ
(1.17)
Further, in an adiabatic gas, p ∝ ργ if γ is the adiatabic index (the ratio of specific heats). The second
term simplifies, so that Bernoulli’s relation for an inviscid adiabatic gas is
1 2
γ p
v +
+ Φg = constant along streamline
2
γ−1ρ
(1.18)
Alternatively, in an incompressible fluid, ρ is constant,
and the second term in (1.16) becomes simply p/ρ.
Thus, for an incompressible fluid, Bernoulli’s relation
is
1 2 p
v + + Φg = constant along streamline (1.19)
2
ρ
D. Work in a Rotating Frame
Euler’s equation effectively expresses force balance. It
must therefore be modified in a rotating system, where
we expect Coriolis and centrifugal forces. Recall your
basic mechanics. We want to transform vectors from
our (inertial) frame to a frame rotating at Ω . Let r be
the usual vector from the coordinate origin to the observation point (polar coordinates), and let R be a vector
from the rotation axis, perpendicular, to the observation
point (cylindrical coordinates). The time derivative of a
general vector, P, transforms as
¶
µ
¶
µ
dP
dP
=
+Ω ×P
dt i
dt r
Thus, velocities and accelerations transform as
vi = vr + Ω × r ;
Ω × vr + Ω × (Ω
Ω × r)
ai = ar + 2Ω
Ω × vr − Ω 2 R
= ar + 2Ω
From this, we can write the incompressible Euler equation in a rotating frame (now dropping the r subscripts):
1
Dv
Ω×v
= − ∇p + g + Ω2 R − 2Ω
(1.20)
Dt
ρ
The two new terms are the centrifugal and Coriolis
forces.
4
E. Dimensional Analysis
Much of the art of fluid mechanics is knowing which
terms in the equations can be thrown away. We are
guided in this by several dimensionless numbers. In this
section, we need “characteristic” values of the velocity
V , the length scale L, the gas pressure p, the density
ρ, and the magnetic field B. In addition, transport coeffients come in: ν is the viscosity, η the magnetic diffusivity, and κ is the thermal conductivity (these quantities will be introduced in more detail later, as they are
needed).
As an example of why we consider these dimensionless numbers, return to the Euler equation, but with the
viscous term added (just now, take my word for it, we’ll
derive it in Chapter 3 and give it another name):
ρ
Dv
= −∇p + ρν∇2 v
Dt
(1.21)
The last term, a second spatial derivative in v, is the
viscous stress term. Consider the magnitude of each of
the three terms:
µ
¶
µ 2¶
µ 2¶
µ
¶
V
V
c
V
O ρ
∼O ρ
; O ρ s ; O ρν 2
t
L
L
L
In writing these estimates, we have introduced the
sound speed, c2s = ∂p/∂ρ (it will appear formally in
Chapter 7). In addition, we have assumed we can find
a characteristic velocity V and a characteristic length
L. Now, we can form the ratio of the first (the inertial
term) to the second (the pressure gradient):
µ 2¶
V
Dv/Dt
(1.22)
∼O
∇p
c2s
and, the ratio of the first term to the third (the viscous
force):
µ
¶
VL
Dv/Dt ∼ O
(1.23)
ν
Thus, the relative importance of the three terms in
the Euler equation are determined by the dimensionless
quantities V 2 /c2s and V L/ν. These two ratios are fundamental in hydrodynamics, and have their own names.
A similar analysis of other terms in these and/or other
equations, as they arise, leads to several other important
dimensionless numbers. I list several of these here.
The two most important for hydrodynamics are:
• the Reynolds number is
Re =
VL
ν
(1.24)
Viscosity ν is addressed and defined in Chapter 3. This
measures the ratio of the inertial term to the viscous
term in the force equation. Viscosity is important at low
Re; flows are laminar, i.e., we can ignore viscosity, at
higher Re. Interestingly, the transition to turbulence occurs at very high Re (even though viscosity is critically
important in turbulence; this is discussed later).
• the Mach number is
v
M=
(1.25)
cs
if cs is the sound speed (shown in §3 to be given by
c2s = ∂p/∂ρ). An important limit is the incompressible, or Boussinesq limit, namely that the fluid density
is constant (from 1.4, this is equivalent to ∇ · v = 0);
this is justified for flows in which M ≪ 1.
The most important scaling numbers for MHD are
• the plasma beta is
β=
8πp
B2
(1.26)
and gives the ratio of the gas pressure to the magnetic
pressure (pB = B 2 /8π).2 In low-beta plasmas, the
magnetic field dominates the dynamics; in high-beta
plasmas the gas pressure dominates and magnetic effects are small.
• the Magnetic Reynolds number is
Rm =
LV
η
(1.27)
if η is the magnetic diffusivity (addressed in Chapter
13). Flows with high Rm have strong coupling of the
B field to the flow; flows at low Rm have only a weak
coupling.
• the Alfven Mach number is
v
MA =
(1.28)
vA
if vA is the Alfven speed, a characteristic signal speed
in a magnetized
√ fluid (shown in chapter 13 to be given
by vA = B/ 4πρ).
In addition, several other dimensionless numbers
appear here and there. Some that we may meet are
• the Prandtl number is
ν
Pr =
(1.29)
κ
and measures the radio of viscous diffusion to thermal
diffusion.
2
The SI expression is β = 2µo p/B 2 .
5
• the Rayleigh number is
Ra =
αg∆T L3
κν
F. Appendix: When can we use hydrodynamics?
(1.30)
and measures the ratio of the bouyancy force to the stabilizing effects of diffusion. α is the thermal expansion
coefficient.
• the Ekman number is
E=
ν
ΩL2
(1.31)
for a system rotating at angular speed Ω. This measures
the ratio of viscous to Coriolis terms in the force equation.
• the Rossby number is
Ro =
V
ΩL
(1.32)
in a rotating system. It measures the ratio of inertial to
Coriolos terms in the force equation.
• the Hartmann number is
Ha = √
BL
4πρνη
(1.33)
and measures the ratio of magnetic forces to diffusive
forces in a MHD fluid. Some authors omit the 4π.
• the Magnetic Prandtl number is
ν
Pm =
η
Finally, we need to consider when we can use the hydrodynamic (or magnetohydrodynamic) model. Fluid
mechanics may seem the obvious choice when we are
concerned with laboratory or terrestrial liquids; they
are “clearly” well described by macroscopic quantities
such as density, pressure, velocity. For gases, and particularly plasmas, however, the relevance of fluid mechanics is slightly subtler. In particular, some astrophysical gases and plasmas are very tenuous: when can
we justify a fluid approach, and when must we consider
the dynamics of individual particles?
The general answer is, HD or MHD can be
used when both the interparticle distance (equal to
(density)−1/3 ) and the mean free path of fluid particles
(molecules, atoms, ions) is small compared to any characteristic scale of the system. That is, these distances
must be small compared to the overall scale of the system, and must also be small compared to the scale of
any gradients (ρ/∇ρ, for instance, is the density gradient scale).
1.
HARD SPHERE COLLISIONS
Thus, we need to look at the mean free path in fluids,
gases and plasmas. To start, we recall the basics of
hard-sphere collisions. If a “gas” of billiard balls, say,
has a number density n and each particle has a random
velocity v and a radius a, we define the collision cross
section,
σ = πa2
(1.35)
(1.34)
and measures the ratio of viscous diffusion to magnetic
diffusion.
From this we find the mean free path (the average distance between collisions),
λ≃
1
nσ
(1.36)
and the mean time between collisions,
References
Much of this discussion is “just from me” – i.e.,
I’ve pulled the topics together “out of my head”. Good
places to go for more discussion include Kundu (who’s
usually quite mathematical); Tritton (who’s much better with words and concepts); or Faber (our text).
For the plasma/Coulomb scattering discussion,
Spitzer’s Physics of Fully Ionized Gases is the classic
reference.
tcoll ≃
1
.
nσv
(1.37)
This last can be inverted to describe the collision rate
per particle, t−1
coll ≃ nσv.
For hard spheres this analysis is straightforward, of
course; they will not interact unless there is a direct
“hit”, and the geometrical cross section is the relevant
one to describe energy exchange. Neutral atoms and
molecules behave similarly, in that they need a very
close hit; their cross sections can be calcalated from basic physics. Typical atomic cross sections ∼ 10−14 cm2 .
6
2.
PLASMAS : THE COULOMB CROSS SECTION
There is another type of encounter which is important in
plasmas: a long-range encounter between two particles
which feel a 1/r2 Coulomb force. The particles never
directly collide with each other; but the long-range scattering allows exchange of energy and momentum, and
thus makes the system act like a fluid.
• Start with a single encounter, in which particle A
(an electron, say) scatters on particle B (a proton, say;
with mp >> me , we can assume the proton stays at
rest. Let the incoming particle have velocity v and mass
me , and let in come in at impact parameter b. We can
solve this problem exactly, from classical mechanics,
and find the deflection angle, θ, and the resultant velocity and momentum changes, ∆p = m∆v. Here, we
will approximate this analysis.
R • The net impulse on the electron will be ∆p =
F(t)dt, integrated over the collision. Now, the force
is strong only when the two particles are close. Since
they are close for a period of time ∆t ≃ 2b/v, we can
approximate F ≃ e2 /b2 and ∆p ≃ 2F b/v. (Since we
know the net deflection is perpendicular to the initial
direction of motion, we can also drop the vector notation). This gives us the net energy gain per collision,
∆E =
(∆p)2
2me
≃
bmin ≃
(1.41)
¡
¢1/2
bmax ≃ λD = kB T /4πne2
(1.42)
(kB is the Boltzmann constant, and T is the temperature). Thus, the best choice of ln Λ clearly depends
on the exact situation one is considering. Luckily, for
our purposes, this is only a logarithmic uncertainty, and
will not be critical for most of our calculations. The
choices above, with typical astrophysical parameters,
give ln Λ ≃ 10 − 20, in almost any diffuse-matter setting.
• Numerically, for a thermal plasma with 12 me v 2 ≃
kB T , the Coulomb cross section becomes,
σc ≃ 7 × 10−13
ln Λ
cm2
2
T4
(1.43)
where T4 = T /104 K; so that
me b2 v 2
4πe4 n
2e4
2πbnvdb =
ln
2
2
me b v
me v
e2
2me v 2
bmax is less straightforward. A common choice is the
Debye shielding length (the scale over which an extra
charge causes charge separation in a plasma):
2e4
We want to extend this analysis, to find the net rate of
energy exchange with the plasma. But the collision rate
of our electron, with particles at impact parameter b is,
(collisions/second) = 2πnbv db, we find the net energy
exchange rate by integrating over all allowed b:
bmax
(clearly, we cannnot integrate from bmin = 0 to bmax =
∞, since the integral in (1.38) would diverge). bmin is
usually taken to be the distance corresponding to maximum energy transfer,
3/2
T4
sec
n ln Λ
(1.44)
T42
cm .
n ln Λ
(1.45)
tcoll ≃ 4 × 104
and
λ ≃ 1 × 1012
¶
bmax
bmin
bmin
(1.38)
• Now, we want to express this in terms of a cross
section:
• A useful short way to remember the Coulomb
cross section is as follows. Similarly to the bmin estimate above, we can define an effective “radius”, aef f ,
by equating potential and kinetic energies:
E
dE
=
= nvσc E
dt
tcoll
1
e2
= me v 2
aef f
2
dE
=
dt
Z
µ
(1.39)
This defines the Coulomb cross section, σc :
σc = 8π
µ
e2
me v 2
¶2
ln Λ
(1.40)
if ln Λ = ln (bmax /bmin ) is defined as the Coulomb
logarithm.
• The Coulomb logarithm depends on the largest
and smallest impact parameters that are important
(1.46)
and then, estimating σc = 2πa2ef f ln Λ. This recovers
the form of equation (1.40), and resembles the hardsphere cross section, (1.35), “with a factor of ln Λ
tacked on”. In extending this to other examples, as
we will do just below, the exact numerical factor that
scales πa2ef f ln Λ cannot be recovered by this method
of guessing; one would have to do a more formal analysis to get the correct order-unity numerical factor for
each cross section.
7
3.
COLLISIONLESS PLASMAS
Finally, a brief comment on a third limit. We saw
above that Coulomb collisions can transfer energy and
momentum between charges in a plasma, thus “tying
the plasma together” and justifying our treating it in
the fluid limit. In many applications, however, the
Coulomb mean free path derived using (1.40) is large
compared to the size of the system. Does this then
mean that we must use a single-particle (kinetic) approach? Not necessarily. Such systems can and do act
like fluids; one example is the solar wind, which has a
very long Coulomb mean free path, and yet is observed
to carry shocks. (We will see later that shocks are mediated by particle-particle collisions; their thickness is
several mean free paths). How can this be? There are
two likely causes:
• Tangled magnetic fields. A charged particle in
a magnetic field undergoes gyromotion. It is thus constrained to move (nearly) along the field line; gyroradii
are typically quite small. If the magnetic field is tangled on a scale small compared to the system size, the
system can be “tied togeter” by this effect, justifying a
fluid approximation.
• Plasma Microinstabilites. A plasma is subject
to a wealth of microinstabilities – in which free energy
(such as that of streaming charges) is converted to wave
energy in the plasma. The plasma waves which are excited involve fluctuating electric and magnetic fields,
which in their turn scatter the plasma charges. This effect can also “tie the system together”; even a low-level
wave energy density can lead to a short mean free path
for particle-wave scattering.
8
2. VISCOSITY AND LAMINAR FLOW: BASICS
Let’s start by looking at some simple fluid flow solutions. We’ll consider steady-state, incompressible
flows. Some very simple solutions can be found when
viscosity is important to the flow. Because viscosous
stresses can, however, get very complicated, we’ll begin with simple examples, then see the general stuff.
A. One-dimensional Laminar Flows
Think about a long, 1D channel such as in Figure 2.1.
The flow is driven by a pressure gradient, p2 < p1 ; and
we assume the velocity must go to zero at the walls (this
is a no-slip boundary). Because the flow is incompressible – has the same density everywhere – the velocity
profile must be independent of x, in order to enable
mass conservation (the same flow rate across any point
within the channel).
y
dx
(p1)
dy
2b
v
Beware: you must be careful when working with
viscosity. The definitions, symbols and even units
of viscosity vary from author to author. I prefer the choice, that ν∇v has the dimensions of a
force per area per gram: and that ρν∇v is a force
per area. Thus, the dimensions of ν are [L]2 /[T].
Many authors (including Faber) include the density; he uses η (dimensions [M]/[L][T]) which is
the same as our ρν. Still other authors use µ where
Faber uses η.
So: putting this back into vector form, we’ve discovered that our governing equation for this simple system
should be1
∇p = ρν∇2 v
But now ... looking back to equation (1.10), our result here suggests that we should add a new term to describe the effects of viscous stresses. Taking this as true
for the moment (we’ll derive it more generally below),
we have the basic equation governing viscous flow:
(p2 )
ρ
x
Figure 2.1. Defining the geometry for this problem. The
flow is driven by a pressure gradient; p2 < p1 . This driver,
combined with the no-slip condition assumed at the walls
and the action of viscous stresses within the flow, leads to
the quadratic velocity profile shown.
Now, think about a little differential volume dxdy,
as shown in the figure. It feels a net force to the right,
due to dp/dx; what balances this force to keep the flow
from accelerating? This must be viscosity: friction on
the little volume due to the shear in the flow. We assume the shear stresses are linearly proportional to the
velocity gradient, transverse to the surface (this is called
a Newtonian flow). The little volume will feel friction
forces on its top and bottom surfaces; the net force on
the volume will be the difference between the forces
on the top and bottom. If any one surface feels a force
ρν(∂v/dy), the net force (per length in the z direction)
on our element is
ρν
µ
∂v
∂y
¶
y+dy
− ρν
µ
∂v
∂y
¶ #
y
dx = ρν
∂2v
dy 2
dx
(2.1)
In this last I’ve assumed dy is small, of course, and also
that ν is independent of y. I’ve also written the constant
of proportionality as ρν, and the constant ν is what I
like to call the coefficient of viscosity; I’ll return to this
shortly.
(2.2)
Compare this back to our first version of the momentum
equation, namely (1.10), which we called Euler’s equation. This version of the force equation, with viscosity
included, is called the Navier-Stokes equation.
In the rest of this chapter, we look at smooth (timesteady) flows in which viscosity plays the dominant
role. This situation is often called laminar flow. Faber
describes laminar flow in a more restricted sense, implying that “the fluid can be treated as an assembly
of laminae of uniform thickness, whose boundaries remain fixed as the flow moves between them”. The more
general, and more usual, definition, uses laminar flow
to mean smooth, non-turbulent flow. In most applications, we note that a real fluid does not slip freely past
a boundary (as was the implicit assumption in potential
flow theory), but rather sticks: we will generally use
no-slip boundary conditions.
1
"
∂v
+ ρ(v · ∇)v + ∇p = ρg + ρν∇2 v
∂t
What does ∇2 a mean, if a is a vector? For Cartesian it’s simple:
´
`
∇ 2 a = ∇ 2 a x , ∇ 2 a y , ∇ 2 az
where
∇2 =
„
∂2 ∂2 ∂2
,
,
∂x2 ∂y 2 ∂z 2
«
is the usual Laplacian operator. For cylindrical and spherical coordinates, ∇2 a is a more complicated form – check the vector
references I’ll put up on the web, or your favorite vectors-infunny-coordinates reference book.
9
B. Steady Flow: Cartesian Applications
In particular, in this section and the next, we consider
steady flows, where gravity is unimportant. Thus, we
are solving
ρ(v · ∇)v + ∇p = ρν∇2 v
(2.3)
and the second derivative gives this equation the propertires of a diffusion equation.
1.
FLOW BETWEEN PARALLEL PLATES
In this section, we consider flow along x̂, with gradients
in the ŷ direction v = (v(y), 0, 0). Thus, the system
(2.3) simplifies still further to
∂2v
∂p
= ρν 2
∂x
∂y
∂p
=0
∂y
(2.4)
Note, we have eliminated the inertial term utterly, by
picking our geometry carefully: so that the gradients
are perpendicular to the velocities. The second of these
equations says that the pressure can vary only with x. In
the first, then, the first term can be only a function of x,
while the second can be only a function of y; thus, both
must be constant. Thus, dp/dx = constant, (2.4) becomes an ODE, and our general solution for v(y) comes
from
ρνv(y) = Ay + B +
y 2 dp
2 dx
open surface allows us to assume the pressure is everywhere equal to atmospheric pressure, so that ∇p is
zero. Keep the Cartesian coordinates aligned with the
flow bed. The basic equation is then
ρν
Other variants, shown in Figure 2.2, will appear in the
homework.
FLOW IN AN OPEN CHANNEL
This comes from Faber. Let the channel have an open
surface, and be inclined to horizontal at an angle α. The
d2 v
= −ρg sin α
dy 2
(2.7)
The boundary conditions are now v = 0 at the lower
surface, and dv/dy = 0 at the upper (free, open) surface, a distance b away. The latter is due to the fact
that shear stresses are continuous at a boundary, and
the atmosphere (having very low viscosity) is assumed
to carry no shear. This solves to
(2.5)
Thus, the transverse velocity profile is a quadratic in y.
Specific examples come from various imposed boundary conditions, as illustrated in Figure 2.2.
One particular case is plane Poiseuille flow, in
which an external pressure gradient drives a flow
through two stationary walls. This is what we introduced in Figure 2.1. The velocity profile here is a centered parabola, the flow being fastest in the middle (as
you’ll see in the homework). The mass flux per unit
width is
Z b
1 3 dp
b
(2.6)
ρv(y)dy = −
Q=
12ν dx
0
2.
Figure 2.2. Plane-flow solutions for various boundary
conditions. From Kundu figure 9.4.
v(y) =
¡
¢
1
g sin α 2by − y 2
2ν
which gives a mass flux,
Z b
1
ρgb3 sin α
ρv(y)dy =
Q=
3ν
0
3.
(2.8)
(2.9)
HELE SHAW FLOW
This describes the means by which the striking twodimensional images used to illustrate potential flow are
made. Inject the fluid through closely spaced parallel
plates, with some obstacle in the middle. We want the
vertical direction (between the plates; call it ẑ) to be
governed by viscosity, while the two horizontal directions (say the flow goes along x̂) can be treated as potential flow. Let the plate separation be b, the scale of
the central obstacle be L, and the flow be driven by a
fixed, externally imposed dp/dx.
First, use our parallel-plate solution to find the discharge rate Q. From this define a characteristic velocity
10
(strictly, the mean over the z-direction):
V =
Q
b2 dp
=−
b
12ρν dx
to get (2.5), we argue here that each side of 2.13 must
be constant, giving a flow profile
(2.10)
v(r) =
(the minus sign just notes that V points opposite to the
pressure gradient). But I can write the right hand side
as the gradient of a scalar:
µ 2 ¶
b
p
(2.11)
V =∇
12ρν
That is, the horizontal velocity can be treated in terms
of a potential, φ = −(b2 p/12ρν). This shows that all
the potential flow apparatus of chapter 3 can be used
here; and, conversely, that flows where this works are
good illustrations for two-dimensional potential flow
solutions.
But then: what does it take to justify this? We must
be able to ignore the viscous forces in the x and y directions, and to ignore the inertial force in the x direction.
That is, we need
∂ 2 vx
∂vx
≪ν 2 ;
vx
∂x
∂z
b2
1
≪
2
L
Re
Thus, the Reynolds number determines the required dimensions of the apparatus.
C. Steady Flow: Cylindrical Applications
We can repeat the exercise for cylindrical geometry;
differences from the planar cases highlight the effects
of geometry. The flow is now taken to be in the z direction, as is customary in cylindrical coordinates.
1.
Figure 2.3. The circular Poiseuille flow solution,
showing the velocity field and also the stress field (τ ; our σ)
From Kundu figure 9.5.
But now, A = 0 to keep v finite as r → 0; and a
no-slip boundary at r = a gives
v(r) =
(2.12)
(2.14)
where A, B are again integration constants, set by the
boundary conditions.
∂ 2 vx
∂ 2 vx
≪
∂y 2
∂z 2
But now: the second condition simply requires that
L ≫ b; while the first requires
V2
νV
≪ 2 ;
L
b
r2 dp
+ A ln r + B
4ρν dz
r2 − a2 dp
4ρν dz
(2.15)
that is, another centered parabolic profile. The mass
flux here is
Z a
πa4 dp
2πvrdr = −
Q=ρ
(2.16)
8ν dz
0
This is called Poiseuille’s Law, or the Hagen-Poiseuille
law.
2.
CIRCULATING FLOW
Another case is flow between two concentric, rotating
cylinders, called circular Couette flow.
PIPE FLOW
Now, consider steady flow in a pipe of radius a. This
is circular Poiseuille flow. The radial component of the
momentum equation again required dp/dr = 0, ie no
transverse pressure gradients. Assuming a very long
pipe, so that things only vary with r, the z-component
of the momentum equation is
µ
¶
dp
dv
ρν d
r
(2.13)
=
dz
r dr
dr
As there is only a z-component to the velocity, we have
dropped the subscript. By the same argument as used
Figure 2.4. The geometry and the velocity field for
circular Couette flow. From Kundu figure 9.6.
Referring to (2.43) and (2.44) in Appendix F, and noting that only vφ 6= 0, the two components of the equa-
11
tion of motion are
vφ2
1 dp
r
ρ dr
¸
·
d 1 d
(rvφ )
0=ν
dr r dr
the shear force points along the interface between the
two fluid parcels). The coefficient ν is the viscosity or
shear viscosity.
=
(2.17)
The general solution for the azimuthal velocity is
B
(2.18)
r
and the pressure gradient, required to offset centrifugal
force, is
µ
¶
B 2
ρ
dp
Ar +
(2.19)
=
dr
r
r
vφ (r) = Ar +
Using boundary conditions vφ (R1 ) = Ω1 R1 and
vφ (R2 ) = Ω2 R2 gives the specific values for A and
B.
This has some interesing limits. One has Ω = Ω2
and R1 = 0, that is a rotating cylindrical tank. This has
vφ = Ωr
(2.20)
showing that the fluid inside goes into solid body rotation. Alternatively, if the outer cylinder is at infinity, with Ω2 = 0, and the inner one has R = R1 and
Ω = Ω1 , then the solution is
ΩR2
(2.21)
r
We’ll see this again in chapter 4 when we work with the
irrotational vortex.
vφ =
D. Viscous Stresses, Generally
OK, we’ve avoided this long enough .. we need to write
down the more general form of the viscous stresses.
This will be a tensor. We start by determining the surface forces on a piece of fluid, due to deformations
(arising from velocity gradients) of the adjacent bits of
fluid.
1.
DO IT PHYSICALLY FIRST
Following Faber here. Consider a planar flow in the x̂
direction, with transverse velocity gradients (along ŷ).
We expect the shear stress to be linear: ν x̂dvx /dy is the
force between adjacent streams in the fluid.2 (Note that
2
This is a BIG assumption – it’s traditional, but not obvious to
me that it should always hold. Look ahead to the effects of a
magnetic field and j × B forces, for a possible counter-example.
Luckily, however, we rarely need to worry about viscous stresses
in MHD applications ....
Figure 2.5. The two-dimensional geometry for the stress
tensor. pi are the normal forces on the fluid square, and sij
are the shear forces. From Faber, Figure 1.2
Life is a bit more complicated, however. Consider
a two-dimensional Cartesian system, where s12 is the
force in direction 2, due to adjacent fluid in direction 1;
and s21 is the force in direction 1, due to adjacent fluid
in direction 2. These two forces must be equal: s12 =
s21 ; if not we would have a net torque on the parcel of
fluid. Thus, we must symmetrize our expression for the
shear force:
¶
µ
∂v2
∂v1
= s3
(2.22)
s12 = s21 = ρν
+
∂x2 ∂x1
where the last equality simply defines s3 , to shorten the
notation.
There are also normal forces on the parcel: due to
the fluid pressure, and also due to the fluid deformation.
Faber approaches this by considering a frame rotation,
into the 45◦ degree primed frame. From force balance,
we find that
1
s′3 = (p1 − p2 )
2
(2.23)
1
′
p1 = (p1 + p2 − 2s3 )
2
and also, from a Taylor expansion, we find that
¶
µ ′
∂v2′
∂v1
−
s3 = ρν
∂x′1 ∂x′2
from which we get
p1 + 2ρν
∂v1
∂v2
∂v3
= p2 + 2ρν
= p3 + 2ρν
(2.24)
∂x1
∂x2
∂x3
12
But now, taking the mean (isotropic) pressure to be
1
(p1 + p2 + p3 )
(2.25)
3
we end up with an expression for the normal force:
µ
¶
∂v1
∂v2
∂v3
2
p1 = p − ρν 2
−
−
(2.26)
3
∂x1 ∂x2 ∂x3
p=
So, generalizing to three dimensions, we have all 9
components (6 independent) of the surface forces acting on this parcel, which we collect as the tensor ~~σ :


p1 s12 s13




(2.27)
σij = − s21 p2 s23 


s31 s32 p3
(the minus sign has been added for consistency with the
usual treatment).
and we note that the Ωij term is 1/2 of the usual vorticity, Ω = ∇ × v. Also, note that the trace of Dij
is
Dmm =
∂v2
∂v3
∂v1
+
+
=∇·v
∂x1 ∂x2 ∂x3
(2.30)
Now, we argue that Σij should depend only on the symmetric part of the velocity gradient, Dij (otherwise,
simple rotation, such as solid body, would lead to a spurious stress term). Thus, requiring linearity, we have in
general
Σij = αijmn Dmn
(2.31)
which involves no fewer than 81 separate α’s. However,
we can make arguments about symmetry and isotropy
to reduce this to the most general form useful for us:
¶
µ
1
Σij = 2ρν Dij − δij Dmm + ρνb δij Dmm (2.32)
3
That is, we have only two α’s – ν, the shear viscosity,
usually just called the viscosity, and νb , the bulk viscosity. Thus, the general stress tensor is
¶
µ
1
σij = −pδij + 2ρν Dij − δij Dmm + ρνb δij Dmm
3
(2.33)
From this, we have the full expression for the vector
surface force per area:
T = eˆk σik
Figure 2.6. The surface forces on a three-dimensional
fluid cube. The notation here uses τ for our σ. The normal
forces are τii , and the shear forces are τij . From Kundu
figure 2.2.
2.
THEN DO IT FORMALLY
We can also approach this more formally (following
Thompson). First, we want to write the surface forces
as a general stress tensor,
σij = −pδij + Σij
(2.28)
Next, write a general velocity gradient in terms of symmetric (Dij ) and antisymmetric (Ωij ) parts:
∂vi
= Dij + Ωij
∂xj
¶
¶ (2.29)
µ
µ
∂vj
∂vj
1 ∂vi
1 ∂vi
=
+
+
−
2 ∂xj
∂xj
2 ∂xj
∂xi
(2.34)
if eˆk is the unit vector in the kth direction.
In many applications the bulk viscosity is ignored.
This is called the Stokes hypothesis; Thompson discusses it nicely. This hypothesis is physically true
(at the kinetic level) only for simple systems – dilute,
> ν is possimonatomic gases. For other fluids, νb ∼
ble. However, in many applications the flows are incompressible, and (from 2.30 or 2.33) the effects of the
bulk viscosity term can be ignored.
13
E. The Navier-Stokes Equation (in Cartesian)
Now, we want to add viscous stresses to the force equation. Remember that we’re still working in Cartesian. We
need to generalize the ∇p term in the Euler equation (e.g., 1.10) to the full stress tensor. With the stress term added,
the force equation is called the Navier-Stokes equation. Recall, g = −∇Φ is the body force (derivable from a
potential). In Cartesian, in index notation, we have the form
¶
µ
∂σki
∂vi
∂vi
= ρgi +
+ vj
(2.35)
ρ
∂t
∂xj
∂xk
and in vector notation, we have the form
µ
¶
1
∂v
2
+ ρ(v · ∇)v + ∇p = ρg + ρν∇ v + ρνb + ρν ∇(∇ · v)
ρ
∂t
3
(2.36)
If Stokes’ hypothesis holds, this simplifies a bit to
ρ
∂v
1
+ ρ(v · ∇)v + ∇p = ρg + ρν∇2 v + ρν∇(∇ · v)
∂t
3
(2.37)
which is still general (allows variation of ρ). If we now assume the fluid is incompressible, ∇ · v = 0, (2.37)
becomes
ρ
∂v
+ ρ(v · ∇)v + ∇p = −ρ∇Φ + ρν∇2 v
∂t
(2.38)
which is the most commonly used form of this important equation, and recovers our “guess” in (2.2).
References
I’ve mostly followed Kundu and Faber here, as well as Thompson (whose discussion of viscosity and stress
tensors I like). For the non-Cartesian forms of the NS equation, Pozrikidis is one good reference.
F. Appendix: Navier-Stokes in other coordinates
It is also worth storing the stress tensor and the N-S equation in curvilinear coordinates (taken from Pozrikidis).
Here, we are using only the shear viscosity, and dropping the bulk viscosity term.
1. C YLINDRICAL COORDINATES
The 6 independent components of the stress tensor are
σrr
σrz
σφz
∂vr
= −p + 2ρν
∂r
µ
¶
∂vr
∂vz
= ρν
+
∂z
∂r
¶
µ
∂vφ 1 ∂vz
+
= ρν
∂z
r ∂φ
σrφ
σφφ
σzz
·
¸
∂ ³ vφ ´ 2 ∂vr
= ρν r
+
∂r r
r ∂φ
µ
¶
2 ∂vφ 2
= −p + ρν
+ vr
r ∂φ
r
∂vz
= −p + 2ρν
∂z
(2.39)
14
and the 3 components of the NS equation are:
vφ2
vφ ∂vr
∂vr
∂vr
∂vr
1
+ vz
+ vr
+
−
= gr +
∂t
∂z
dr
r dφ
r
ρ
µ
∂σzr
1 ∂(rσrr ) 1 ∂σrφ σφφ
+
+
−
∂z
r ∂r
r ∂φ
r
∂vφ
∂vφ
∂vφ vφ ∂vφ vr vφ
1
+ vz
+ vr
+
+
= gφ +
∂t
∂z
∂r
r ∂φ
r
ρ
vφ ∂vz
∂vz
∂vz
1
∂vz
+ vz
+ vr
+
= gz +
∂t
∂z
∂r
r ∂φ
ρ
µ
µ
¶
∂σzφ
1 ∂(r2 σrφ ) 1 ∂σφφ
+ 2
+
∂z
r
∂r
r ∂φ
∂σzz
1 ∂(rσrz ) 1 ∂σφz
+
+
∂z
r ∂r
r ∂φ
(2.40)
¶
¶
(2.41)
(2.42)
Alternatively, we can write the NS equations out explicitly for the case of constant viscosity:
vφ2
vφ ∂vr
∂vr
∂vr
∂vr
+ vz
+ vr
+
−
∂t
∂z
dr
r dφ
r
· 2
µ
¶
¸
∂ vr
∂ 1 ∂(rvr )
2 ∂vφ
1 ∂ 2 vr
1 ∂p
+ gr + ν
+
− 2
+ 2
=−
ρ ∂r
∂z 2
∂r r ∂r
r ∂φ2
r ∂φ
(2.43)
∂vφ
∂vφ
∂vφ vφ ∂vφ vr vφ
+ vz
+ vr
+
+
∂t
∂z
∂r
r ∂φ
r
· 2
µ
¶
¸
∂ vφ
1 ∂p
∂ 1 ∂(rvφ )
1 ∂ 2 vφ 2 ∂vr
=−
+ gφ + ν
+
+
+ 2
ρr ∂φ
∂z 2
∂r r ∂r
r ∂φ2
r ∂φ
(2.44)
vφ ∂vz
∂vz
∂vz
∂vz
+ vz
+ vr
+
∂t
∂z
∂r
r ∂φ
· 2
µ
¶
¸
∂ vz
1 ∂
∂vz
1 ∂ 2 vz
1 ∂p
+ gz + ν
+
r
+ 2
=−
ρ ∂z
∂z 2
r ∂r
∂r
r ∂φ2
(2.45)
2. S PHERICAL P OLAR C OORDINATES
The 6 independent components of the stress tensor are
σrr
σrφ
σθφ
∂vr
= −p + 2ρν
∂r
¸
·
∂ ³ vφ ´
1 ∂vr
+r
= ρν
r sin θ ∂φ
∂r r
·
¸
³
vφ ´
sin θ ∂
1 ∂vφ
= ρν
+
r ∂θ sin θ
r sin θ ∂φ
σrθ
σφφ
σθθ
·
¸
∂ ³ vθ ´ 1 ∂vr
+
= ρν r
∂r r
r ∂θ
¶
µ
∂v
ρν
φ
= −p +
+ vr sin θ + vθ cos θ
2
r sin θ
∂φ
µ
¶
2 ∂vθ
2vθ
= −p + ρν
+
(2.46)
r ∂θ
r
and the 3 componts of the NS equation are:
vθ2 + vφ2
vφ ∂vr
∂vr
∂vr
vθ ∂vr
+vr
+
+
−
∂t
∂r
r ∂θ
r sin θ ∂φ
r
µ
¶
2
σθθ + σφφ
1 1 ∂(r σrr )
1 ∂(σθr sin θ)
1 ∂σφr
+gr +
+
+
−
ρ r2
∂r
r sin θ
∂θ
r sin θ ∂φ
r
(2.47)
vφ ∂vθ
cot θ
∂vθ
vθ ∂vθ
vr vθ
∂vθ
+vr
+
+
+
− vφ2
∂t
∂r
r ∂θ
r sin θ ∂φ
r
r
¶
µ
2
σφφ
1 ∂(σθθ sin θ)
1 ∂σφθ
σrθ
1 1 ∂(r σrθ )
+
+
+
−
cot θ
+gθ +
ρ r2
∂r
r sin θ
∂θ
r sin θ ∂φ
r
r
(2.48)
15
∂vφ vθ ∂vφ
vφ ∂vφ vφ
∂vφ
+vr
+
+
+
(vr + vθ cot θ)
∂t
∂r
r ∂θ
r sin θ ∂φ
r
µ
¶
1 ∂σφφ
1 1 ∂(r2 σrφ ) σrφ 1 ∂σθφ 2 cos θ
+
+
+
σθφ +
= gφ +
ρ r2
∂r
r
r ∂θ
r
r sin θ ∂φ
(2.49)
Once again, we can write these out explicitly:
2
2
vθ + vφ
vφ ∂vr
∂vr
∂vr
vθ ∂vr
+vr
+
+
−
∂t
∂r
r ∂θ
r sin θ ∂φ
r
µ
¸
·
¶
2vr
2 ∂vθ
2 ∂vφ
1 ∂p
2
+ gr + ν ∇ vr − 2 − 2
+ vθ cot θ − 2
=−
ρ ∂r
r
r
∂θ
r sin θ ∂φ
(2.50)
vφ ∂vθ
cot θ
∂vθ
vθ ∂vθ
vr vθ
∂vθ
+vr
+
+
+
− vφ2
∂t
∂r
r ∂θ
r sin θ ∂φ
r
r
·
¸
1 ∂p
2 cos θ ∂vφ
∂v
2
v
r
θ
2
=−
+ gθ + ν ∇ vθ + 2
− 2 2 − 2 2
ρr ∂θ
r ∂θ
r sin θ r sin θ ∂φ
(2.51)
∂vφ vθ ∂vφ
vφ ∂vφ vφ
∂vφ
+vr
+
+
+
(vr + vθ cot θ)
∂t
∂r
r ∂θ
r sin θ ∂φ
r
·
¸
vφ
∂p
2 cos θ ∂vθ
1
2 ∂vr
+ gφ + ν ∇2 vφ − 2 2 + 2
+ 2 2
=−
ρr sin θ ∂φ
r sin θ r sin θ ∂φ
r sin θ ∂φ
(2.52)
In this set, for f some scalar function,
∇2 f =
1 ∂
r2 ∂r
and this is the end of this chapter.
µ
¶
µ
¶
1
∂f
1
∂f
∂
∂2f
r2
+ 2
sin θ
+ 2 2
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
16
3. POTENTIAL FLOW
which becomes, explicitly,
In the first chapter we wrote down fairly general (partial
differential) equations describing fluid flow. We now
must look for solutions. In this chapter we consider
one of the traditional methods, potential flow, which describes incompressible flows in which the velocity field
has no curl.
A. Setups and Assumptions
Our first set of solutions consider flows with no vorticity: ω = ∇ × v = 0. In addition, we will generally assume the flows are incompressible. Why are
such restrictive solutions interesting? First, we can find
analytic solutions: so let’s see where they lead. Second, these solutions have some relation to reality . . .
because vorticity turns out to be a (nearly) conserved
quantity, and hard to generate away from boundaries:
we will prove this later. Thus, flows which start out
irrotational tend to stay this way.
1.
VELOCITY POTENTIAL
For irrotational flows, the fact that ∇ × v = 0 allows
the velocity to be written as the gradient of a scalar potential:
v = ∇φ
(3.1)
Explicitly, this is
vx =
∂φ
;
∂x
vy =
∂φ
∂y
(3.2)
Further, when the flow is incompressible – which is the
case in all examples we will see — we have ∇ · v = 0,
so that
∇2 φ = 0
(3.3)
That is, the velocity potential satisfies Laplace’s equation. But this is great if we are traditional analytic
physicists: we can use all the (glorious) solution methods worked out in electrostatics. This will be our general approach – picking examples where we can find solutions to Laplace’s equation, based on what instinct we
have from analytic E&M. Once we have found φ(x, y),
the velocity solution comes immediately from (3.1).
2.
STREAM FUNCTION
If the flow is two-dimensional, we can also define another useful function. This is the stream function, ψ.
Define that
v = −∇ × (ψẑ)
(3.4)
vx =
∂ψ
;
∂y
vy = −
∂ψ
∂x
(3.5)
(Why is this possible? From 3.4 it’s clear that ∇·v = 0.
Thus, this works for incompressible flows.) The stream
function has several useful properties.
• It’s related to the velocity potential; we can get
one from the other:
vx =
∂φ
∂ψ
=
;
∂x
∂y
vy =
∂φ
∂ψ
=−
(3.6)
∂y
∂x
• ψ also satisfies Laplace’s equation; we can solve
for ψ rather than φ if we wish.
• Streamlines are tangent to the local velocity v.
A streamline obeys
vy dx = vx dy
(3.7)
(compare 1.1). Thus, the differential of ψ along
some path is
dψ =
∂ψ
∂ψ
dx +
dy
∂x
∂y
(3.8)
Combining (3.7) and (3.8) shows that dψ = 0
along a streamline. We can therefore use the function ψ to label streamlines – as in the examples.
• Finally, using (3.6), it follows directly that
∇φ · ∇ψ = 0, so that streamlines and equipotential lines cross each other at right angles. This
again says that v is tangential to the contours of
constant ψ – so the value of ψ provides a label for
the streamlines in the flow.
3.
OTHER COORDINATES
As long as the problem is two dimensional and incompressible, we can define φ and ψ functions.
Plane Polar Coordinates
The velocity is connected to the potential by
vr =
∂φ
;
∂r
vθ =
1 ∂φ
r ∂θ
(3.9)
(Note that we are using θ for the physicist’s customary angle φ here, to avoid confusion with the velocity
potential.) The incompressibility condition is
1 ∂
1 ∂vθ
(rvr ) +
=0
r ∂r
r ∂θ
(3.10)
From this we can define a stream function:
vr =
1 ∂ψ
;
r ∂θ
vθ = −
∂ψ
∂r
(3.11)
17
Cylindrical Coordinates, Axisymmetric
The velocity is connected to the potential by
vr =
∂φ
;
∂r
vz =
∂φ
∂z
(3.12)
The incompressibility condition is
1 ∂
∂vz
+
(rur ) = 0
∂z
r ∂r
(3.13)
From this we can define a stream functionn:
vz =
1 ∂ψ
;
r ∂r
vr = −
φ = Ux ;
1 ∂ψ
r ∂z
(3.14)
1.
Spherical Polar Coordinates, Axisymmetric
The velocity is connected to the potential by
vr =
∂φ
;
∂r
1 ∂φ
r ∂θ
vθ =
This describes radial flow away from a source (or inwards to a sink if C < 0). Note that ∇ · v 6= 0 at the
origin, and that the potential is undefined there. Integrating the mass flux around a source or sink shows us
that ρC describes the amount of mass per time being
added at the source, per length in the z direction (recall
that this is a 2-D problem in 3-D).
Another basic building block is that of a constant
flow of speed U in the x direction, represented in Cartesian by
We can combine these two building blocks (and work
in plane polar):
C
ln r
2π
C
vr = U cos θ +
2πr
φ = U r cos θ +
The incompressibility condition is
1
r2 sin θ
∂ψ
;
∂θ
vθ = −
1 ∂ψ
r sin θ ∂r
(3.21)
vθ = −U sin θ
(3.16)
to get what is called flow past a half body. We see why,
by noting that the corresponding stream function is
From this we get a stream function:
vr =
(3.20)
SOURCES AND SINKS , FLOW PAST HALF BODY
(3.15)
1 ∂ ¡ 2 ¢
1 ∂
r vr +
(vθ sin θ) = 0
r ∂r
sin θ ∂θ
v = ∇φ = U x̂
(3.17)
ψ = U r sin θ +
C
θ.
2π
(3.22)
B. Two-dimensional Planar Problems
Consider Laplace’s equation in two dimensions, in
plane polar coordinates (r, θ).
µ
¶
∂φ
1 ∂2φ
1 ∂
2
r
+ 2 2 =0
∇ φ=
r ∂r
∂r
r ∂θ
We know the possible general solutions:
φ = constant ;
φ ∝ rn sin(nθ) ;
φ ∝ rn cos(nθ);
φ ∝ ln r ;
φ ∝ r−n cos(nθ) ;
φ ∝ r−n sin(nθ); Figure 3.1.
(3.18)
And, of course, any superposition of these is a solution.
So, we proceed by finding useful superpositions . . .
One important building block will be a simple
source or sink:
Cθ
C
; φ=
ln r
2π
2π
C
vr =
; vθ = 0
2πr
ψ=
Potential flow past a two-dimensional “half
body” (represented mathematically by a source function).
The boundary streamline is labelled by ψ = m/2 (that’s
C/2 in our notation); the shape of this streamline defines
the shape of the half body. S shows the stagnation point in
front. From Kundu figure 6.9
The flow clearly has a stagnation point: if it’s at position (a, π), then v(a, π) = 0 requires a = C/(2πU ).
Now, the value of the stream function passing through
this point – its label – is
(3.19)
ψs = U a sin π +
C
C
π=
2π
2
18
and the equation of the streamline passing through this
point is found by taking ψ(r, θ) = ψs = C/2:
U r sin θ +
C
C
θ=
2π
2
(3.23)
This stagnation streamline separates the flow into an
upstream region (v → U x̂ at ∞), and the region inside the streamline, a region with a smooth nose, called
a half body. We can, therefore, put a physical body with
this shape into a flow, and keep the same external solution.
Once we have the velocity solution, we can find the
drag on the half body – the net force the fluid exerts on
the body (or vice versa). As long as we ignore friction
(i.e., viscosity), the net force on the body is just the
(vector) integral of the pressure over the surface of the
body. That is, for this simple case, we expect the drag
force to be
Z
(3.24)
FD = − pn̂dS
S
if n̂ is the normal to the surface S, and the integral is
taken over the surface of the body. Bernoulli’s relation
for an incompressible fluid (1.19) can be written as
1
1
p + ρv 2 = p∞ + ρU 2
2
2
(3.25)
where we have evaluated the constant at “infinity”, that
is far upstream. Work this out: the (normalized) excess
pressure,
Cp = 2(p − p∞ )/ρU 2
turns out to be positive at the front of the body, zero at
θ = 113◦ , and negative along the sides. You can show
from this, by setting up the integrals, that the net pressure, integrated over the surface of the body, is zero:
there is no drag in this system.
2.
Figure 3.2.
A doublet: a source and a sink which
approach each other, keeping the product of their strength C
and separation ǫ constant. The streamlines become circles,
whose centers lie on the y-axis and which are tangent to the
x-axis at the origin. From Kundu figure 6.8
Now: combine this with a uniform stream flow, φ =
U x and ψ = U y: combining and putting in polars, with
a2 = µ/U , gives
¶
µ
a2
cos θ ;
φ=U r+
r
¶
µ
a2
vr = U 1 − 2 cos θ ;
r
¶
a2
ψ=U r−
sin θ
r
µ
¶
a2
vθ = −U 1 + 2 sin θ
r
µ
(3.27)
But, now, this shows that ψ(a, θ) = 0: the streamline
ψ = 0 is a circular cylinder of radius a. Thus, flows
inside this cylinder have no influence on flows outside;
we can again replace this doublet with a physical cylinder, and find the same flow outside.
FLOW PAST A CYLINDER
In another example, we start with a doublet: a paired
source and sink, of equal strength C, separated by ǫ;
let them approach each other while the product of their
separation and their strength stays finite, Cǫ/π → µ.
The potential and stream function, from Kundu, are
φ=
x2
µx
;
+ y2
ψ=−
x2
µy
+ y2
(3.26)
The streamlines, given by ψ = constant, are circles
whose centers lie on the y-axis, and which are tangent
to the x-axis at the origin.
Figure 3.3. Potential flow past a circular cylinder,
described by the ψ = 0 streamline. The inside of the
cylinder is represented mathematically by a doublet. From
Kundu figure 6.11
Finally, we can again work out the drag. The pres-
19
sure difference
Cp =
FLOW AROUND A SPHERE
v2
2(p − p∞ )
=
1
−
= 1 − 4 sin2 θ
ρU 2
U2
(3.28)
so, again, is positive at the front and negative at the back
. . . so that, once again, there is no net pressure drag on
this cylinder.
C. Axisymmetric 3D Problems
Axisymmetric 3D problems are also amenable to this
type of solution. Spherical polar coordinates are appropriate here. The general solutions to Laplace’s equations for this geometry are
φ =constant ;
φ ∝ rn Pn (cos θ) ;
(3.29)
φ ∝ r−(n+1) Pn (cos θ)
which are just your favorite (axisymmetric) Laplace’s
solutions in spherical polars. The velocity potentials
and stream functions obey
vr =
1 ∂ψ
∂φ
= 2
;
∂r
r sin θ ∂θ
1 ∂ψ
1 ∂φ
=−
vθ =
r ∂θ
r sin θ ∂r
(3.30)
As with flow past a cylinder, this adds a uniform stream
to a doublet which opposes the stream.
A Doublet, in this geometry, has the potential and
stream function,
φ=
m
cos θ ;
r2
ψ=−
m
sin2 θ
r
(3.34)
The velocity solutions for flow around a sphere are
straightforward. We have, then
m
cos θ ;
r2
m
1
ψ = U r2 sin2 θ − sin2 θ
2
r
φ = U r cos θ +
(3.35)
From this, we see that ψ = 0 for θ = 0, π at any r
(that is, the x-axis), and also for r = a = (2m/U )1/3
(any θ). Thus, again, we have a spherical surface, flows
inside of which have no effect on the outer stream flow.
Again, then, we can work out the pressure/drag on
this system. The velocity components are
·
³ a ´3 ¸
vr =U 1 −
cos θ ;
r
¸
·
(3.36)
1 ³ a ´3
sin θ
vθ = −U 1 +
2 r
so that the pressure is
We can run through some familiar-looking solutions.
¶
µ
9
1 2
2
p(a, θ) = p∞ + ρU 1 − sin θ
2
4
POINT SOURCE AND STREAM FLOW
These are our two most basic. A point source has the
potential and stream function,
φ=−
Q
;
4πr
ψ=−
Q
cos θ
4π
(3.31)
where Q (units:
is the discharge rate of the
source. A uniform stream flow has the potential and
stream function,
1
ψ = U r2 sin2 θ
2
(3.32)
The velocity solutions for these two basics are straightforward. It is worth noting that the sum of these two
gives the velocity field for an axisymmetric half body,
vr = U cos θ +
Q
;
4πr2
vθ = −U sin θ
So, once again, this is symmetric in θ, leading to zero
drag on the sphere. Finally, let me refer you to Faber for
a good discussion of real-world effects in this problem,
§4.7.
A LINE SOURCE
m3 /s)
φ = U r cos θ ;
(3.37)
(3.33)
Here, we describe a mass source which is distributed in
a line; let it have a source
R function k per length, so that
its total source is Q = kdx.
Each differential bit of the line source has potential
and stream functions,
dφ = −
kdξ
;
4πr
dψ = −
kdξ
cos α
4π
(3.38)
if ξ is the running variable along the length of the
source.
We also let R = r sin θ be the vertical distance to the observation point, and r1 =
¡ 2
¢1/2
r + a2 − 2ra cos θ
be the radial distance from
20
(r, θ)
r
θ
0
Figure 3.4.
α
ξ
a
The geometry of a line source, which extends
from ξ = 0 to ξ = a.
ξ = a (the end of the source) to the observation point.
The net stream function is then
Z a
Z a
k
k
dψ = −
ψ(r, θ) =
cos α(ξ)dξ =
(r−r1 )
4π 0
4π
0
(3.39)
where the last step is obtained by straightforward geometrical identities and changes
R of the integration variable. [Exercise: what is φ = dφ?]
D. d’Alembert’s “Paradox”
We have now seen several examples of flow past a solid
body, described by potential flow. In each case we
found that the net force exerted by the flow on the body
is zero. Thus, there is no drag in a potential flow around
a body. This result turns out to hold for any body, symmetric or not.
Does this sound wrong? We know experimentally
that a drag force does exist when a body is immersed in
a fluid flow. This apparent contradiction, between the
theory and experiment, became known as d’Alembert’s
paradox. As you might suspect, the solution of the paradox is that the theory is incomplete (no, we don’t say
“wrong”). In particular, we have ignored two important
pieces of the physics here. One, is that these solutions
can have nonzero velocity at a solid surface: check Figure 3.1 or 3.3. This is naive: any real fluid has a finite,
if small, viscosity. But viscosity will force to a no-slip
boundary condition. This leads to a viscous boundary
layer, which will modify the solutions close to the surface, and add tangential stress which contributes some
drag. Two, the flow behind the body may very well be
turbulent – as this viscous boundary separates from the
body and connects into the wake. We will see later that
this also contributes to a drag force. Both of these ideas
are illustrated in Figure 3.5.
We can also carry out a general drag calculation, for
a finite body, that illustrates this argument. We could
work out the pressure in the fluid, everywhere on the
Figure 3.5. Illustrating the nature of the flow field around
an arbitrarily shaped body. From Currie Figure 9.1.
surface of the body, and then integrate it over the surface. Or we can try a different approach.
P
U
p
Q
0000000
1111111
1111111
0000000
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
0000000
1111111
o
o
u(y)
y
R
S
x
Figure 3.6. Momenteum balance of flow past an arbitrary
body. The upstream flow has constant speed Uo and
pressure po . The downstream velocity may be a function of
y; in this sketch it is lower immediately behind the body (as
due to a wake). Note, the body need not be spherical, that
was just easy for me to draw... Following Kundu figure 4.7.
Consider in figure 3.6 the region of fluid within the
rectangle SRQP; the net drag on the body within this
region is just the rate of change of momenteum flux:
FD = Ṗout , i.e. the net outflow of x-momentum from
the region. The outflow through PS and QR is easy to
calculate:
Z L
PS
Uo ρUo dy = −LρUo2 ;
Ṗ
=−
o
(3.40)
Z L
2
QR
u(y) dy
Ṗ
=ρ
0
Now: there must be mass outflow, Ṁ , through the
sides, SR and PQ (this is necessary to conserve mass
if u(y) < Uo , as shown). Mass conservation requires
Z L
PQ
SR
u(y)dy
(3.41)
LρUo = Ṁ
+ Ṁ + ρ
O
But if the sides are far enough away from the rock that
the x velocity ≃ Uo there, then the rate of outflow of
21
x-momentum is
Ṗ P Q + Ṗ SR = ρUo
Z
0
L
(Uo − u(y))dy
(3.42)
Finally, we can collect all of this, to find the net drag on
the body: FD = Ṗ out = Ṗ P S + Ṗ QR + Ṗ P Q + Ṗ SR .
This gives,
FD = Ṗ
out
=ρ
Z
L
O
u(y)(Uo − u(y))dy
(3.43)
And .. referring back to our simple example of potential
flow past a cylinder, note from (3.27) that the downstream velocity field → U as r → ∞. That is, there
is no slow-down of the flow in this idealized case; and
thus no drag. Drag comes from viscosity and boundary
effects which generate a wake behind the body.
References: I’ve mostly followed Kundu here.
22
4. VORTICITY
Integral: the magnitude of the circulation is the velocity integrated around a closed path:
I
Z
(4.2)
K = v · dl = (∇ × v) · dA
For God’s Sake Let us sit upon the ground
and tell sad stories
Of vortex filaments.
How some have been ill-posed, some singular,
Some poisoned by their self induction,
some core size killed,
Some haunted by the mathematics they
have involved.
A
where Stokes’ theorem is used in the second equality.
As we noted above, K can be taken as a vector, with
direction given by the sense of the flow, using the right
hand rule. We can think of K as a “flux” of vorticity;
it measures the amount of ω passing through some surface bounded by dl.
Let’s look at some examples. Solid body rotation is
described by vθ = Ωr;
All murderous.
1 ∂
(rvθ ) = 2Ω ;
r ∂r
Z 2π
vθ rdθ = 2πr2 Ω
K=
For within the swirling motion that
rounds the mortal circulation
Of a vortex
Keeps futility his court,
And there the non-linearity sits
Scoffing at his state and grinning at his
theories
Allowing him a breath, a little scene to
linearize, compute and fill with approximations
And then at last he comes and with a
little inconsistency bores through the
costly hopes and
ω=
0
so solid body rotation has a net vorticity, and a net circulation.
An Irrotational vortex comes from potential flow.
Pick vθ = C/r; then rvθ = constant, so that
ω = 0, r 6= 0 ;
Z 2π
vθ rdθ = 2πC
K=
(4.4)
0
Farewell . . .
Shakespeare, Richard II, Act 3, Scene 2
“translated” by H. C. Yuen, quoted in
Vortex Dynamics, P. G. Saffman
A. Vortex Kinematics
Vorticity is a large topic, and we can only treat it lightly
here. We start with kinematics. We saw an example in
Chapter 3 of a flow field with a net circulation: in which
the line integral of the velocity around a closed loop is
finite. Such a flow has a net circulation. We generalize
this here, with differential and integral definitions.
Differential:
(4.3)
vorticity is the curl of the velocity field:
ω =∇×v
(4.1)
We can think of vortex lines – just like magnetic field
lines or streamlines, they are lines which are everywhere tangent to the local vorticity. Also we will run
into vortex tubes, regions of high vorticity (picture a
tornado; that is an extreme example).
Thus, circular streamlines do not imply that the flow
has a net vorticity everywhere. In this example ω 6= 0
only at the origin; elsewhere the flow is curl-free. The
infinite vorticity at the origin, however, is enough to
give this vortex a net circulation.
The Rankine vortex is a more physical extension of
the irrotational one. A real vortex must have a rotational
core (we can’t have vθ → ∞). We can approximate this
with vθ ∝ r, r ≤ R (the radius of the core of the vortex); and vθ ∝ 1/r for r > R. Thus, the Rankine vortex
has uniform vorticity in its core, and zero elsewhere.
B. Vortex Dynamics
We will only need this result in the nearlyincompressible limit. That is, we take ∇ · v ≃ 0, but
formally allow ∇ρ 6= 0. We start with the incompressible form of the Navier-Stokes equation, from (2.38),1
1
∂v
+ (v · ∇)v + ∇p = −∇Φ − ν∇ × (∇ × v) (4.5)
∂t
ρ
1
The viscous term is written slightly differently, by the magic of
vector analysis.
23
To proceed, take the curl of (4.5). Using the vector relation v · ∇v = (∇ × v) × v + 12 ∇v 2 , we get the
dynamical equation for vorticity:
Kundu proves this as follows. Start with Euler’s equation, (1.10) (note we aren’t worried about viscosity right now):
Proof #1
1
Dv
= − ∇p + ∇Φ
Dt
ρ
ω
∂ω
1
ω × v) = ν∇2ω + 2 ∇ρ × ∇p (4.6)
+ ∇ × (ω
∂t
ρ
or2
ω
1
Dω
= ω · ∇v + ν∇2ω + 2 ∇ρ × ∇p
Dt
ρ
(4.7)
A possible further step is to consider the barotropic
limit, in which the pressure is only a function of the density [p = p(ρ)] and vice-versa, for instance in an adiabatic gas with p/ργ = constant. In that case, ∇ρ k ∇p,
so their cross product is zero and the equations simplify,
to
ω
∂ω
ω × v) = ν∇2ω
+ ∇ × (ω
∂t
(4.8)
ω
Dω
ω · ∇)v + ν∇2ω
= (ω
Dt
(4.9)
and
Yet another possible step is to work in a rotating
frame. Referring back to chapter 1, the same analysis
gives here,
ω
1
Dω
ω + 2Ω
Ω) · ∇v + ν∇2ω + 2 ∇ρ × ∇p (4.10)
= (ω
Dt
ρ
C. Conservation of Circulation: Kelvin’s
Theorem
A basic result here is Kelvin’s theorem: in an inviscid, barotropic flow with conservative body forces, the
circulation K around a closed curve, moving with the
fluid, is a constant of the motion. This means that vortex lines can be thought of as “frozen into” the fluid:
if you stretch or compress a piece of fluid, any vortex
tube running through it must also be stretched or compressed.
This is worth two slightly different proofs, as follows.
2
If we want to eliminate the nearly incompressible assumption,
we simply need to start with (2.37) and carry out the full vector
curl, to go from v to ω ; students with time on their hands are
welcome to do this.
Now, take the formal derivative of K:
I
I
DK
Dv
D
=
· dl + v ·
dl
Dt
Dt
Dt
(4.11)
(4.12)
But using Euler’s equation, and noting that ∇p · dl =
dp, the difference in pressure between two adjacent
points:
I
I
I
dp
D
DK
= ∇Φ · dl −
+ v·
dl (4.13)
Dt
ρ
Dt
Now: the body (gravity) force, being from a potential, is conservative; so the first term goes to zero. If
the density is only a function of pressure, ρ = ρ(p) –
that is if the fluid is barotropic – then dp/ρ is a perfect differential, so that the second integral (around a
closed curve) also goes to zero. Finally, the third term
involves D(dl)/Dt = dv, and thus v · D(dl)/Dt =
D(v 2 /2)/Dt, also a perfect differential, so the third
term also vanishes. Thus, K is a constant of the motion in this flow.
I have not seen this in a text, but it follows
directly from magnetic field flux freezing arguments
(which will be presented in a later chapter). Here, we
keep the viscosity term around, but retain the incompressible, barotropic limit. That means we are again
using (4.5) as the governing equation, and (4.6) derived
from it, but now assuming ∇ρ × ∇p = 0. The formal
derivative of K can now be written
¸
Z ·
Z
ω
∂ω
∂K
=
− ∇ × (v × ω ) ·dA = ν∇2ω ·dA
∂t
A ∂t
(4.14)
In expanding out the K integral, we have allowed for
intrinsic changes of ω with time, and also for motion
of the boundary of the surface, A. But now the result
is clear: if ν → 0, K → constant. Furthermore, a
finite viscosity will lead to eventual dissipation of the
circulation.
Proof #2
The Helmholtz theorems are related; they can be derived from the same assumptions that we used for
Kelvin’s theorem. The theorems are:
• Vortex lines move with the fluid.
24
• The strength of a vortex tube – the circulation – is
constant along its length.
• A vortex tube cannot end within the fluid. It must
either end at a solid boundary, or form a closed loop (a
vortex ring).
• The strength of a vortex tube is constant in time.
D. The Magnus Effect
Let’s return to potential flow. Consider, again, flow past
a cylinder, but this time let the cylinder rotate. This
situation leads to a net force on the cylinder.
To be specific, consider a potential, and stream function, given by
φ=
K
θ;
2π
ψ=−
K
ln(r/r0 )
2π
(4.15)
(where r0 is an arbitrary constant to fix units in the
log), which recovers the velocities vr = 0 ;
vθ =
K/2πr and streamlines r = constant, as we expect.
(Note, the r term in the log in the expression for ψ must
contain a scaling parameter to make it dimensionless;
this will not matter as we only take derivatives of ψ).
Thus, this potential-stream pair describes counterclockwise flow (if K > 0) about the origin. Or, put in a
minus sign to make clockwise flow, and add this to our
previous solution for irrotational flow around a cylinder:
¶
µ
K
a2
θ;
cos θ +
φ=U r+
r
2π
(4.16)
µ
¶
a2
K
ψ=U r−
ln r
sin θ −
r
2π
We recall a is a length scale, corresponding to the radius of the cylinder. This potential/stream pair has the
velocity field,
¶
µ
a2
vr = U 1 − 2 cos θ
r
(4.17)
µ
¶
a2
K
vθ = −U 1 + 2 sin θ +
r
2πr
H
Note that
this
velocity
field
has
a
net
circulation:
v·
R
dl = vφ rdφ = K, if the integral is taken around any
closed curve that encloses the r = 0 axis.
Now, at the surface of the cylinder, vr = 0 and
vθ = −2U sin θ −
K
2πa
so that the sign of K – the intrinsic circulation associated with the r ≤ a region – determines whether the
Figure 4.1.
Potential flow past a circular cylinder for
different values of the circulation Γ (K in our notation).
Note the location of the stagnation points S depends on the
ratio of circulation to flow speed U . From Kundu figure 6.14
vθ flow outside agrees with K or with the ambient flow
U . We also see from (4.17) that the flow is no longer
symmetric above and below the cylinder; the flow simply moves faster “over the top”. Figure 4.1 illustrates
various possibilities. At low K values, there are two
stagnation points, at the surface of the cylinder. At high
enough K, there is one stagnation point, well below
the cylinder. The streamline passing through this stagnation point contains a region of fluid which remains
separated from the outer flows; it simply circulates and
never reaches large distances away.
And now: there is a net surface pressure. We can
find it as usual from Bernoulli, and show that it contributes a net lift:
Z 2π
p(a, θ) sin θadθ = ρU K
(4.18)
L=−
0
We can remember the direction of the force by noting
that the lift L in this example is upwards. If we define
the direction of the circulation K as along the axis of
the cylinder, in the sense given by the right hand rule,
then the direction of the lift is L ∝ U × K. This allows
a reasonable analogy with basic EM: a current I in an
external magnetic field B feels a force F ∝ I × B.
It turns out that this result – L = ρU K – holds for
any two-dimensional shape, not just a true cylinder. A
rotating body feels a lateral force: this is the Magnus
effect. Why is this different from the potential flow
solutions of chapter 3? One answer would simply be,
“Bernoulli”; the asymmetric velocity above and below
the rotating object leads to an asymmetric pressure, thus
a net lateral force. But one could also ask how a rotat-
25
ing body develops the circulation which is needed for
the Magnus effect. The answer here is, again, viscous
effects at the boundary of the cylinder. Any rotating
body must have a no-slip surface; this will generate a
small layer of fluid rotating with the body.
E. Vortex lines and their behavior
Vortex lines demonstrate very interesting behavior. In
these notes we only summarize; the literature is extensive here.
• Free vortex lines are linear (but not necessarily
straight) structures, with a net circulation K, which is
constant along the line and also a constant of the motion
(from Kelvin’s theorem). By Kelvin’s theorem, a loop
in the fluid which at one point encloses the core (r → 0)
will have a finite circulation K; it must retain that K
value no matter how the fluid is distorted in later flow.
Thus, a vortex line must move with the fluid; the fluid
is effectively attached to the vortex, particularly to its
core (about which the line integral defining K is finite).
• A vortex sheet is made up of an infinite number of
vortex filaments placed side by side, with all filaments
rotating in the same sense. This sheet generates a tangential velocity discontinuity, as in Figure 4.2. Such
structures are important in understanding the flow over
aircraft wings.
Figure 4.2.
Figure 4.3.
A line vortex (A) near a wall, and its image
(B). From Kundu figure 5.13
• Paired vortices interact. Two nearby vortex
lines, with opposite K values, will drift together in a
direction perpendicular to their separation. If the two
lines have parallel K values, they will circle around
each other – or twist into a helix if one end of each is
somehow tied down.
An idealized vortex sheet. From Kundu
figure 5.15
• The behavior of vortex lines can be very intriguing. They move under their own power. A vortex line will move in a transverse flow, in a direction
U × K. (Recall the Magnus effect). A bent vortex line
will move in a direction K × n̂, if n̂ is the direction
of its radius of curvature R; Batchelor shows its speed
∼ K/4π.
• Images are useful in calculating the behavior of
vortex lines near boundaries (remember your images
in electrostatics?). A wall, such as in Figure 4.3, is a
streamline, and must have zero normal velocity. This
means we can add an imaginary image vortex, behind
the wall and paired with the real one, to understand the
motion of the vortex line.
Figure 4.4.
A vortex ring approching a wall. From
Kundu figure 5.14
• Vortex rings are formed when a vortex line connects back on itself. A vortex ring will move itself along
through the background fluid, at a speed ∼ K/4πb if
b is the radius of the ring. (Any two opposite points
around the ring act like paired, anti-parallel K lines.
Alternatively, this is a case of well-defined curvature of
a single vortex line.) One vortex ring will spread out as
it approaches a wall. Two rings will move around and
through each other – Faber shows this in his figures.
• Vortex lines can reconnect. Viscosity breaks the
conservation of circulation, and allows adjacent vor-
26
tex lines or rings to reconnect, thereby changing their
topology.
F. Generation of Vorticity
We argued above that circulation is conserved; this was
Kelvin’s theorem. How, then, can a fluid which starts irrotational, ever gain vorticity? The answer, once again,
is the combination of viscosity and boundary effects.
To illustrate, I combine a figure from Kundu with the
discussion from Batchelor.
Figure 4.5.
Top view of a vortex pair generated by
moving the handle of a spoon through a cup of coffee. From
Kundu figure 5.12
Start with a quiescent cup of coffee, turn your
spoon upside down, and move its handle perpendicular to itself through the cup. Vortices will be
generated, as shown. Why? Even though there
is a formal no-slip condition at the surface of the
handle, there must be a finite tangential velocity
somewhere close to the handle. In this problem,
it’s at the edges. As Batchelor says, “the motion
that would be generated from rest in the absence
of diffusion of vorticity across the boundary . .
.is accompanied by a nonzero tangential relative
velocity at the boundary. Since the no-slip condition requries the tangential component of relative velocity to be zero at each point of the solid
boundary, however small the viscosity may be, the
vorticity in the flow is infinite at the boundary.”
Thus: the need to go from a no-slip boundary to
a tangential slip in the coffee, some tiny distance
away, generates vorticity local to the handle; diffusion (as per 4.6) takes care of the rest.
References
Vorticity is a large, interesting and complex topic.
I’ve mostly followed Faber and Tritton (for the words)
and Kundu (for the math); but there’s also good material in Shivamoggi, not to mention a whole book by
Saffman (quite mathematical), and several useful articles in Annual Review of Fluid Mechanics over the
years.
27
5. LAMINAR FLOW: MORE APPLICATIONS
We continue the previous discussion with more examples and applications from here and there. We start
with a simple discussion of flow in a (quasi) twodimensional, rotating system – all sorts of fun things
happen when you’re rotating. We then look at two more
mathematically involved examples of viscous flows.
A. Geostrophic Flow
rock is not very disturbed by the rock’s motion (that is,
the flow will vary with height, above the rock). Now
consider the same thing but with a rotating water tank.
The Taylor-Proudman result says that the flow at all
heights above the rock must be the same. This produces
a Taylor column: the water above the rock forms a vertical column which is fixed relative to the rock; water
not over the rock flows around this column, as though
it were a solid object.
ATMOSPHERIC CIRCULATION
What happens in a rotating system? Because our important application is to terrestrial (geophysical) flows, we
can work in the incompressible limit. Referring back to
chapter 1, the general equation of motion comes from
(1.20), but now has viscosity added, as per Chapter 2:
One striking example of our first point is the atmospheric circulation around a pressure extremum:
the circulation is counterclockwise (seen from above)
around a low, and clockwise around a high – as shown
in the figure.
Dv
1
Ω × v + ν∇2 v (5.1)
= − ∇p + g + Ω2 R − 2Ω
Dt
ρ
High
pressure
We consider the limit which is interesting for terrestrial
problems: steady (or nearly steady) flow, in which the
advective term (v·∇v) and the viscous term (ν∇2 v) are
Ω × v). In addismall compared to the Coriolis force (2Ω
tion we ignore the centrifugal force and gravity terms.1
We then have a situation in which the equation of motion simplifies to
Ω × v + ∇p = 0
2ρΩ
(5.2)
This is known as geostrophic flow; such flows have interesting properites, as follows.
• The velocity is perpendicular to the pressure gradient. Thus, isobars are also streamlines; the flow is
along lines of constant pressure. This is in marked
contrast to non-rotating systems, where we tend to
think of pressure gradients driving the flows, or (from
Bernoulli’s equation) think of pressure variations along
streamlines.
• The flow must be two-dimensional. That is: take
Ω × v) = 0. But if Ω k ẑ,
the curl of (5.2), to get ∇ × (Ω
say, this requires that dv/dz = 0. The flow cannot depend on z. This is known as the Taylor-Proudman theorem, and will be important in our discussion of Rossby
waves, below. It has one striking consequence: the Taylor column. Think about a rock on the bottom of a tank
of water. You tie a string to the rock and pull it through
the water. If the tank is static, the water well above the
1
Purists can note that both of these can be expressed as gradients
Ω × (Ω
Ω × r) = −∇(Ω2 R2 ) (in the
of a scalar: g = ∇Φg , and 2Ω
notation of §1.D.). Thus they can be subsumed into the pressure
term – where they will be only small corrections.
Low
coriolis
pressure
coriolis
Figure 5.1. Illustrating geostrophic flow around high and
low pressure centers. Following Kundu Figure 13.4.
The jet stream is a related effect. To be specific, think
about the (north) polar jet stream. The atmosphere at
the top of the world is cold, while that at lower latitudes
is warm; the hot and cold air masses meet in a region
called the polar front. The horizontal pressure gradient
across the polar front generates transverse winds, again
by the Coriolis balance of (5.2); this large-scale circulation, from west to east, is called the polar jet stream. It
typically lies at about 10 km altitude, has average speed
∼ 100 km/s, but can reach maximum speeds 300-400
km/s. The northern hemisphere also has a sub-tropical
jet, where the temperate mid-latitude air meets the hot,
tropical air. These two jets are by no means steady –
they wander about, and sometimes split, depending on
the local thermodynamics of the atmosphere. Some of
the wiggles and fluctuations in the jet streams are the
result of fluid instabilities – which we’ll discuss later in
the course.
ROSSBY WAVES
These waves exist only in particular geometries in a rotating system. For a simple version, consider a rotating layer of fluid, bounded by two planes which are not
quite parallel. As in Figure 5.2, let the height of the
fluid (parallel to the rotation axis ẑ) vary slightly. We
know from the Taylor-Proudman theorem that the fluid
flow must be independent of z. A “geostrophic” flow
28
in this geometry would have to follow contours of constant depth h. Consider, then, a small radial displacement of a column of the fluid, to a region with h + δh.
This violates the T-P theorem; the column will try to
return to its initial position. But it has some inertia, so
it will overshoot. We thus have an oscillation: this is
one version of a Rossby wave.
Rossby waves exist in the atmosphere, and in the
ocean; they are found to be low frequency ω < Ω
and long wavelength. The restoring force for terrestrial waves is due to variations in the depth of the atmosphere parallel to Ω (here meaning the earth’s rotation).
In the atmosphere they are easily observed as the largescale meanders of the mid-latitude jet stream. In the
ocean they are harder to find, being very small amplitude (several cm) and very long wavelength (hundreds
of km), but now detected in satellite data.
The boundary conditions are: v = 0 at z = 0 (no-slip
at the solid surface), and v = U x̂ as z → ∞. This
system solves to
h
i
vx = U 1 − e−z/δ cos(z/δ)
(5.4)
vy = U e−z/δ sin(z/δ)
with δ 2 = ν/Ω. Thus, the velocity vector rotates
through a spiral, while it increases from zero at the
> δ.
ground to ∼ U x̂ at z ∼
• Free boundary. Our driver here is the wind stress
at the ocean surface, again in the x direction; call it τ x̂.
We can ignore pressure gradients in this case; thus our
equations are, simply,
d 2 vx
dz 2
d 2 vy
2Ωvx = ν 2
dz
−2Ωvy = ν
Ω
α
h
Figure 5.2. The geometry for a simple version of Rossby
waves. The curved arrow marks the deviation of the upper
plate from parallel. Following Tritton Figure 16.13.
THE EKMAN LAYER
Coriolis forces also do interesting things at boundaries.
The two applications relevant to geophysical flows are
solid boundaries (such as land) and free boundaries
(such as the ocean). In each case, the combination of
a driving wind (think of geostrophic flow) and viscosity within the boundary layer results in a transverse flow
within that layer. In this section I present an overview;
details will appear in the homework.
• Solid boundary. At some high altitude, let the
wind velocity be U x̂. Thus, (5.2) reduces at high altitudes to
1 dp
2ΩU = −
ρ dy
(note dp/dx = 0, right?). Within the boundary layer
we must keep the viscous terms. Here, (5.1) becomes
d 2 vx
dz 2
d 2 vy
1 dp
2Ωvx = ν 2 −
dz
ρ dy
−2Ωvy = ν
(5.3)
(5.5)
Take z = 0 at the ocean surface, and negative below.
Our boundary conditions now are v → 0, z → −∞,
and dvy /dz = 0, dvx /dz = τ /ρν, z = 0 (refer back to
the definitions of stress in chapter 4). These solve to
³ z π´
vx = Vo ez/δ cos − +
δ
4
(5.6)
³ z π´
z/δ
vy = −Vo e sin − +
δ
4
√
where Vo = τ /ρ 2Ων, and δ is the same as above.
Again, the velocity vector rotates through a spiral; note
that it is at a 45◦ angle to ~τ at the ocean surface.
Both cases have similar consequences. A prevailing wind or flow results in a transverse flow within the
Ekman boundary layer (this is Ekman transport). Mass
conservation then forces a net vertical motion (why?)
into or out of the layer (this is Ekman pumping). In
the atmosphere this is related to updrafts in low pressure regions (thus storm formation); in the ocean this
is related to phenomena such as upwelling of cold subsurface water in regions with a steady prevailing wind
direction.
B. Viscous Flow: Time-Dependent Problems
Viscosity is dissipative: we expect a viscous flow to
decay with time. In this section we consider solutions
of the basic equations, (4.5) or (4.6), but here omitting
gravity and the advective terms. Thus we want solutions of
ρ
∂v
+ ∇p = ρν∇2 v
∂t
(5.7)
29
and of
ω
∂ω
= ν∇2ω
∂t
(5.8)
∇2
The
term makes these “diffusive” DE’s, and we can
use standard methods for their solution. Let’s do this
by example.
1.
SIMILARITY METHODS IN A DIFFUSION
EQUATION
Here’s one good approach to solving diffusion equations. Take a Cartesian system, ignore the pressure gradient, and take v = v(y, t)x̂ as a simple geometry. The
basic equation is then
∂v
∂2v
=ν 2
∂t
∂y
(5.9)
To set up a similarity transform, we note that there is
no physical scale applied to this problem (no edges of
finite length, no rocks
√ in the flow). We also notice that
the quantity η = y/ νt is dimensionless. We therefore
guess that the (normalized) solution, v(y, t)/U can be
expressed as a function of η alone, call it f (η). Now do
some chain rules (you should check this!)
∂2v
U
= η 2 f ′′
2
∂y
4
(5.10)
(here, f ′ = df /dη, and etc). Putting these into (5.9),
our DE becomes simpler, and can be integrated by simpler methods:
Z
′2
′′
′
f + 2ηf = 0 ; f (η) = A e−η dη ′ + B (5.11)
Uη ′
∂v
=−
f ;
∂t
2 t
∂v
U
= ηf ′ ;
∂y
2
and the integration constants A, B (and/or limits of the
integral) are chosen for the boundary conditions of the
problem.
2.
SMOOTHING OUT A VELOCITY JUMP
Consider planar flow with a sharp jump in the velocity.
At t = 0, take
vx = +U ,
y>0
= −U ,
y<0
Viscosity will clearly try to smooth out this jump. We
again have (5.9) as the basic equation; and (from the
last section), we can find its solution, with reasonable
boundary conditions vx → ±U, |y| → ∞):
Z y
U
2
v(y, t) =
e−u /4νt du
1/2
(πνt)
0
¶
µ
y
= U erf √
4νt
(5.12)
The last expression simply notes that this defines the
error function, erf(x). Thus, the |y| → 0 region obeys
vx ∝ y, and the outer has
√ vx → ±U , as required; the
transition occurs at y ∼ 4νt. Thus, viscosity spreads
the transition region out with time.
3.
FLOW ABOVE AN OSCILLATING PLATE
Another variant is an infinite, flat plate which moves
back and forth parallel to itself. The basic equation
for fluid above the plate is again (5.9), and the boundary conditions are v(0, t) = U cos ωt, and that v stays
bounded as y → ∞. To solve this, we try a separable
solution, v(y, t) = eiωt f (y). Note, we are looking for
the “steady” solution, established after initial transients
have gone away. Plugging this test solution in and doing algebra, we find
r ¶
µ
√
ω
−y ω/2ν
cos ωt − y
(5.13)
v(y, t) = U e
2ν
The cosine term here represents a signal propagating
away in the y direction, while the exponential is a decay
in the y direction. Thus, the flow resembles
p a damped
transverse “wave”, with wavelength 2π 2ν/ω in the
y direction, propagating
√ in the y direction, with ampli−y ω/2ν
.
tude falling off as e
4.
IRROTATIONAL VORTEX DECAY
A third application is the time-dependent evolution of
the irrotational vortex. Switching to cylindrical coordinates, we have
¸
·
∂ 1 ∂
∂v
=ν
(rv)
(5.14)
∂t
∂r r ∂r
The interesting solution of this equation, for a system
which initially obeys v = K/2πr everywhere, is
´
K ³
2
1 − e−r /4νt
(5.15)
v(r, t) =
2πr
Thus: we find v ∝ r for r → 0 (that is the core of
the vortex); v ∝ 1/r for r → ∞ (the outer part of
the vortex); and the transition
√ occurs at a radius which
grows with time, as r ∝ 4νt. Thus, viscous dissipation causes the core of the vortex to spread. Some
details of this may appear in the homework.
30
C. Creeping Flow
Now, consider very viscous flow; flow at very low
Reynolds number. This is called creeping flow.2 In
creeping flow we assume the inertial term v ·∇v can be
dropped compared to the viscous and pressure gradient
terms: this means the flow speed is very slow. We continue to assume steady state and incompressible flow.
The basic equation is, then,
∇p = −ρν∇ × ω = ρν∇2 v
(5.16)
Taking the curl of this, recalling ω = ∇ × v is the
vorticity, we have
∇2 ω = 0
(5.17)
Now, we consider flow around a sphere of radius a,
placed in a stream flow with asymptotic velocity U . We
want to find the drag force between the sphere and the
flow. The solution is long but straightforward ... here
goes.
First, we need to find the velocity field. Pick polar
coordinates so that θ = 0 corresponds to the direction
of downstream flow. The only nonzero component of ω
is then
·
¸
1 ∂(rvθ ) ∂vr
−
(5.18)
ωφ =
r
∂r
∂θ
But this is axisymmetric flow, so we can use a stream
function (as in Chapter 2); in terms of ψ, the vorticity
is
·
µ
¶¸
1
1 ∂
1 ∂2ψ
1 ∂ψ
ωφ =
+ 2
(5.19)
r sin θ ∂r2
r ∂θ sin θ ∂θ
Putting this into (5.17), we get a fourth order equation
for ψ:
·
∂2
sin θ ∂
+ 2
2
∂r
r ∂θ
µ
1 ∂
sin θ ∂θ
¶¸2
ψ=0
(5.20)
This is now something that can be solved. Our boundary conditions are
ψ(a, θ) = 0
dψ/dr(a, θ) = 0
(5.21)
The first is vr = 0 at the surface (recall vr ∝ dψ/dθ,
and we want vr = 0 for all θ); the second is vθ = 0
at the surface; and the third is uniform flow at infinity.
Choosing a separable solution, ψ(r, θ) = f (r) sin2 θ
(to match the distant boundary), we find the solution is
¶
µ
a3
1 3a
2
2
(5.22)
−
+
ψ(r, θ) = U r sin θ
2 4r 4r3
From the stream function, we can find the velocity components:
µ
¶
3a
a3
vr = U cos θ 1 −
+
2r 2r3
(5.23)
µ
¶
3a
a3
vθ = −U sin θ 1 −
−
4r 2r3
We’ll need these in the next step.
Next, we want to find the drag force on the sphere
– which is the integral of the stress over the surface of
the sphere. Separating out Σij as the non-pressure component of the stress tensor, the component of the drag
force in the U direction3 is
FD = [−p cos θ + Σrr cos θ − Σrθ sin θ]r=a (5.24)
For the pressure, we go back to (5.16) and solve for p
(taking zero pressure at infinity, or solving for the overpressure):
p(r, θ) = −
(5.26)
Putting this into the expression for FD , multiplying by
4πa2 and integrating over θ, gives the drag
FD = 6πρνaU
3
For once the name makes sense!
(5.25)
(Note that we couldn’t use Bernoulli’s law here; why?)
Thus, the pressure is highest at the forward stagnation
point, and lowest (negative in fact) at the rear stagnation point. For the rest of the stress tensor, we use the
velocity solution, from (5.23), to get
µ
¶
3a
3a3
∂vr
= 2νU cos θ
− 4
Σrr = 2ν
∂r
2r2
2r
·
¸
³
´
∂ vθ
1 ∂vr
3νU a3
Σrθ = ν r
+
=−
sin θ
∂r r
r ∂θ
2r4
1
ψ(∞, θ) = U r2 sin2 θ
2
2
3aρνU cos θ
2r2
(5.27)
Compare our earlier discussion in Chapter 3, in and around
(3.24).
31
Thus, we get a finite drag due to the viscosity. This is
Stokes’ law.
References
I’ve mostly followed Kundu, who does the more
mathematical examples. He also has a good chapter
on geophysical fluid dynamics. Tritton is also good for
geostrophic flows – as always his discussions are helpful (not just pure math).
32
6. BASICS OF COMPRESSIBLE FLOW
Up to now we have assumed incompressible flows. We
now extend to the more general case, when the density is allowed to vary. This is necessary to work with
(i) wave propagation; (ii) hydrostatic atmospheres, over
scales large compared with the scale height; (iii) flows
in which the fluid speed is comparable to (or larger
than) the sound speed.
1. CONSTANT GRAVITY: THE EXPONENTIAL
ATMOSPHERE
One example is hydrostatic balance in a fixed (external)
gravitational field g. The most obvious application of
this is the earth’s atmosphere. The basic equation is:
∇p = ρg
Or in a vertically stratified medium, such as earth’s atmosphere, this is
A. Some useful thermodynamic quantities
Since density changes are accompanied by temperature
changes, we need to remember a bit of thermodynamics. I store some terms and definitions here, in no particular order (some of the definitions are circular..). When
specific expressions are needed, I work with an ideal
gas.
• Equation of state. p = nkB T = ρRT ; kB is the
Boltzmann constant, a fundamental constant. R is the
gas constant, which varies by composition. It’s related
to the “fundamental” gas constant R by R = R/M , if
M is the molecular/atomic weight of the gas in question. Or if you prefer microphysics, R is related to the
Boltzmann constant by R = kB /m, if m is the mass
per particle.
• Internal energy and enthalpy. Let e be the internal energy (per mass): we define the enthalpy as
h = e + p/ρ. For a simple ideal gas we have p = nkT ;
e = (3/2)kT /m (for a monatomic ideal gas, if m is the
mass per particle – add kT /m per accessible degree of
freedom for other ideal gases); also ρe = p/(γ − 1);
and ρh = γp/(γ − 1).
• Specific heats and adiabatic index. The SH’s
are defined by de/dT = cV (constant volume) and
dh/dT = cP (constant pressure). The adiabatic index can be defined as γ = cP /cV , the ratio of specific
heats. We also have cp − cV = R; cv = R/(γ − 1);
cp = γR/(γ − 1).
• Degrees of freedom and adiabatic index. Let f be
the number of d’s of f, so that the mean KE per particle
is (f /2)kB T . This connects to the adiabatic index by
γ = (f + 2)/f . Typical values for γ are γ = 1.4 for the
atmosphere; γ = 5/3 for a monatomic ideal gas; and
γ = 4/3 for a relativistic gas (such as a photon gas).
B. Hydrostatics: gaseous atmospheres
The effects of compressibility must be considered in
most situations involving gases (or plasmas) in static
equilibrium in a gravitational field.
(6.1)
dp
= −ρg
dz
(6.2)
and g is constant over any relevant scale in this application. We can directly define a scale height from this. If
T is also constant with height, the pressure solution is
p(z) = po e−z/H
(6.3)
where H = RT /g = kT /mg is the scale height.
Another example of this, perhaps less obvious,
is the interstellar medium in our galaxy. For nonastronomers: our galaxy is a flat disk, composed of stars
and gas rotating about a common center. The thickness of the disk is controlled by the local mass density
(which produces a constant gravity, g), and the temperature (plus random motions) of the gas or stars in the
disk. The vertical structure of the disk is controlled by
the same equation, (6.3) ... just on a much larger scale.
2. VARIABLE
SPHERE
GRAVITY:
THE
ISOTHERMAL
A different HSEq system is a self-gravitating sphere.
This might be a naive picture of a star: held together by
its own gravity, and supported against gravitational collapse by its internal energy. The basic HSEq equation,
(6.1) still holds; in spherical geometry it is
GM (r)
dp
= −ρ
dr
r2
(6.4)
where M (r) is the mass inside r. Alternatively, this can
be written
dp
= ρ∇Φg
dr
(6.5)
where Φg is the gravitational potential, defined through
g = −∇Φg . The mass, density and Φg are connected
via
Z r
4πr2 ρdr ;
M (r) =
0
µ
¶
(6.6)
d
1
2 dΦ
2
r
= 4πGρ
∇ Φg = 2
r dr
dr
33
Putting this into (6.5) and differentiating with r, the basic equation becomes
µ
¶
d r2 dρ
G
4πr2 ρ
(6.7)
=−
dr ρ dr
RT
The solutions of this are less straightforward. First, we
note that a basic scale length appears:
¶
µ
9RT 1/2
(6.8)
ao =
4πGρo
if ρo is some characteristic density, say that at the origin. (the numerical constants appear to simplify things
later on.) We expect physical solutions of (6.7) to involve lengths scaled to ao .
Consider an isothermal gas; these are the simplest
solutions. One solution of (6.7) is the simple power
law, ρ ∝ 1/r2 . The divergence as r → 0 keeps this
from being an interesting solution, however. More interesting physical solutions must be found numerically,
starting with ρ(r = 0) = ρo and working out. These
solutions do, indeed, have a turnover at r ≃ ao ; at large
radii they do approach ρ ∝ 1/r2 . (These solutions also
have problems, for the total mass M (r) diverges ∝ ln r
as r → ∞. Physically, the problem is that the gas cannot be maintained isothermal everywhere.)
Numerical solutions of this equation are called polytropes, or Lane-Emden solutions, for polytropic index
n = 1/(γ − 1). These are physically well-behaved, in
that the density is finite at the center and falls faster than
1/r2 at large r.
3.
REALITY: NONISOTHERMAL ATMOSPHERES
Just a note of caution here. Both of the preceding examples assumed a constant temperature throughout the
atmosphere. This naive assumption allowed us to find
(nearly) analytic solutions. More realistically, however,
we expect the temperature to be a function of position,
controlled by the thermodynamics of the system. For
earth’s atmosphere, T (z) is determined by the combined actions of solar radiation coming in at the top,
all the chemistry and energy transfer effects within the
atmosphere, and radiative losses back into space. For a
star, T (r) is determined by nuclear energy generation
in the star’s core, radiative energy transfer within the
star, and eventual radiative losses into space.
4.
ADIABATIC ATMOSPHERE
Another analytic model of an atmosphere assumes the
gas is adiabatic: that is, as it compresses or expands
(while satisfying HSEq), it heats or cools accordingly.
This is useful in analyzing the stability of an atmosphere to convection (which we’ll do immediately below).
To develop this idea, go back to basic HSEq, (6.1):
dp/dz = −ρg. But now, assume the gas is adiabatic,
so that
µ ¶(γ−1)/γ
µ ¶1/γ
T
p
p
ρ
;
(6.10)
=
=
To
po
ρo
po
We can use these to relate dT /dz to dp/dz for an adiabatic atmosphere:
Figure 6.1.
Density (and projected density, Σ) solution
for a self-gravitating isothermal sphere. The dotted line is
the asymptotic ρ ∝ 1/r2 power-law solution. From Binney
& Tremaine figure 6.7
Self-gravitating spheres can also be modelled as adiabatic gases. Take the equation of state as p = Kργ .
The hydrostatic equation, (6.5), and the differential
form (6.7), become
Kγργ−2
dΦ
dρ
= −ρ
;
dr
µdr
¶
1 d
2 dΦ
r
= 4πGρ
r2 dr
dr
(6.9)
1 dT
γ − 1 1 dp
=
;
T dz
γ p dz
1 dρ
1 1 dp
=
ρ dz
γ p dz
(6.11)
From here, assuming p = ρRT (ideal gas) and using
cp − cv = R (refer back to §6.1) we get the condition
for an adiabatic atmosphere:
g
(γ − 1) mg
dTad
=− =−
dz
cp
γ
kB
(6.12)
where cp is the isobaric specific heat, and m is the mass
per particle. This clearly gives a linear temperature
drop with altitude z; g/cp ≃ 10◦ C/km for typical atmospheric conditions. The vertical temperature gradient is
often caled Γ = dT /dz; and is sometimes called the
“lapse rate”.
34
C. Convective Stability
So ... we now have a description of hydrostatic atmospheres. But, are they stable? One of the most important instabilities, for the earth’s atmosphere or for a star,
is convection. We need to understand whether the atmosphere is stable, or unstable, to convection – that is,
whether or not convection will develop spontaneously.
To picture the situation, here’s how Shore puts it:1
Picture a duck sitting calmly on a pond. . . If we
say that the bird is bouyant, we mean that if we
depress him a bit by pushing from above, he will
bob back to the surface and, ignoring his agitation,
bounce up and down for awhile [this is called neutral bouyancy.]. . .If we have one, on the other
hand, who is not well preened and therefore not
waterproof, and [we] push down on him, he may
sink. Now think of a blob which is hotter than
its surroundings. It will begin to rise, since we
already know that its density will be lower than
that of the [surrounding] medium, and it will thus
be bouyant. If it remains underdense, it will continue to rise – we call this an instability. It will
continue to rise until it reaches a level at which it
is neutrally bouyant again. On the other hand, if
the blob is pushed down, and if it remains overdense, it will sink until it reaches a point at which
the density again allows for stable balance.
We would like to find a condition to tell if the situation is stable or unstable. To get there, think about
some blob again: assume it starts at some vertical position z, with density and pressure in balance with its
surroundings (i.e, the “outside”, which usually refers to
the atmosphere). Thus: it starts at ρin = ρout = ρ1 and
Tin = Tout = T1 . Now, raise it some distance ∆z; let
the conditions in the blob at z + ∆z be labelled by “∗”.
For the blob, we assume it evolves adiabatically. Thus,
it reaches a new density and temperature,
µ ¶
dT
= T1 +
∆z
dz ad
(6.13)
If we want to evaluate the z-derivatives in (6.13), we
need to know something about the specific situation.
A common assumption is that the blob rises slowly
enough to remain in pressure balance with its surroundρ∗in
1
µ ¶
dρ
∆z ;
= ρ1 +
dz ad
∗
Tin
An Introduction to Astrophysical Hydrodynamics (Academic
Press) 1992, ch. 9.
ings. If this is the case, we will have
µ ∗ ¶1/γ
pin
∗
∗
∗
;
pin = pout ; ρin =ρin
pin
µ ∗ ¶1/(γ−1) (6.14)
pin
∗
Tin = Tin
pin
*
in
no (z+dz), T o(z+dz)
n
g
dz
nin
n (z), T (z)
o
o
Figure 6.2. A cartoon illustrating the buoyant/convective
instability. Imagine a blob starts in balance with its
surroundings. It is then displaced vertically, some dz;
during this rise it (i) remains in pressure balance with its
surroundings, and (ii) remains adiabatic. How does its
density at z + dz compare to the density outside? The
answer to this question determines the stability of the
atmosphere.
The surroundings, however, are not necessarily adiabatic: they have some other dT /dz and dρ/dz values (specified by the situation – for instance the heating/cooling balance for the outer layers of a star, or the
earth’s atmosphere). So, we can determine stability or
instability by asking whether, when the blob has risen
this ∆z, it is at a higher or lower density than the surrounding atmosphere. In the first case it will sink again
(and thus the atmosphere is stable); in the second case it
will keep rising (and thus the atmosphere is unstable).
Thus, our condition for instability becomes condition
on the external density gradient. If we assume the blob
remains in pressure balance with its surroundings, we
also have a condition on the atmospheric temperature
gradient. Thus, the atmosphere is buoyantly unstable if
µ ¶
µ ¶
µ ¶
µ ¶
dρ
dT
dT
dρ
<
;
>
dz ad
dz atm
dz ad
dz atm
(6.15)
Because we usually consider situations with dρ/dz <
0, for instance a hydrostatic atmosphere, the condition
for instabilty is often written in terms of absolute values:
¯ ¯
¯ ¯
¯ ¯
¯ ¯
¯ ¯
¯ dT ¯
¯ dT ¯
¯ dρ ¯
¯
¯
¯ ¯ > ¯ dρ ¯
; ¯ ¯ < ¯¯ ¯¯
(6.16)
¯ dz ¯
¯ dz ¯
dz ad
dz atm
ad
atm
35
Thus: if the outside (atmospheric) temperature changes
too rapidly with altitude, the atmosphere is convectively
unstable. An underdense blob will continue to rise, and
an overdense blob will sink.
1.
ADIABATIC ATMOSPHERE
Referring to our discussion of adiabatic atmospheres,
above, we recall that the real atmosphere needs to be
compared to the ideal, adiabatic case in order to determine convective (in)stability. This gives a condition
for instability, written in terms of either the pressure or
temperature gradients:
¶
µ
1 dρ
1 T dp
dT
1 1 dp
>
;
1−
>
(6.17)
γ p dz
ρ dz
γ p dz
dz
Note that both p and T drop with z in any hydrostatic
situation; so both sides of these inequalities involve
negative quantities.
2.
while the ambient density has changed as
ρ(z + dz) = ρ(z) +
POTENTIAL TEMPERATURE
The adiabatic gradient can be important in the convective stability of the atmosphere. Another useful quantity here is the potential temperature. Consider a parcel of atmosphere, that starts in local balance at some
(p, T ). Take it adiabatically to some other point, with
local pressure ps (typically sea level pressure). The
temperature this parcel reaches at point s is called its
potential temperature – called θ – which is given by
µ ¶(γ−1)/γ
p
d
T dθ
T =θ
=
(Tad − T )
;
ps
θ dz
dz
(6.18)
But this is useful, because the gradient of θ depends
on the difference between the actual and adiabatic gradients in the atmosphere; thus θ is a handy tool when
one’s thinking about convection. Specifically, from the
results above, we see that the atmosphere is convectively stable if dθ/dz > 0, and unstable if dθ/dz < 0.
3.
Figure 6.3.
Vertical variation of the (a) actual and (b)
potential temperature in a typical terrestrial atmosphere.
The straight lines are the adiabatic Tad (z) structure. From
Kundu figure 1.9.
BRUNT- V ÄIS ÄL Ä FREQUENCY
Now, let’s return to the blob (or the duck). If the atmosphere is convectively stable, the blob (or duck) will
simply bob up and down. Its easy to show this is simply harmonic motion, and to find its characteristic frequency.
To proceed, picture the earth’s atmosphere. Move
the blob vertically by some ∆z, so that it reaches a new
density ρ∗in . The blob’s new density, still assuming an
adiabatic displacement, will be
ρ∗in
1 ρ dp
= ρ1 + ∆ρ∗ = ρ(z) +
dz
γ p dz
(6.19)
dρ
dz
dz
(6.20)
(Note, I’ve now dropped the subsript “out”). Thus, the
bouyant force (per unit volume of the blob/duck) is
µ ¶¸
·µ ¶
dρ
dρ
−
∆z
(6.21)
Fbuoy = g
dz ad
dz
Alternatively, using the definition of potential temperature, this becomes
Fbuoy =
g dθ
∆z
θ dz
(6.22)
So: the buoyant force ∝ ∆z – this is clearly simple harmonic motion. The equation of motion (still per mass)
can be written
1
d2 ∆z
= Fbuoy = N 2 ∆z
2
dt
ρ
(6.23)
and this last implicitly defines the Brunt-Väisälä frequency, N , frequency,2
µ ¶¸
·µ ¶
dρ
dρ
g
2
−
(6.24)
N =
ρ
dz ad
dz
If the right hand side is positive (which is the same thing
as saying the atmosphere is convectively stable – refer
back to 6.15) , the blob simply oscillates at the B-V
frequency. If the RHS is negative (if the atmosphere is
unstable), the displacement ∆z grows exponentially.
2
Caveat to the student: very few authors leave N 2 in this form
– they express it in terms of θ, dT /dz, specific heats, and what
have you.
36
D. Energetics of Compressible Flow
find
We also need an equation for energy conservation. We
consider two forms of energy: kinetic energy density,
v 2 /2, and internal energy density, e, both defined per
unit mass.
R
The net energy in our volume V is V ρ(e+ 12 v 2 )dV .
The net rate of change of this energy, from intrinsic
changes and from flows is
¸
·
Z
∂
1 2
ρe + ρv dV +
2
V ∂t
· µ
¶¸
(6.25)
Z
1 2
∇ · v ρe + ρv
dV
+
2
V
This net energy change must be accounted for by (a)
work done by the external force, f ; (b) work done by
the external pressure; (c) direct energy gains or losses,3
which we collect as H.4 These three energy-change
factors are
Z
Z
Z
ρf · vdV −
pn̂ · vdA +
HdV
(6.26)
V
A
V
We can use Gauss’ law to express the pressure work
term as a volume integral, and can derive one version
of the differential energy conservation law:
¸
· µ
¶¸
·
1
1
∂
ρe + ρv 2 +∇ · v ρe + p + ρv 2
∂t
2
2
= ρf · v + H
(6.27)
Note that the enthalpy h = e + p/ρ appears naturally in
the second term on the LHS.
At this point, we need to look at alternative forms
of (6.27). The forms we derived for mass conservation (1.4) and momentum conservation (1.10 or 2.2)
are pretty standard. However, there does not seem to
be one standard form for the energy conservation equation; rather, one uses the form that works best in a given
application. Therefore, at the expense of a little algebra,
we will look at several alternate forms of (6.27).
First, we simply separate out the ∂ρ/∂t and ∇·(ρv)
terms in (6.27), using the continuity equation (1.4), and
3
4
Examples of direct heating include resistive dissipation (of a current), or such things as cosmic ray heating (relevant in astrophysics); direct losses are most commonly by radiation
We will treat viscous dissipation below. We could also include
thermal conductivity, which provides a separate means of energy
transport out of or into the volume; it brings in second derivatives, and we will not need it here.
ρ
∂
∂t
µ
¶
µ
¶
1
1
e + v 2 +ρv · ∇ e + v 2
2
2
(6.28)
= ρg · v − ∇ · (pv) + H
This one alternate form that we will use again. We can
isolate the rate of change of e, by subtracting v·(the
momentum conservation equation) from (6.28), giving
ρ
∂e
+ ρv · ∇e = −p∇ · v + H
∂t
(6.29)
In this expression, we can see that the rate of change of
the internal energy depends explicitly on compression
work (“pdV ” work), and on the net heating and cooling
rates.
Yet another common form of the energy equation
uses the Lagrangian derivative. Writing ∇ · v in terms
of the density derivatives, and collecting the p and ρ
derivatives separately, we get
µ ¶
D
p
= (γ − 1)H
(6.30)
Dt ργ
which is the last of our alternate forms of the energy
equation. This last form allows us to consider a couple
of important limits.
• The first is the adiabatic limit. If H = 0, so that
there is no net gain or loss of energy to the system,
(6.30) shows that
p
= constant
ργ
(6.31)
which is the usual adiabatic law (the consequence
of there being no gain or loss of heat from a system).5
•The second limit is the isothermal limit. A good
many astrophysical calculations assume T = constant, which simplifies things enormously. From
(6.22), we note that
p∇ · v = H
(6.32)
is the condition that must be satisfied if T (or e) is
constant.
It might be comforting to prove that we can extract
Bernoulli’s relationship from this formalism, in addition to our earlier derivation from the force (Euler’s)
equation. We assume, again, H = 0. We use the expression for De/Dt, and expand out the ∇ · (pv) term,
using the continuity equation. We also assume the system is in a steady state, so that ∂/∂t = 0; thus, D/Dt
37
measures the rate of change of a quantity, due to its motion through a region in which the flow field changes.
We get,
µ
µ ¶
¶
1 2
D p
D
e+ v +ρ
= ρg
(6.33)
ρ
Dt
2
Dt ρ
cluding viscosity, equation (2.2); multiply it by vi and
interchange dummy indices, to get an expression for the
time-change of kinetic energy:
Now, using g = ∇Φ, Bernoulli’s relationship becomes
Subtract this from the full energy equation (also written
in Cartesian), and symmetrize the tensor term (in the
second step), to get two forms of the energy equation
including viscosity:
1
p
1
e + v 2 + + Φ = h + v 2 + Φ = constant (6.34)
2
ρ
2
which does, indeed, recover the Bernoulli relation that
we derived above from momentum conservation. We
recall, again, that this law holds along any one streamline in the flow.
References
I mostly follow Thompson for the basic energy
conservation and dissipation analysis. The isothermal
sphere discussion follows Binney & Tremaine (Galactic Dynamics); the terrestrial atmosphere discussion
leans on Kundu and Tritton.
ρ
D v2
= σik,i uk + ρgk vk
Dt 2
De
dvk
= −p
+Σik Dik
Dt
dxk
(6.36)
This brings back the (symmetrized) deformation tensor
in Cartesian:
¶
µ
∂vj
1 ∂vi
(6.37)
+
Dij =
2 ∂xj
∂xi
ρ
De
= σik vk,i = σik Dik ;
Dt
ρ
and introduces what I’ll call the dissipation function:
D = Σij Dij . Referring back to (2.37) and (2.38),
it’s normal to assume νb = 2ν/3, which simplifies the
stress and dissipation tensors. In this limit, still Cartesian, the energy equation becomes
∂e
∂e
∂vk
ρ + ρvi
+p
= ρν
∂t
∂xi
∂xk
E. Appendix: Viscous Dissipation
Well, that was so much fun, let’s do it again: in Cartesian to be explicit. Start with our force equation, in-
(6.35)
µ
∂vk
∂xk
¶2
(6.38)
To close, I write out explicitly the dissipation functions for all 3 coordinate systems.
Cartesian:
D=
where, as above,
¤
2ρν £
(D11 − D22 )2 + (D22 − D33 )2 + (D33 − D11 )2
3
¢
¡ 2
2
2
+ ρνb (D11 + D22 + D33 )2
+ 4ρν D12
+ D13
+ D23
1
Dij =
2
µ
∂vj
∂vi
+
∂xj
∂xi
¶
(6.39)
(6.40)
Cylindrical:
where
¢
¡ 2
2
2
2
2
2
+ ρ(νb − 2ν/3)(∇ · v)2
D = 2ρν Drr
+ Dθθ
+ Dzz
+ 2Drθ
+ 2Dθz
+ 2Dzr
Drr
Dzz
∂vr
;
=
∂r
∂vz
;
=
∂z
Dθθ
Dθz
µ
¶
1 ∂vθ
vr
1 1 ∂vr
∂ vθ
=
+
; Drθ =
+r
r ∂θ
r
2 r ∂θ
∂r r
µ
¶
µ
¶
1 ∂vθ
1 ∂vz
∂vz
1 ∂vr
=
+
+
; Drz =
2 dz
r ∂θ
2 ∂z
∂r
(6.41)
(6.42)
38
Spherical is even better (recall that θ is the polar angle and φ is the azimuthal angle):
¢
¡ 2
2
2
2
2
2
+ ρ(νb − 2ν/3)(∇ · v)2
D = 2ρν Drr
+ Dθθ
+ Dφφ
+ 2Drθ
+ 2Dθφ
+ 2Dφr
(6.43)
where
1 ∂vφ vr
vθ cot θ
1 ∂vθ
vr
∂vr
; Dφφ =
+
+
; Dθθ =
+
∂r
r sin θ ∂φ
r
r
r ∂θ
r
µ
¶
µ
¶
1 ∂vr
∂ vφ
1 1 ∂vr
∂ vθ
1
+r
; Drθ =
+r
=
2 r sin θ ∂φ
∂r r
2 r ∂θ
∂r r
µ
¶
1 sin θ ∂ vφ
1 ∂vθ
=
+
2
r ∂θ sin θ r sin θ ∂φ
Drr =
Drφ
Dθφ
and this is the end of this chapter.
(6.44)
39
7. SIGNAL PROPAGATION
In this chapter we introduce a very important basic tool:
the speed at which a disturbance propagates through
a fluid. Because the disturbance we’re considering is
compressive – think of a locally overpressured region –
one example is a sound wave. Thus the signal speed is
called the sound speed: we’ll see that it’s given by
c2s =
∂p
∂ρ
(7.1)
First we’ll derive this important speed, then look at how
causality can dramatically change the nature of supersonic flows.
Thus, δv > 0 if δρ > 0; the passage of a compression
wave leaves behind a fluid moving in the direction of
the wave. Now, apply momentum balance: the net force
on some control volume, from the pressure difference,
equals the rate of change of momentum in that volume.
That is,
p − (p + δp) = ρcs [(cs − δv) − cs ]
so that, again to first order in things small, we have
δp ≃ ρcs δv
Combining (7.2) and (7.3), we get a condition on the
wave speed (to allow mass and momentum balance):
A. Sound Waves and the Signal Speed
This is worth two derivations; one physical, and another
one more mathematical.
SOUND WAVES : A PHYSICAL APPROACH
We can also demonstrate that cs is the speed of a travelling wave. Let some perturbation (δρ, δp, δT ) be moving at some cs . Ahead of the wave the fluid has v = 0;
behind the wave the fluid has δv, in the same direction
as the wave motion.
(7.3)
c2s =
δp
δρ
(7.4)
Thus, we have one justification of our expression (7.1)
for cs .
SOUND WAVES : A FORMAL APPROACH
We can also use a more mathematical approach, deriving a formal wave equation by linearizing our two basic
equations, continuity & momentum:
∂ρ
+ ∇ · (ρv) = 0
∂t
(7.5)
(assuming no body forces)
ρ
∂v
+ ρ(v · ∇)v = −∇p
∂t
(7.6)
Our question is, at what speed does a small, compressional disturbance in the gas travel? We start with a
uniform, static unperturbed state, described by ρo , po
and vo = 0. We add small perturbations, ρ1 , p1 and v1 .
If we put these into the mass and momentum equations,
(7.5) and (7.6), noting that the “o” terms are constant in
space and in time, we get
Figure 7.1. Propagation of a sound wave: (a) into a still
fluid; (b) a stationary wave. From Kundu figure 15.1. Note,
u here is the same as v in the text.
Mass conservation at the wave front, in a frame moving
with the wave front, gives
and to lowest order in things small, this gives
δρ
ρ
and
(ρo + ρ1 )
∂v1
+ (ρo + ρ1 )v1 · ∇v1 = −∇p1
∂t
In these equations, we now drop terms which are second order in the perturbed quantities, and we write
∇p = (∂p/∂ρ)∇ρ. This gives us
ρcs = (ρ + δρ)(cs − δv)
δv ≃ cs
∂ρ1
+ v1 · ∇ρo + (ρo + ρ1 )∇ · v = 0
∂t
(7.2)
∂ρ1
+ ρo ∇ · v 1 = 0
∂t
(7.7)
40
B. Why is the sound speed important?
and
ρ0
∂v1
+
∂t
µ
∂p
∂ρ
¶
∇ρ1 = 0
(7.8)
Two approaches are possible here.
• Method # 1. This is more physically intuitive.
From this pair of equations we can, for instance, eliminate v1 , to get a second order DE in ρ1 :
µ ¶
∂ 2 ρ1
∂p
=
∇2 ρ1
(7.9)
2
∂t
∂ρ
This, of course, has travelling wave solutions: the disturbances travel at a speed given by
µ ¶1/2
∂p
(7.10)
cs =
∂ρ
This recovers our guess in (7.1).
• Method # 2. This is more formal, and a simple example of a general method which is the standard
approach when the situation is more complex. (This
will be useful in the homework for instance). Refer
back to equations (7.7). They are linear in the perturbations, which allows us to use Fourier techniques.
That is, we can consider one simple perturbation, say
ρ1 , v1 ∝ ei(k·x−ωt) ; and we know that any arbitrary
perturbation can be expressed as a sum of these (ω, k)
waves.
With this, taking (7.1) as a definition (only that, for
the moment), and dropping down to a 1D perturbation
to simplify, equations (7.7) and (7.8) become
−iωρ1 + ρo ikv1 = 0
1. WHEN CAN WE ASSUME INCOMPRESSIBLE
FLOW ?
In the first five chapters, we assumed incompressible
flow, and noted that this is a good approximation when
the flow speed is much less than the sound speed. To
start, we justify this assumption. We begin with the
continuity equation:
∇ · (ρv) = 0
From these we immediately find the dispersion relation:
(7.12)
Thus, we have verified that we have non-dispersive
waves propagating at speed cs . This again verifies our
guess in (7.1).
The derivative in (7.1) is usually taken assuming the
disturbance is adiabatic, so that p/ργ is constant, and
c2s = γp/ρ. This corresponds, physically, to the heating and cooling times for the perturbation to be long
compared to the wave travel time. An alternative choice
which is sometimes used, is to consider isothermal perturbations: T = constant – which is the limit in which
the local cooling and heating times are short compared
to the wave travel time. In this case, c2s = p/ρ.
(7.13)
This reduces to the incompressible condition, ∇·v = 0,
if (in 1D Cartesian for simplicity),
v
∂ρ
∂v
≪ρ
;
∂x
∂x
vδρ ≪ ρδv
(7.14)
Now, the momentum/Euler equation becomes,
vδv ≃
1
δp
ρ
(7.15)
If we now use the sound speed, from (7.1), we have
δp ≃ c2s δρ, and thus
v 2 δv
δρ
≃ 2
ρ
cs v
(7.16)
We thus find that the density changes are negligible –
and thus we can work in the incompressible limit –
when
v2
= M2 ≪ 1
c2s
(7.11)
−iωρo v1 + ikc2s ρ1 = 0
ω 2 = c2s k 2
I’m storing two important ideas in this section.
(7.17)
Here, we have introduced the Mach number, M =
v/cs .
2.
THE IMPORTANCE OF CAUSALITY
We have just demonstrated that the sound speed is the
speed at which information propagates; it is also critical
to the dynamics of the flow.
There are important difference between subsonic
and supersonic flows. Subsonic flows can be thought of
as quasi-hydrostatic. That is, the flow field is strongly
influenced by pressure gradients which are determined
by conditions a long distance away (such as at boundaries).
Supersonic flows, however, are quasi-ballistic. This
distinction is due to the fact that information travels at a
finite speed, the speed of a simple sound wave. Both 1D
41
and 3D cartoons, in Figures 7.1 and 7.2, can illustrate
this point. Pressure gradients have only a limited range
of influence, and conditions far away have little or no
effect on a solution locally. We’ll see that supersonic
flows can (and usually do) contain discontinuous jumps
in the flow properties (shocks). They can violate our
subsonic intuition, for instance a supersonic flow in a
diverging channel will accelerate (as we’ll see below).
perturbation
Undisturbed flow
cs
cs
reverse waves
forward waves
Figure 7.2.
Physical illustration of simple waves. The
information that the flow has been “whacked” at point a,
propagates by simple sound waves, moving at speed cs
relative to the fluid in the pipe. Following Thompson figure
8.6
ct
s
vt
vt
cst
M=0
M<1
M>1
Figure 7.3. Mach’s construction for the propagation of a
disturbance. Consider a point source of sound (Thompson
suggests a bumblebee) in a moving medium. If the source
and flow are stationary, the sound propagates spherically
from the source. If the source/flow are moving subsonically.
the motion only distorts the spherical wavefronts. If,
however, the motion is supersonic. all disturbances are
confined to a Mach cone; an observer located outside of this
cone does not receive any information about the bee. The
opening angle of the cone is called the Mach angle:
sin µ = 1/M.
Some authors use the following classifications,
based on the Mach number M = v/cs (I follow Kundu,
for instance):
< 0.3 everywhere; can ignore
• incompressible: M ∼
density variations due to pressure changes.
< M < 1 everywhere. Need to be
• subsonic: 0.3 ∼
∼
careful with density fluctuation (now above the 10%
level), but no shock waves in the flow.
< M < 1.2, say; M is “around
• transonic: 0.8 ∼
∼
unity”. Shock waves appear, increasing drag. These are
the hardest flows to analyze analytically, as the nonlinear terms in the governing equations are important, but
the simplifying effects of M > ∞ flow can’t be used
yet.
< M < 3, typically. Shock waves
• supersonic: 1 ∼
∼
are usually present. Analysis is easier because information propagates along well-defined directions in (x, t)
space, called characteristics.
> 3, say. Qualitatively the same as
• hypersonic: M ∼
supersonic flow, but some interesting new effects – such
as strong heating and ionization of boundary layers –
come into play. Shocks can be analyzed in the strongshock limit (chapter 9).
C. Weak Waves and Causality
How can we use the signal speed – the sound speed – to
understand a flow? One way to approach this, following
Currie, is to consider “weak waves”. Specialize to a
1D system, and let co be the undisturbed value of cs .
Equations (7.7) and (7.8) become
∂ρ1
∂v1
∂v1
∂ρ1
+ ρo
= 0 ; ρo
+ c2s
=0
∂t
∂x
∂t
∂x
In the previous derivation, cs can be a function of density, and thus can vary within the wave. Here, we simplify by taking cs to be a constant, co . But now: because ρo is assumed constant, and because vo = 0, we
can write
∂ρ
∂v
∂v
∂ρ
+ ρo
= 0 ; ρo
+ c2o
=0
(7.18)
∂t
∂x
∂t
∂x
Compare these to the originals, (7.7) and (7.8): we have
quietly removed the bothersome, nonlinear terms. This
is valid only as long as we continue to assume a small
perturbation – a “weak wave”. We can now divide the
first of (7.18) by ρo , the second by ρo co , and add and
subtract, to get
¶
¶
µ
µ
ρ
ρ
v
v
∂
∂
+
+
+ co
=0 ;
∂t co ρo
∂x co ρo
¶
¶
µ
µ
ρ
ρ
v
v
∂
∂
−
−
− co
=0
(7.19)
∂t co ρo
∂x co ρo
These now have the form of a total (Lagrangian) derivative (compare equation 1.5). Thus, we have an important result:
µ
¶
v
ρ
+
= constant on x − co t = constant
co ρ o
(7.20)
and
¶
µ
ρ
v
−
= constant on x + co t = constant
co ρ o
(7.21)
An alternate form of (7.20 and 7.21) can be found with
a bit of algebra. Using (7.7) and ρ1 ≪ ρo , it’s easy to
express ρ/ρo in terms of p/po and γ, to write
µ
¶
v
1 p
J+
=
+
= constant
c0
co γ p o
(7.22)
on x − co t = constant
42
and
J−
=
c0
µ
1 p
v
−
co γ p o
¶
= constant
(7.23)
on x + co t = constant
The interpretation is simple for (7.20) or (7.22). The
lines x = ±co t are the loci of forward and backward
propagating sound waves. Our results say that the quantities on the left of equations (7.21) and (7.23) are constant along these trajectories.
Note that the above derivation is only valid for weak
shocks, working in a frame where the fluid speed is
much less than the sound speed. However, these conditions may easily be relaxed. In a moving fluid, one finds
that the characteristics are (not surprisingly) subject to
a Galilean transformation by speed v. The invariants,
and associated characteristics, become
¶
µ
Z
∂ρ
+
= constant
J = v + cs
ρ
(7.24)
on x − (cs + v)t = constant
J
1 p
1 p1
v
+
=
co γ p o
γ po
v
1 p
1
−
=−
co γ p o
γ
and
These contain enough information to solve the system.
We find that
and
−
the low-pressure gas. The information that this has happened can only travel at the sound speed, co . Thus, a
compression wave will move to the right, and an expansion wave to the left, both at co . The problem: what
is the velocity and pressure of the gas everywhere, as a
function of time?
Figure 7.4 shows an (x, t) diagram for this system,
with the two wave paths as solid lines. Gas ahead of
the forward wave must be undisturbed: it has p =
po , v = 0. Similarly, gas behind the reverse wave has
p = p1 , v = 0. What of the middle region? Consider an observation point P (xp , tp ). Two wave lines
intersect P : one forward wave which originates from
(x < xp , t = 0), and a reverse wave which originates
from (x > xp , t = 0). But now: we know v = 0 everywhere at t = 0, so we can evaluate the constants in
(7.22, 7.23) for each of these wave lines. For the reverse
one, we have J − /c0 = −1/γ; and for the forward one,
we have J + /c0 = p1 /γpo (verify this for yourself!).
Thus, at point P we know that
µ
¶
Z
∂ρ
= v − cs
= constant
ρ
(7.25)
on x + (cs − v)t = constant
J±
In this form,
are known as the Riemann invariants,
and they are valid even for strong shocks and arbitrary
fluid speeds. Note that density changes may now be
large, so cs may be a function of position. Thus, the
characteristics may be curved. In addition, note that
for supersonic flow, both characteristics will be in the
same direction (compare figure 7.3 for M > ∞, on the
horizontal line). Details (and the derivation) will follow
in chapter 10.
D. Two examples of simple waves
Two standard examples show how this analysis can be
used. In each, the pressure changes are assumed to be
small, so that cs can be replaced by the constant c0 , and
we can use the weak form of the invariants.
v=
co p 1 − p o
;
2γ po
p=
p1 + po
2
(7.26)
Thus: the region inbetween the two waves has a uniform pressure, equal to the mean of p1 and po ; and it
moves to the right at a uniform speed, proportional to
the initial pressure difference.
diaphragm
p1
p0
t
x = -ct
x = ct
(3)
P
(3)
(2)
(1)
x
1.
SHOCK TUBE
Consider a gas confined to a 1D tube. A diaphragm at
x = 0 separates high-pressure (p1 , x < 0) gas from
low-pressure ( p0 , x > 0) gas. At t = 0 the diaphragm
breaks, and the high-pressure gas begins to expand into
Figure 7.4. Illustrating the geometry of the
ruptured-diaphragm problem, in the x,t plane. The two solid
lines are the forward and reverse waves which start at the
origin; the two dotted lines are the waves which intersect
the observation point P. Following Currie Figure 11.2.
43
PISTON PROBLEM
U
Consider another 1D tube filled with gas; at t = 0 a
piston is moved into the gas, at velocity U . How does
the gas respond?
In Figure 7.5 we show this problem in the (x, t)
plane. Once again, we expect the gas to be undisturbed
(pressure po , zero velocity) ahead of the forward-wave
which starts from the piston at t = 0. Also again, consider an observation point P (x, t). It connects the the
forward x-axis by a reverse wave; we can evaluate the
constant J − /c0 , because we know v = 0, p = po at
the x-axis. The point P also connects to the piston by a
forward wave. We cannot evaluate J + /c0 yet, however;
we know the velocity vp = U , but not the pressure, pp ,
at the piston. We thus need a third wave, another reverse one which connects the piston to the x-axis. We
thus have three algebraic equations:
1 pp
1
U
−
=−
co γ p o
γ
1 p
U
1 pp
v
+
=
+
co γ p o
co γ p o
v
1 p
1
−
=−
co γ p o
γ
(7.27)
This now solves the system: we can find pp , and then
solve for conditions in the gas between the piston and
the forward wave:
v=U ;
3.
p
U
=γ +1
po
co
(7.28)
WAVES AT BOUNDARIES
Finally, a comment about wave reflection at boundaries.
Waves reflect in like manner off solid boundaries. This
means that at a solid wall, a compression wave reflects
as a compression wave, and an expansion wave reflects
as an expansion wave. Free boundaries are different;
waves reflect in an unlike manner. That means that expansion waves reflect as compression waves, and vice
versa. The reason is that along a solid boundary, the
flow direction is the boundary condition, while off a
free boundary the pressure is the boundary condition.
References
Good references here include Thompson, Currie,
Kundu and Faber.
(piston)
t
x=Ut
P
x=ct
(v=U)
2.
x
(v=0, p=p0)
Figure 7.5. Illustrating the geometry of the piston
problem, in the x,t plane. The two solid lines are the locus
of the piston (assumed to be close to vertical; U ≪ co ), and
the forward wave starting from the origin. The dotted lines
are the two waves which intersect the observation point P ,
and the third (reverse) wave used to connect the piston to
the x > 0 axis. Following Currie Figure 11.5.
44
8. ONE-DIMENSIONAL STEADY FLOW
We found in Chapter 7 that the sound speed, cs , is the
speed at which a signal propagates in a fluid. Because
of this, there are important differences between subsonic and supersonic flow.
Although most flows are (at least!)
twodimensional, we can learn a lot from the simpler case of
1D flows. These are often called channel flows; think
of a firehose, the Alaska pipeline, or a wind tunnel. If
you’re an astrophysicist, you can think about jets (from
protostars, or accretion disks around black holes). Another astrophysical application is to spherically symmetric flows (such as the solar wind), which are effectively one-dimensional.
A. Two-point Connections in Steady Flow
One approach, especially in the engineering-type literature, is to connect flow properties “here” to those
“there”.
It’s useful in this game to remember the Adiabatic
Relations. Steady flows are often assumed to be adiabatic, with a specific heat γ. This gives us useful scaling
laws:
¶
µ
µ ¶(γ−1)/γ µ ¶γ−1
csf 2
pf
ρf
Tf
=
=
=
(8.1)
T
cs
p
ρ
In the above set of expressions, the subscript “f” can
refer to any fiducial point (“there”) at which we know
T , cs , p or ρ; the unsubscripted quantities then refer to
“here”.
To be specific, now, consider 1D flow in a channel,
or pipe, of cross-section A. To sound more formal, let
“here” be region 1, and “there” be region 2 (or vice
versa, of course). We will want to apply two conservation laws.
v
1
v
region 1
known
1
γ p 1 2
+ v = h + v 2 = constant
γ−1ρ 2
2
as
(8.4)
(recall h = e + p/ρ is the enthalpy). This becomes
1
1
h1 + v12 = h2 + v22
2
2
(8.5)
Energy conservation with heat addition sometimes appears in 1D problems. In this case, Bernoulli is
modified as
1
1
h1 + v12 + q12 = h2 + v22
2
2
(8.6)
if some quantity q12 of heat (per gram) is added between points 1 and 2. This could describe, for instance,
to a turbojet, in which ignition of a (hopefully tiny!)
amount of fuel adds heat to the flow, and accelerates
it (if the flow is subsonic – more on that in the homework).
Two specific reference points in the flow are sometimes useful.
The stagnation point is just that, a point on a streamline where the velocity goes to zero. The subscript
“o” may be used here. One refers, for instance, to ho ,
the “stagnation enthalpy”. This is often connected to a
“reservoir” – picture a large tub of gas, from which a
flow emerges through a nozzle.
Some useful relations here start from
c2s +
γ−1 2
v = c2so
2
(8.7)
(this is just Bernoulli again, right?), which can also be
written,
2
(8.8)
From this, the ratios ρ/ρo , p/po , come simply from the
adiabatic relations (eqns 8.1).
region 2
Mass conservation has
reservoir
outflow
"o"
(8.2)
In our current application, this becomes
ρ1 v1 A1 = ρ2 v2 A2
otherwise
To
γ−1 2
=1+
M
T
2
Figure 8.1. A simple 1D “channel”, which has
cross-section area A. We want to connect conditions in
region 1 (ρ1 , T1 , p1 , v1 ) to those in region 2 (ρ2 , T2 , p2 , v2 ).
ρvA = constant
Energy conservation,
Bernoulli, has
(8.3)
Figure 8.2. A 1D flow from a “reservoir” out into the
universe. The gas in the reservoir is assumed to have zero
velocity, and carries subscript o: its conditions are
ρo , po , To .
45
The sonic point is another useful reference – the
point where v = cs . The subscripts “*” or “s” are used
here. The sonic and stagnation points can be connected
simply:
T∗
2
=
;
To
γ+1
¶(γ/γ−1)
2
;
γ+1
¶1/γ−1) (8.9)
µ
2
ρ∗
=
ρo
γ+1
p∗
=
po
µ
Mass conservation can be used to derive another useful
relation, between the sonic point and any general point
in the flow with area A and Mach number M:
µ ¶2
µ
¶ γ+1
A
2
γ − 1 2 γ−1
1
(8.10)
+
M
=
A∗
M2 γ + 1 γ + 1
Figure 8.3. Nozzles and diffusers in subsonic and
supersonic regimes. A converging nozzle accelerates
subsonic flow, and decelerates supersonic flow. A diverging
nozzle decelerates subsonic flow, and accelerates supersonic
flow. From Anderson Figure 5.3.
• At supersonic speeds, however, (M > 1), a decrease of area leads to a decrease of speed, and
conversely. Thus, a supersonic nozzle must be divergent; and a supersonic diffuser myst be convergent. The reason for this can be noted from
the force equation, which can be written dρ/ρ =
−M2 dv/v so that, for supersonic flows, the density decreases faster than the velocity increases;
so the area in an accelerating flow must increase,
in order to keep the product Aρv constant.
B. One-dimensional channel flow
Another approach is to consider the structure of the flow
throughout the channel. In particular, consider flow in
a channel of area A; work in the limit where the flow is
1D, that is, ignore cross-channel variation. We’re particularly interested in the effect that variations in A have
on the flow properties.
The steady continuity and momentum equations are
1 dρ 1 dv
1 dA
+
+
=0
ρ dx v dx A dx
(8.11)
and
ρv
dv
dp
+
=0
dx dx
(8.12)
Using c2s = dp/dρ, these combine as
¡
M2 − 1
¢ 1 dv
1 dA
=
v dx
A dx
2.
A SMOOTH TRANSITION ?
What about connecting these? It is possible to accelerate a flow smoothly through the sound speed, by a
proper combination of converging and diverging channels. Figure 8.2 shows an example of this. By considering (8.13), we can see that such flow must reach M = 1
at the throat (where dA/dx = 0). Such a configuration
is called a supersonic nozzle, a converging-diverging
nozzle, or a de Laval nozzle.1
(8.13)
which is the basic, governing equation.
1.
NOZZLES AND DIFFUSERS
This leads to a striking result: the behavior of the flow
in a channel depends on whether it is subsonic or supersonic. In particular, compare the flows in converging
and diverging channels:
• At subsonic speeds (M < 1), a decrease of
area increases the flow speed. A subsonic nozzle
(which accelerates the flow) must therefore have a
convergent profile, and a subsonic diffuser (which
decelerates the flow) must have a divergent profile. Compressibility does not change this qualitative behavior.
Figure 8.4.
Transonic flow in a a convergent-divergent
nozzle. From Thompson Figure 6.3
1
The last name is for Carl de Laval, a Swedish inventor who developed such a nozzle, in 1883; he was best known for developing
a high-speed turbine for driving a cream separator. Astronomers
may be familiar with the name as a model for the formation of
radio jets by pressure confinement in active galactic nuclei.
46
We must note, however, that M = 1 is not necessarily reached at the throat. Other combinations of
initial/boundary conditions allow other flows. Equation
(8.13) also allows smooth solutions where the velocity
v is an extremum: everywhere subsonic flow can have a
maximum of v at the throat, and everywhere supersonic
flow can have a minimum. Figure 8.5 illustrates this.
We also apply energy conservation, assuming an adiabatic shock:
γ p2 1 2
γ p1 1 2
+ v1 =
+ v
γ − 1 ρ1 2
γ − 1 ρ2 2 2
Now, these equations can be used to express the three
post-shock quantities, ρ2 , v2 , and p2 , in terms of their
pre-shock counterparts. Working through the algebra
(see chapter 9) gives the normal shock jump conditions:
2
γ−1
1
ρ1
=
+
2
ρ2
γ+1 M γ+1
2γM2 − (γ − 1)
p2
=
p1
γ+1
v2
2
γ−1
1
=
+
v1
γ + 1 M2 γ + 1
Figure 8.5. More convergent-divergent nozzles, in which
the velocity does not reach M = 1 at the throat. From
Kundu figure 15.7
3.
NORMAL SHOCKS
Most flows that go between subsonic and supersonic
aren’t so smooth, however (in fact it takes very special
conditions to achieve the smooth, de Laval transition of
Figure 8.4). More typically, mismatched variations in
channel area and/or external pressure lead to shocks in
the flow. Chapter 9 is devoted to shock properties; but
we also need some short discussion here.
In brief, a shock is a discontinuous transition in the
flow. If you force a supersonic flow to change suddenly
(compared to the sound travel time), a shock forms at
the right place and strength to accomodate the change.
Here we’ll look at normal shocks, which means
shocks normal to the flow.2 Put yourself in a frame
in which the shock is at rest; let “1” be conditions upstream (flow coming into the shock) and “2” be conditions downstream.
We apply mass and momentum conservation across
the shock. These two conditions become
ρ 1 v1 = ρ 2 v 2
(8.14)
ρ1 v12 + p1 = ρ2 v22 + p2
(8.15)
and
2
No, there are no “abnormal” shocks, sorry.
(8.16)
(8.17)
Once you have these, the jumps in T and cS can also be
derived from the usual adiabatic and ideal gas relations.
In a nutshell: a shock decelerates the flow (v2 < v1 ).
The density must increase, due to mass conservation
(ρ2 > ρ1 ). The “lost” kinetic energy goes to heat: T2 >
T1 and p2 > p1 .
And a caveat: equations (8.17) hold for the rest
frame of the shock. In many applications,however, the
shock moves through the fluid. If this is the case, you’ll
need to remember to do a Galilean transform to the
shock-rest frame before applying the jump conditions.
4. 1D
FLOWS WITH INTERNAL SHOCKS
Now, let’s return to 1D flows, and consider what happens if smooth flow is not reached.
The pressure jump maintained across the nozzle can
determine the flow possibilities. From (8.12), the sign
of dp is opposite to the sign of dv; accelerating flow
has a pressure drop, and vice versa. Consider, say, a
nozzle connecting an input region of pressure po , and
an exit at pe . What happens? (I follow Anderson in this
discussion; refer to Figure 8.6.)
(a) For pe = po , there is no flow of course. Let pe
drop a bit (say to pa in the figure): the flow is completely subsonic, with highest M and lowest p at the
throat.
(b) At some specific value of pe , the flow reaches
sonic at the throat. It remains subsonic everywhere
else, however (pb is not small enough to allow a smooth
sonic transition); this is called choked flow.
(c) As the exit pressure is further reduced, a region
of supersonic flow occurs downstream of the throat.
Because it can’t continue so all the way to the exit, a
47
normal shock forms in the diverging nozzle; the flow
slows down to subsonic at the shock.
(d) For some specific value of pe , the shock is located exactly at the exit. The fully isentropic, subsonicsupersonic flow pattern now exists throughout the entire
duct, except at the exit.
flow. In particular, a smooth transition from subsonic to
supersonic flow is possible if the gas stays hot enough
(extended heating sources are required).
The basic solution is due to Parker. Consider a
steady, spherical outflow. Mass conservation in this
case is ρvr2 =constant; or,
1 dρ 1 dv 2
+
+ =0
ρ dr v dr
r
(8.18)
while the momentum equation becomes in this case
(noting that gravity from the central star is important),
ρv
dv dp
GM
+
= −ρ 2
dr
dr
r
(8.19)
Writing dp/dr = c2s dρ/dr, these two equations combine to give the basic wind equation,
¶
µ
2c2 GM
c2s dv
= s − 2
(8.20)
v−
v dr
r
r
This does not have analytic solutions over the whole
range of r. It must be solved numerically; examples are
shown in Figure 8.7. However, as with our analysis of
channel flow, we can learn a lot by simple inspection of
(8.20).
Figure 8.6.
The effect of the downstream pressure on a
convergent-divergent nozzle. See text for details. From
Kundu figure 15.11
(e) As pe is reduced further, the normal shock is replaced by oblique shocks emanating from the edge of
the nozzle. This is overexpanded nozzle flow.
(f) Finally, there is one value of pe that is just right,
exactly matching that required by the isentropic flow
solution. No shocks of any kind exist here; the flow
exits quietly into the post-nozzle region.
(g) At still lower values of pe , expansion waves will
emanate from thet edge of the nozzle. This is an underexpanded nozzle flow.
C. Spherical stellar wind flow
A different example of one-dimensional “channel” flow
is spherical outflow from a central mass: a stellar wind.
Consider a spherical expansion, driven against gravity
by a steady mass and energy source at the origin. In this
case, the gravity plays the role of the area in channel
Figure 8.7. Possible solutions of the wind equation,
(8.20). Initial or boundary conditions determine which of
these solutions will exist in a particular system. The heavy
lines show the two interesting solutions, which pass through
a sonic point (at rs ) and connect smoothly between r → 0
and r → ∞. The outflow solution represents a stellar wind,
and the inflow solution represents steady, spherical
accretion. From Frank, King & Raine, Figure 2.1.
(i) The left hand side contains a zero, at v 2 = c2s . If
we want to consider well-behaved flows, that is to say
those in which the derivative dv/dr does not blow up,
then the right hand side of (8.20) must go to zero at the
same point. This defines the condition that must be met
at the sonic point:
v 2 = c2s
at
r = rs =
GM
2c2s
(8.21)
48
Whether or not a particular flow satisfies this condition
depends on the starting conditions, such as with what
velocity and temperature it left the stellar surface, and
also what the boundary conditions at large distances
are. If it does not start in such a way to satisfy this condition, it either stays subsonic (corresponding to finite
pressure at infinity), or cannot establish a steady flow.
(ii) The solution beyond the sonic point depends on
the temperature structure of the wind. The only solutions with dv/dr > 0 for r > rs are those for which
c2s (r) drops off more slowly than 1/r; it is only these
for which the right-hand side stays positive. In the case
of an isothermal wind, with c2s = constant, (8.20) can
be solved in the limit r >> rs :
v 2 (r) ≃ 4c2s ln r = constant
(8.22)
Thus, the wind will be supersonic, by a factor of a
few, as r → ∞. The question of how the solar wind
manages to stay nearly isothermal is not solved; it is
probably due to energy transport by some sort of waves
(MHD or plasma waves, for instance) which are generated in the photosphere and damped somewhere far out
in the wind.
(iii) Inside the sonic point, the gravity term will
dominate the right hand side of (8.20). Thus, solutions
with dv/dr > 0 , and v 2 < c2s , will obey
dv
GM
≃− 2
dr
r
This equation looks as if gravity is driving the wind out!
This unlikely-looking result comes from the fact that
the flow is nearly subsonic in this region; therefore, the
dp/dr term in (8.19) – which actually drives the wind
out – is nearly equal to the gravity term.
What happens at the outer edge of such a wind? Remember that the wind is not flowing into vacuum, but
rather into some external medium (call it the ISM, for
interstellar medium: this theory has been developed to
describe the solar wind). The pressure in the wind is
dropping with radius (because the density is dropping,
right?). When the wind pressure is close to the ambient pressure, the wind must slow down. In fact, we
expect a two-shock structure. At this outer boundary
of the wind, we expect some sort of shock transition,
since the wind is supersonic. Past this shock, the hot,
shocked wind-gas will expand into the ISM (at about
its own sound speed, to start); as long as this expansion
is supersonic relative to the ISM, the expanding hot gas
will drive a “snowplowed” shell of ISM, and a second
shock, out into the ISM.
A cartoon of this region, at some point in time,
would be that in Figure 8.8. Let region “a” be the wind;
v
S2
C
S1
a
b
c
d
11111111111111
00000000000000
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
R(t)
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
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00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
00000000000000
11111111111111
Figure 8.8.
Cartoon of the structure of a stellar wind,
and its interaction with the ISM. Left: the shock structure
within the wind. Right: The outer shell of dense,
snowplowed ISM. From Dyson & Williams figures 7.3 and
7.4.
S1 be the inner shock; region “b” be the wind-gas which
has been through the shock; C be the contact surface between the wind and the ISM; region “c” be the shocked
ISM; and S2 be the outer shock (moving into the ISM).
We expect S1 to be an adiabatic shock (since the wind
is probably hot and low density, and thus will have a
long cooling time); region “b” will contain hot, shocked
2
3 mvwind
∼ several × 107 K (noting
wind, with Tb ∼ 16
kB
1
that m = 2 mp is the mean mass per particle if region
“b” is fully ionized). The outer shock will probably be
isothermal, since the ISM is denser and cooler than the
wind. Thus, the shocked ISM will be in a thin shell,
containing all of the original ISM that lay between S2
and the star.
We can’t say anything about the details of the shock
transitions until we know more about shocks ... and
that’s next.
References
The basic 1D flow material can be found in several places; I’ve followed Thompson, Kundu and Anderson, mostly. The more specialized spherical-wind
applications are found in, for instance, Frank, King &
Raine, Accretion Power in Astrophysics; and Dyson &
Williams, Physics of the Interstellar Medium.
49
9. SHOCKS IN FLUID FLOW
Next, assume we have formed a shock, and consider
how it affects the fluid moving through it. We idealize
the shock as an infintesimally thin discontinuity in the
flow. This allows us to use conservation laws to determine the “jumps” in macroscopic quantities across the
shock (but of course it does not allow us to say anything
about the fluid flow within the shock).
A. Jump conditions
To start, recall our three basic equations – conservation
of mass, momentum, energy – written in conservative
form. Here, we ignore body forces, viscosity, and heating/cooling losses. The basic equations, then, are:
∂ρ
+ ∇ · (ρv) = 0
∂t
(9.1)
(from 1.4);
∂
(ρv) + ∇ · (ρvv + P) = 0
∂t
(9.2)
from (1.13), recalling P is the pressure tensor; and
µ
¶
· µ
¶¸
∂
1
1
ρe + ρv 2 + ∇ · v ρe + p + ρv 2
=0
∂t
2
2
(9.3)
(from 6.28). Comment: we are ignoring viscosity (it
makes this job so much easier ...). You remember that
viscosity depends on the second derivative of the velocity field. It follows that viscosity is critical to the
internal structure of the shock; but not to the jump conditions.
We use these conservation laws to find jump conditions. We work in a frame moving with the shock,
and assume a steady flow in that frame, so ∂/∂t = 0.
(Other frames require a Galilean transform to this shock
frame). General notation: the incoming (upstream)
quantities are labelled “1”, the outgoing (downstream)
quantites are labelled “2”, and [[A]] = A2 − A1 is the
jump in A across a boundary. Let n̂ be the vector normal to the boundary, t̂ be a unit vector tangential to the
boundary, and work in a frame in which the shock is
stationary. Our conservation laws become, then,
• Mass flux:
[[ρv · n̂]] = 0
(9.4)
which describes continuity of mass flux.
• Momentum flux (note P is a trace-only tensor if
we ignore viscosity:)
[[ρv(v · n̂) + pn̂]] = 0
(9.5)
is the basic momentum flux. One useful consequence
of this (resembling Bernoulli’s equation) comes from
dotting (again) it with n̂:
££
¤¤
p + ρ(v · n̂)2 = 0
(9.6)
We can also dot (9.5) with t̂, and use (9.4), to show
££
¤¤
(v · t̂) = 0
(9.7)
that is, the component of velocity in the shock plane
does not change.
• Energy flux (recall ρe + p = γp/(γ − 1):)
¶
¸¸
··µ
γ
1 2
ρv +
p v · n̂ = 0
(9.8)
2
γ−1
B. Normal shocks
We first consider normal shocks, for which v k n̂. In
this case, the system above reduces to the following.
Continuity gives
ρ 1 v1 = ρ 2 v2
(9.9)
Momentum conservation gives
ρ1 v12 + p1 = ρ2 v22 + p2
(9.10)
The energy jump condition, factoring out ρv from each
side, can be written in the case of an adiabatic shock,
γ p2 1 2
γ p1 1 2
+ v =
+ v
γ − 1 ρ1 2 1
γ − 1 ρ2 2 2
(9.11)
Now, these equations can be used to express the three
post-shock quantities, ρ2 , v2 and p2 , in terms of their
pre-shock counterparts. One way to do this is to take
the density jump as the basic variable: X = ρ2 /ρ1 .
One can then eliminate the ratios p2 /p1 , and cs2 /cs1 ,
from the equations, deriving a quadratic in X:
£
¤
¡
¢
X 2 2 + M2 (γ − 1) −2X 1 + M2 γ +M2 (γ + 1) = 0
and the positive solution of this gives the interesting solution. Writing out the ratios, we get the jump conditions:
2
γ−1
1
ρ1
=
+
2
ρ2
γ+1 M γ+1
p2
2γM2 − (γ − 1)
=
p1
γ+1
v2
2
γ−1
1
=
+
2
v1
γ+1 M γ+1
(9.12)
50
(Jumps in T and cS can also be derived). We can also
find a condition for the change in the Mach number:
M22 =
M21 + 2/(γ − 1)
2γ/(γ − 1)M21 − 1
(9.13)
(In this last expression I have explicitly put ’1” and “2”
subscripts on the Mach number; when omitted, M is
generally taken to refer to the upstream value.) These
expressions (9.12) & (9.13) are simple enough to evaluate. Nonetheless, most fluid books in the engineering
tradition present shock tables – tabular forms of the solutions as functions of M.
STRONG SHOCK LIMIT, NORMAL SHOCKS
When M >> 1, these equations simplify, to
ρ1
γ−1
=
;
ρ2
γ+1
γ−1
v2
=
;
v1
γ+1
p2
2γM2
=
;
p1
γ+1
γ−1
M22 =
2γ
(9.14)
In particular, if we consider an ideal gas with γ = 5/3,
and let M become large, we find ρ2 /ρ1 = v1 /v2 = 4.
This is the strong shock limit. Also in this limit, the
temperature jump is T2 /T1 = 5M2 /16, giving T2 =
3mv12 /kB , and M2 = 0.447; the upstream kinetic
energy is converted to internal energy in an adiabatic
shock.
C. Oblique shocks
Now, consider oblique shocks; in which the upstream
velocity v makes an angle β < π/2 with the shock surface. We know, from (9.7), that the tangential velocity
component u = v cos β does not change in the shock.
Thus, the normal upstream velocity, w = v sin β, controls the jump conditions. It follows from this that an
oblique shock exists only if
M1 sin β ≥ 1
This expression defines the Mach angle, µM : sin µM =
1/M. (Check back to chapter 7 for another usage of the
Mach angle). We expect, and will find, that w must decelerate in the shock; so v bends towards the shock face
(away from the shock normal), due to the deceleration
of the normal component of the velocity.
The jump conditions for an oblique shock can be
found from the normal jump conditions, (9.12) & 9.13),
by using w1 sin β in place of w1 . Alternatively, looking
ahead to MHD shocks, we can write the jump conditions explicitly in terms of components:
[[ρu]] = 0 ;
[[v]] = 0 ;
££
¤¤
p + ρw2 = 0
··
¸¸
1 2
h+ w
=0
2
Figure 9.1.
Downstream conditions (solutions of 9.12,
9.13) as a function of the shock Mach number. Top, for
γ = 1.4 (useful for air); bottom, for γ = 5/3 (useful for a
monatomic gas, such as ionized hydrogen). The differences
in γ are apparent in the
(9.15)
It is also useful to define the deflection angle: δ =
− [[β]] = β − β2 . This is the amount by which the
flow bends towards the shock (or away from the shock
normal). Geometry gives us, directly,
tan β2
tan(β − δ)
u2
=
=
u1
tan β
tan β
51
mach number,
w
2
M22 sin2 (β − δ) =
δ
w1
(γ − 1)M2 sin2 β + 2
2γM2 sin2 β + (1 − γ)
(9.18)
Note that both M2 > 1 and M2 < 1 are possible.
β
1.
TWO POSSIBLE DEFLECTIONS
We can also use the jump conditions to solve for the
post-shock angles:
tan β2 = tan β
β
u
2 w
2
(9.19)
M2 sin2 β − 1
M2 (γ + cos 2β) + 2
(9.20)
and
v
2
tan δ = 2 cot β
1
u
v
(γ − 1)M2 sin2 β + 2
(γ + 1)M2 sin2 β
2
Figure 9.3 shows numerical solutions for δ(β, M), and
Figure 9.4 shows a cartoon.
β
1
w
1
Figure 9.2. Oblique shock geometry. Top in terms of the
incoming flow; w1 is the incoming velocity, at angle β to
the shock; w2 is the outgoing velocity, and δ is the
deflection angle, through which the flow bends at the shock.
Bottom, the same thing with the shock rotated, and w
divided up into components perpendicular and parallel to
the shock face. This follows Faber, Figure 3.11, but note,
our notation differs from his. In particular, we have
subscripts 1 and 2 for upstream and downstream; he has
unprimed and primed, respectively.
and the jump conditions give
−
tan δ
[[w]]
=
v
cos2 β (1 + tan β tan δ)
(9.16)
We can use the jump conditions to find the usual
post-shock quantities. If we start by specifying β, we
can find relations analogous to (9.12 & 9.13)
(γ + 1)M2 sin2 β
ρ2
=
ρ1 (γ − 1)M2 sin2 β + 2
p2 2γM2 sin2 β γ − 1
=
−
p1
γ+1
γ+1
2
u1
(γ + 1)M sin2 β
=
u2 (γ − 1)M2 sin2 β + 2
(9.17)
(remembering that M refers to the upstream value,
M1 .) We can also find an expression for the post-shock
Figure 9.3. Oblique shock solutions. The dashed lines
are the strong shock branch, the solid lines are the weak
branch, and the heavy dotted line shows the maximum
deflection angle δmax . From Kundu figure 15.15. Note, σ in
this figure is β in our notation.
This contains important results. First, the incoming
angle, β, corresponding to a given deflection, δ, is double valued.1 That is: for any specified deflection angle,
1
This analysis gives us analytic expressions, (9.17, 9.19 and 9.20),
for the post-shock conditions as functions of M and β. Alternatively, in shock tables, one may find M and δ specified;
then there are two sets of post-shock conditions, corresponding
to strong and weak shocks.
52
there are two possible shock angles (relative to the incoming flow). A given δ can come from (a) a strong
shock, corresponding to high β values and giving subsonic postshock flow; and also a weak shock, from low
β values and, generally, supersonic postshock flow. The
flow will generally find the weak shock solution if it
can.
strong
shock
M
weak
shock
θ
Figure 9.4. Weak and strong shocks. The geometry (θ) of
the wall forces the flow to bend through θ; the mach number
M1 (and flow properties) determine whether forced bend
will happen through a weak shock, a strong shock, or
neither.
A second important result from this analysis is that
there is a maximum deflection angle possible for any
incoming flow. This connects to the the existence of
detached shocks – that’s what happens when the flow
can’t bend enough at an obstacle to form an attached
shock. Figure 9.5 illustrates this.
attached
shock
1
detached
shock
M
HIGH MACH NUMBER LIMIT
Finally, as one would expect, oblique shocks also have
a M ≫ 1, “strong-shock” limit. The limits are just
the same as for a normal shock, (9.14), if the normal
velocity w is substituted for the full velocity v. The
strong-shock limit of the deflection angle is still doublevalued, and can be found from the high-M solutions in
Figure 9.3.
D. The Weak Shock Limit
1
M
2.
1
Figure 9.5. Attached vs. detached shocks. Top, the
opening angle of the wedge is less than the maximum
deflection angle of the flow; an attached shock forms.
Bottom, the opening angle of the wedge is larger than the
maximum possible deflection of the flow; a detached shock
forms.
The weak shock limit describes shocks at oblique angles (close to the Mach angle) to the upstream flow, and
thus shocks in which the velocity and pressure jumps
across the shock are small. These “mach wave” shocks
are important in bending supersonic flows.
For weak shocks, we can treat [[p]] /p1 as a small
parameter. (Notation: Faber calls this σ, and does not
necessarily assume it is small. I’m following Thompson here, in the weak-shock case.) To avoid notation
confusion, here we let
Π=
[[p]]
γp1
(the γ factor simplifies the algebra later), and we assume Π ≪ 1. With this, we can expand the jump conditions in Π, as
[[u]]
[[ρ]]
≃Π;
≃ −Π
ρ2
cs1
γ−1
[[cs ]]
(9.21)
≃
Π
cs1
2
γ+1
γ+1
M1n ≃ 1 +
Π;
M2n ≃ 1 −
Π
4
4
Thus, both the incoming and downstream Mach numbers are close to unity. Seen in the lab frame, the shock
is propogating at nearly the sound speed (as one would
expect for a low-amplitude wave). We can also find the
deflection and the shock angle, for weak oblique shocks
¡ 2
¢1/2
M −1
δ≃
Π
2
(9.22)
¶
µ M
Π
γ+1
β − µM ≃ 1 −
4
(M2 − 1)
Thus, the flow deflection is small, and the shock angle
(relative to the incoming flow) becomes the Mach angle, in the weak shock limit.
PRANDTL - MEYER FUNCTION
The relation between the deflection, δ, and the Mach
number, M, turns out to be a perfect differential (as we
53
This can be written in terms of M: using
will show). The integral form of this function will be
useful later.
Rearranging the weak-shock relations, we find to
lowest order in [[u]],
¡ 2
¢1/2
M −1
[[u]]
δ≃−
2
M
cs1
dw
dcs
dM
=
=
;
M
w
cs
wdw +
cs dcs
=0
Γ−1
where the factor Γ = (γ + 1)/2 for a perfect gas. This
gives
(9.23)
This is useful now. Since only the normal velocity
changes at the shock, we have w12 − w22 = u21 − u22 ;
and to first order in the jumps, w1 [[w]] ≃ u1 [[u]]. Thus,
with u1 /v ≃ sin µM = 1/M,
dw
dM/M
=
w
1 + (Γ − 1)M2
1 [[u]]
[[w]]
=
w1
M2 cs1
and, thus, we have expressed the deflection angle as
athe perfect differential. This defines the PrandtlMeyer function, δ(M) :
(9.26)
Thus, we can rewrite (9.23) as
δ
[[w]]
=
2
w1
(M − 1)1/2
¢1/2
M2 − 1
dM
dδ =
2
1 + (Γ − 1)M M
¡
(9.24)
In the limit of infinitesimal strength, the oblique shock
becomes a Mach wave, and with δ → dδ, [[w]] → dw,
¡
¢1/2 dw
dδ ≃ ± M2 − 1
w
δ(M) =
µ
γ+1
γ−1
¶1/2
This can be integrated formally. Taking the constant of
integration so that δ(1) = 0, the function describing the
deflection angle, in the weak shock limit, as a function
of total upstream mach number is
(9.25)
−1
tan
"µ
γ+1
γ−1
While we have only considered weak shocks, with M−
1 ≪ 1, we can formally look at the full range of this
function. In particular, when M → ∞, the PM angle δ
has a limit,
"µ
#
¶
γ + 1 1/2
π
−1
(9.29)
δmax =
2
γ−1
and for γ = 5/3, this happens to be just δmax = π/2.
References
I’m taking this mostly from Thompson, Kundu and
Faber.
(9.27)
¶1/2
¡
2
M −1
¢1/2
#
¡
¢1/2
− tan−1 M2 − 1
(9.28)
54
10. ONE-DIMENSIONAL UNSTEADY
FLOW
In chapter 7 we introduced simple waves which carry
information about some disturbance in the fluid (such
as a piston moving or a sudden “hit”). In Chapter 9 we
studied the local structure of shock fronts. In this chapter we return to the arguments of chapter 7, formalize
them and allow for waves that are not weak. That is,
we will formally introduce characteristics to solve unsteady flow problems. One important application of this
is to shock formation and propagation.
A. Shock Formation: The Physical Picture
In Chapter 4, we treated sound waves as smallamplitude perturbations, and found that they propagate
at a characteristic speed, cs , which depends only on
the temperature of the fluid. In reality, however, any
such wave has higher density at its “center”, relative to
its surroundings, and has a smooth variation of density
with position. Figure 10.1 shows a discretized version
of this.
Figure 10.2.
Schematic stages in the development of a
wave which steepens into a shock. Note that all of the
profiles contain the same area. From Faber figure 3.6.
SHOCK THICKNESS
Physically, shocks cannot be infinitely thin surfaces,
although this is how they are treated mathematically.
In practice, the fluid variables cannot change instantaneously; there must be a smooth transition between upstream and downstream. This defines a shock thickness,
which is determined by viscosity, that is by interparticle
collisions.
We can find a scaling argument for this: consider
the balance between inertial and viscous forces, which
must apply at the shock:
d2 v
dv
≃ν 2
(10.1)
dx
dx
This suggests the shock thickness, ∆, is determined by
∆ ∼ ν/v. But, the viscosity can be found from kinetic theory as ν ∼ λhvi, where λ is the mean free path
and hvi is the mean particle speed. Thus, noting that
hvi ∼ cs (check this out for an ideal gas if you don’t
believe it!), we would estimate ∆ ∼ λ. This makes
some physical sense: the information that the flow conditions have changed cannot be communicated over a
distance shorter than a collision mean free path.
This is not the entire picture, however. An ionized
gas – a plasma – can support shocks which are much
thinner than their collisional (Coulomb, cf.chapter 1)
mean free path. The details of how these shocks work
are still being argued about. There seems to be agreement, however, that these shocks are supported by microturbulence in the plasma: these small-scall fluctuations transfer the energy and momentum from upstream
ρv
Figure 10.1.
Density profiles for layers of higher
density, propagating to the right. From Faber figure 3.5.
But the higher central densities lead to a higher central pressure and temperature (assuming an adiabatic
wave which is the most likely). Thus the central signal speed will be higher than that at the edges; therefore the center will tend to overtake the leading edge. If
the wave amplitude is large enough, an initially smooth
waveform will be significantly distorted, and the wave
will “break”. That is, the wave profile will become
triple valued at its leading edge; this leads to the reforming of the edge in a shock discontinuity (as sketched in
Figure 10.2).
55
to downstream, and convert ordered to random energy,
just as interparticle collisions do in a shock in a neutral
gas. The width of a collisionless shock is typically a
fundamental plasma scale, such as the ion gyroradius or
the length derived from the plasma frequency, ∼ cs /ωp .
Thus, this function F is simply a constant times the local sound speed. Now, the pressure derivatives can be
written in terms of F , as
∂F ∂p
1 ∂p
∂F
=
=
∂t
∂p ∂t
ρcs ∂t
∂F ∂p
1 ∂p
∂F
=
=
∂x
∂p ∂x
ρcs ∂x
B. The Method of Characteristics
In chapter 7 we introduced weak waves – waves of such
small amplitude that the local conditions (density, pressure) are not significantly disturbed. These waves propagate at a constant speed, co , thus their loci in an (x, t)
diagram are straight lines.
Our goal in this section is to extend this approach to
the general case, rather than limiting ourselves to smallamplitude or subsonic flows. To do this, we first find the
equations that describe the characteristics. Start with
the 1D momentum and continuity equations (in a uniform channel, no A terms, and with no external body
forces):
∂v
1 ∂p
∂v
+v
+
=0
∂t
∂x ρ ∂x
(10.2)
and
∂ρ
∂ρ
∂v
+v
+ρ
=0
∂t
∂x
∂x
(10.3)
But this latter can be written, using ρ = ρ(p) and dp =
c2s dρ, as
1 ∂p
v ∂p
∂v
+
+ cs
=0
ρcs ∂t
ρcs ∂x
∂x
(10.4)
Now, these last two can be added and subracted to form
two new equations,
µ
¶
∂v
1 ∂p
1 ∂p
∂v
±
+ (v ± cs )
±
= 0 (10.5)
∂t ρcs ∂t
∂x ρcs ∂x
As Thompson notes, it appears doubtful that this is any
simplification, but it is nonetheless. Define a new funciton F = F (p, s) (where s is the specific entropy),
which satisfies
Z p
dp
(10.6)
F =
po ρcs
where po is some (useful) reference state, and the integration is carried out at constant entropy. For an adiabatic perfect gas, the F integral can be done straightforwardly:
Z
Z
Z
dp
dρ
2dcs
2
F =
= cs
=
=
cs
ρcs
ρ
γ−1
γ−1
(10.7)
so that (10.5) can be written as
·
¸
∂
∂
(v ± F ) = 0
± (v ± cs )
∂t
∂x
(10.8)
At this point, each equation (the “+” and “-” ones) contain only one derivative operator:
D+
∂
∂
=
+ (v + cs )
;
Dt
∂t
∂x
D−
∂
∂
=
− (v − cs )
Dt
∂t
∂x
(10.9)
so that our two equations are, finally,
D+
(v + F ) = 0 ;
Dt
D−
(v − F ) = 0
Dt
(10.10)
These equations turn out to be quite useful. We must
interpret them physically, as follows. From (10.9) and
(10.10), we see that the quantities
2
cs
γ−1
2
=v−F =v−
cs
γ−1
J+ = v + F = v +
J
−
(10.11)
are constant along the lines described (and labelled) by
C+ :
−
C
:
dx
= v + cs ;
dt
dx
= v − cs
dt
(10.12)
These lines are called characteristic lines. They describe the paths of signals (sound waves) travelling forward or backwards in a flow, at the local sound speed.
This, the quantity v +F is constant for an observer travelling at velocity v + cs , or on any plane perpendicular
to the x axis which moves at v + cs (that is, with a forward, or positive, sound wave in a flow with velocity v.
this is the plus characteristic, or the C + wave). Similarly, the quantity v − F is is constant for an observer
travelling at velocity v − cs (that is, a backward or negative sound wave; the minus characteristic or the C −
56
regions reached by no characteristic.
1.
PISTON MOTION IN A CHANNEL
This may be made more concrete by the example of
piston withdrawal, or advance, in a 1D channel. Let the
piston position be described by X(t); withdrawal has
dX/dt < 0 and advance has the opposite sign.
Figure 10.3.
Characteristic lines, along which
J + = v + F and J − = v − F are constant. The solution at
point d is uniquely determined by the initial data at points a
and b. From Thompson figure 8.3
wave). These quantities, J ± = v ± F , are called the
Riemann invariants.
But these characteristics allow the simple solution
of an initial value problem (conceptually; in practice
numerical integration may be required). Refer back to
Figure 10.4, which assumes we have a time-dependent
problem in which initial values are specified over the
range a < x < b. Consider point d in that figure. It is
connected by a plus (forward direction) characteristic to
point a, and a minus (backward direction) characteristic
to point b. That means the quantity Ja+ = v +F at point
d is the same at point a (since C+ connects the two
points), and the quantity Jb− = v − F is the same at
point d and point b (since C− connects
¡ them). Thus,
¢ we
know the solution uniquely: v = 21 Ja+ + Jb− ; cs =
¢
γ−1 ¡ +
Ja − Jb− .
4
Figure 10.5.
The geometry and characteristics
associated with smooth piston withdrawal. PP labels the
path of a particle starting at a particular point on the x axis.
From Thompson figure 8.17.
We first notice that C − characteristics come in from
ahead, from the undisturbed flow region (which has v =
0, cs = cs,o . On these lines,
v(x, t) −
Going further, referring to Figure 10.4, we can see
the extent of the solution to this problem. The solution v(x, t) is completely determined by the initial values in the region enclosed by the triangle abd. Outside
this region, it is partly influenced in regions where one
characteristic reaches; and it is totally undertermined in
(10.13)
This relation allows us to find v from cs,o , or vice versa.
Now, the C + characteristics connect the “influenced
flow region” to the piston motion. On these lines,
v=
Figure 10.4.
Regions of influence for the same
initial-data problem as in Figure 10.4. From Thompson
figure 8.4
2
2
cs =
cs,o
γ−1
γ−1
dX
;
dt
cs = cs,o +
γ − 1 dX
2 dt
(10.14)
In this latter, we have noted that a typical straight C + ,
which propagates into a uniform region, characteristic
carries nearly constant cs and v values. Thus, on C + we
can identify v with dX/dt at the piston (assuming slow
enough withdrawal that the fluid doesn’t separate); and
we have used (10.13) as well. We note that the inverse
slope of the C + characteristics is v+cs = cs,o +dX/dt.
As this quantity becomes smaller, the characteristics
become more and more vertical. We also note that the
second expression in (10.14) shows that cs < cc,o for
a withdrawing piston; the expansion of the gas cools
it.These relations (10.13) and (10.14) effectively solve
the problem; they specify the values along each characteristic, and thus all regions of (x, t) space are addressed.
57
Piston advance can be treated in the same way, taking dX/dt > 0. The difference is that a shock forms
ahead of the piston. The analysis above carries through
just the same – we again note that v+cs = cs,o +dX/dt
is the inverse slope of the C + lines. But now, with
dX/dt > 0, the slope decreases moving along the path
of the accelerating piston: the characteristic lines must
cross. As we will see next, a shock forms at the point
where they first intersect.
ary region ahead of the wave. On these characteristics,
J − = Jo− = constant everywhere, so that
J− = v −
and thus
2
2
cs = −
cs,o ;
γ−1
γ−1
(10.15)
2
(cs − cs,o )
γ−1
(10.16)
v=
Thus, fluid within compression regions (cs > cs,o )
moves in the +x direction, the direction of travel of
the wave. In addition, any portion of the C + wave must
move with speed
v + cs =
Figure 10.6.
The geometry and characteristics
associated with smooth piston advance. From Thompson
figure 8.21.
2.
γ+1
2
cs −
cs,o
γ−1
γ−1
(10.17)
Both of these connections verify the monotonic increase of wave speed with wave amplitude (cs − cs,o )
which gives rise to the distortion of the wave into a
shock. We can visualize this from characteristics also.
The slope of each characteristic, in a (t, x) diagram,
is the inverse of the local sound speed – so regions
of higher overdensity have higher cs , and thus lower
slopes. Thus, characteristics from some region will intersect at some later time – shown as point s in the figure.
CONNECTION TO SHOCK FORMATION
We argued in §10.1 that simple waves can steepen into
shocks; this is based on the local propagation speed is
an increasing function of the local density; so overdense
regions move faster, and overtake the less dense regions
ahead.
We can treat this more formally using characteristics. Let the initial wave be traveling in the +x direction (a forward, C + , wave), into stationary gas with
po , ρo , cs,o values far ahead of our disturbance.
Figure 10.8. A characteristic diagram (wave diagram),
showing characteristics intersecting at what becomes a
shock. From Thompson figure 8.12
Figure 10.7.
The geometry, and progressive distortion,
of the steepening waveform described by characteristics in
Figure 10.9. From Thompson Figure 8.10.
Now, the region around the disturbance includes C −
characteristics which come in from the uniform station-
When is this time? In general, two nearby pieces
of fluid have similar values of the quantity u = v +
cs ; say they vary by δu (take this to be positive, just
to pick a specific case). Referring to Figure 10.7, we
see that the shock first forms, then, at the point where
this derivative dx/du is a minimum – which occurs at
the steepest point in the wave (that is along the leading
edge, as in Figure 10.2).
Note, after the shock forms, our method of characteristics must be extended to include jumps at the shock
itself.
58
3.
TRAFFIC SHOCKS
Finally, let’s consider a different approach: characteristics and shocks in traffic flow.
Figure 10.9.
A traffic shock; from Thompson figure 7.3.
That is: consider the flow of cars along a one lane
road (no passing!). We can take ρ as the density of cars
(per length), v as their speed, and q = ρv as the traffic
flow (what are the units here?). Conservation of cars is
expressed by
∂ρ ∂q
+
=0
∂t
∂x
(10.18)
But this is, of course, just the continuity equation. Now,
traffic-flow people simplify this by assuming the flow
can be taken as a function only of the density: q =
q(ρ).1 This allows the basic equation to be written
∂ρ
∂ρ
+ c(ρ)
=0
∂t
∂x
(10.19)
if c(ρ) = ∂q/∂ρ.
Now, we know how to treat the equation (10.19):
the solution ρ is constant along the characteristic lines
dx
= c(ρ)
dt
(10.20)
Why? Consider the case when the “total derivative” is
zero:
∂ρ dx ∂ρ
Dρ
=
+
=0
Dt
∂t
dt dx
But this is just equation (10.18) if the dx/dt term is
given by (10.20). Thus, (10.19) shows that ρ is constant
along the “path” in (x, t) space given by (10.20). That
is, the lines on which ρ is constant, are simple waves,
straight lines in the (x, t) plane.
Typically, the problem is set up in terms of initial
conditions: ρ(x, 0) = f (x), that is some function f (x)
1
describes the car density at t = 0. But each point in the
(x, t) plane can be mapped back, along its local characteristic, to a starting point xo . Thus, if things are
single-valued, the solution can be found directly from
ρ(x, t) = f [xo (x, t)]. (If things are multivalued, that
is characteristics from more than one xo pass through a
given (x, t), then we find that a shock develops. Examples of both of these are given in the homework.
A typical choice assumes the velocity is v(ρ) = V (1 − ρ/ρo ),
ρ < ρo . That is, a linear relation between density and speed, up
to some maximum density ρo at which gridlock sets in and no
further motion occurs.
References
I’m mostly following Thompson here, also Faber on
shock development. Traffic shocks can also be found as
examples in PDE books.
59
11. TWO-DIMENSIONAL STEADY FLOW
We have studied transonic, one-dimensional flows, both
time-steady and not. We now move to 2D, steady flows.
There are two mathematical approaches that can be useful: linearized potential flow, and characteristics.
A. The Nature of Steady, two-dimensional flow.
As we have noted already, subsonic and supersonic
flows behave very differently. A formal way to demonstrate this is through an extension of potential-flow
techniques to a general, compressible flow. Consider a
steady, irrotationa, inviscidl flow. Because ∇ × v = 0,
we choose a velocity potential, v = ∇φ. The continuity and Euler equations for steady flow are
∇ · (ρv) = 0 ;
1
(v · ∇)v + ∇p = 0
ρ
(11.1)
Now, we note the useful fact, still for steady flow
(whence?):
1
v · ∇p = −c2s ∇ · v
ρ
We can combine these results to get a second order
equation in the velocity potential:
£
¤
1
∇φ · ∇(∇φ)2 = c2s ∇2 φ
2
(11.2)
To be more explicit, we can pick Cartesian coordinates,
and write
∂φ ∂φ ∂ 2 φ
∂2φ 2 2 ∂2φ 2 2
(cs −φx )+ 2 (cs −φy ) = 2
(11.3)
2
∂x
∂y
∂x ∂y ∂x∂y
These equations (11.2) and (11.3) are general, but not
very useful as they are horribly nonlinear in φ.
We can simplify it, however, by linearizing. In this
section I describe one particular version of this: twodimensional Cartesian flow. That is, consider a simple,
unperturbed flow – for instance, a uniform flow U in
the x̂ direction. Add a thin body, such as an airfoil,
nearly aligned with the x-axis. This will perturb the
flow slightly; we can specify the velocity potential to
describe the perturbed flow. Going through the algebra (check Thompson, or Schreier, for the details), the
linearized potential equation becomes
¡
U 2 − c2s
2
¢ ∂2φ
2∂ φ
=
c
s
∂x2
∂y 2
(11.4)
But now, we note a significant difference for M > 1
and M < 1 flow. For subsonic flow, M < ∞, (11.5)
is essentially Laplace’s equation, (3.3); so we can solve
this linear problem with our potential flow techniques
of chapter 2. We also note that such flows are totally
governed by their boundary conditions.1
However, if M > 1, the nature of equation (11.5)
changes. In particular, it becomes a wave equation; and
has as the general solution,
φ(x, y) = F (x − βy) + G(x + βy)
(11.6)
where β 2 = M2 − 1, and F , G are any function. This
is getting back into the regime of characteristics – the
flow is determined only by those regions nearby from
which information can propagate.
We can restate this using some results from formal
PDE theory. Equation (11.5) is a second order
¡ PDE; its¢
nature depends depends on the sign of the 1 − M2
term. If M < 1, the equation is elliptic – and does not
have any real characteristics. If M > 1, the equation is
hyperbolic, and does have real characteristics. We can
exploit those characteristics in the solution of steady,
two-dimensional problems; which is the point of this
chapter.
A final note: this breakdown into elliptic vs. hyperbolic does not depend on the linearization; Schreier
demonstrates that the full potential system, (11.2) and
(11.3) also changes type at M = ∞.
B. Signal Propagation in Flows
If you don’t care for such a mathematical approach, we
can try a more physical one. Consider the Mach construction (due to Ernst Mach in an 1887 paper on supersonic projectiles). Let some object emit a steady signal
while moving at some speed v through a fluid. (This
signal might, for instance, be the simple fact that the object is moving and thus disturbing the local fluid). The
signal propagates at cs relative to the fluid. If the object
is moving subsonically, the signals “converge” ahead
of the object, and “diverge” behind it: this leads to the
familiar Doppler effect. Note, all of space eventually
receives the signals. The situation is different however,
if the object is moving supersonically. In this case the
spherical wavefronts emitted as the object moves, define an (upstream) cone of influence: fluid inside this
cone can receive information about the motion, while
Or, in terms of the Mach number,
1
¡
¢ ∂2φ ∂2φ
+ 2 =0
1 − M2
∂x2
∂y
(11.5)
Remember all those problems you solved in electrostatics, for
the electric potential, where the boundary conditions determined
everything?
60
fluid outside cannot. The opening angle of this cone is
sin µM =
cs
1
=
v
M
(11.7)
which is called the Mach angle. The surface bounding
the range of influence of the object is called the Mach
surface or characteristic surface; in two dimensions it
becomes the Mach lines, or simply characteristics.
corner. Thus then tend to lean “forward”. Minus characteristics come from ahead of a given point in the flow;
at any point the intersection of plus and minus characteristics determines θ and M (the latter through the
value of δ(M)). In particular, the flow angle and speed
are constant along any given characteristic. Thus, the
flow must follow the bending of the Mach lines. This
geometry forces the streamlines to diverge, so the flow
must expand. Expanding a supersonic flow accelerates
it – just as in chapter 7.
By the way: where does the energy for the acceleration come from?
Figure 11.1. Repeating Figure 7.3, showing Mach’s
construction for the Mach cone and Mach angle. Note in
this figure, the flow is moving to the right relative to the
object; alternatively one can envision the object as moving
to the left. From Thompson figure 5.2.
C. How Does Supersonic Flow Turn a Corner?
Now, we illustrate the use of characteristics in steady,
supersonic flow with one example: flow around a corner.
Before we hit the math, however, let’s think about
the physics. Consider supersonic flow along a wall, and
let the wall have a corner. Remember that information
about the corner cannot propagate very effectively upstream, so the incoming fluid cannot “know” very easily
about the corner. Just how, then, does the flow turn the
corner?
To be specific, think about flow around a wall which
turns a corner away from the flow, as in Figure 11.2b.
Since we’re working with supersonic flow, we can think
back to the channel flow of Chapter 7, in particular flow
into an expanding channel. The flow must expand, and
being supersonic it must also speed up. This can be
seen in terms of characteristics (also called Mach lines
in this context). In the next section we will show that
the Riemann invariants in this situation are given in
terms of two angles: θ, the angle of the flow relative
to some axis (say the wall before it turns), and δ(M,
the Prandtl-Meyer angle defined in chapter 9. The invariants are θ ± δ(M). Our boundary (“initial”) condition must be along the wall: the flow (streamline) must
parallel the wall, at the wall. Thus, the characteristics
must be straight lines which originate from the wall.
Now, the plus (forward) characteristics have values of
θ which go more and more negative, moving along the
Figure 11.2.
Flow turning corners, with streamlines and
plus (forward) characteristics (Mach lines) shown. The left
diagram shows flow into a bending wall. The streamlines
come closer together as they follow the wall around, so the
flow must compress and decelerate. A shock forms where
the characteristics intersect. The right picture shows the
opposite case, flow around a corner. The characteristics
diverge, as does the flow; this expansion results in an
acceleration. From Kundu figure 15.18
Now, consider the converse case: flow “into” a wall
which bends, as in Figure 11.1a. The geometry forces
the flow to be compressive, and thus (still being supersonic) it must slow down. Here, however, the cornerturning takes place through a set of shocks; it cannot
be smooth. We can see this by drawing in characteristics again. The nature and angle of the characteristics
follows the same rules as above; the angle of the wall
sets the angles of the Mach lines which arise from it.
But now, it’s apparent that these lines must intersect: a
shock will form. (Formally: think about characteristics
intersecting. If “adjacent” lines carry different values
of θ ± δ, the values of θ and/or δ will not be uniquely
defined at the intersection. That is not good. The fluid
will respond by setting up a shock – an infinitely thin
(well, nearly) jump between “front” and “back” properties.
By the way: when the flow decelerates, where does
the energy go?
61
D. One example: Prandtl-Meyer flow
Now, let’s do this mathematically. A specific example of flow around corners is the case of flow around
a sharp corner (as in Figure 3.14 of Faber). The flow
expands through a “fan” of Mach lines centered at the
corner, called the Prandtl-Meyer expansion fan. The
mach number increases along a streamine through the
fan, while the pressure falls along that streamine. Each
mach line is inclined at µM to the local flow direction.
Thus, supersonic flow turns via a sequence of standing
oblique shocks.
This can be described in detail with characteristics.
We could formally start with the second order PDE,
(11.5), and find its characteristics. However, Thompson
(his chapter 9) has an alternative approach which might
be more illuminating. Following him, we search for
characteristics by starting with the basic, 2D, steadyflow equations:
∇ · (ρv) = 0 ; ∇ × v = 0
µ 2¶
v
ρ∇
+ ∇p = 0
2
(11.8)
We write these in streamline coordinates: s is the distance along the streamline, at local angle v̂; n̂ is the
local normal, and θ is the flow angle (relative to some
reference direction). Figure 11.2 illustrates the geometry. The zero-curl equation becomes
∂v
∂θ
+v
=0
∂n
∂s
(11.9)
which is the first equation of motion. For the moment,
let’s work in cylindrical coordinates. The continuity
equation becomes, in these coordinates,
∂θ
sin θ
1 ∂ρ 1 ∂v
+
−
=−
ρ ∂s v ∂s ∂n
r
(11.10)
∂v
1 ∂p
+
=0
∂x ρ ds
(11.11)
These last two combine, using dp = c2s dρ, as
¡ 2
¢ 1 ∂v
∂θ
sin θ
+
=
M −1
v ∂s ∂n
r
the two equations are
√
M2 − 1 ∂v
∂θ
+
=0
tan µM
v
∂n
∂s
√
M2 − 1 ∂v
∂θ
tan µM sin θ
+ tan µM
=−
v
∂s
∂n
r
(11.13)
But the first term in this second equation is ∂δ/∂s, if
δ(M) is the Prandtl-Meyer function, defined in (9.27)
above. Putting δ(M) into (11.13) where we can, and
adding or subtracting the two, gives us our characteristic equations:
¶
µ
tan µM sin θ
∂
∂
(θ + δ(M)) = +
+ tan µM
∂s
∂n
r
µ
¶
∂
tan µM sin θ
∂
(θ − δ(M)) = −
− tan µM
∂s
∂n
r
(11.14)
This is the first equation of motion we want. Next, the
momentum equation (in the s direction) becomes
v
Figure 11.3.
(a) Streamline coordinates for the
characteristic solution. The flow velocity is u, at angle θ to
the x-axis; the distance along a streamline is s; n is the
vector normal to the streamline. (b) Two Mach waves, m+
and m− , intersecting a streamline, (c) A geometrical way to
remember the Mach angle. From Thompson figure 9.6
(11.12)
which is the second equation of motion.
Now we use (11.9) and (11.11) to look for characteristics. In terms of the mach angle,
p
tan µM = 1/ M2 − 1
These can be interpreted as follows. (Compare equations (10.10) in our earlier discussion of characteristics.) At any point along the streamline, two characteristics (a.k.a. Mach waves) can be found. They head
out at angles ±µM relative to v, which defines the local streamline. (These are the analogs of C ± in chapter
10).
But the left hand side of equations (11.14) are just
the derivatives of θ ± δ(M) along the lines m± . To
see this, recall that (by the chain rule) the derivative of
some function F along m+ is
∂F ds
∂F dn
dF
=
+
dm+
∂s dm+
∂n dm+
62
But, ds/dm+ = cos µM and dn/∂m+ = sin µM ; so
that
µ
¶
∂F
∂F
dF
= cos µM
+ tan µM
dm+
∂s
∂n
m− .
A similar result obtains for derivatives along
Thus, the two characteristic equations for steady, twodimensional flow can be written,
sin µM sin θ
d
(θ + δ(M)) = +
dm+
r
(11.15)
d
sin µM sin θ
(θ
−
δ(M))
=
−
dm−
r
Now, there are more general ways to treat characteristic
problms, if the derivative of some function along a characteristic is not zero.2 For now, let’s switch back from
cylindrical to plane geometry: that means take r → ∞
in (11.15). This gives
θ + δ(M) = constant
on
m+
on
−
(11.16)
θ − δ(M) = constant
m
and these equations are good analogs of (10.10). Thus,
we have our characteristics (m± ) and our invariants
[θ ± δ(M)].
Figure 11.4.
The geometry used in the example of a
Prandtl-meyer solution. The wall makes a sudden turn by
angle ∆ away from the initial flow direction. m−
characteristics starting from the sudden turn, are illustrated.
From Thompson figure 9.13
Finally, then, we can use this formalism to describe
Prandtl-Meyer flow around a corner. To pick an example, consider upstream flow at M1 , meeting a corner
that turns by some angle ∆. The flow angle θ has a discontinuity at the corner; the interesting characteristics,
m− , diverge from here.
Now, the entire flow field is covered by characteristics m+ which originiate in the uniform, upstream region 1 (where θ → 0): they have
θ + δ(M) = δ1
2
We didn’t bother with this in chapter 10, feel free to bug a mathematician if you’re interested.
everywhere. The other characteristics have
θ − δ(M) − 2θ − δ1
m−
on
where δ1 = δ(M1 ). From this, we get
on m− :
θ = constant
and
(11.17)
δ(M) = δ(M)1 = θ = constant
Thus, each m− characteristic is a straight line, with
constant δ(M) and θ values everywhere along it. There
will then be a first intersting m− characteristic, on
which θ = 0◦ and δ(M) = δ1 = δ(M1 ); and a last
intersting one, on which θ2 = −∆, and δ2 = δ1 + ∆;
the final M value can be found from this, through the
Prandtl-Meyer function.
References
Mostly from Schreier and Thompson, here.
63
12. SIMILARITY SOLUTIONS
Similarity solutions can be a powerful method when
they apply. Consider an example: a point explosion
going off in a uniform atmosphere (such as a supernova
explosion in a uniform interstellar medium). We expect
a spherical shell will be driven out into the medium.
How does this shell expand with time (what is the law
for its size, R(t)?), and what is the gas structure within
the shell?
We can skip ahead if we realize that there are only
two scaling parameters in the problem: the total energy
E, and the ambient density ρo . We expect the shell radius must involve only these two parameters, and the
time t since the explosion. If we further guess that the
expansion goes as a power law right (this turns out to
be right; if we had made a bad guess our later analysis would show up the inconsistency), there is only one
way to form a distance out of energy, density and time.
That is, R(t) ∝ (E/ρ)1/5 t2/5 . This must be the form
of the dymamical law for the outer shell. We can skip
further ahead: the internal structure of the expanding
bubble must also be described in terms of this scaling
law. For instance, r must appear only in combination
as r/R(t); velocity only as v/Ṙ(t); and so on.
Putting this scaling into the basic dynamical equations looks rather horrible, but actually leads to a very
useful simplification of the analysis. This is best done
by example, and I present three in what follows. You
will see that the art of similarity solutions lies in understanding the problem, physically, beforehand, so that
you can make a good guess as to the scaling laws and
similarity variables. As often in solving physics problems, hindsight is a great help.
A. Blast Waves: the Sedov-Taylor Solution
We start with our initial example, a spherical blast wave
resulting from a point explosion.
Dump a quantity E of energy, instantaneously, into
a perfect-gas atmosphere with density ρo . The energy
appears in the gas as kinetic and internal energy, with a
strong psherical shock propagating away from the origin. Let Vs is the shock velocity, and ρs , ps and vs are
the conditions just behind the shock. The jump conditions (9.12) for a strong shock become
2
ρo Vs2
γ+1
γ+1
ρo
ρs =
γ−1
2
vs =
Vs
γ+1
ps =
(12.1)
We see from this that the only constant dimensional
quantities which enter the problem are E and ρo ; these
along with r and t consitutue the independent variables
of the problem. (The dependent variables are v, p, ρ).
We want to put the basic equations into a dimensionless form. The standard approach in similarity solutions
is to search for a dimensionless variable, η, which rules
the problem. In general we want
η = (constant)rλ t−µ
(12.2)
with λ, µ and the constant to be determined from the
specific problem. Next, we set up scalings of the other
physical quantities:
v = rt−1 ṽ(η) ;
p = r−1 t−2 p̃(η) ;
ρ = r−3 ρ̃(η)
(12.3)
Derivatives of all the quantities must also be transformed. For instance,
µη d
∂
→−
;
∂t
t dη
∂
λη d
→−
∂r
r dη
(12.4)
(Think: why?). When this is done, the PDE’s from the
basic equations are transformed into ODE’s. If there
are no independent physical scales in the problem (for
instance, does the ambient density have a length scale?
If so then we can’t solve the problem this way), then
these ODE’s can be solved once and for all time, giving
us the solution to the problem.
Returning to the blast wave problem, the arguments
in the introduction give us the dimensionless combination of the basic variables:
³ ρ ´1/5 r
o
(12.5)
η=
E
t2/5
and this is the similarity variable for this problem. The
outer shock must then be labelled by some value of η,
say ηo ; so that the shock position as a function of time
is
µ ¶1/5
E
t2/5
(12.6)
Rs (t) = ηo
ρo
That is, we expect the radius of the outer shell to be
close to our dimensional analysis guess, (12.6), but we
can’t be sure of the exact coefficient yet. From (12.6),
the shell/shock velocity it
2 Rs
2
Vs (t) =
= ηo
5 t
5
We will find that ηo ≃ 1.
µ
E
ρo
¶1/5
t−3/5
(12.7)
64
We can go further and determine the gas structure
inside the outer shock. We work with dimensionless
forms of the basic variables:
5(γ + 4) u
4
r/t
25(γ + 1)
p
p̂ =
8
ρo r2 /t5
γ−1 ρ
ρ̂ =
γ + 1 ρo
at the shock front. The gas flow behind the shock is, as
usual, described by the basic three equations:
v̂ =
∂v 1 ∂p
∂v
+v
+
=0
∂t
∂r ρ ∂r
∂v
∂ρ 2ρv
∂ρ
+ρ
+v
+
=0
∂t µ ∂r
∂r¶ r
∂
p
∂
ln γ = 0
+v
∂t
∂r
ρ
(12.8)
and each of these are only functions of η. (Check back
with equations 12.3). The boundary conditions of the
problem are
v̂(ηo ) = 1 ;
p̂(ηo ) = 1 ;
ρ̂(ηo ) = 1 . (12.9)
After a decent bit of algebra, these can be written in
terms of the “hatted” variables:
dv̂
dp̂
1
ρ̂ (2v̂ − γ − 1)
+ (γ − 1)
=
[ρ̂v̂(5γ + 5 − 4v̂) − 4(γ − 1)p̂]
dη
dη
2η
¶
µ
3v̂
γ + 1 1 dρ̂ dv̂
+
=−
v̂ −
2
ρ̂ dη dη
η
d
p̂
1 5(γ + 1) − 4v̂
ln γ =
dη ρ̂
η 2v̂ − (γ + 1)
But now, the bottom line: these equations are coupled
ODE’s, in the basic variable η. They can be integrated
directly (that means numerically), for a given value of
γ. This is the Sedov-Taylor solution, whose behavior is
shown in Figure 12.1.
The value of ηo is also found numerically, from conservation of energy. That is, the total energy must satisfy
¶
Z Rµ
1
1 2
ρv +
p 4πr2 dr
(12.12)
E=
2
γ
−
1
0
and in dimensionless terms, this is
Z 1
¡ 2
¢ 4
32πηo5
ρ̂v̂
+
p̂
ξ dξ
1=
25(γ 2 − 1) 0
(12.10)
(12.11)
Figure 12.1. (Numerical) solutions for the gas structure
inside the shock, as a function of the dimensionless variable
η = r/rsh , where rsh ≃ (Et2 /ρo )2/5 . From Shu, Figure
17.3.
(12.13)
where ξ = η/ηo . Numerical evaluation, after the p̂(η)
and v̂(η) solutions are found, gives ηo – which turns
out to be close to unity (γ = 1.34 ⇒ ηo = 1.000;
γ = 1.40 ⇒ ηo = 1.0033; γ = 1.67 ⇒ ηo = 1.153).
Finally, we should note that this Sedov solution does
not apply over the entire life of the explosion. It describes the early phases, in which the blast energy E is
entirely contained in the explosion (Shu calls this the
energy conservation phase). At later times, radiative
losses become important,1 the outer edge of the explo1
Recall that radiative energy losses depend only on the microphysics, that is the density and temperature; they thus have a
characteristic time scale which is not included in the analysis of
this problem
65
sion gets denser and becomes a dense, thin shell. Past
this point the expansion is governed by momentum conservation rather than energy conservation. . . which is
another topic that we won’t address here.
B. Prandtl-Meyer flow, revisited
In Chapter 11 we slogged through the characteristicbased solution to this problem; now, let’s repeat the
analysis using similarity methods. Here, we do not have
an easy identification with a physical scale (there is no
“radius of the shell” to consider). Instead, we guess that
our fundamental variables must be angles (or, equivalently, the two cartesian coordinates (x, y) must only
appear as the ratio x/y). Figure 12.2 has the geometry,
in today’s notation.
(the corner) to the point (r, ψ) under consideration. The
irrotational equation becomes
1 dv
dθ
+
cot(ψ − θ) = 0
dψ v dψ
(12.15)
and, using the useful connection,
µ ¶
µ ¶
dρ
dh ∂ρ
dv ∂ρ
ρv dv
=
=v
=− 2
dψ
dψ ∂h s
dψ ∂h s
c dψ
(12.16)
(where we used the chain rule; energy equation; and
Gibbs relation, dh = T ds + vdp, to find (∂ρ/∂h|s =
ρ/c2 ). With this, the continuity equation becomes
cot(ψ − θ)
¢ 1 dv
dθ ¡ 2
+ M −1
=0
dψ
v dψ
(12.17)
Equations (12.15) and (12.17) describe the system, as
two ODE’s in ψ.
Now, we can solve these two. Isolating dθ/dψ and
then eliminating it, we get
p
cot(ψ − θ) = ± M2 − 1)
(12.18)
Figure 12.2.
Geometry for the Prandtl-Meyer solution.
From Thompson figure 10.1
The basic equations are, still, the three in (11.8).
Rewriting them in polars, we have
∂vr
− vψ = 0
∂ψ
∂
ρvψ = 0
∇ · (ρv) = ρvr +
∂ψ
v2
h+
= constant
2
∇×v =
(12.14)
(where we have used the definition of enthalpy h, from
Chapter 6). We now note that there is no characteristic length scale in the problem, either in (12.14)
nor in our boundary conditons (upstream quantities
→ p1 , v1 , ρ1 , θ = 0; downstream θ → θ2 , the angle
of the corner). Thus, the spatial variabiles x, y must
appear only in the dimensionless form y/x. We pick
the polar angle ψ = tan−1 (y/x) as the fundamental
variable in the solution. This means there must be no
dependence on the radial variable r.
Now: write vr = v cos(θ − ψ), and vψ = v sin(θ −
ψ). Recall, θ is the angle of the velocity vector, relative
to the x axis, and ψ is the polar angle from the origin
But, since the RHS is the cotangent of the mach angle,
we can write
ψ − θ = µM
(12.19)
This says that a radial line from the origin is a Mach
line, ie the characteristic m− . Using (12.18)), the equations of motion reduce to
1 dv
dθ p 2
+ M −1
=0
dψ
v dψ
This has the formal solution,
Z p
dv
θ+
M2 − 1
= constant
v
(12.20)
(12.21)
But, the integral on the RHS is, again, the PrandtlMeyer function (check equation 9.27). So, we have finally,
θ + δ(M) = δ(M1 )
(12.22)
This solves the problem: the bending angle θ2 is
known, from the boundary conditions. Thus, the expansion reaches δ(M2 ) = δ(M1 ) − θ2 , which allows
one to find M2 . Check back to chapter 11 – you can
verify that this gets us to the same place as the characteristic analysis we used there.
66
C. Gravitational Collapse: the Shu Solution
A nice third example of this is the gravitational collapse
of a self-gravitating cloud – such as a protostar. Frank
Shu has developed this area extensively, and I take this
(quite basic) analysis from his book.
What is the physical picture? Let’s start with a
static, isothermal sphere; take a = (kT /m)1/2 as the
sound speed. For this argument, let’s consider the simple version with ρ(r) ∝ 1/r2 , and ignore the unphysical divergeance at the origin. Imagine some perturbation, at t = 0, that makes the inside of the clour (protostar) slightly overdense. This will lead to an “insideout” collapse of the protostar. As the expansion starts,
for r ∼ 0, we do not expect the outer regions to know
about the expansion at all; an expansion wave will move
outwards, at a speed a, as the collapse proceeds. This
suggests a similarity variable, x = r/at. If we assume there is no important core (ignore the core discussed in (6.8) and following), then there is no physical
scale length in the problem, and we can apply similarity
methods.
Pick our basic equations, then. It is convenient to
rewrite the continuity equation (1.4) in terms of mass
shells, taking M (r) as the mass within radius r:
∂M
= 4πr2 ρ ;
∂r
∂M
∂M
+v
=0
∂t
∂r
(12.23)
For isothermal flow, the force equation is
∂v
a2 ∂ρ GM
∂v
+v
=−
−
∂t
∂r
ρ ∂r
r
ρ(r, t) =
(12.25)
v(r, t) = av(x)
Putting these into (12.23) and (12.24) gives, after some
algebra, an expression for the mass function,
m(x) = x2 a[x − v(x)]
Figure 12.3.
Self-similar solutions for Shu’s collapse
model. The solid line describes the density; the dotted line
the infall velocity; and the dashed line, the mass function.
From table 18.1 of Shu.
Applications and extensions of this may be discussed in class or in the homework (depending on
Jean’s mood and time).
(12.24)
We convert to similarity variables:
α(x)
4πGt2
a3 t
M (r, t) =
m(x)
G
These are the heart of this problem (compare the sets
12.11, or 12.15 and 12.17). Shu presents numerical solutions, which are illustrated in Figure 12.3.
(12.26)
and the two coupled ODEs,
·
¸
£
¤ dv
2
2
= (x − v)a −
(x − v)
(x − v) − 1
dx
x
·
¸
£
¤ 1 dα
2
(x − v)2 − 1
= α − (x − v) (x − v)
α dx
x
(12.27)
References
The Sedov solution is the classical example here; I
followed Thompson and Shore, but it’s also in many
other references. I took the Prandtl-Meyer solution
from Thompson, and the Gravitational Collapse from
Shu’s book.
67
13. FOUNDATIONS OF MHD
In this chapter, we extend our fluid treatment to address
current-carrying fluids in magnetic fields. The extra
force term, from the j × B Lorentz force, adds – shall
we say – an interesting richness to the problem. I am
using several references for this part of the course, as I
have not found one single book which covers the material usefully. Thus, some of my references come from
labratory plasma work (confinement, stability, toroids),
and others come from space physics and astrophysics
(basic equations, flows, reconnection).
A. Remember your E&M?
We start with the full Maxwell:
∇×B=
4π
1 ∂E
j+
c
c ∂t
∇·B=0
1 ∂B
∇×E=−
c ∂t
(13.1)
where ρq is the charge density. A couple of simplifications are common in MHD. First, note that we usually
work with highly conductive fluids, that cannot support a finite space-charge density (why??). Thus, the E
fields present are those induced, which are O(vB/c) ≪
B in subrelativistic flows (our limit in this course).
Thus, ∂E/∂t ∼ O[(v/l)(vB/c)] = O[(cB/l)(v 2 /c2 )],
i.e. small, and we can also drop the displacement current in most of what follows. With these approximations, we can write the “reduced Maxwell equations:
4π
j
c
∇·B=0
∇×E=−
1 ∂B
c ∂t
We can also interpret (13.4) by noting that E′ is the
field felt by an observer (or an electron) moving with
the fluid element.
2.
∇ · E = 4πρq
∇×B=
In our reduced MHD, we can write 4πj = c∇ × B.
Now go to a frame moving at v (relative to the lab);
and keep v ≪ c to avoid worrying about relativity. In
that frame there will be a simple Ohm’s law, j′ = σE′ .
But we know the fields transform as
´
³
´
³
v
v
E′ = E + × B ; B′ = B − × E
c
c
(13.3)
(recall γ ≃ 1 here). Now, we know that any charge density ρq in the lab frame will contribute an extra current
ρq v in the comoving frame ... however with E small,
and v also small, any extra j′ is (v 2 /c2 ). We therefore
have j = j′ , and thus in the lab we can write
³
´
v
j = σE′ = σ E + × B
(13.4)
c
(13.2)
In a high density fluid collisions usually dominate particle motions, so that a scalar conductivity σ and the
vector form of Ohm’s law, (13.4), is enough. Occasionally, however, we need to extend the description to
consider microphysical effects in a low density plasma.
In particular, applying a E field to a magnetized plasma
will result in currents parallel and perpendicular to the
background field Bo . The details are in the Appendix;
here, just note that the Generalized Ohm’s law can be
written as
j = σo Ek + σ⊥ E⊥ + σH n̂ × E
1.
CONDUCTIVITY AND OHM ’ S LAW: I
To close the (reduced) system we need to connect fields
to currents. In most situations we can assume a vector
Ohm’s law, which in the absence of inductive effects
would be written j = σE, with scalar electrical conductivity σ. How do we include inductive effects here?
(13.5)
Here, Ek is the component of E along B; j, E⊥ is the
component of E across B; and b̂ is the unit vector along
B. The first term uses the normal (collisional) conductivity; the second uses the Pederson conductivity; and
the last (usually the smallest) uses the Hall conductivity.
∇·E=0
which will govern almost everything we do in this
course.
CONDUCTIVITY AND OHM ’ S LAW: II
B. Field Evolution: Induction Equation
How does the field evolve? We use (13.4), but consider
only the inductive part;1 and we work with ”reduced”
Maxwell (13.2):
E=−
1
c
v
×B+
∇×B
c
4πσ
(13.6)
Why? because we don’t expect any unshielded free charge to
stick around ... right?
68
Alternatively, using vector algebra and defining the
quantity η = c2 /4πσ, this becomes
∂B
= ∇ × (v × B) + η∇2 B
∂t
(13.7)
which is often called the induction equation. Two comments on this equation.
• Compare this to equation (4.8), which describes
the evolution of vorticity, in the barotropic limit. They
are just the same: and so we should expect the magnetic
field and the vorticity to behave similarly in hydrodynamic situations. One example is flux freezing, which
we derive immediately below: it’s the MHD analog of
Kelvin’s theorem. Another example is the existence of
magnetic flux ropes: a field subject to fluid forces tends
to bunch into linear, high-field regions (think of a tornado or a vortex line).
• Equation (13.7) has two important limits, as follows. When we worked with fluids, we noted that the
behavior of a system is sensitive to the ratio of the advective to dissipative terms. We expressed this ratio in
terms of the Reynolds number. The induction equation
admits a similar separation of effects. We take the magnetic Reynolds number
Rm =
LV
η
(13.8)
where L, V are characteristic length and velocity scales
of the problem. This quantity again measures the radio
of inertial/advection, to dissipative, terms.
1.
IDEAL LIMIT: FLUX FREEZING
An imporant limit in MHD is the limit when Rm ≫ 1,
called ideal MHD . This is the case of a highly conducting plasma. Here, the induction equationk (13.7),
reduces to
∂B
≃ ∇ × (v × B)
∂t
(13.9)
Now, (13.7) allows us to write this, generally, as
Z
dΦB
η∇2 B · dS
(13.13)
=
dt
S
Thus, as η → 0 (the perfect conductivity limit, an
“ideal” plasma), ΦB becomes a constant of the motion. That is, the magnetic flux through a loop which
is “attached to the plasma”, stays constant as that loop
is stretched (or compressed) by the flow field. The
mean magnetic field in that loop therefore decreases (or
grows), in proportion to the change of loop area. This
is Flux Freezing.2
Magnetic Field Lines? (A discussion from
Schmidt...) The concept of a magnetic line of
force is an abstraction. IN general no identity can
be attached to these lines (they cannot be labelled
in a varying field), nor can we speak of “motion”
of field lines. In a perfect conductor, however, the
concept of field lines becomes meaningful, due to
flux freezing.
Consider a material line in the fliud (say a chain
of labelled droplets, or particles painted pink), defined by intersecting material surfaaces. Choose
these surfaces everywhere tangential to B at t =
0. The flux through both surfaces is therefore zero
to start, and their intersection defines a field line at
that point. Flux freezing guarantees that these surfaces continue to ΦB = 0 at any later time. Thus,
their intersection continues to define a field line,
in fact the same field line – it has become identifiable; labelling the material (painting it pink) has
labelled the field line, and the local fluic velocity
v(x, t) is also the velocity of that section of the
field line. The field line is attached to – “frozen
into” – the fluid.
2.
while Ohm’s law becomes
v
E+ ×B≃0
c
The first term describes intrinsic changes in B, while
the second describes advection of the field through the
boundary. Now, Stokes’ theorem rewrites this as
¶
Z µ
dΦB
∂B
=
− ∇ × (v × B) · dS (13.12)
dt
∂t
S
(13.10)
Consider, now, a closed curve (C) bounding a surface
(S) which is moving with
R the plasma. The magnetic
flux through S is ΦB = B · dS; and its rate of change
can be written
Z
I
∂B
dΦB
B · (v × dl)
(13.11)
=
· dS −
dt
S ∂t
C
RESISTIVE LIMIT: FLUX ANNIHILATION
In a fluid with finite conductivity, flux freezing no
longer holds. We can explore this by going to the other
limiting case, when Rm ≪ 1. This is diffusive limit.
If we simply ignore the advection term, equation (13.7)
becomes
∂B
= η∇2 B
(13.14)
∂t
2
Compare Kelvin’s Theorem, from chapter 4; you will see that the
math is identical, as is the nature and interperation of the result.
69
This describes the effect of Ohmic dissipation on the
magnetic field; note that it is a standard diffusion equation.
Diffusion of field lines? We know how solutions
to (18.4) behave: an initial field will decay on a
timescale ∼ L2 /η. Some authors discuss this in
terms of field line “diffusion” or “slippage” out
of the fluid. Remember that the density of field
lines is related to the strength of the field; so a
lower density of field lines, with time, should correspond to field lines “diffusing” out of the field.
In particular, when η is finite, field lines are no
longer tied to parcels of the plasma; some authors
talk of field lines “moving through” the plasma in
dissipative regions.
C. Fluid Equations: Lorentz force
The effect of the B field on the force equation is
straightforward. We simply add the Lorentz force to
the momentum equation:
Dv
j
ρ
= −∇p + × B
(13.15)
Dt
c
Note, I have ignored viscosity here, as well as any external forces (such as gravity). Now: expand out the
Lorentz force as
1
j
×B=
(∇ × B) × B
c
4π
(13.16)
1
1
2
(B · ∇)B
= − ∇B +
8π
4π
This is an important breakdown of the Lorentz force;
it demonstrates that the field exerts a magnetic tension and a magnetic pressure on the fluid. The first
term in (13.16) represents the gradient of a scalar pressure, pB = B 2 /8π. It appears in the momentum
equation parallel to the fluid pressure....you can think
of trying to compress a magnetic field, parallel to itself, with the field resisting the compression (“fighting back”). The second term in (13.16) is non-zero
only if the field varies parallel to itself. A simple illustration is a curved field line. The curvature means
there is a current flowing along the field line; the j × B
force points inwards (relative to the curvature). Thus,
curved field lines “want to straighten out”...Some authors combine both effects by describing magnetic field
lines as “elastic bands within the fluid”, which resist
being stretched: either pushed together, or pulled transverse to their length.
D. Fluid Equations: Energetics
We need two items here: first recall the energetics of
the fields, by themselves; then their effects on the energetics of the fluid.
1.
ENERGETICS OF THE E AND B FIELDS
To remind yourself of this...go to Jackson.3 Keep all of
the terms – the full Maxwell’s equations. Consider the
quantity E · j: it expands as
4πj · E = cE · (∇ × B) − E ·
∂E
∂t
But now, use the vector identity ∇·(E×B) = B·(∇×
E) − E · (∇ × B), and Maxwell again, to get
c
∂E
∂B
∇ · (E × B) + E ·
+B·
4π
∂t
∂t
Now: we identify the the energy density in the fields, as
uE = E 2 /8π; and uB = B 2 /8π. And we integrate the
equation above over some volume V , with a surface S,
giving
Z
Z
d
−
j · EdV =
(uE + uB ) dV
dt
V
V
Z
(13.17)
c
(E × B) · dS
+
S 4π
−j · E =
Thus: the first term is the rate of change of field energy
in V . The second term – involving the Poynting flux,
S = (c/4π)(E×B) – is the flow of EM energy through
the surface. The last term, then, is the work done by
the fields on the sources in the volume (and it can have
either sign ... thinking about driving vs. dissipation .. I
think). This can also be written in differential form:
d
(uE + uB ) + ∇ · S = −j · E
(13.18)
dt
which looks much like any other conservation equation.
2.
ENERGETICS OF THE FLUID
Schmidt has a nice approach: I’ve redone it in cgs,
keeping all the terms. To start, dot v into the force
equation (13.15):
ρv ·
1
Dv
+ v · ∇p = v · j × B
Dt
¸
· c
(13.19)
1
1 ∂E
×B
=
v · (∇ × B) −
4π
c ∂t
where I’ve used the full Maxwell on the right. With
some algebra, the terms on the LHS can be written:
¸
·
∂v
+ (v · ∇)v
ρv·
∂t
¶
µ
¶
µ
(13.20)
∂ 1 2
1 2
=
ρv + ∇ ·
ρv v
∂t 2
2
3
Preferably the 1st edition, where he works in in cgs.
70
and
v · ∇p =
γ
1 ∂p
+
∇ · (pv)
γ − 1 ∂t
γ−1
(13.21)
just the same as the first part of the course. Now, consider the last term, using vector algebra to reorganize
each term:
where we’ve used p ∝ ργ in this last; otherwise these
are both general, for compressible fluids. So far, this is
·
¸
1
1 ∂E
1
∂E
1
v · (∇ × B) −
(v × B) ·
× B = − (v × B) · (∇ × B) +
4π
c ∂t
4π
4πc
∂t
(13.22)
Now ... use v × B/c = j/σ − E; and E · (∇ × B) = B · ∇ × E − ∇ · (E × B); expand out and collect terms, to get
µ
¶
1 2
1
∂E
∂B
c
v
· (j × B) = j −
E·
+B·
−
∇ · (E × B)
(13.23)
c
σ
4π
∂t
∂t
4π
Collecting everything, we get
·
¸
·
¸
∂ 1 2
1
1
γ
c
1
1 2
2
2
ρv +
p+
(E + B ) + ∇ · ρv v +
pv +
E × B = − j2
∂t 2
γ−1
8π
2
γ−1
4π
σ
This is still fully general; it combines the EM and fluid
terms in a conservative form... We can now follow the
usual MHD approach and drop E 2 in the ∂/∂t term (as
Schmidt does). Also note that Parker writes the Poynting flux, in the MHD case, as S = B × (v × B)/4π =
v⊥ B 2 /4π, explicitly assuming ideal and an inductiononly E field.
References
Once again, much of this basic material is “just from
me”. Possibly useful references might be Priest (who
has a very good MHD introduction), also Woods and
Davidson. I followed Schmidt for the energetics analysis, and cannot for the life of me remember where I first
found the Generalized Ohm’s law arguments (probably
Chen’s plasma book? ... sorry).
(13.24)
tween collisions, tcoll , or its inverse the collision rate,
νcoll = 1/tcoll , as follows. Consider a free electron, in
a plasma, subjected to an external electric field E. The
net force on the particle can be estimated,
Fnet ≃ eE −
∆p
∆t
(13.25)
where ∆p/∆t is the mean rate of momentum change
per collision. But if the charges have a net drift velocity
vD , we can estimate ∆p/∆t ∼ me vD /tcoll ; then, in a
steady state we have Fnet ≃ 0, so that
eE ≃ nvνcoll
(13.26)
and the drift velocity must be vD = eEtcoll /me . Next,
we can use this in the (static) Ohm’s law, to relate the
conductivity to the drift velocity:
j = ne evD = σE
(13.27)
where the second equality defines σ. Collecting everything, we end up with
E. Appendix I: Conductivity notes
1.
σ=
ωp2
ne e2
=
τcoll
me νcoll
4π
(13.28)
COLLISIONAL CONDUCTIVITY
First, simple collisional (parallel) conductivity, σ. We
can find σ simply, in terms of the mean time be-
In the last expression the plasma frequency is used:
ωp2 = 4πne2 /m. The simplest physics is when the collision time is determined by Coulomb collisions in the
71
plasma, as in chapter 1 (check §1.F). Alternatively, the
collisions may be with microturbulence in the plasma
... in which case the collision rate usually has to be put
in by hand (rather than calculated from first principles).
2.
CROSS - FIELD CONDUCTIVITY
Consider a general situation in which E and B exist;
there will be Ek and E⊥ components, relative to B. We
can anticipate the three terms in the Generalized Ohm’s
law, as follows: (1) collisional conductivity, giving a
field-aligned current proportional to Ek , as usual; (2)
quasi-collisional conductivity, giving a cross-field current proportionao to E⊥ ; and (3) a transverse current,
perpendicular to both E and B, based on single-particle
E × B drift.4 Note that this drift is independent of particle charge and mass .. but we get a net current because the collision frequencies are not. There are two
approaches, macro and micro.
• macroscopic. Now: just consider one species,
still, and extend (??eqn13.26) as
´
³
v
(13.29)
e E + × B + mvνcoll = 0
c
Retain the definition j = nev; so that
mνcoll
1
j + j × B = −eE
ne
nc
(13.30)
Because we have cross products, things get complicated
.. choose the form,
j = σo Ek + σ⊥ E⊥ + σH n̂ × E
(13.31)
Put this into (13.30), do the algebra and collect the
terms5 We find,
ne2
mνcoll
ν2
σ⊥ = σo 2 coll 2
νcoll + Ω
νcoll Ω
σH = σo 2
νcoll + Ω2
σo =
5
dvx
= qEx + qBvy
dt
dvy
= qBvx
m
dt
dvx
= qEz
m
dt
m
with solutions
vx = vxo cos Ωt +
(13.32)
Quick physics here: a single particle undergoes gyromotion
around the local magnetic field, at a frequency Ω = eB/mc. If
there is also a perpendicular E field, the center of the gyro-orbit
shifts during one orbit. Following the particle’s trajectory, it’s
easy to show that the orbit-center moves at a steady drift speed,
∝ E × B.
Remember the definitions: ωp2 = 4πne2 /m, and Ω = eB/mc;
that’s how the frequencies ωp and Ω come into the expression.
vyo + qEx
sin Ωt
qB
vyo + qEx
qEx
(13.33)
cos Ωt − vxo sin Ωt −
qB
qB
qEz
vz = vzo +
t
m
vy =
(This shows the combination of gyromotion and E ×
Bdrift). Now, include collisions. Let νcoll be the collision frequency, and assume collisions occur randomly
in time. The probability of a collision is e−νcoll t n and
the probability that a particle will escape a collision in
time t, t + dt is given by νcoll e−νcoll t dt. Inbetween collisions the trajectories of (13.33) are followed. Thus,
the drift speeds are
hvx i = A
where Ω is the gyrofrequency, as usual. These are the
collisional, Pederson and Hall conductivities.
4
• microscopic. Alternatively, consider single particle motion. The particles undergo gyromotion around
B; they undergo an E × B drift; and they suffer collisions which disrupt these ordered motions. To illustrate (I take this from Park), let B = (0, 0, B) and
E = (Ex , 0, Ez ). The equations of motion, in the absence of collisions, are
Z
νcoll e−νcoll t (Ex /B) sin Ωtdt
Ex νcoll Ω
=
2 + Ω2
B νcoll
Z
hvy i = A νcoll e−νcoll t (Ex /B)(cos Ωt − 1)dt
Ex
Ω2
=−
2
B νcoll + Ω2
Z
hvz i = A νcoll e−νcoll t (Ex /B) sin Ωtdt
=
Ez
Bνcoll
(13.34)
R
if the normalizing factor A−1 = νcoll e−νcoll t dt. Finally, use this in the definition of current density, j =
72
nqhvi. We get
ne Ωνcoll
Ex
B ν 2 + ω2
ne Ω2
jy = −
Ex
B ν 2 + ω2
ne Ω
jz =
Ez
B νcoll
jx =
(13.35)
Compare this to (13.32); you can see we get the same
answers.
where, in SI, η = 1/µo σ is the magnetic diffusivity;
note that it differs from the electrical resistivity, which
is 1/σ.
Note of caution: just to confuse us all, the usual
cgs definition is η = 1/σ, and lightspeed terms
are carried through the equations explicitly. In
addition, some authors working in SI also take
η = 1/σ. Sigh . . . be sure to check the definitions in each different book you use.]
3. FORCE EQUATION , MAGNETIC TENSION AND
PRESSURE
F. Appendix II: Do It in SI
Both SI and cgs are used in the literature, which makes
it confusing to go between sources. For reference, I
store the main equations in SI.
1.
MAXWELL IN
SI:
∇ × B = µo j +
∇·B=0
∇×E=−
∇·E=
2.
The force equation is
Dv
= −∇p + j × B + F
ρ
Dt
with the separation of terms, as
1 ∂E
c2 ∂t
j×B=−
∂B
∂t
(13.36)
INDUCTION EQUATION
If we eliminate E from Maxwell # 1, Maxwell # 3 and
(13.3), we have
(13.37)
4.
ENERGY EQUATION
After dropping the charge-based E, but keeping the
Poynting flux explicitly:
·
¸
·
¸
∂ 1 2
1
1 2
γ
1
1
1
ρv +
p+
B + ∇ · ρv 2 v +
pv + E × B = − j 2
∂t 2
γ−1
2µo
2
γ−1
µo
σ
and that’s the end of this chapter, folks.
(13.39)
As disscussed in §13.C, the first term represents the gradient of a scalar pressure, which in SI ispB = B 2 /2µo .
The second term is non-zero of B varies parallel to its
length, and represents magnetic tension.
1
ρq
ǫo
∂B
= ∇ × (v × B) + η∇2 B
∂t
1
1
∇B 2 + (B · ∇)B
2µo
µo
(13.38)
(13.40)
73
14. SIMPLE MHD EQUILIBRIA
How many ways can you construct a steady magnetic
field? When you have done that, can it confine a
plasma? There are several common applications.
A. Potential Fields
Now and then we can simplify by considering static
magnetic fields in vacuum – that is fields arising from
a current which is confined to some finite spatial region. A typical example is the magnetic field of the
sun, above the solar surface.
Go back to Maxwell: in a current-free region, the
magnetic field satisfies ∇×B = 0. That means it can be
found from a scalar potential, Φm . Because ∇ · B = 0
always, the potential (and field) satisfy
∇2 Φm = 0 ;
B = −∇Φm
(14.1)
But this is Laplace’s equation, and has well-known solutions, usually expressed as series expansions. For instance, go back to your favorite E&M book. In spherical geometry the general solution is straightforward (if
long), and can be expresed in terms of spherical harmonics:
Φm (R, θ, φ) =
∞
X
l
X
l=0 m=−l
(14.2)
h
i
alm rl + blm r−(l+1) Plm (cos θ)eimφ
and for axisymmetry, this simplifies to
i
Xh
Φm =
al rl + bl r−(l+1) Pl (cos θ)
(14.3)
dynamics, as the fluid equations do not provide for intrinsic static equilibria (unless a gravitational field is
included). The plasma equations do, in principle, provide for self-confinement: if the plasma pressure can
just balance the Lorentz forces from the fields. Most
commonly, flows are ignored, as is resistivity. The general condition for equilibrium is, then,
j
× B = ∇p
c
(14.5)
This is, of course, can be solved, subject to the constraints
∇·B=0;
∇·j=0;
j=
c
∇×B
4π
(14.6)
(The second relation holds in steady state, right?). A
system which satisfies (14.5) also obeys
j · ∇p = 0 ;
B · ∇p = 0
(14.7)
That is, constant-pressure surfaces are also “magnetic
surfaces” and “current surfaces”: B and j lines lie in
constant-p surfaces.
Now... (14.5), with its auxilaries (14.6) and (14.7),
is “all” that is needed for laboratory confinement. We
just have to solve it, and then test for stability. In
these notes I confine myself to infinitely long plasmas
in cylindrical geometry. The basic equation, (14.5), becomes
Ã
!
2
2
Bφ2
d Bφ + Bz
dp
+
+
=0
(14.8)
dr dr
8π
4πr
l
where Pl is the Legendre polynomial. The coefficients
al , bl or alm , blm , are determined by the boundary conditions. Working in cylindrical geometry, with coordinates (r, φ, z), is a bit less general, because we have to
choose particular boundary conditions. One choice of
solution is
X
Φm =
[cn Jn (kmn r) + dn Yn (kmn r)] einφ±kmn z
m,n
(14.4)
where Jn and Yn are Bessel functions, and the eigenvalues kmn are chosen by looking at zeros of the Bessels
(for instance see Jackson’s E&M).
B. Plasma Confinement
Plasma confinement is the fundamental problem for
laboratory plasmas. There is no real parallel in fluid
There are several common applications in the literature.
1.
THETA PINCH
The simplest is a field which is everywhere parallel
to the axis of the cylinder. For instance (following
Woods), set up azimuthal currents in the wall of a cylinder which contains plasma; opposing currents are induced in the plasma, and the resulting axial magnetic
field, Bz , can’t easily penetrate the plasma. Instead,
the magnetic force compresses or “pinches” the plasma,
until a pressure balance is reached.
To quantify this: if B = (0, 0, Bz ), then confinment
must have Bz rising outside of the plasma. The equation is simple:
d
dr
µ
B2
p+
8π
¶
=0
(14.9)
74
so that we have a simple, radial pressure balance. For
instance, one possible equilibrium (illustrated in the figure) is
One possible equilibrium (illustrated in the figure) is
Bφ (r) =
−r 2 /a2
p(r) = po e
h
i
2
2
Bz (r) = Bo 1 − βo e−r /a
(14.10)
with βo = 8πpo /Bo2 .
Such a system is effectively confined by the box
(which carries the φ-currents). We will see that it is
stable; however losses out the ends are severe, so that it
isn’t particularly useful for laboratory confinement.
Bo r
1 + r2 /a2
(14.12)
(exercise for the student: what are the corresponding pressure and current density profiles?) This type
of pinch can be self-confining.R Note that the curr
rent within radius r is I(r) = 0 2πjz rdr, and from
Maxwell the B field is Bφ (r) = 2I(r)/rc. Using these
and the pressure balance condition (14.11),
Z a
1 2
I
(14.13)
prdr =
2 a
4πc
0
where Ia is the current in the entire pinch (out to radius
a), and we’ve assumed p(a) = 0. (To the student: can
you derive this?). Thus, the plasma can self-confine if
it carries the right current. This type of pinch is attractive, in that particles don’t escape out the ends, and the
current is carried by the plasma itself. However, this
configuration turns out to be seriously unstable, thus is
also of little practical interest.
B
j
z
3
j
B
2
1
z
2.5
0
0
1
2
3
r
Figure 14.1. Above, the geometry of a linear theta pinch;
note the B field is axial and the current azimuthal. Below,
qualitative equilibrium profiles for a theta pinch. Following
Freidberg figure 15.1, 15.2
2
1.5
1
2.
BENNET PINCH OR Z PINCH
0.5
The other limit is a purely azimuthal field: B =
(0, Bφ , 0), so that magnetic tension that confines the
plasma. The most interesting case is that in which the
plasma confines itself by carrying just the right net current Io . The basic relation is
Ã
!
Bφ2
Bφ2
d
(14.11)
p+
=−
dr
8π
4πr
0
0
1
2
3
r
Figure 14.2. Above, the geometry of a linear pinch: the
field is azimuthal and the current is axial. Below, qualitative
solutions for a linear pinch. Following Freidberg figures
15.4, 15.5
75
3.
GENERAL SCREW PINCH
A combination of these two cases is clearly possible,
in which both Bz and Bφ are non-zero. The general
equilibrium is, then, (14.8). One example, which may
be considered in more detail in the homework, comes
from Freidberg: consider
Bφ =
r
2I
2
c r + a2
(14.14)
where I is the plasma current and a is a scale length.
Let the pressure and Bz field be related by
Bz2 (r)
=
Bo2
− λp(r)
The simplest approach here is to pick a simple form
for α, usually constant, and look for analytic solutions.
Note one simple consequence: taking the divergeance
of (14.16), we have
∇ · (αB) = (B · ∇)α = 0
so that α must be constant along field lines; that is, B
lies on surfaces of constant α; these are magnetic flux
surfaces. If we take α = constant everywhere, in fact,
(14.16) becomes
¡ 2
¢
∇ + α2 B = 0
(14.15)
where Bo is the externally applied field and λ is a constant, determined experimentally.
A mixed pinch of this type turns out to be more stable than the simple Z pinch, due to its axial field. However it retains the problem of losses out the ends. The
next step might be to bend the ends of the screw pinch
together, to form a toroid. The simplest form of that is
a tokamak; many, many trees have been lost to suppor
the books and papers analyzing this geometry. It turns
out still to suffer serious instability problems. You can
go one step further by twisting the cylinder, about its
axis, before connecting the ends...that gives you a stellarator. I have no intention of pursuing the specialized
geometry that these machines require, in these notes.
j×B=0;
∇ × B = αB
(14.16)
where α is, thus far, some general function. These are
force-free fields; they are discussed in lab plasma literature and occasionally in astrophysics. Now, note
that we cannot say there is no plasma pressure; there
must be a net current, by definition, so there must be
some matter. Rather, we are saying the plasma pressure is small – very small – compared to the magnetic
forces. This approximation holds well for some laboratory plasma configurations (see “Taylor relaxation”,
later in the course). Astrophysically, it is often applied in the solar corona, where pb ≫ pgas ; and can
be applied to other idealized, strongly magnetized systems (such as pulsar magnetospheres and, possibly, radio jets).
(14.18)
Now: the mathematically astute among you will recognized immediately that this is the vector Helmholtz
equation. (The rest of you – like me – can refer, for
instance, to Morse & Feshbach chapter 13). It has wellunderstood, if complicated, solutions. Here, I quote a
couple of useful solutions without deriving them.
1.
CYLINDRICAL GEOMETRY
The system becomes (following Priest)
d
dr
Ã
Bφ2 + Bz2
8π
!
+
Bφ2
4πr
=0
(14.19)
dBz
= αBφ
−
dr
C. Force-Free Fields
Another track considers solutions to the MHD force
equation when both gravity and plasma pressure are
negligible. In this case, we must have
(14.17)
But this has Bessel function solutions:
Bφ (r) = Bo J1 (αr) ;
Bz (r) = Bo Jo (αr) (14.20)
as originally found by Lüst & Schlüter (1954). These
solutions describe helical fields, lying on coaxial cylindrical magnetic surfaces; the pitch of the helices increases going away from the axis.
2.
SPHERICAL GEOMETRY
Chandrasekhar & Kendall (1955) found the general solution, in terms of spherical harmonics and spherical
Bessel functions. The axisymmetric version of this
(m = 0 in the general Ylm term) is
ψ=
X
An r−1/2 Jn+(1/2) (αr)Pn (cos θ)
n
which yields the following
(14.21)
76
Br =
X
Cn n(n + 1)r−3/2 Jn+(1/2) (αr)Pn (cos θ)
n
h
i dP
n
Cn −nr−3/2 Jn+(1/2) (αr) + αr−1/2 Jn−(1/2) (αr)
dθ
n
X
dPn
Bφ = −
Cn αr−1/2 Jn+(1/2) (αr)
dθ
n
Bθ =
X
These solutions describe nested, toroidal flux surfaces;
the field lines are again helices on the flux surfaces.
3.
NON - LINEAR FIELDS
A third interesting case has uniformly twisted solutions,
and does not assume α = constant. In cylindrical geometry, a field line is traced in (φ, z) by
rdφ
dz
=
Bφ
Bz
(14.23)
Now, a field whose line-twist is independent of r must
satisfy
Bφ
= br
Bz
Bφ d
dBz
+
(rBφ ) = 0
dr
r dr
(14.26)
Combining this with b = constant gives this solution
Bo
;
1 + b2 r 2
Bφ(r) =
Bo br
1 + b2 r 2
(14.27)
This has the property that field lines at different radii
are twisted through the same angle; so that the whole
tube is twisted like a solid body.
4.
In Chapter 6 we addressed the question of hydrostatic
equilibrium: the static support of a fluid, by its own
pressure, in a gravitational field. When we add a magnetic field, the static balance becomes much more complex:
∇p = ρg +
(14.25)
Requiring that this field also be force-free, that is has
j × B = 0, gives
Bz =
conditions, to keep things well behaved. For cylindrical solutions, a physical boundary at some outer radius
makes sense (say from external pressure or a wall). For
the spherical solution just discussed, we can also impose a wall (as in the lab), or can also assume the α
constant is non-zero only inside of some R. This requires that the usual magnetic boundary conditions be
met across r = R, and keeps the currents (sources)
confined to a finite region.
D. Gravitational Equilibrium I
Thus, the twist of a field line between 0 and z is
Z
Z z
Bφ
dz
(14.24)
ϕ(z) = dφ =
0 rBz
Bz
(14.22)
BOUNDARIES
These three solutions illustrate a general result: there
are no force-free fields for which j ∝ ∇ × B is confined to a finite volume (as we would expect for a finite
source region), and for which B is O(r−3 ) at infinity.
Thus, we need to impose physical boundaries, or other
j
×B
c
(14.28)
Recall the gravitational acceleration can be written
in terms of a potential, g = −∇Φg . The threedimensional nature of the Lorentz force, j × B, makes
this a much more complicated problem if we treat it
fully generally. Let’s avoid that and find some simple
approaches.
1.
PLANAR GEOMETRY: THE GALAXY
Let’s start with a nice and simple case, originally due to
Parker.
Consider a simple planar system, such as the plane
of the galaxy. Let the magnetic field be horizontal,
and fully mixed with the gas. Picking ẑ as the vertical direction, that means we take B = By (z)ŷ; and
describe the gas by density ρ(z), pressure and sound
speed p(z) = c2s ρ(z), and gravitational potential Φg (z).
The simplest assumption we can make is that the
ratio of gas to magnetic pressure is constant at all altitudes: B 2 /8πp = αo = constant. Magnetostatic bal-
77
1. “ CONVECTIVE ”
g
B
Figure 14.3. A cartoon of the vertical structure of the
magnetic field and gas structure in the disk of the galaxy.
The grey scale is meant to represent the local density, which
decays vertically (presumably as an exponential if g is
constant). from Shu figure 23.3.
ance, then, can be written
p dΦg
dp
=− 2
(1 + α0 )
dz
cs dz
(14.29)
INSTABILITY
Think back to the hydrostatic atmospheres in chapter 6.
There, we considered a small vertical displacement of
a “blob” (that is, gas parcel), and compared its density
after displacement to the local, external density. We
found that a hydrostatic atmosphere is convectively unstable – the blob will keep rising – if the atmospheric
temperature decreases more slowly with height than a
fiducial value (the adiabatic gradient).
Now, let’s do an analogous calculation with a magnetized atmosphere. To simplify we choose an isothermal atmosphere; that would be stable according to our
analysis in chapter 6, but it can be unstable here. Assume the magnetic field lines are horizontal; displace a
magnetized parcel vertically by some δz, without bending the field lines. 1 The parcel starts at the conditions
of the ambient atmosphere: ρin = ρo and Bin = Bo .
Let the parcel density and field change by δρin and
δBin . By flux freezing and mass conservation, with
this geometry, we know for the parcel that
which has the solution,
·
¸
Φg (z)
p(z) = p(0) exp −
(1 + αo )c2s
·
¸
Φg (z)
By (z) = By (0) exp −
2(1 + αo )c2s
δBin /Bo = δρin /ρo
(14.30)
Thus, the field as well as the gas must be concentrated
toward the central plane of the galaxy. If we take a
constant gravitational acceleration, then Φg (z) = go |z|,
and we recover the expected exponential dependence of
the pressure (and the field).
Looking to useful tools ahead, we can also find the
vector potential of this field. Recall that B = ∇ × A in
general. In this geometry, A only has a component in
the x direction, A = Ax (z)x̂, and By (z) = dAx /dz.
In this simple geometry, it’s easy to “invert” the solution
for By (z) and show that
Ax (z) = −sgn(z)
2
(14.31)
(1 + αo )2 By (0)
go
·
¸
go |z|
× exp −
2(1 + αo )c2s
Note that the magnetic field lines lie in surfaces of constant Ax .
E. Magnetic Bouyancy
The solution above has a difficulty: it is unstable. Magnetic fields tend to be buoyant, thus want to rise against
gravity. There are several approaches to this situation,
depending on the problem one wants to solve. We’ll
work through two of them.
(14.32)
Now, we again assume the parcel remains in pressure
2 /8π = p2 +
balance with its surroundings: pin + Bin
o
2
Bo /8π. This translates to
δpin +
Bo
Bo
δBin = δpo +
δBo
4π
4π
(14.33)
But now, if we assume the temperature doesn’ change,
we get two results.
First, we know that the parcel will continue to rise
if δρin < δρo . As long as Tin is not too different from
Tout , we see from (14.33), that δBin > δBo is required
for instability. That makes sense: a larger B in the risen
parcel requires a smaller ρ, to stay in pressure balance;
thus, bouyancy. It’s more useful, however, to get our
instability condition in terms of the atmospheric structure. Using (14.32) and (14.33), again with the condition that δρin < δρo , we get the condition for bouyant
instability:
µ ¶
δρo
d Bo
δBo
<
;
<0
(14.34)
Bo
ρo
dz ρo
Thus: an isothermal atmosphere can be unstable if the
magnetic field strength falls off faster than the density
does.
1
So this is really displacing a “flux tube”, a cylindrical object containing field and plasma.
78
This analysis can be extended by considering adiabatic changes of the parcel density and temperature,
as in chapter 6; we may explore this in class or in the
homework.
2.
SCALE
OF
UNSTABLE
PARKER I NSTABILITY
PERTURBATIONS :
Now, let the magnetic structure bend as it rises. The
gas is free to flow along the lines, and thus will accumulate in the “valleys”; the “tops” of the bent field lines
will thus be lighter than the surrounding gas, and will
rise due to bouyant forces. This instability turns out to
be important for long wavelenghts: only large perturbations are unstable.
To see this, first consider an isolated flux tube, such
as might give rise to a sunspot. Let the external gas have
a density scale height H = kB T /mg (refer back to
chapter 6 for hydrostatic equilibrium with no magnetic
field). If the magnetic field is confined in the flux tube,
and the intial state is in pressure balance, we have the
internal/external balance,
pi +
B2
= pe
8π
(14.35)
buoyancy force will overcome the tension, leading to
instability. Comparing these two forces, we find instability occurs if
g∆ρ >
B2
;
4πλ
λ > 2H
(14.37)
We can make the same argument for the magnetic
bouyant (in)stability of a flat flux sheet (as in the planar
model of the galaxy, above). Our static solution gives a
new scale height:
p
ρ
B2
(14.38)
=
= 2 = e−z/Λ
po
ρo
Bo
¡
¢
2 /2 /g, in direct extension of the
where Λ = c2s + vA
nonmagnetized scale height from above (check back to
Chapter 6). Let the flux sheet be raised some small ∆z,
with the gas sliding down into the “valleys” again. If the
perturbation has horizontal scale λ, its effective curvature radius is, from simple geometry,
r≃
λ2 2
16 ∆z
The density at the top of the perturbation is now (take
“o” to be at the height of the undisturbed sheet)
¶
µ
∆z
−∆z/H
(14.39)
ρi (∆z) ≃ ρo e
≃ ρo 1 −
H
(Note there is no magnetic pressure supporting the gas
inside the bent flux sheet). The change in external density does know about magnetic pressure however:
µ
¶
∆z
−∆z/Λ
ρ3 (∆z) ≃ ρo e
≃ ρo 1 −
(14.40)
Λ
Figure 14.4. The geometry of a sub-surface flux tube
before it erupts from the sun due to bouyancy, and its
possible post-eruption state; from Tajima &Shibata figure
3.17.
Assuming the gas inside is at the same temperature,
it must be at lower density than the outside. Thus leads
to a bouyancy force,
Fbuoy = g (ρe − ρi ) = g∆ρ =
B2
8πH
(14.36)
Say, now, that the flux tube is bent upwards, locally,
with a radius of curvature λ. If λ is short, magnetic
tension will pull the tube back towards its initial position, giving a stable system. If, however, λ is long, the
If we again compare bouyancy to the restoring force of
magnetic tension, we find the instability condition for
this case:
λ2 >
16Λ2
(1 + 1/β)2
(14.41)
if β = 8πp/B 2 is the ratio of gas to magnetic pressure.
References
The general cylindrical MHD equilibria are discussed in several books; Priest and Bateman are good
sources. Moffatt discusses the force-free solutions. I
took the Magnetic Buoyancy discussion from Shu.
79
15. NOT-SO-SIMPLE EQUILIBRIA
Most magnetostatic systems are a bit more complex
than the simple ones in Chapter 14. In this chapter we’ll
look at two important methods. First we’ll look at ways
to describe MHD equilibria, by reducing vector DE’s to
scalar DE’s (through flux functions, described below).
Then we’ll look at a process by which an MHD system,
left to itself, will “relax” to its own equilibrium (this is
called Taylor relaxation).
equation. To do so we must introduce flux functions. In
this section we’ll work in Cartesian coordinates; later
we’ll use a similar approach in cylindrical.
z
B
A. Flux functions I: Gravitational Equilibrium
Revisited
da
x
To start, think back to the last chapter, which we ended
with simple, planar equilibria. What if we can’t assume
the field is planar? Our basic magnetohydrostatic equation,
1
−∇p + ρg + j × B = 0
c
(15.1)
is now a more complicated vector equation, and hard to
solve. becomes harder to solve. How can we proceed to
search for an equilibrium (stable or unstable)? A common procedure introduces what is called a flux function
– that’s a scalar function that can be used to find the
vector B field.1
Before we start here, you should check back to the
stream function, ψ, that we introduced in chapter 3.
Some points about that ψ are good analogs for what
we’re doing here, as follows.
• The stream function is a scalar that can be used
to find the vector v field.
• Solutions for ψ(r) are generally found by solving some sort of DE.
• The stream function ψ “labels” streamlines.
From this, it follows that δψ = ψB − ψA (the
difference in ψ values on two streamlines A and
B), is proportional to the fluid flow between those
streamlines. (You can work this out if you don’t
believe me.)
So, now let’s do some algebra. We’re working in Cartesian geometry here.
1.
DEFINE THE FLUX FUNCTIONS
To start, pick a two-dimensional problem: let the field
B = B(x, z) = (Bx , 0, Bz ). A common approach
allows us to reduce the vector equation (15.1) to a scalar
1
This approach may look intimidating, but people do it because
it’s almost always easier to solve one scalar equation than three
inter-related equations for vector components.
Figure 15.1. Geometry for Cartesian problem.
The
R
magnetic flux through the square is ΦB = B · da. Using
our definition (15.2),
per unit
R length in the y
R and working
R
dz
=
dΨ. See the text
direction, ΦB = Bx da = ∂Ψ
∂z
for more details.
If the B field has only two components, we may
choose its vector potential to have only a single component, A = (0, Ψ, 0). We then have
Bx =
∂Ψ
∂Ψ
; Bz = −
; B = ∇Ψ × ŷ
∂z
∂x
(15.2)
It’s easy to show that the magnetic field lines lie on surfaces of constant A. We can see this from (15.2), by
simply noting that B is perpendicular to the gradient of
A; thus it must lie in constant-Ψ surfaces. Alternatively,
we can consider the differential of Ψ along a field line:
dΨ =
∂Ψ
∂Ψ
dx +
dz = −Bz dx + Bx dz
∂x
∂z
(15.3)
But, a field line obeys
dx
dz
=
Bx
Bz
(15.4)
Thus, (15.3) shows that dΨ = 0 along a field line.
Thus, the scalar function Ψ labels magnetic surfaces
(or magnetic field lines in a 2D representation of the
geometry).
The function Ψ can also be called a flux function. To
see why, recall the integral form of the defining equation for A:
Z
I
ΦB = B · da = A · dl
(15.5)
80
For our geometry, consider the magnetic flux through
a surface taken in the (x, z) plane (think of a unit
length in y to make a reasonable mental picture, so
da = dydz → dz). This is just
Z z
∂Ψ
dz = Ψ(z) − Ψ(0)
(15.6)
ΦB =
o ∂z
Thus: if we take Ψ(0) = 0, then Ψ(z) measures the
magnetic flux contained “within” (below in this case)
the surface of constant Ψ.
2.
APPLY THEM : SOLAR MAGNETIC ARCHES
Now: put this formalism into the magnetostatic equation, (15.1). We note several useful facts.
Fact 1. The associated current is easy to find:
j = (c/4π)(∇2 Ψ)ŷ
(15.8)
Details here: The ∇2 Ψ term is a scalar function, which
multiplies the vector ∇Ψ. We have taken an important step: we have reduced our system to one scalar
unknown, Ψ, rather than the vector unknowns B and j.
Fact 2. What is the pressure distribution? Consider
some field line, locally B, at a direction θ to the vertical.
The component of (15.1) parallel to B is
dp
− ρg cos θ;
ds
dp
= −ρg
dz
(15.9)
if ds = dz/ cos θ is the differential distance along the
field line. The second equality just converts from ds
to dz, and shows that normal hydrostatic equilibrium
obtains along the field line. This can be written,
−
p(z) = po (Ψ)e
Rz
o
dz/H(z)
(15.10)
height.3
if H(z) = kB T (z)/mg is the local scale
I
have noted the explicit dependence of po on the local
2
3
j · ∇p ≃ 0 .
Thus, the magnetic field and current lie nearly in surfaces of constant pressure. Deviations occur only on
> H(z) (for an example see the homework).
scales ∼
Fact 3. we can consider the full force equation as a
system to be solved for the function Ψ(z) (and thus the
field structure). The gravitational and pressure terms
can be combined as
2
which can also be written,
0=−
B · ∇p ≃ 0 ;
ρ∇Φg + ∇p = −e−Φg /cs ∇q
You can find this in your favorite EM book, or derive it
directly from ∇ × (∇ × A).2 From (15.2) and (15.1),
the components of the Lorentz force are
·
¸
j
1 ¡ 2 ¢ ∂Ψ ¡ 2 ¢ ∂Ψ
×B=
, 0, ∇ Ψ
∇ Ψ
(15.7)
c
4π
∂x
∂z
1 ¡ 2 ¢
j
∇ Ψ ∇Ψ
×B=
c
4π
value of Ψ at the footpoint of the line (for instance
the surface of the sun). This result also shows us that
surfaces of constant pressure nearly coincide with field
lines. For scales ≪ H(z), the pressure is nearly constant, and hydrostatic balance becomes j × B ≃ c∇p.
Dotting this with B or with j gives
BIG NOTE: ∇2 Ψ = ∂ 2 Ψ/∂x2 + ∂ 2 Ψ/∂z 2 is a scalar, with
simple form in Cartesian.
Compare to the scale height in Chapter 6 for a uniform T .
(15.11)
where, in the last step, we have introduced the usual
gravitational potential, given by g = −∇Φg , and have
2
defined the function q = peΦg /cs . Combining this with
(15.8), we can write the full force equation, (15.1), as
−
1
2
(∇2 Ψ)∇Ψ = e−Φg /cs ∇q
4π
(15.12)
This balance requires that ∇Ψ and ∇q be parallel to
one another; so that surfaces of constant Ψ must also
be surfaces of constant q. It follows that we can pick
q = q(Ψ), i.e. that q can be written as a function of Ψ.
Given this, our basic equation turns into
2
∇2 Ψ = −4πe−Φg /cs
dq
dΨ
(15.13)
This is an important result: we have reduced the system
from a general vector equation, (15.1), to a quasilinear
PDE in Ψ(x, z). If we can pick q(Ψ) (for instance, constant or linear or some other simple function), we can
find solutions for Ψ(x, z) by standard PDE methods.
An example of this will appear in class or in the
homework.
B. Flux Functions II: Grad-Shafranov Equation
Another version of flux functions is applied to plasma
confinement problems which can be reduced to two independent variables (such as an axisymmetric cylinder
problem). Again, the general equilibrium condition, in
the absence of gravity, is
j
× B = ∇p
c
(15.14)
81
Once again, it follows that j · ∇p = 0 and ∇p · B = 0.
That is, B and j are normal to ∇p; each of B and j lie
in constant-p surfaces. These are called magnetic surfaces or flux surfaces; they contain the field lines. Most
interesting equilibria consist of sets of nested magnetic
surfaces. Once again, we will label these surfaces with
flux functions, ψ.4 We will have ψ = constant in a flux
surface; and we can write p = p(ψ), so that p(ψ) is
constant on the surface given by ψ = constant.
z
B
One function, ψ, is enough if the field has no φ component. For the more general case we need a second
scalar function. Consider the total current J, within
surface S which is planar and lies perpendicular to the
z axis (refer back to Figure 15.2 for such a surface):
Z 2π Z r
Z
jz rdφdr
(15.17)
j · da =
J=
S
0
0
But now, since rjz = ∂(Bφ r)/4π∂r in axisymmetry,
we can write
c
(15.18)
J = Bφ r
2
This is our second surface function, J(ψ), the axial current crossing through the surface S. It pairs with p in
the analysis, both functions being constant on a ψ surface.
r
2.
APPLY THEM : THE G - S EQUATION
To start, we ¡can get¢ the toroidal current density from
jφ = (c/4π) ∇2 A φ :
Figure 15.2. Geometry for the axisymmetric cylindrical
problem. The magnetic flux through the circle,
R
perpendicular to the z-axis, is ΦB = B · da = Bz 2πrdr.
Using (15.16), this becomes ΦB (r) = 2πrAφ = 2πψ(r, z)
(you should derive this!). See the text for more details.
1.
DEFINE THE FLUX FUNCTIONS
We need to introduce two scalar functions. The first is
the φ- component of the vector potential, which can be
related to ψ, the poloidal flux function
ψ(r, φ) = Aφ r
(15.15)
(Figure 15.2 shows the basic geometry for the simpler,
axisymmetric case. As with the Cartesian case, the
scalar function ψ(r) labels the flux within a circle of
radius r. It follows, as before, that the poloidal field
components can be written
¶
µ
ψ
φ
⇒
Bpol = ∇ × φ̂
r
(15.16)
1 ∂ψ
1 ∂ψ
Br = −
;
Bz =
r ∂z
r ∂r
4
Yes, the notation has changed slightly ... its conventional to
use (lower case) ψ for flux functions in non-cartesian problems
(which tend to be laboratory plasmas, in cylindrical or toroidal
geometries).
·
µ
¶¸
c ∂2ψ
d 1 ∂ψ
c
jφ = −
+r
= − ∆∗ ψ
2
4πr ∂z
∂r r ∂r
4π
(15.19)
where we have followed convention and defined
·
µ
¶¸
1 ∂2ψ
d 1 ∂ψ
∗
∆ ψ=
+r
(15.20)
r ∂z 2
∂r r ∂r
to shorten the notation. From this we can write the general B field in terms of two scalars J and ψ:
¶
µ
1 ∂ψ 2J 1 ∂ψ
,
,
(15.21)
B= −
r ∂z cr r ∂r
Similarly, the total current density is
µ
¶
1
1 ∂J c ∗ 1 ∂J
−
j=
, ∆ ψ,
2π
r ∂z 2
r ∂r
(15.22)
These surface functions, J and ψ, can be used to
reduce the equilibrium condition, (15.14), to one DE
rather than three. As we noted abive, because both the
pressure p and the axial current J are surface functions,
we can write p = p(ψ), and J = J(ψ). Our most
useful relation is the r-component of (15.11):
1
∂p
= (jφ Bz − jz Bφ )
∂r
c
But now, since p = p(ψ), we can write ∂p/∂r =
(dp/dψ)(∂ψ/∂r). We can then cancel the ∂ψ/∂r
82
terms, insert the full expression for jφ , and get one form
of the G-S equation:
· 2
µ
¶¸
1 ∂J 2
1
∂ 1 ∂ψ
∂ ψ
∂p
=0
+
+
+r
∂ψ 2πr2 c ∂ψ 4πr2 ∂z 2
∂r r ∂r
(15.23)
This is the simplest interesting version of the G-S equation, outside of that in Cartesian coordinates, that I
know. (Most are set up for toroidal coordinate systems,
to be relevant in the lab, and are truly horrible ..) If
p(ψ) and J(ψ) are known functions of ψ, then this is a
PDE in the scalar flux function, ψ. Once we solve this
for ψ, everything else we need is known (check 15.19,
15.21, 15.22).
3.
EXAMPLE : SIMPLE PINCHES
To solve the G-S equation, one chooses functional
forms for dp/dψ, and dJ/dψ, then finds solutions of
(15.23).
As a very simple example, let’s assume J(ψ) = 0,
and look for a solution independent of z (think of
plasma confinement in a simple pinch geometry). The
G-S equation becomes
µ
¶
1 ∂ 1 ∂ψ
∂p
=0
(15.24)
+
∂ψ 4πr ∂r r ∂r
But, we know Bz = (1/r)(dψ/dr); so this equation is
just
µ
¶
d
Bz2
p+
=0
(15.25)
dr
8π
(does that look familiar?). A slightly more complicated
solution is the screw pinch. We again assume the equilibrium is independent of z, but retain J(ψ) 6= 0. If we
multiply the G-S equation by (4πr2 )(dψ/dr), we get
J dJ
∂p Bz dBz
+
+
=0
dr
4π dr
4πr2 dr
(15.26)
But because J = cBφ r/2, the G-S equation becomes
dp
d
+
dr dr
Ã
Bφ2 + Bz2
8π
!
+
Bφ2
4πr
=0
(15.27)
which recovers our earlier result for cylindrical plasma
confinement.
The previous example was overkill; we did not need
to go through the G-S equation to recover the result.
Most situations are more complex, however, with z as
well as r dependence. Searches for a solution start by
assuming an analytic form for p(ψ) and J(ψ).5 Most
realistic p(ψ) and J(ψ) choices requires numerical solution of the equation; this is the current approach in
the literature. However, some analytic solutions were
found early on, and are still useful for gaining insight.
You will see an example in class or in the homework.
C. Helicity and Taylor relaxation
Now, a different track ... return to the force-free fields
which we discussed in §14.3. There, we presented some
simple solutions; in this section we discuss why they
might be interesting.
This subject was motived by laboratory plasmas,
which exhibit self-organization. These plasmas, in tin
cans, tend to evolve spontaneously towards a small
number of preferred states, which are independent of
the initial conditions of the system. The structure of
these preferred states depends only on a few global parameters, such as total magnetic flux, and on the geometry of the system (say, cylindrical or toroidal). Examples are the reversed field pinch (RFP), the spheromak,
and the tokamak (each of which have particular stable
field configurations, that I’m not going to describe in
detail here). The process by which the plasma evolves
from an arbitrary initial condition to its long-lived final state is called plasma relaxation. Taylor (1974, and
later) suggested that this is a variational process – in
which the system evolves toward a state of minimum
energy, but constrained by one or more invariants. To
describe this we need two mathematical tools.
1.
HELICITY
First, we must define helicity. For a magnetic system,
the helicity is defined by the integral,
Z
H = B · AdV
(15.28)
where the integral is over some relevant closed volume
V : that of the system, or that of some particular flux
surface, depending on the application. Let this volume be bounded by a surface S. Helicity can only
be uniquely defined in a closed system: one in which
the magnetic field lines do not cross the surface. To
see this, recall that the vector potential A is only defined up to an arbitrary gradient (the choice of which
is called the gauge). Consider some gauge transform;
5
This sounds arbitrary, yes; but think even of simple solutions
we’ve seen to (15.25) or (15.27). Those problems also start by
“assume we know the behavior of at least one of the functions”,
then proceed to find the behavior of the other(s).
83
let A → A′ = A + ∇χ, where χ is some arbitrary
function. The effect of this on the helicity is
H′ − H =
Z
v
∇χ · BdV =
Z
S
χB · n̂dS
• How can we understand helicity? Moffatt &
Tsinober (1992) give two useful illustrations. For
one illustration, consider a circular magnetic flux
tube, in which each field line is a circle running
parallel to the axis of the tube. The helicity of
this state is (clearly?) zero. Let the magnetic flux
of this tube be ΦB . Imagine now that we cut the
tube at any section, twist it through an angle 2πγ,
and reconnect. Each field line in the tube is now
a torus knot; and any pair of field lines is now
linked. This new system has finite helicity.
• Or, consider a second, more complicated example. Two magnetic flux tubes, topologically linked
through each other. The integral in (15.28) simplifies to two line integrals, around the axes of each
tube:
I
I
H = κ1
A · dl + κ2
A · dl
C2
But now, the first integral is equal to the flux of
field through a surface spanning loop C1 , that is
±κ2 ; and similarly for the second integral. Thus,
the net helicity is H = ±κ1 κ2 .
Having defined helicity, we now need to introduce two
important conservation laws. First, helicity is an invariant in an ideal, closed MHD system. Second, the minimum magnetic energy available to a system at a fixed
(invariant) value of the helicity corresponds to that of a
force-free field. Here are the proofs for each of these.
2. I NVARIANCE OF THE HELICITY
First, we want to establish that H is an invariant in ideal
flow. The general rate of change can be written,
¶
∂A
∂B
B·
+A·
dV
∂t
∂t
V
·
µ
¸
¶
Z
∂A
∂A
−∇ · A ×
· ∇ × A dV
=
+2
∂t
∂t
∂H
=
∂t
E+
(15.29)
Thus, H is uniquely defined only if B · n̂ = 0 (that is,
if there is no component of B across the boundary; n̂ is
the surface normal vector.)
C1
To work with ∂A/∂t, we need to make a choice of
gauge. Recall the general expression,
Z µ
(15.30)
1 ∂A
= −∇Φ
c ∂t
if Φ is the scalar electric potential. Note, further, that
E + v × B = 0 in an ideal system. If we work now in
a gauge where ∇Φ = 0, we can show H is constant.
• method 1. One way to proceed (following Priest)
is in terms of conditions on A. In an ideal system, we
have
∂B
= ∇ × (v × B) ;
∂t
∂A
= v × B . (15.31)
∂t
The first relation in (15.31) comes from the basic induction equation, for an ideal (η → 0) system. The second
comes from “uncurling” B in the first, and choosing a
gauge with Φ = 0. From this, (15.30) can be written,
¶
Z µ
Z
∂A
∂H
B · (v × B)dV
=−
A·
· n̂dS + 2
∂t
∂t
S
V
(15.32)
But the second term is clearly zero. Thus we find that
H is invariant in systems for which A is held constant
on the boundary.
• method 2. An alternative proof comes from
Biskamp, who works with conditions on B and v. Still
using the fact that B · E = 0, we rewrite (15.30) as
Z
∂H
=
A · ∇ × (v × B)dV
∂t
ZV
(15.33)
[(A · v)B − (A · B)v] · n̂dS
=
S
From this last, we find that H is invariant if B · n̂ =
v · n̂ = 0; that is, if neither B nor v connect across the
boundary.
Both of these methods show that H is constant for
an ideal, closed system (highly conductive and not connected to the outside universe).
3.
THE MINIMUM ENERGY STATE
A second important relation is the fact that when the
magnetic energy,6
Z
1
W =
B 2 dV ,
(15.34)
8π
6
This is traditional notation ... W instead of E ... sorry, folks –
JAE.
84
is minimized subject to the constraint H = constant,
the resultant field is force free.
We prove this using a Lagrange multiplier technique. That is, we consider small perturbations to a
magnetic system, and search for the state which satisfies
δ (W − αH) = 0
(15.35)
In particular, following Priest, consider small perturbations which ahve δA = 0 on S (the surface; again a
fixed A thereon), and δB = ∇ × δA. The change in
W is then (linearizing and subtracting αH),
Z
8πδW = 2B · δB − α (δA · B + A · δB) dV
(15.36)
Algebra allows us to rewrite this as
Z
8πδW = ∇ · (−2B × δA + αA × δA) dV
Z
(15.37)
+ 2 (∇ × B − αB) · δAdV
But the first term again can be written as a surface integral, which vanishes due to δA = 0 on S. We are left,
then, with the second term:
Z
1
(∇ × B − αB) · δAdV
(15.38)
δW =
4π
But this says that we have a minimum energy state if
the field is force-free:
δW = 0
if
∇ × B = αB
in which the Bessel-function solutions (14.20) are seen.
It has also been suggested to occur in solar flares (which
involved a sudden, dramatic release of energy), and
even in radio jets (where it has been used to predict the
magnetic field structure). We should note, however, that
applications to astrophysics, while very tempting, are
hampered by the lack of rigid boundaries. The derivations above made heavy use of the tin-can boundary that
we have in the lab.
How does this work? How can magnetic energy be
dissipated while magnetic helicity is not? There is still
a fair amount of discussion about this; the sense of the
literature is that this works because magnetic helicity
decays much more slowly than magnetic energy does.
Following Bellan, think about the dimensions of magnetic energy vs. magnetic helicity: W ∼ B 2 L3 , while7
H ∼ ABL3 ∼ B 3 L4 , where L is the characteristic linear dimension of the system (or the scale of the Fourier
component which holds most of the power). Now, let
B be decreased (for instance by resistive dissipation);
the total energy decays ∝ L3 , while the helicity decays
∝ L4 . Thus, the smaller the scale, the bigger is the ratio
of energy loss to helicity loss.
We can do this a bit more formally,following Orlanti & Schnack. Magnetic energy decays due to
resistivity, as
Z
dW
= −η j2 dv
(15.40)
dt
while helicity decays as
Z
dH
= −2η j · BdV
dt
(15.39)
Thus: we have shown that the state of minimum
energy, energy, given a specified value of helicty, is a
force-free state.
4.
(to get this last, we started with the definition,
(15.28), used dA/dt = −cE, and Ohm’s law).
But now, if we
P think in terms of Fourier components, B = k bk eikr for a set of wavenumbers
k, then we have
TAYLOR RELAXATION
This result has an interesting application in the lab (and
presumably elsewhere). Lab plasmas are observed to
“relax” spontaneously. If a plasma starts in some initial state (specified by the experimenters, say), it goes
through a turbulent state (which develops due to instabilities in the initial configuration), and ends up in a
final, longer-lived configuration. Taylor hypothesized
that this corresponds to the plasma dissipating magnetic energy (through self-generated turbulence), while
maintaining its initial value of helicity (as determined
by the initial configuration). Its final state should then
be a force-free state, as in (14.16). This is indeed observed, most famously in an experiment called the Reverse Field Pinch, which has cylindrical geometry and
(15.41)
X
dW
k 2 b2k
→ −η
dt
(15.42)
k
and
X
dH
kb2k
→ −2η
dt
(15.43)
k
Thus, again, at a given k, the energy decay ∝ k 2 ,
while the helicity decay ∝ k.
7
Remember how A connects to B.
85
So, finally, to close the arguments: turbulence, and
thus turbulent dissipation, is dominated by small scales
(large k values). We therefore expect the energy to dissipate more rapidly than the helicity does – thus supporing Taylor’s hypothesis (and its apparent validation
in the lab).
References
I’ve taken the flux-function material and solutions
from a mixture of Priest, Bateman and the plasma literature. The helicity and relaxation discussion comes
partly from me, and partly from references such as Moffatt and Biskamp, as well as the more specialized books
by Bellan, and Orlanti & Schnack.
86
16. MHD EFFECTS IN FLUID FLOWS
We now have the basics: we know, in principle, how
B fields interact with fluid flows. They exert magnetic
tension and pressure on the fluid; the fluid, in return,
modifies and/or determines the local B field, through
inductive effects (that is, the v × B EMF creates local
currents). In this chapter we visit some applications.
A. Magnetic damping and stirring
Magnetic fields can have some unexpected effects on
fluid flow – if you set things up right, they can either
accelerate or decelerate the flow. Here are a couple of
brief examples; we may discuss them more in class.
1. M AGNETIC DAMPING .
Magnetic fields can decelerate a flow. For a concrete
example, let’s say you try to drive a flow (maybe a jet)
across a pre-existing B field. Qualitataively, we know
that the flow will initially try to “stretch” the field lines;
the resultant magnetic tension will resist – and decelerate – the flow. More quantitatively, the EMF caused
by the flow, v × B, will generate a transverse current j;
this will give a “backwards” j × B force which will try
to decelerate the flow.
B
2. M AGNETIC STIRRING .
Magnetic fields can also induce motion in a fluid. One
simple example is a magnetic field rotated (at some angular frequency Ω) around a fluid that is initially stationary (think about external electromagnets rotating
around the cylinder).
Ω
B
Figure 16.2. Illustrating magnetic damping. A fluid sits
in a long cylinder (seen end-on). A magnetic field B
pervades the fluid, and is rotated at some Ω.
v
j <−> v x B
Figure 16.1. Illustrating magnetic damping. Sent a jet,
velocity v, into a transverse B field. The resultant EMF,
v × B, will drive a current: which way will the Lorentz
force act? Will it accelerate or retard the flow?
The energetics are simple. From Ohm’s law, j =
σ (E + (v/c) × B). If the flow is set up with no freecharge E (for instance by keeping all boundaries at the
same potential), then we find that the rate of work done
on the flow, by the Lorentz force, is
1
j2
1
v · (j × B) = − j · (v × B) = −
c
c
σ
1
D
= v · (j × B) − ∇ · (pv)
Dt
c
How does this cause motion? Think about this in a
frame in which the B field is at rest; in this frame the
fluid initially has an azimuthal motion vphi = Ωr. But
this gives us an EMF,
ΩrB
v
×B→
(16.4)
c
c
This induces a current, as usual (in which direction??),
which leads to a Lorentz force
j
σ
× B → 2 ΩrB 2
(16.5)
c
c
Think through the directions: this force will accelerate
the flow, in the direction of motion of the rotating B.
B. Channel flow with MHD: Hartmann flow
(16.1)
Connect this to the force equation (which we dot with
v to get an energy equation):
v·
Integrate this over a finite box (without losses through
the sides):
Z
Z
d
1 2
1
j 2 dV
(16.3)
ρv dV = −
dt
2
σ
Thus, we find that the energy lost from the flow, due
to the Lorentz force, is just balanced by ohmic losses –
which of course appear as heat in the flow: (Compare
the more general energy equation, at the end of Chapter
13 – this is a simple application of the same physics.)
(16.2)
Here’s a classic example of MHD effects on a fluid
flow; think of a low-Rm , laboratory-type channel flow.
We already saw pure-HD versions of this, in Chapter 9.
Here, we revisit this situation with MHD effects added.
We’ll find that MHD effects can either drive the flow (a
“pump”, or drive a transverse current (a “generator”),
depending on the cionfiguration of the experiment.
87
1. T HE BASIC SETUP
We get
The setup is: steady, viscous, incompressible, planar
flow between plates, and across a B field. As in the
case of Poiseuille flow, which we saw in Chapter 3, the
flow here is maintained against friction by a pressure
difference applied across the ends of the channel. The
presence of a magnetic field will, however, change the
nature of the solution.
The basic equations are
∂b
=0
∂z
∂ 2 v Bo ∂b
∂p
ρν 2 +
−
=0
∂z
4πµ∂z ∂x
¶
∂
b2
p+
=0
∂z
8π
Bo v + η
Play algebra on these, eliminate b, and find an equation
for v:
d2 v
Bo2
dp
− ρν 2 −
v=0
dx
dz
4πcη
∇×E=0
∇·v =0
(16.6)
η∇ × B = E + v × B
∇p − ρν∇2 v =
1
(∇ × B) × B
4π
(16.7)
(16.8)
Because dp/dx is constant, we have a differential equation which can be solved for v(z). If we apply no-slip
boundary conditions, the solution can be written
µ
¶
Haz
v(z) = A cosh Ha − cosh
(16.9)
l
The first one is from Maxwell, with ∂B/∂t = 0; the
third is from the steady-state induction equation (think
of “uncurling” ∂B/∂t = 0. The other two should be familiar to you. Once again, ν is the kinematic viscosity,
η = c2 /4πσ is the diffusivity, and σ is the conductivity.
where A is an integration constant (which can be determined from the problem), and the Hartmann number
has been defined as
1111111111111111111111111
0000000000000000000000000
0000000000000000000000000
1111111111111111111111111
0000000000000000000000000
1111111111111111111111111
0000000000000000000000000
1111111111111111111111111
0000000000000000000000000
1111111111111111111111111
How does this compare to the non-MHD channel
flows in Chapter 2? Well, the velocity at the center of
the flow is vo = v(z = 0) = A(cosh Ha−1). The form
(16.9) converts to the usual parabolic velocity profile,
as B → 0 (that is as Ha → 0:
¶
µ
z2
(16.11)
v(z) → vo 1 − 2
l
b
dp/dx
B
v
z
x
1111111111111111111111111
0000000000000000000000000
0000000000000000000000000
1111111111111111111111111
0000000000000000000000000
1111111111111111111111111
0000000000000000000000000
1111111111111111111111111
0000000000000000000000000
1111111111111111111111111
Figure 16.3.
The geometry for Hartmann flow: a viscous
conducting flow between parallel plates with a transverse B,
and also a component b k v. Following Somov figure 15.4
Figure 16.3 shows our setup. Let the plates lie at
z = ±l, with the flow along x̂. Things only vary with
z (so the driving ∇p = dp/dx) is constant along x, as
before). The imposed field is B = (0, 0, Bo ). We look
for induced fields b = (b(z), 0, ) and v = (v(z), 0, 0).
In general, we allow a constant transverse electric field,
E = Eo ŷ; but (you should be able to convince yourself
that) if we hold the boundaries at a fixed potential, we
will have E = 0 inside the flow.
2.
THE SOLUTION
To proceed, ignore any E field (for reasons given
above), and specify the system (16.6) to one dimension.
Ha =
lBo
(4πρνη)1/2
(16.10)
(Compare the channel-flow solutions in chapter 2). Alternatively, as Ha gets large, the solution flattens:
h
i
(16.12)
v(z) → vo 1 − e−Ha(l−z)/l
This is a much flatter velocity profile, with an exponential boundary layer close to the walls.
Note that everything else you might want for the solution – current, b, etc – can be found from the basic
equations once you have the v(z) solution.
3.
AN MHD GENERATOR OR AN MHD PUMP ?
What determines the on-axis flow, vo ? Combine (16.9)
with Maxwell, to find the current density in the duct:
µ ¶2
jy Bo
Ha
∂p
Haz
=
− ρνA
(16.13)
cosh
c
∂x
l
l
88
This integrates to give the total current, I =
IBo
∂p
= 2l
− 2ρνA
c
∂x
µ
Ha
l
¶
Haz
sinh
l
R
jy dz:
(16.14)
(Note, this latter is an implicit solution for the velocity
amplitude A).
Thus, the Hartmann flow speed depends on the applied pressure gradient and the Lorentz force. This allows two regimes of operation. (1) If the flow is maintained by an external pressure gradient, the duct operates as an MHD generator, driving the transverse current I (this is useful if you somehow complete the circuit external to the duct..). (2) Alternatively, one can
apply an external EM driver to create the transverse current I; interaction of this current with Bo gives rise to
a Lorentz force that makes the plasma move along the
duct, that is creates an MHD pump. This may be discussed further in class.
This is a quite different application, derived from space
plasmas... think of the earth’s ionosphere (which
is partly ionized, partly neutral) and magnetosphere
(which is fully ionized).
Picture: two plasma slabs, the top fully ionized, the
bottom an ion-neutral mix, separated by some lowerdensity plasma. A vertical magnetic field threads both.
Now let the top layer move at some vo ; if it is close to
ideal, flux freezing will pull the field lines along, and
thus exert a transverse force on the lower layer. The
system is governed, of course, by the momentum equation:
Dv
= −∇p + j × B − ρνcoll v
Dt
1.
PURE MHD
The spirit here is, ignore currents and E fields; use
Maxwell to convert j to B. We thus write
j × B ≃ −∇
Bo2
1
+
(B · ∇)B
8π
4π
(16.16)
But now, we ignore gradients in the magnetic pressure (they are small), and estimate the second term as
(B · ∇)B ≃ (By Bo /H)ŷ. We thus have two force
B
H
11111111111111111111
00000000000000000000
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
H 11111111111111111111
00000000000000000000
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
00000000000000000000
11111111111111111111
y
x
v
1
Figure 16.4.
Two plasma slabs, in relative motion. The
top slab moves at vo , and is tied by B to the bottom slab.
The slabs each have height H, which is also the inter-slab
spacing. The slabs are infinite in the y direction, and have
finite extent Lx in the x direction. Following Cravens
Figure 8.19.
equations, for the top and bottom slabs:
By Bo
Dvo dp
+
=−
Dt
dy
4πH
By Bo
Dv1
=
− ρ1 νcoll v1
ρ1
Dt
4πH
ρo
(16.17)
The first equation describes the “drivers”: either
Dvo /Dt or dp/dy must be finite, in order to drive the
system. The second equation describes the response of
the lower slab; clearly a steady state is possible there if
By Bo /4πH ≃ ρ1 νcoll v1 – in fact this will determine
v1 , the velocity of the lower slab.
But: what determines By , i.e. the curvature (and
tension) of the field lines? If we stick to the pure-MHD
approach, we answer this using the induction equation
(13.7). That is,
(16.15)
We allow a ∇p term, and have added a “friction” force,
ρνcoll v; assuming some collisional coupling term νcoll .
In addition we anticipate a Lorentz force, j × B.
We can analyze the system in two ways.
o
z
C. Magnetic Coupling of Two Plasma Slabs.
ρ
v
H
∇ × (v × B) + η∇2 B = 0
By
vo Bo
+η 2 =0
H
H
(16.18)
where the second equation is again from scaling/dimensional analysis. We can thus estimate By ≃
Bo Hvo /η; and use this in (16.17) to solve the full system. NOTE that we must still pick the appropriate resistivity η for the bottom slab – Cravens argues for crossfield Pederson conductivity, using the same ion-neutral
collision rate νcoll that provides the frictional force on
the lower slab.
2.
CURRENT- BASED APPROACH
The above is a perfectly good solution...but Cravens
notes that simple dimensional analysis may not be good
enough when By is small (and presumably numerical
89
j⊥o H = j⊥1 H, or j⊥0 = j⊥1 (because we’ve chosen the two slabs to have the same thickness). We can
therefore combine the results above, as
µ
¶
Dvo dp
1
+
ρo
(16.21)
v1 = −
ρ1 νcoll
Dt
dy
methods may have trouble). We can also gain insight
by repeating the analysis in terms of the currents and
electric fields induced by the motion. We expect two
cross-field currents, Pederson and Hall... The Pederson
current is the interesting one.1 The Lorentz force in the
upper slab acts to decelerate it; that in the lower slab
accelerates the slab:
which allows, again, the possibility of the two drivers,
Dvo /Dt and dp/dy.
To finish this, we still need to find j⊥o . Consider
the top slab: if it is collisionless, we expect E =
−v × Bo = −vo Bo x̂ in the top slab. But also, because this is a steady-state, one-dimensional problem,
we have ∇ × E = 0, thus E must be the same in the
bottom slab (as well as in the intervening space). It follows that the Bo lines “map” the field from one slab
to the other: the magnetic field lines are equipotentials.
We can, therefore, use Ohm’s law (for the Pederson current) to find j⊥o = j⊥1 = σ⊥ vo Bo ..thus finishing the
problem.
top : j × B ≃ −j⊥o Bo ŷ
bottom : j × B ≃ j⊥1 Bo ŷ
(16.19)
The two momentum equations become
Dvo dp
+
= −j⊥o Bo
Dt
dy
Dv1
ρ1
= j⊥1 Bo − ρ1 νcoll v1
Dt
ρo
(16.20)
We can again find an equilibrium solution for the lower
slab, which has v1 = j⊥1 Bo /ρ1 νcoll .
J
1
0
J
v
0
3.
_
z
z
y
B
J ||
1
0
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J_
000000000000000000
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000000000000000000
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000000000000000000
111111111111111111
000000000000000000
111111111111111111
000000000000000000
111111111111111111
v
1
Β
J
||
1111111111111111111
0000000000000000000
0000000000000000000
1111111111111111111
0000000000000000000
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0000000000000000000
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0000000000000000000
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0000000000000000000
1111111111111111111
J_
0000000000000000000
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0000000000000000000
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0000000000000000000
1111111111111111111
Figure 16.4. Two views of the problem, emphasizing the
induced currents. Left, same view as in the first figure;
induced j × B Lorentz forces as shown. The top slab carries
j⊥ out of the page, the bottom has j⊥ into the page. The
induced E is into the page throughout. Right, the same
situation, but seen along the y axis. The complete induced
current loop can be seen, with j⊥ within the slabs (not
labelled with subscripts due to limitations of Xfig), and
field-aligned jk connecting the slabs. Following Cravens
Figure 8.19.
Now: how do we find the current densities? First,
consider current conservation, ∇ · j = 0 in a steady
state. The Pederson current flows along x̂, and the
slab is finite in that direction. Thus, the cross-field
currents in the plasma slabs must be connected by
field-aligned currents inbetween the slabs, as shown in
the figure. In addition, currrent conservation requires
1
the Hall current is there, in principle infinite in extent, so without
gradients, and not interesting for this problem.
x
ENERGETICS AND EQUIVALENT CIRCUIT
Finally, consider the energetics of this example. We can
think of j · E as a “local” source of EM energy. In this
example, jo ·E > 0 while j1 ·E < 0. Thus, we can think
of the top slab as an MHD generator, which produces
energy that is dissipated in the lower slab, the resistor.
The powerRper unit length generated (top) or dissipated
(lower) is j · Edxdz ≃ Lx Hvo Bo j⊥o . This is also,
nicely, the same as the powerRtransferred from top to
bottom by the Poynting flux: (E × B)4π −1 dxdy ≃
vo Bp By Lx /4π (check: you can verify this is the same
result). We can also describe this problem in terms
of the total circult and a global resistance...let E ≃
vo Bo Lx be the EMF generated (volts); the resistance
of the lower slab is R = Lx /σ⊥ HLy (ohms); and the
current I = /R (amps), as it should be.
References
I’ve taken the magnetic damping/stirring material
mostly from Davidson. My Hartmann flow discussion
follows Woods & Somov. The plasma-coupling discussion follows Cravens.
90
17. WAVES AND SHOCKS IN MHD
Nonmagnetized flows have one important characteristic signal speed – that’s the sound speed, as we saw in
Chapter 7. When we add a B field, things get more
interesting ... there is more than one possible “signalcarrying” wave.
A. MHD Waves
In Chapter 7 we found that a density perturbation propagates at a speed cs = (∂p/∂ρ)1/2 . This is a critical quantity in fluid dynamics – the speed at which information can propagate in a gas. For a magnetized
plasma, the situation is (not surprisingly) more complex. There are two characteristic waves, Alfven and
magnetosonic. To start, we carry out a linear analysis
to discover what types of perturbations propagate, and
at what speeds.
1.
where c2s = γkB To /m is the square of the (unperturbed) sound speed.
These equations (17.3) can be combined, algebraed, and eventually made into one equation for the disturbance velocity, v1 :
∂ 2 v1
= c2s ∇(∇ · v1 )
∂t2
= [∇ × (∇ × (v1 × Bo ))] ×
v1 (r, t) = v1 ei(k·r−ωt)
ω 2 v1 =c2s k(k · v1 )
p = po + p1 ;
ω 2 v1 = c2s k(k · v1 )
(17.6)
(17.7)
Dotting this with k, and assuming k · v1 6= 0 (i.e.the
perturbed velocity has some component parallel to the
wavevector), this gives
ω = cs k
(17.1)
(17.8)
for the dispersion relation. This should make you feel
good – we have recovered sound waves.
v = v1
B = Bo + B1
Bo
4πρo
Now, finally, we’re in position to analyze basic wave
modes for this system.
As a prelude, note that if Bo = 0, (17.6 reduces to
Linearize these in the usual way. That is, take
ρ = ρ o + ρ1 ;
(17.5)
our linearized equation reduces to
+ [k × (k × (v1 × Bo ))] ×
Dρ
+ ρ∇ · v = 0
Dt
1
Dv
= −∇p +
(∇ × B) × B
ρ
Dt
µ4π ¶
D
p
=0
Dt ργ
∂B
= ∇ × (v × B)
∂t
∇·B=0
(17.4)
If we now assume a plane-wave solution,
BASIC STRUCTURE : LINEAR ANALYSIS
Start with a set of basic, unperturbed equations:
Bo
4πρo
2.
(17.2)
where the unperturbed (“zero” subscript) state is taken
as uniform and homogeneous. (The more general case
of background structure, won’t be addressed here).
With this, we get
∂ρ1
+ (v1 · ∇)ρo + ρo (∇ · v1 ) = 0
∂t
1
∂v1
= −∇p1 +
(∇ × B1 ) × Bo
ρo
∂t
4π µ
¶
∂p1
2 ∂ρ1
+ (v1 · ∇)po − cs
+ (v1 · ∇)ρo = 0
∂t
∂t
∂B1
= ∇ × (v1 × Bo )
∂t
(17.3)
∇ · B1 = 0
ALFVEN WAVES
These are waves in which the magnetic field dominates.
It exerts the restoring force; fluctuations in the plasma
density and pressure are either exactly zero, or unimportant. We can start by guesstimating the likely wave
speed. Recall that waves in an elastic wire propagate
due to the tension; as the field lines in a plasma exert a
tension Bo2 /4π, one might expect a wave speed
vA =
Bo
(4πρo )1/2
(17.9)
This is the Alfven speed, and it is, indeed, a useful scaling speed for waves in a magnetized plasma. We can
also note directly that ∇ · B = 0 ⇒ k · B1 = 0; so
that the magnetic field perturbation must be normal to
the wavevector.
91
angle terms explicit letting θ be the angle between k
and Bo ), and letting b̂o be the unit vector along Bo , we
have the Alfvenic wave dispersion relation:
Following Priest, we now specify to the magnetizedonly limit, that is, dropping the po terms from (17.3).
This is called the cold-plasma limit, and means we’re
ignoring any internal-pressure effects (wo we no longer
have sound waves). Doing this, rewriting to make the
h
i
ω2
2
2
v
=
k
cos
θv
−
(k
·
v
)k
cos
θ
b̂
+
(k
·
v
)
−
k
cos
θ(
b̂
·
v
)
k
1
1
1
o
1
o
1
2
vA
Dotting this with bˆo gives bˆo · v1 = 0; thus, waves in
this limit just have the perturbed velocity normal to the
ambient magnetic field. Dotting this with k gives
¢
¡ 2
2
k · v1 = 0
(17.11)
ω − k 2 vA
an Alfvenic disturbance described by
v1 = −
This has two separate solutions, as follows.
B0
mag tension
v
1
B1
B
mag tension
mag tension
Figure 17.1. Schematic of (shear) Alfven waves; the
perturbed B1 and v1 terms are perpendicular to the
background field Bo . Following Cravens Figure 4.16.
• (shear) alfven waves. This is the case of an incompressible perturbation:
∇ · v1 = 0 ;
k · v1 = 0
(17.12)
which is, of course, one solution to (17.11). These are,
thus, transverse waves (particle motion transverse to k,
B1 ⊥ Bo ). Using this, (17.10) gives
ω = kvA cos θ
(17.13)
for Alfven waves (sometimes called shear Alfven
waves). These waves have a phase speed vA cos θ,
which agrees with our simple argument for waves propagating exactly along Bo . Putting this solution back
into the linearized equations (17.3), the first two show
that there are no density or pressure perturbations: ρ1 =
p1 = 0. So, the plasma just moves back and forth with
B1 , without any compressive effects. (The same equation sshow that v1 = B1 / (4πρo )1/2 , so that v1 and B1
are in the same direction).
An interesting aside is that these solutions also describe finite-amplitude waves. Most wave solutions require small-amplitude perturbations (as we began with
here). However, with some algebra one can verify that
(17.10)
B1
(4πρo )1/2
;
|Bo + B1 | = constant
(17.14)
satisfies the full equations, 17.1), without the need to
linearize. One important consequence of these waves,
Priest notes, is that finite-amplitude waves do not tend
to steepen, and so dissipate must less readily than other
wave modes.
• compressional alfven waves The second solution
to (17.11) is
ω = kvA
(17.15)
which describes compressional Alfven waves. For these
waves, the linearized equations (17.3) show that v1 is
normal to Bo , and lies in the (k, Bo ) plane. It therefore
has components both along and transverse to k, and so
gives rise to density and pressure fluctuations.
3.
MAGNETOSONIC WAVES
When the gas pressure is dynamically comparable
to the magnetic field, the wave nature is different.
We might expect the wave speed to be a mixture of
compressive effects (through cs ) and magnetic effects
(through vA ).
B
1
k
v1
Bo
Figure 17.2.
Schematic of magnetosonic waves,
illustrating a compressive wave propagating at right angles
to the background Bo . Following Cravens figure 4.17.
Returning to (17.6) and keeping the pressure and
compresssive terms, the ruling equation here is
92
h
i
ω2
2
2
ˆo + (1 + c2 /v 2 )(k · v1 ) − k cos θ(bˆo · v1 ) k
v
=
k
cos
θv
−
(k
·
v
)k
cos
θ
b
1
1
1
s
A
2
vA
Dotting this with k and bo , in turn, gives two useful
relations:
¢
¡ 2
2
2
(k · v1 ) = k 3 vA
cos θ(bˆo · v1 )
−ω + k 2 c2s + k 2 vA
(17.17)
and
k cos θc2s (k · v1 ) = ω 2 (bˆo · v1 )
(17.18)
If k · v1 = 0, we recover the Alfven wave solution
(17.13). If this isn’t zero, we can combine these two relations to find the dispersion relation for magnetosonic
waves:
¢
¡
2 4
2
+ c2s vA
k cos2 θ = 0 (17.19)
ω 4 − ω 2 k 2 c2s + vA
For forward-pointing (ω/k > 0) waves, there are two
distinct solutions:
¢1/2
¢ 1¡ 4
1¡ 2
ω2
2
4
2
=
cs + vA
cs + vA
− 2c2s vA
cos 2θ
±
2
k
2
2
(17.20)
The minus sign in (17.20) gives what is known as
the slow mode; the plus sign gives the fast mode. The
Alfven speed lies inbetween these two wave speeds;
thus the Alfven wave is occasionally called the intermediate mode.
These two magnetosonic modes may be thought of
(as Priest notes) as a sound wave, modified by the magnetic field, and a compressional Alfven wave, modified
by the plasma pressure. If the B field becomes small
(vA → 0), the slow mode disappears, and the fast wave
becomes a sound wave. If the gas pressure becomes
small (cs → 0), the slow mode disappears again, and
the fast wave becomes a compressional Alfven wave.
4.
VALIDITY OF MHD WAVE THEORY
An important limitation on this MHD wave theory is
that it assumes simple, collective particle motion. At
higher frequencies, the single-particle motions in mixed
magnetic and electric fields cannot keep up with the
driving wave. The wave modes become more complicated as a result, and require a plasma-physical approach (following individual particles, or at least individual species). The common usage is that we can treat
waves by MHD methods for wave frequencies below
Ωi = eBo /mi c, the ion gyrofrequency.
(17.16)
B. MHD Shocks; Jump Conditions
In Chapter 9, we found that density perturbations –
sound waves – will steepen and can develop into
shocks. Just as there are more than one signal-carrying
type of MHD wave, there are more than one type of
MHD shock.
Also in chapter 9, we applied conservation laws
to determine the jump conditions at hydrodynamic
shocks. We will follow the same path here, and we will
find that there is more than one soltuion to the jump
conditions – more than one type of shock. In particular,
magnetosonic waves are compressive, and can steepen
into shocks; the two MS modes, fast and slow, connect
to two types of MHD shocks. Alfven waves, not being
compressive, do not steepen into shocks (although there
are formal discontinuities that one can find, associated
with Alfven waves.)
We work in a frame in which the shock is at rest; this
makes everything steady state. We also ignore dissipation; it matters within the shock, but here we idealize to
an infintely thin jump. We use to the notation of Chapter 9, that is [[A]] is the jump in A across a boundary.
The unit vectors are n̂, normal to the boundary, and t̂,
tangential to it. We start with the basic equations, and
for each write down the jump across the shock.
• Maxwell. We know ∇ · B = 0; this
[[B · n̂]] = 0 ;
Bn = constant
(17.21)
is the continuity of the normal component, Bn , of the
B field. (You recall that the jump in Bt can be finite if
there is a surface current).
• Mass flux. We have, in a steady state, ∇ · (ρv) =
0, so that
[[ρv · n̂]] = 0 ;
ρvn = constant
(17.22)
is the continuity of mass flux.
• Induction. In steady state, ∇ × (v × B) = 0.
Across the boundary, this gives
[[(v × B)t ]] = 0 ;
[[vn Bt ]] = Bn [[vt ]]
where we’ve used Bn =constant in the last.
(17.23)
93
• Momentum flux.
··
B · n̂ B 2
ρv(v · n̂) + pn̂ − B
n̂
−
4π
4π
µ
¶¸¸
=0
(17.24)
is the basic momentum flux. A useful alternative to this
(resembling Bernoulli’s equation) comes from dotting
it with n̂:
··
¢2
1 ¡
(B · t̂)2 − (B · n̂)
p + ρ(v · n̂) +
8π
2
¸¸
=0
(17.25)
We can also scalar multiply with t̂ to get
··
Bn Bt
ρvn vt −
4π
¸¸
C. Perpendicular (Normal) Shocks
=0
(17.26)
These last two can be turned into more useful forms if
we isolate the (conserved) mass flux, ρvn :
[[p]] +
ρ2 vn2
• Parallel Shocks have the shock normal parallel
to B. These are no different from unmagnetized
shocks. The fluid can flow along the B field without hindrance. There is no v × B term to maintain
an EMF, nor is there a current (as B is constant by
assumption).
• Perpendicular (Normal) Shocks have the shock
normal perpendicular to B. These are the simplest
cases of magnetized shocks.
• Oblique Shocks are, as before, the most complicated. My presentation follows Priest, and
Woods; the latter is the most thorough. There does
not seem to be an extensive literature on oblique
MHD shocks – the complexity of the results may
be why.
·· ¸¸
1 ££ 2 ¤¤
1
Bt = 0
+
ρ
8π
That is, let Bn = 0, but allow non-zero vn . A useful
fact: from (17.28), we know [[vt ]] = 0 in this case. We
can then, always, transform to a frame with vt = 0;
this simplifies things (and I drop the subscript on vn ).
The jump conditions now become, for perpendicular
shocks,
ρ1 v1 = constant
(17.27)
B1 v1 = constant
(17.31)
B12
= constant
8π
1 2
B2
v1 + h1 + 1 = constant
2
4πρ
and
ρ1 v12 + p1 +
ρvn [[vt ]] −
Bn
[[Bt ]] = 0
4π
(17.28)
• Energy flux. There are several useful forms here.
A basic form is
··
1 2
γ
1
ρv v +
pv −
(v × B) × B
2
γ−1
4π
¸¸
=0
(17.29)
As before (chapter 6), we can define h = e + p/ρ =
γp/(γ − 1)ρ, with e = p/(γ − 1)ρ; and again use continuity of Bn and mass flux, to write this as
··
v2
B2
vn2
+h+ t + t
2
2
4πρ
¸¸
Bn
[[vt · Bt ]]
4π
(17.30)
As before, the nature of the shock depends on its
orientation relative to the flow; and also, in this case,
on its orientation to the B field. We can think of three
cases.
ρvn
=
where “constant” means the quantity is the same upstream (subscript 1) and downstream (subscript 2) of
the shock.
The geometry is illustrated in the Figure.
B
v
1
1
000
111
111
000
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
000
111
B
2
v
2
S
Figure 17.3. A plane perpendicular shock; the magnetic
field is parallel to the shock front, and perpendicular to the
shock normal.
Again, let’s call the compression ratio X = ρ2 /ρ1
(where “2” is downstream and “1” is upstream”), and
94
let the upstream B field be scaled by b = B 2 /4πρc2s =
2 /c2 . The various equations can be written as,
vA
s
ρ2
ρ1
v1
v2
B2
B1
p2
p1
=X
=X
(17.32)
=X
= γM
2
µ
X −1
X
¶
+1+
¢
γb ¡
1 − X2
2
where the compression ratio is the solution of
£
¤
b (γ − 2)X 2 − γX
£
¤
= X (γ − 1)M2 + 2 − (γ + 1)M2
Solving (17.33) shows us that a perpendicular, magnetized shock is compressive only if the upstream flow
(relative to the shock) is above the fast magnetosonic
2 /c2 ; the upspeed. That is, X ≥ 1, if M2 ≥ 1 + vA
s
2 . Compared to
stream flow must satisfy v12 ≥ c2s1 + vA1
the HD solutions (Chapter 7), the effect of the B field is
to reduce the density jump X below its hydrodynamic
value (since the flow energy can go to magnetic as well
as thermal energy). At the same time, the pressure
and temperature jumps are larger than for the B = 0
case. These shocks have the same limiting behavior as
M → ∞: the density and field jumps asymptote to
(γ + 1)/(γ − 1); while the pressure and temperature
jumps rise without bound.
D. Oblique Shocks
(17.33)
Solutions to this are shown in Figure 17.4. (Details:
doing the algebra on equations (17.32), and eliminating
p2 /p1 , gives a cubic equation in X. I’ve factored out
(X − 1) (which is not zero in any interesting case), to
get this quadratic. Note also typos in two of Priest’s
equations have been fixed.)
MHD shocks at general angles can be more complex, of
course. To specify the geometry, let x̂ be the direction
along the shock normal, and ŷ be in the plane of the
shock, and in the plane of the B field.
1.
DO IT GENERALLY
The general jump conditions (or conservation laws) are
as follows. For mass
ρvx = constant
3
(17.34)
(where “constant” means has the same value on both
sides of the shock). The ∇ · B Maxwell-eqn becomes
2
Bx = constant
(17.35)
Two components of the force equation become
1
p+
B 2 Bx2
−
+ ρvx2 = constant
8π
4π
(17.36)
and
0
ρvx vy −
-1
5
10
15
20
Figure 17.4.
Numerical solution of the jump conditions,
17.11, for unmagnetized shocks and for two different values
2
of b = vA
/c2s . See text for discussion.
Bx By
= constant
4π
(17.37)
The induction equation becomes
vx By − vy Bx = constant
(17.38)
And, finally, the energy equation becomes
¶
µ
¶
µ
B·v
1 2 B2
B2
+ vx ρe + ρv +
vx p +
− Bx
= constant
8π
4π
2
8π
(17.39)
95
[[vx ]] − m [[ν]] = 0
These equations can, in principle, be solved for the
general shock jump conditions. But they are long and
complex, so most authors simplify by working in a useful reference frame.
2.
n [[vx ]] + [[p]] +
FIND A USEFUL REFERENCE FRAME
(17.42)
1
hBy i [[By ]] = 0
4π
(17.43)
1
Bx [[By ]] = 0
4π
(17.44)
At this point, it is very helpful to transform to a frame
moving along the shock face, at a speed
v1y
B1y
= v1x
B1x
(17.40)
Such a transform can always be found, if Bx 6= 0. In
this frame, the plasma velocity is parallel to the B field
on both sides of the shock; which in turn implies that
E = 0 on both sides.
This transform simplifies the situation, but the algebra is still complex. It also turns out that three solutions are possible: fast shocks (which develop from
fast-mode MS waves); intermediate shocks (which develop from Alfven waves); and slow shocks (which develop from slow-mode MS waves). I’ve seen two different analytic approaches, & will summarize both here.
3.
n [[vy ]] −
NOW SOLVE THE SYSTEM : I
¤¤
1 ££ 2
vx + vy2 + [[h]] = 0
2
and
Bx [[by ]] − m [[ν]] hBy i − mhνi [[By ]] = 0)
µ
(17.41)
c2s
2 ;
vA
[[By ]]
;
ǫ=
B1
S=
[[p]]
p1
1S
1
χ=
P+ ǫ
γ ǫ
2
P =
ρ2
χ + sin θ1
=1+ǫ
ρ1
1 + χ sin θ1
(17.49)
(17.47)
In these variables, the jump conditions combine to a
quadratic (taking γ = 5/3 explicitly):
¶
¶
µ
5
2
2
ǫ sin θ1 − 1 + S1 χ − (ǫ + 2S1 sin θ1 ) − 0
2 sin θ1 − ǫ χ − 2
3
6
• method # 1. which can be solved directly for χ, given
choices of S1 and θ1 (that is, choices for the upstream
magnetic field ratio and the flow angle). Slow shocks
correspond to χ < 0, and fast shocks to χ > 0. The
more usual ratios can be found from the χ solution:
(17.46)
To solve this system, Woods defines θ1 as the angle
between B1 and the shock normal; andintroduces the
“useful intermediate variables”
To follow convention, and shorten the notation a bit,
we define ν = 1/ρ as the inverse density (“specific volume”). Two important constants, across the shock, are
m = ρv = v/ν, the normal mass flux; and Bx , the
normal field component. Recall [[A]] = A2 − A1 is the
jump in A; and hAi = 21 (A1 + A2 ) is the mean value
of A. The jump conditions are, now,
[[Bx ]] = 0
(17.45)
5ǫ
p2
=1+
p1
3S1
µ
¶
1
χ− ǫ
2
(17.48)
(17.50)
and
2
sin θ1
vx1
ǫ
= 1+
(χ + sin θ1 ) +
(χ + sin θ1 )
2
2
cos θ1
cos2 θ1
bx1
(17.51)
Woods presents a long discussion of the details of this
system and some solutions.
96
4.
NOW SOLVE THE SYSTEM : II
5.
Alternatively, the conservation laws, (17.41)-(17.46),
can be written as
ρvx = constant
Bx = constant
ρvv =
Ã
p+
By2
8π
!
n̂ −
Bx
B = constant
4π
(17.52)
1 2
(v + vy2 ) + h = constant
2 x
(recalling that h = γp/(γ − 1)ρ is the enthalpy).
To solve these, we can (following Priest) again use
the compression ratio, X = ρ2 /ρ1 . Keeping our defi2 = B 2 /4πρ (all evaluated
nitions, c2s = γp/ρ, and vA
upstream), the jump conditions become
v2x
v1x
v2y
v1y
ρ2
ρ1
B2y
B1y
BACK TO THE PHYSICS
Whichever solution method one uses, three solutions of
these equations can be found.
One has X = 1, is called an intermediate or Alfven
wave, and isn’t of great interst as it isn’t compressive.1
In the intermediate wave, the tangential field component is simply reversed in sign, so the overall field rotates without changing magnitude; hence the name.
The other two solutions are compressive, with X >
1, and are slow and fast shocks. The slow shock has
2 , and also has B < B . That is, the magnetic
v12 ≤ vA1
2
1
field bends towards the shock normal in a slow shock.
Similarly, in the slow shock the parallel flow is slowed
down, v2y < v1y . Conversely, the fast shock has v12 ≥
2 , and so B > B . Thus, the field bends away
XVA1
2
1
from the shock normal in a fast shock. The parallel flow
is speeded up in the fast shock, v2y > v1y .
1
X
2
v 2 − vA1
= 21
2
v1 − XvA1
=
=X
(17.53)
2
v 2 − vA1
= 21
2 X
v1 − XvA1
p2
(γ − 1)Xv12
=X+
p1
2c2s1
(a)
µ
X2 − 1
X2
¶
where θ is the angle between the upstream magnetic
field and the shock normal. The compression ratio X is
now the solution of
¢
¡ 2
2
F1 (X, θ)
v1 − XVA1
(17.54)
1 2 2 2
+ vA1 v1 sin θXF2 (X) = 0
2
where
F1 (X, θ) = Xc2s1
1
+ v12 cos θ [X(γ − 1) − (γ + 1)]
2
−
2
XvA1
((γ
(c)
(d)
(e)
There are two particular special cases of these. The
slow solution becomes a switch-off shock when v1 =
vA1 . In this limit, the tangential field vanishes behind
the shock, B2y = 0. The fast solution becomes a
switch-on shock when v1 = X 1/2 vA1 . In this limit, the
upstream field is parallel to the shock normal: B1y = 0.
(17.55)
and
F2 (X) = (γ + X(2 − γ)) v12
(b)
Figure 17.5.
Left, the changes in magnetic field
direction that are caused by the three types of oblique MHD
shock waves; (a), slow mode shock, in which B bends
toward the shock normal; (b) intermediate wave, in which B
“flips”; and (c) fast shock, in which B bends away from the
shock normal. From Priest figure 5.6. Right, the two special
cases of switch-off (d) and switch-on (e) shocks. From
Priest figure 5.7.)
(17.56)
+ 1) − X(γ − 1))
Presumably this equation could be solved numerically
(picking values of cos θ of M = v1 /cs1 and also of
vA1 /cs1 ). However, I have not seen such solutions presented, nor have I felt the need to work them out myself.
References
The two best references I know for MHD waves and
shocks are Priest and Woods. I’ve followed both of
them here.
1
Apparently it was the first one found, and was at that point of
interest for laboratory plasmas; so I gather from Bazer & Ericson.
97
18. MAGNETIC RECONNECTION
Magnetic reconnection is an innocent-sounding name
for an important, and initially unintuitive, process. It
can be defined as the breaking, and topological rearrangement, of B field lines. That means, it violates
what we’ve learned about magnetic flux conservation –
which says that the B lines are “frozen into” the plasma.
Magnetic flux is conserved only when resistive dissipation can be ignored; reconnection is an effect of localized, intense dissipation regions. It results in magnetic
energy being converted to plasma energy (both heat and
bulk motions).
Because reconnection allows field lines to be rearranged, and to “move through” a plasma, it can “connect” regions of plasma which had previously been
magnetically isolated. To be specific: think of space
plasma, where the electrical conductivity is very high
(all those free charges can move where they want to),
and therefore the flux freezing limit is a good one. This
means that the plasma particles can freely move and
mix along field lines, but cannot do so across the field.
A particle always remains tied to the same field line as
it moves with the plasma flow. Think, then, about two
initially separate plasma regions which come into contact with one another.1 It follows that the two plasmas
will not mix; rather, a thin boundary layer will form
between them, which keeps them quite separate. (The
location of the boundary layer is, of course, determined
by pressure balance and large-scale fluid dynamics.) In
general, the magnetic fields on either side of the boundary layer will have different strengths and orientations,
so that a current sheet must exist within the boundary
layer.
Thus, flux freezing leads inevitably to the prediction that in plasma systems space will be divided into
separate cells, which contain magnetized plasma from
different sources, separated from each other by thin current sheets. Based on what we’ve seen up to now, the
plasmas from these separate domains should be able to
mix only very slowly, as resistivity dissipates the current and magnetic flux. Looking ahed to (18.5), this
should happen on the (usually very slow) timescale,
τdiss ∼ L2 /η. But: the answer is much more interesting. A mixture of dynamics and resistivity allows
the topology of the magnetic field to change, and the
plasmas to mix, on a much shorter timescale. This process is called magnetic reconnection, and quite a culture
1
To be specific, think of the solar wind where it encounters the
earth’s magnetosphere. Both of these are diffuse, magnetized,
flux-frozen plasmas.
has evolved around it. In this chapter I’ll try to give an
overview of this culture, both the simple treatments often found in the literature, and an impression of what
might realy happen.
Space examples are often quoted – the interaction
of the solar wind with the earth’s magnetosphere, or
the rapid energy release that creates a solar flare. But
reconnection is also important in laboratory plasmas,
where it underlies the “Taylor relaxation” process we
talked about in Chapter 15; and it is critical to the existence of the dynamos that are responsible for the earth’s
and sun’s magnetic fields.
A. Advection vs. dissipation of magnetic field
To set the stage, think back to the induction equation,
∂B
= ∇ × (v × B) + η∇2 B
∂t
(18.1)
The first term on the right-hand side describes inductive
effects of the plasma flow on the B field; the second
term describes resistive diffusion or dissipation. The
ratio of the two terms is given by the magnetic Reynolds
number,
Rm =
Lv
η
(18.2)
where the magnetic diffusivity η = c2 /4πσ and σ is the
electrical conductivity. It’s worth noting, here, that we
can “uncurl” (18.1), to get
E=
η
v
×B+ ∇×B
c
c
(18.3)
– and remember from Maxwell that ∂B/∂t = −c∇ ×
E; so there is also an E field in the system when we
have either advection (v × B) or gradients (∇ × B).
Now, envision a region of adjoining, magnetically
isolated regions, as in Figure 18.1, and think about how
it might evolve.
B
x
j
B
Figure 18.1. The setting for a reconnection event – the
oppositely directed B field separates plasma in the regions
above and below the current sheet (j, located at x ≃ 0).
98
1.
RESISTIVE LIMIT: FIELD DECAY ON AXIS
First, consider the highly resistive case, when Rm ≪ 1.
We have simple diffusion,
∂B
= η∇2 B
∂t
(18.4)
As we saw earlier, we expect the field close to the x = 0
line to decay, and the current sheet to broaden, with
time (the first is also described as B lines “diffusing
through the plasma”), This basic equation has wellknown solutions; I’m storing one in the Appendix to
this chapter. But we don’t need the full solution to
guesstimate the timescale for this process. If the xextent of the system is initially ∼ L, then from (18.4)
we can directly estimate the diffusion time:
τdif f ≃
L2
η
τA ≃
∂B
= ∇ × (v × B)
∂t
(18.6)
and we talk about magnetic flux freezing. In the context
of Figure 18.1, envision the plasma flowing towards the
current sheet, with some velocity Vo . Because of flux
freezing, the B lines will be carried with the plasma,
in towards x = 0. Thus, B will grow, close to the xaxis (the field lines must bunch up as they’re carried inwards). I’ve also put one analytic solution to this problem in the Appendix to this chapter. Once again, we
can also estimate a (dynamic) timescale for this process, without knowing the details:
τdyn
L
≃
Vo
Our next task, then, is to combine advection and diffusion. When we keep both terms in (18.1), we lose the
chance at simple analytic solutions. Instead, we’ll work
with dimensional and scaling arguments2
1.
WHY DO FIELD LINES “ BREAK ”?
To understand how resistivity can “break” and “reconnect” field lines, think about the geometry in Figure
18.2. We know resistivity is important in regions of
high current density - such as the central region (around
OP). If we set up this geometry and waited awhile, the
central magnetic field would decay as the current layer
supporting them is dissipated. This would deplete the
magnetic pressure in this region. The plasma above
and below this region would be pushed inwards by its
own pressure, bringing a in a fresh supply of field (and
plasma).
V
i
A
E
V
o
(18.7)
D
F
B
Its also worth noting that this process can’t continue
forever – at some point the magnetic energy will build
up enough to exert a back pressure and shut down the
flow ... unless this energy is dissipated as heat which
can escape the system. That means that we have to incorporate both advection and dissipation ... which leads
us to reconnectoin.
But first, one more timescale. We’ll find below that
the inflow velocity Vo is something to be solved for, not
(18.8)
B. Steady, 2D Reconnection
ADVECTION LIMIT: FIELD GROWTH ON AXIS
Alternatively, consider a highly conductive case, when
Rm ≫ 1. Now, the first term in (18.1) dominates;
L
vA
where vA is evaluated at large distances from the current sheet.
We might expect that τA would be the best we could
do to annihilate magnetic flux – simply drive the field
into the dissipation region. We’ll see below that things
aren’t this easy – that τA underestimates the timescales
(or overestimates the reconnection rate). The inflow
must be regulated by the dissipation rate ... so the final
answer for the reconnection timesscale will be somewhere between τA and τdif f .
(18.5)
This limit destroys magnetic energy, and heats the
plasma – but it doesn’t, by itself, change the magnetic
topology.
2.
a parameter of the problem .. and so people usually
prefer to estimate an Alfven time:
C
O
V
o
P
C’
B’
E’
Vi
A’
F’
D’
Figure 18.2. Illustrating a sinple reconnection geometry;
see text for discussion. Following Choudhuri figure 15.2.
2
This is the true state of the field, folks .. it’s either these scaling
arguments, or numerical simulations.
99
Now, look at this in more detail (I’m following
Choudhuri’s discussion here). The field lines ABCD
and A’B’C’D’ move inwards, with velocity vin . Eventually the BC and B’C’ parts of the field lines decay
away. The AB part of the field line is moved to EO,
and the A’B’ part of that field line is moved to E’O.
Thus, these “fragments” of two original field lines now
make up one new field line, EOE’. And similarly, the
parts CD and C’D’ eventually make up a new field line,
FPF’. Thus, “cutting and pasting” of field lines (otherwise known as reconnection) takes place in the central
region. And, of course, there must be plasma flow away
from the region – sideways in this cartoon – to conserve
mass.
2.
SWEET- PARKER MODEL
We can be more quantitative about this geometry, and
find simple scaling laws to describe this situation. The
model I’m describing here is Sweet-Parker reconnection, and is illustrated in Figure 18.2.
v
in
annihilating magnetic field and energy (where does it
go?).
• Next, use the induction equation. In a steady state,
it is ∇ × (v × B) = η∇2 B, which gives (by dimensional/scaling analysis)n
ηB
vin B
≃ 2 ;
l
l
vin ≃
η
l
This is saus that we can keep the diffusion region steady
if the rate at which flux is brought in is equal to the rate
at which it is annihilated. We can also note that there
must be an E field, as shown in the figure, to maintain
the current sheet. From Maxwell, we get
j = σE ; ∇ × B =
4π
B
4π
j⇔
≃
σE
c
l
c
v
out
L
v
out
(18.13)
Along and within the current sheet (call that the ŷ direction), there is no v × B force, so only the pressure
gradient accelerates the flow. We have then,
v
in
Figure 18.3. Geometry of Sweet-Parker reconnection.
The current sheet (grey shaded area) has transverse width l
and lateral extent L The input velocity is vin and the output
velocity is vout . The induction E ∝ v × B is normal to the
system (into/out of the paper), as is the current in the current
sheet. Quantities far away from the current sheet are
labelled with subscript o. Following Cowley figure 5.5.
(18.12)
where σ is again the electrical conductivity.
• Thirdly, consider force balances. In the vertical direction, we note that B → 0 at the center of the current
sheet, and that p → pmax there (its maximum value).
Pressure balance in this direction therefore requires
Bo2
≃ pmax − po
8π
l
(18.11)
ρvy
∂vy
∂p
≃−
;
∂y
∂y
2
vout
≃ pmax − po
2
(18.14)
Thus, the outflow speed must be
2
2
vout
≃ vA
=
Bo2
4πρ
(18.15)
Our analysis is strictly two-dimensional and steady
state. I’ll assume the flow is incompressible – that it
stays at constant density (which turns out to be a good
approximation if vin and vout are both subsonic. There
are three important steps here.
• First, mass conservation requires
OK: we can combine these three parts to get our
main results, namely, the inflow velocity ahd thickness
of the dissipation layer:
vin L = vout l
To connect with the literature, we can also rewrite our
results in terms of Rm:
(18.9)
2
vin
≃
From this, directly, conservation of magnetic flux gives
Bo vin = Bout vout
(18.10)
Thus: the outflow speed must be significantly faster
than the inflow speed (because we’re assuming l ≪ L).
The outflow field is much smaller than the inflow; we’re
2
vin
=
vA η
;
L
l2 ≃
ηL
vA
2
vAi
B2
L2
2
; Bout
= o ; l2 =
Rm
Rm
Rm
(18.16)
(18.17)
Thus, when Rm ≪ 1 – which we expect for the highly
conductive plasmas in our examples – reconnection is
indeed slow.
100
3.
ENERGETICS
Short notes here: where does the energy go? When
you work through the energetics, you’ll find that the
incoming kinetic energy is small (because the inflow
speed is slow); most of the incoming energy is electromagnetic (Poynting flux). But also, magnetic energy is
dissipated in the current sheet; this drives a plasma outflow. One can show that the outgoing kinetic energy (in
the plasma flow) is equal to half of the incoming EM
energy (question: where does the other half go?)
numberes, any “rate” ∼ 1/(Rm)1/2 is called “slow reconnection”.)
Thus, Sweet-Parker reconnection is faster than just
ohmic decay; but observed reconnection events proceed
much faster than this simple model predicts. People
have, therefore, spent a lot of time trying to invent faster
versions of this model. Some are as follows ..
C. Speed up the reconnection?
I’m just putting brief comments here; this section
would be far too long if I attempted a full review of
the arguments.
1.
Figure 18.4. Numerical solution for Sweet-Parker-type
reconnection. This simulation (presumably) started with a
field geometry like that in Figure 18.1, and continued until it
reached a steady state, as shown. The upper figure shows
magnetic field lines; the lower one shows velocity vectors.
You can see that a slow flow in from the top and bottom has
changed the field topology and driven “jets” of plasma out
the sides. From James Drake, Maryland.
4.
This model has gotten a lot of attention in the literature, but does not seem to be supported by recent
work. Several people approached the slowness problem in the Sweet-Parker model by changing assumptions about the geometry; the Petschek model (Figure
18.35) is the most generally discussed of these models.
The point of the Petschek model, and its relatives, is to
keep the lateral extend of the diffusion region small: if
the reconnection region occupies only a small fraction
of the system size, then the scaling arguments above
give an inflow rate faster than what the Sweet-Parker
model predicts.
v
1
0
0
1
0
1
0
1
0
1
0
1
This analysis gives us, finally, the spontaneous reconnection rate; the rate of slow inflow that allows thinge
to go steadily.
Many authors describe the vin solution as a “reconnection rate”, despite the imprecision of this term. Or,
they use its Alfven Mach number (defined in terms of
the asymptotic Bo ) as a rate:
vin
1
≃
vAo
(Rm)1/2
(18.18)
We might better talk about a “reconnection time”, as
a geometric mean of the (slow) diffusion time and the
(much faster) dynamic time, τdyn = L/vAi :
τrec ≃
L
≃ (τdif f τdyn )1/2
vin
11
00
00 E
11
11
00
00
11
00
11
00
11
00
11
00
11
00
11
(18.19)
Clearly this type of reconnection is “medium fast”:
τdyn < τrec < τdiss . (Or, in terms of Alfven mach
B
B
111 v
000
v
A
1111
0000
TIMESCALES AND RATES
MA,in =
PETSCHEK RECONNECTION
A
111
000
000
111
000
111
000
111
000
111
000
111
11
00
00 E
11
1
0
0
1
0
1
0
1
0
1
0
1
v
Figure 18.5.
Sketch of the field and flow configuration
in the Petschek reconnection model. A small central
diffusion region, of scale L∗ , bifurcates into two,
standing-wave current sheets (standing slow-mode shocks)
in the downstream flow; again shown shaded. Following
Cowley Figure 5.6.
In Petschek’s original model,3 he argued that two
slow-mode standing shocks exist in the outflow region,
and they determine the lateral size, L∗ . Similar, more
general, solutions to this were also given by Sonnerup
(1970). This class of solutions is often quoted in the
literature as the answer to the slowness problem. These
3
Note that this is not A. Petschek who used to teach at Tech; I
believe it is a relative.
101
solutions assume that the diffusion region will adjust
itself to respond to the external boundary (driving) conditions, in doing so making L∗ ∼ l and thus making the
reconnection rate nearly independent of η.
Do Petschek-type mechanisms work? It looks at
present as if the Petschek mechanism is not obtained
in real situations. Several numerical simulations (e.g.,
Biskamp 1986, Ugai 1995) have looked for steady
reconnection solutions, and found that simple, twodimensional systems will reach the Sweet-Parker solution for low inflow rates. At higher rates, however,
the simulations find a transition to unsteady behavior,
rather than a transition to the Petschek mode. In addition, recent lab work which was configured explicitly
to measure reconnection physics (cf.review by Kulsrud,
1998) seems to support the Sweet-Parker model.
2.
COMPRESSIBLE FLOW
Another suggestion is that the plasma in the dissipation region (DR) is compressed. If this is the case, then
(18.9) is replaced by vin ρo L ≃ vout ρDR vout . This will
increase vin relative to vout – its a plausible idea, but
I’m not aware of any lab tests yet. This will be investigated in the homework.
3.
ANOMALOUS DIFFUSION
This is my personal favorite, and is a good example of
how plasma microphysics can be important in macroscopic situations. That is: the electrical resistivity σ,
which sets the size of the η term, is traditionally calculated assuming Coulomb collisions are the mechanism for energy transfer and dissipation. (Check the
appendix of Chapter 1 for Coulomb scattering, and the
back of Chapter 13 for conductivity discussions). The
critical point is that electrical conductivity is regulated
by the collision rate of the current-carrying electrons:
σ=
j
nevD
ne2
=
≃
τcoll
E
E
m
(18.20)
where τcoll is the electron collision time.
Now, Coulomb collisions are the “hands-off”
particle-particle collisions that result from the longrange nature of the electrical force. They must happen
in any ionized plasma – but they are slow (at least in
low-density plasmas, such as space or the lab), and thus
not very effective at dissipating a current. Are Coulomb
collisions the only dissipation mechanism? The answer is, almost certainly not. If the plasma is turbulent on microscopic scales (which is very likely to be
the case), and if the turbulence involves charge separation (ditto), we expect the plasma to contain fluctuating
electric fields (again on microscopic scales; they won’t
affect the macroscopic dynamics that we’re studying in
this course).
Plasma physics details: the most common origin
of microturbulence is streaming of charges relative to the background plasma. If the streaming
speed is high enough – typically
p greater than the
plasma thermal speed, vth ∝ kT /m – the system is unstable, and some of the streaming kinetic
energy is turned into microturbulence. The most
common way to set up relative streaming is with
a high current density. As above, j = nevD ; a
higher j leads to a higher drift velocity vD , and
when vD > vth streaming instabilities occur. But
the current density is controlled by external conditions – for instance by the scale over which B
changes significantly, in the simple reconnection
models. Thus, thin current sheets ⇒ high current
density ⇒ microturbulence.
Once microturtubulence is created, stochastic electric
fields will scatter the plasma particles; if the plasma turbulence is strong enough (and very weak levels will do)
this turbulent scattering will be much more important
than single-particle interactions. Because this is a nonlinear process (at what amplitude does the turbulence
saturate?), we can’t write down collision times from
first principles. One common approach (way to get
around this) argues that we can get an upper limit to the
resistivity (lower limit to τcoll is to set τcoll = 2π/ωp ,
if ωp2 = 4πne2 /m is the square of the plasma frequency (in cgs units). This is used in (18.20) to estimate
anomalous conductivity or anomalous resistivity.
D. Other approaches
Reconnection is a very active field these days. I’m just
putting a few short comments here; once again, these
notes would be very long if I tried to review the entire
field.
1.
SPONTANEOUS RECONNECTION
The simple models assumed a steady state, with some
mass inflow to balance the flux annihilation. Thus they
are “driven” in a general sense. But the plasma will find
its own way; oppositely directed magnetic fields across
a thin current sheet will spontaneously reconnect. This
is called a tearing instability; and typically results in
formation of magnetic islands in the current sheet. Figure 18.6 illustrates this behavior.
This instability can be studied analytically; the math
is long and horrendous, and I’ll put it in a later chapter.
It can also be studied with numerical simulations; an
102
example of that will also be in that later chapter.
Figure 18.6. Effect of a perpendicular plasma
displacement in a neutral sheet, leading to field compression
for the ideal case (η = 0; top figure) and to field disruption
and island formation for the resistive case (η 6= 0; lower
figure). Following Biskamp Figure 4.4.
2.
DRIVEN RECONNECTION
This is a different approach. It’s worth remembering
that the 2D models above are all “passive”: put two misaligned B fields together and wait to see how quickly
they reconnect. But nature does not always work this
way. One can envision a situation in which the two
anti-parallel magnetic structures are driven together, by
large-scale flows in the system (one example of this is
MHD turbulence, in which different parts of the plasma
move in random directions, at a speed set by the energetics of the turbulence). In this case one (this one at
least) expects that the inflow speed (vin ) in our notation above) will be set by the large-scale flow (say the
turbulent speed). How can the reconnection site adjust
to this? Some authors (e.g. Parker) suggest that the
internal structure of the reconnection layer – its thickness, density or resistivity (set by microscale turbulence
therein) – will adjust as necessary. I personally tend to
agree with them.
3.
NON - STEADY RECONNECTION
Here’s another personal impression: who says steadystate reconnection is relevant to any natural situation?
That is: the arguments above show that steady reconnection models must be forced, sometimes rather
severely, to connect with what we think is occuring
in nature. But examples of non-steady reconnection
events are easy to find:
• Reconnection in solar flares. Flares are seriously
transient events; the large amount of energy released is
believed to be due to very fast annihilation of magnetic
field, in a reconnection event. Flares have motivated
much of the work on steady-state models; however as
an outsider I suspect time-dependent, patchy, localized
reconnection must be taking place.
• Reconnection at the magnetopause, where the solar wind hits the earth’s magnetic field. Spacecraft observations suggest this is very patchy, localized and
time-dependent. Picture, for instance, magnetic flux
tubes being carried along in the solar wind; and let one
of them impact the magnetopause, which also has its
field bunched into ropes. This process can allow solar
wind plasma, and field, to penetrate into the magnetosphere.
There are many numerical simulations of dynamic
reconnection in the literature; I’ll bring some to class.
4.
THREE - DIMENSIONAL RECONNECTION
Another observation: only rarely can a reconnection
event be well described by a two-dimensional analysis. My last example in fact assumed this – because
the intersection of two magnetic flux ropes is clearly a
three-dimensional process. Going to 3D is challenging,
and work is only starting here (helped significantly by
recent advances in resistive MHD codes). If I can I’ll
bring some recent images to class.
References
Some of this is “just from me”, from my reading
in the field. I’ve taken the basic Sweet-Parker and
Petschek material from Priest; and the two formal solutions in the Appendix from Priest & Forbes.
E. Appendix
1.
DIFFUSION - ONLY SOLUTION
Refer back to Figure 18.1, so that we have a 1D problem, B = (0, B(x, t), 0); and take the figure as an initial condition
B = Bo , x > 0 ; B = −Bo , x < 0
(18.21)
The time evolution of the system is governed by
∂2B
∂B
=η 2
∂t
∂x
(18.22)
103
There are scads of different ways to solve this equation.
One uses a Green function:
Z
B(x, t) = G(x − x′ , t)B(x′ , 0)dx′
(18.23)
and for this initial condition, the Green function is
G(x − x′ , t) =
1
′ 2
e−(x−x ) /4ηt
(4πηt)1/2
(18.24)
From this we get the desired solution, in terms of an
error function:
µ
¶
x
2Bo
B(x, t) = √ erf √
π
4ηt
(18.25)
Z x/√4ηt
2Bo
−u2
e du
= √
π 0
2.
ADVECTION - ONLY SOLUTION
Refer again to Figure 18.1, but this time specify a velocity field which describes inflow towards x = 0, but
also allows flow out the sides (to conserve mass):
y
x
(18.28)
vx = −Vo ; vy = Vo
a
a
This satisfies ∇ · v = 0 (incompressible); and has
streamlines xy = constant.4 As an initial condition,
pick a B field which increases going away from x = 0,
and changes sign at x = 0:
x
(18.29)
B(x, t = 0) = Bo cos
a
Figure 18.8 shows this geometry. If we ignore diffusion, our governing equation is
∂B
x ∂B
Vo
− Vo
=
B
∂t
a ∂x
a
(18.30)
B(x,t)
B
o
y
t
x
t
x
−Bo
Figure 18.7. Solution for the magnetic field as a function
of time, “diffusing” away from a current layer at x = 0. The
arrow shows the direction of increasing time; solution
details are in the text. Following Priest & Forbes figure 3.1.
This has the expected behavior, as illustrated in Figure 18.7: the B field decays with time, near the current sheet, which itself spreads with time. One can also
show that the total current stays constant:
Z
cBo
c ∂B
; J = jdx =
(18.26)
j=
4π ∂x
2π
and that the magnetic energy decays with time:
Z
Z
duB
B ∂B
B ∂2B
=
dx =
η
dx
dt
4π ∂t
4π ∂x2
¶
Z µ
∂B 2
η
dx
=−
(18.27)
4π
∂x
¶
Z µ
Z 2
η
4π 2
j
=−
j dx = −
dx
4π
c
σ
(where we’ve integrated by parts in the second line, and
used basic Maxwell in the third).
Figure 18.8. Solution for the magnetic field as a function
of time, being advected into a current layer at x = 0. The
dotted lines are velocity streamilines; the magnetic field
builds up close to x = 0 as time goes on. Solution details
are in the text. Following Priest & Forbes figure 3.3.
Priest and Forbes present a cute solution to this.
Consider the characteristic lines (in the (x, t) plane)
x(t) = x∗ e−Vo t/a
(18.31)
(clearly x∗ is the initial value of x, on a given velocity
line, at t = 0). Along these lines, B obeys
dB
∂B dx ∂B
=
−
dt
∂t
dt ∂x
x ∂B
Vo
∂B
− Vo
=
B
=
∂t
a ∂x
a
4
(18.32)
Check back to Chapter 1 for the definition of streamlines, and
how to find them from the velocity field.
104
(the first line is just chain rule, and the second line notices connects back to eqn 18.30). But now it’s easy:
along the characteristic lines, (18.31), the solution is
³x ´
∗
(18.33)
B(x, t) = Bo eVo t/a cos
a
All we have left to do is write this in terms of (x, t)
everywhere:
³x
´
B(x, t) = Bo eVo t/a cos
eVo t/a
(18.34)
a
This shows that the field accumulates – grows in
strength – near the x = 0 axis; in fact it concentrates
there (closer and closer to x = 0 as time goes on.
However, this clearly isn’t a physical solution, because if B grows exponentially, magnetic pressure will
react back on the flow.
105
19. FLUID INSTABILITIES
Fluids, magnetized or not, are subject to a wide variety of instabilities. Many so-called equlibrium configurations, that is steady-state solutions of the basic
equations (or configurations that we might set up in
the lab), are violently unstable. That is, once set up,
they rapidly evolve away from their initial state. Sometimes these instabilities are destructive (such as currentdriven instabilities of plasma pinches), and sometimes
they move the system to another quasi-steady state
(such as convection in the atmosphere or on the sun).
In this chapter we consider a few of these, both hydrodynamic and magnetohydrodynamic. In each case,
we are interested in a criterion for, and a description of,
instability. When – for what initial configurations – is
a system unstable? How fast, and on what scales, does
the instability grow? The answers to these questions
require some lengthy math. We will see two different
approaches. One is modal analysis, in which a small
perturbation is studies in terms of its Fourier components. In this type of analysis, the system is determined
to be unstable if the frequency of the perturbation is
found to have imaginary components. Another method
is the energy method, in which one determines whether
the potential energy of an equilibrium system is a maximum (unstable), or a minimum (stable).
Often the physical insight on why the instability
goes, arrives only after the analysis has been gone
through. In these notes, I will try to focus on the physics
driving each instability, and present only a summary of
the mathematical analysis, rather than reproducing every step (with credit to each author who presents it in
more detail).
A. Buoyancy and Thermal Convection
To motivate things, recall one example we’ve already
seen. In chapter 6 we discussed convective stability,
and in chapter 14 we talked about the effects of a
magnetic field. Let’s revisit this briefly, aiming at the
physics, but with an eye to the more formal stability
analysis about to come.
In §6.2, we considered a small vertical displacement
of a parcel of fluid (air, say) in a static atmosphere. We
expect, physically, that this parcel will be stable against
the displacement if it has a higher density than its surroundings, once displaced; it will just bob up and down.
The frequency of that motion is the Brunt-Vaisala frequency, N (as derived in §6.2). On the other hand, if
the blob when displaced has a lower density than its
surroundings, it will be unstable, and continue to rise.
In this case, convection will develop.
To find the conditions under which this instability
occurs, we argued that the rising blob is adiabatic, and
also stays in pressure balance with the outside atmosphere. We found, somewhat heuristically, that the stability of the blob depends on how rapidly the density of
the atmosphere drops compared to the adiabatic case.
If we didn’t have this physical argument in our head,
we could have done it more formally, by assuming the
blob oscillates as eiN t , and exploring what N is. Check
the result (6.15): the stability criterion is in this equation. If N 2 > 0, the blob is stable; while if N 2 < 0, the
displacement grows exponentially and the system is unstable. But the formal result, (6.15), shows that stability
depends on the sign of dT /dz − (dT /dz)ad .
Thus: stability or instability is determined by the
temperature structure of the atmosphere – which is determined by external conditions (how we set the problem up). In this case we could make a physical argument to lead us to the stability criterion; and a formal
analysis found the same result. In more complex cases,
the exact form of the physical argument is not always
evident ... but the mathematical approach, analogous to
the eiN t method here, will lead us to the answer.
In the rest of this chapter I’ll treat two well-known
fluid instabilities analytically, with and without magnetic fields. Hold on to your hats ...
B. The Rayleigh Taylor Instability
1.
THE PHYSICS
This instability arises when a heavy fluid is supported
on top of a light one, in a gravitational field. This is
clearly unfavorable energetically; if the two fluids can
change places, the system will have a lower potential
energy. This is just what the RT instability does; any
tiny perturbation of the interface between the two fluids, allows “fingers” to develop – in which each fluid
penetrates into the other.
ρ
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g
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Figure 19.1.
Development of the Rayleigh-Taylor
instability. The fluid on top has higher density, and gravity
points downward. Mixing due to the instability produces
fingers that eventually invert the density distribution.
We will work this in terms of a fluid interface; but
any density gradient inverted compared to the local
gravity gives the same results. In addition, the acceleration does not need to come from gravity; a deceleration
106
of a light fluid into a heavy one (as in an explosion in a
dense atmosphere) is also subject to this instability.
2.
THE MATH
Our basic approach here is modal analysis. That is: first
we linearize, assuming small perturbations, so that we
can drop terms which are second order in these perturbations. We have done this before – in deriving sound
waves and MHD waves, for instance. Our goal is to find
the time evolution of these perturbations: do they grow
with time (which is an instability), or simply oscillate
(which says we have wave solutions, but not instabilities). In order to do this, we do a modal analysis. We
assume that any perturbation can be treated as a sum of
Fourier components (which is legal if the defining equations are linear in the perturbations – right??). We can
then focus on the evolution of one such Fourier component – for instance (19.2), below.
My analysis here follows Chandrasekhar, Hydrodynamic and Hydromagnetic Stability. He works through
a modal analysis, and we will pick up on some of his
results. Recall our formal derivation of sound waves,
in Chapter 4: we carried through a linear analysis, expanding the variables v, p, ρ in (unperturbed) + (small
perturbation) terms. Our basic equations are those
of mass conservation, incompressibility of the unperturbed state, and momentum conservation. Geometry: take g = gẑ; assume the unperturbed state had
v = 0, and only has gradients on the ẑ direction. and let
the perturbed velocity components be δv = (u, v, w).
The linearized equations – for the perturbations δρ, δp,
keeping only lowest order “small” terms, are:
∂
∂u
= − δp
∂t
∂x
∂
∂v
= − δp
ρ
∂t
∂y
∂w
∂
ρ
= − δp − gδρ
∂t
∂z
∂u ∂v ∂w
+
+
=0
∂x ∂y
∂z
dρ
∂
δρ = −w
∂t
dz
ρ
(19.1)
Now, assume each perturbed quantity has the form (a
“mode”)
δf (x, y, z, t) = δf (z)ei(kx x+ky y+ωt)
(19.2)
We will generally treat k as given, and want to find the
nature of ω. In particular: if ω is real, (19.2) simply
represents waves or oscillations – so that the system is
stable. However if ω has a negative imaginary part, then
(??) represents a perturbation with grows exponentially
with time: the system is unstable. Let k 2 = kx2 +ky2 , and
denote z-derivatives by Df = df /dz (In this and the
next section only; Chandra’s notation). These equations
then become
ikx δp = −iωρu
iky δp = −iωρv
Dδp = −iωρw − gδρ
(19.3)
ikx u + iky v + Dw = 0
iωδρ = wDρ
These can be used to show
k 2 δp = −iωρDw
(19.4)
and
Dδp = −iωρw +
g
wDρ
iω
(19.5)
These two now combine – eliminating δp between them
– as
D (ρDw) =
gk 2
wDρ + ρwk 2
ω2
(19.6)
But now, this is an equation for w(ω, k), because Dρ
is specified in the problem. Picking w = 0 on the (distant) boundaries of the system, this becomes a Sturmian
characteristic value problem for ω 2 . One can show,
then, that ω 2 > 0 if Dρ < 0 everywhere; and conversely, ω 2 < 0 if Dρ > 0 somewhere in the fluid.
That is: a density structure which decreases with ẑ is
stable; while a density structure which increases with ẑ
is unstable.
The next step is to find out the growth rates, and
whether all wavevectors are unstable. To do this, we
simplify to the case of two uniform fluids, separated by
a horizontal interface at z = 0. Now, away from this
boundary, (19.6) becomes
D2 w = k2 w
(19.7)
which has simple exponential solutions. Picking the
ones which stay well-behaved at z → ±∞, we have
away from the boundary
w(z) = Ae−kz ;
z>0
(19.8)
kz
w(z) = Ae
;
z<0
107
To connect solutions through the boundary, we need
jump conditions. Define
∆s f = f (zs + 0) − f (zs − 0)
that is, the jump in f from the below some surface zs to
above that surface. Here zs = 0. Applied to (??), we
find the jumps in p and ρ are related by
∆o δp =
g
ws ∆ o ρ
ω
(19.9)
Manipulation of the system (??) also gives
2
k ∆o δp = −∆o (−iωρ)
(19.10)
Finally, these combine to give
∆o (ρDw) =
k2
gw∆o ρ
ω2
(19.11)
But in this system, ∆o ρ = ρ2 − ρ1 (upper = lower values); and Dw = ±kw (- for upper, + for lower). The
result (19.11) thus becomes
¶
µ
ρ2 − ρ 1
2
(19.12)
ω = −gk
ρ2 + ρ1
This is our result: ω 2 < 0 – the system is unstable –
if ρ2 > ρ1 . This recovers what we guess from potential
energy arguments, at the start. We also find that all
2
wavenumbers are unstable
√ (that is, have ω < 0), with
a growth rate |ω| ∝ k. This last result is true for
small perturbations; however when the instability starts
to grow, larger-scales dominate the system.
This analysis could also have included surface tension at the interface: when included, one finds that surface tension stabilitizies high wavenumbers (small spatial scales).
3.
THE MAGNETIZED CASE
What happens if we add a magnetic field? We expect
little effect from a vertical field, as it doesn’t hamper
vertical fluid motions. This turns out to be the case –
the stability conditions are not affected. It does have
some effect on the growth rate, however (note that a
vertical perturbed velocity must also involve horizontal, cross-field flows – by the continuity equation). The
more interesting case is a horizontal magnetic field: we
expect this to affect the stability as well as the growth
rates, and it does.
In this section I still follow Chandra. Consider a
uniform B = B x̂, with perturbed field
δB = (bx , by , bz ) =
kx
Bv
ω
(19.13)
(Why is it proportional to the perturbed velocity?) The
perturbed momentum equations become
iρωu = −ikx δp
B
(ikx by − iky bx ) − iky δp
4π
B
g
iρωw =
(ikx bz − iDbx ) − iky δp + wDρ
4π
iω
(19.14)
iρωv =
Now: using (19.11) to convert from b components to v
components, we can show
¶
µ
kx2 B 2
(ikx v − iky u) = 0
(19.15)
iρω +
iω 4π
which requires kx v = ky u. Using this result, and further algebra, we find
iρωDw = −k 2 δp
(19.16)
and
¢
B 2 kx2 ¡ 2
g
2
D
−
k
w = −Dδp + wDρ
2
4πk iω
iω
(19.17)
And now, we can look for the new results. First,
combine (19.16) and (19.17) to get
iρωw −
¢
gk 2
B 2 kx2 ¡ 2
2
D
−
k
w
=
wDρ
4πk 2 ω 2
ω2
(19.18)
which is the generalization of (19.4) for the nonmagnetic case.
From this, first, we note that if kx = 0, which corresponds to propagation perpendicular to the field, (??)
just reduces to (19.6): cross-field perturbations have no
effect on the RT instability. The ones with kx 6= 0,
however, do change the results. We can repeat the nonmagnetized analysis . . . the jump condition (19.11)
generalizes to
D (ρDw)−ρwk 2 −
∆o (ρDw) −
k2
B 2 kx2
∆
(Dw)
=
gw∆o ρ (19.19)
o
4πω 2
ω2
Again taking the case of two uniform fluids separated
by an interface, the result (19.12) becomes
µ
¶
ρ2 − ρ 1
2B 2 kx2
2
ω = −gk
(19.20)
+
ρ2 + ρ1
4π (ρ2 + ρ1 )
Thus: a horizontal magnetic field stabilizes high frequency perturbations. Instability (ω 2 < 0) requires
gk (ρ2 − ρ1 ) >
2B 2 kx2
4π
(19.21)
108
Finally, we note that this analysis assumed an incompressible fluid. That simplifies the algebra but isn’t
required for the physics. Extending to the compressible
case (as done by Shivamoggi, Theory of Hydromagnetic
Stability), one finds that compressibilty has a stabilizing effect, in that it reduces the growth rates in both the
magnetized and unmagnetized cases, compared to their
incompressible analogs.
The basic perturbed equations are
ρ
C. The Kelvin Helmholtz Instability
Well, that was so much fun, let’s do it again. The other
well-known instability of an interface is the Kelvin Helmoltz Instability.
1.
∂u
dU
∂
∂u
+ ρU
+ ρw
= − δp
∂t
∂x
dz
∂x
∂v
∂
∂v
+ ρU
= − δp
ρ
∂t
∂x
∂y
∂w
∂
∂w
+ ρU
= − δp
ρ
∂t
∂x
∂z
∂u ∂v ∂w
+
+
=0
∂x ∂y ∂z
∂
dρ
∂
δρ + U δp = −w
∂t
∂x
dz
We again assume the perturbation has the form, (19.2).
The system (??) becomes
THE PHYSICS
This instability arises at a velocity shear. Consider two
fluids, or two pieces of the same fluid, in relative motion. For instance, think of a horizontal interface; fluid
below is at rest, and fluid above is moving parallel to
the interface. (This might describe a situation in the atmosphere, with high-velocity winds going past a lowvelocity cloud layer). When the interface is displaced
upwards, it forces the overlying flow to deviate over
the perturbation. This means the pressure there drops
(Bernoulli, remember?). The boundary thus feels a lift.
But of course the situation is mirror symmetric across
the interface; the downward displaced boundary feels
a downward “lift”. As the boundary follows these perturbations, it is sheared by the flow into which it penetrates, and rollw up into a vortex sheet.
ikx δp = −iνρu − ρwDU
iky δp = −iνρv
(19.23)
Dδp = −iνρw
ikx u + iky v + Dw = 0
iνδρ = −wDρ
where we have defined the “shifted” frequency, ν =
ω + kx U . From these we can find
iρνDw − iρkx wDU = −k 2 δp
(19.25)
(compare equations 19.4 and 19.5). These combine
(eliminate δp) to give
v
2
D [ρνDw − ρkx wDU ] − k 2 ρνw = 0
v
(19.24)
and
ik 2 Dδp = ρk 2 νw
1
Figure 19.2. The origin of the Kelvin-Helmholtz
instability. Shear at an interface between two fluid layers in
relative motion generates a vortex sheet; pressure
differences (think of the Bernoulli effect) lead to instability,
and eventual mixing of the two fluids.
2.
(19.22)
THE MATH
We again follow Chandrasekhar. The methods are similar to those presented for the RT instability, but the geometry is slightly more complicated. We add a horizontal flow U = ux̂ to the unperturbed system, but we
now ignore gravity.
(19.26)
and from this, our useful jump condition , at an interface zs , is
∆s [ρνDw − ρkx wDU ] = 0
(19.27)
(what quantities must be held constant across an interface? how did we throw out some of the terms in ???)
We again apply this to two uniform fluids. Above
and below the boundary, we again have (19.7), as all
else is constant. But now, the ratio w/(ω + kx U ) must
be continuous across the boundary (why?). Thus, (19.8)
is replaced by
w(z) = Aν2 e−kz ;
kz
w(z) = Aν1 e
;
z>0
(19.28)
z<0
109
U; we don’t expect cross-field flows to differ from the
unmagnetized case.
The modal equations become
(where the subscripts 2,1 refer to (top),(bottom) flow
speeds). Our characteristic equation is simpler here.
Taking ρ1 = ρ2 = ρ, we can derive
(ω + kx U2 )2 + (ω + kx U1 )2 = 0
(19.29)
But this has only complex roots (if kx 6= 0). The
quadratic solves to
1
i
ω = − kx (U1 + U2 ) ± kx (U1 − U2 )
2
2
ikx δp = −iνρu − ρwDU
B
(ikx by − iky bx )
4π
B
(ikx bz − Dbx )
Dδp = −iνρw +
4π
iky δp = −iνρv +
(19.30)
Thus, any such perturbation is unstable (ω has an
imaginary part), no matter how small the difference
U1 − U2 may be.
iνδB = ikx Bv + bz DU
As Chandrasekhar quotes Helmholtz: “Every perfect geometrically sharp edge by which a fluid
flows must tear it asunder and establish a surface
of separation, however slowly the rest of the fluid
may move”.
iνδρ = −wDρ
From these, after a fair bit of algebra, and again using
the “shifted” frequency ν = ω+kx U , we find (compare
19.16 and 19.17)
Equation (19.30) shows that higher k’s grow the
fastest when the perturbation is small. When it becomes
finite, however, this analysis breaks down; experiment
shows that a characteristic and large scale will dominate
the instability.
3.
ρνDw − ρkx wDU = ik 2 δp
THE MAGNETIZED CASE
Finally, we can again consider the effect of a magnetic
field. We expect the interesting case will be when B k
2
2
ik Dδp = ρk νw +
B
kx2
2
4π
·
D
(19.31)
ikx u + iky v + Dw = 0
(19.32)
and
µ
Dw
ν
¶
¸
µ
¶
2
k2 w
DU
3B
−
− kx
D
w
ν
4π
ν2
(19.33)
These two combine as
D (ρνw − ρkx wDU ) +
B
kx2
2
4π
·
D
(compare 19.16). Specializing once more to two uniform fluids separated by a boundary at z = 0, our jump
condition is
∆o (ρνDw) =
B
kx2
2
4π
∆o
µ
Dw
ν
¶
(19.35)
From this, going back from ν to ω, we get the charac-
µ
Dw
ν
¶
µ
¶
¸
2
DU
k2 w
3B
D
w
−
− kx
ν
4π
ν2
(19.34)
teristic equation:
(ω + kx U2 )2 + (ω + kx U1 )2 = kx2
2B 2
4π
(19.36)
The roots of this are
·
¸1/2
2
kx2
1
2
2 B
−
(U1 − U2 )
ω = − kx (U1 + U2 )± kx
2
4πρ
4
(19.37)
110
And so, finally, we have our result: a B field aligned
with the flow will stabilize the KH instability at all
wavelengths if
(U1 − U2 )2 ≤
4B 2
4πρ
(19.38)
Finally, again, one can also do this for compressible fluids. As with RT, one finds that compressibility is stabilizing; it reduces the growth rate, in both the
unmagnetized and magnetized cases, compared to our
incompressible calculation here.
References
As in the text, I’ve followed Chandrasekhar, Hydrodynamic and Hydromagnetic Stability for the hydro,
and Shivamoggi, Theory of Hydromagnetic Stability),
for the MHD.
111
20. IDEAL MHD INSTABILITIES
We now turn to MHD instabilities. They are critical to
fusion plasmas, where one wants to confine the plasma
and keep it stable long enough for fusion to take place.
Thus, an extensive literature has developed on ideal
(non-resistive) instabilities of MHD equilibria.
A. Overview; Energy Methods
The history of this field comes, of course, from the need
to keep plasmas around in the lab long enough for them
to fuse (Bateman has a good discussion of this period;
Figure 20.1 illustrates the basic pictures). In early experiments, a current was driven through a column of
ionized gas, in an effort to heat and confine it. These
early experiments found that the plasma column would
spontaneously pinch itself off in a process that came to
be known as the m = 0 sausage instability. This instability can easily be stabilized by adding an axial magnetic field to the plasma column (requiring an azimuthal
current). However, this configuration introduced a new
instability, in which the plasma column twists into a helical or corkscrew shape; this is the m = 1 kink instability. Higher m modes may be considered, of course;
m = 2 is shown in the figure.
1.
ENERGY METHODS
MHD instabilities often involve more complex geometries than the fluid instabilities we have considered.
Thus, a full modal analysis which describes relevant
(macroscopic) perturbations is often too ugly to consider. A different approach is available here: the energy
method. This analysis allows us to determine if a system is stable or unstable; it does not, however, determine growth rates or, necessarily, details of the nature
of the unstable perturbation.
To illustrate this, consider a simple 1D system, with
a potential energy curve W (x), connected to a force
F = −dW/dx. Equilibrium will be an extremum of
W (x), say xo ; but as in Figure 20.2, it can be either a
stable (minimum of W ) or unstable (maximum of W )
extremum.
We can test this by considering a small displacement, say from xo to x. With this, the change in the
potential energy is
µ 2 ¶
1
2 d W
δW = W (x) − W (xo ) ≃ (x − xo )
2
dx2 o
(20.1)
In this last step we have used dW/dx = 0 at the extremum. Thus, the sign of δW determines the stability. In particular, if we consider some displacement δx
W
W
x
x
(unstable)
(stable)
Figure 20.1.
Stable and unstable potential energy
curves, in a simple one-dimensional illustration. Let the
energy curve W (x) have an extremum at some value of x:
dW/dx = 0. If the energy W is a minimum here, the
system is stable; a small perturbation will return to this
point. If, however, W is a maximum there, the system is
unstable; a small perturbation will evolve away from this
point.
and find δW < 0, we know that we have an unstable
system. Given a slight push, such a system will move
away from its initial equilibrium state, towards a state
of lower energy; thus it is unstable. Conversely, if we
can slow δW > 0 for all possible perturbations, we
know the system is stable.
2.
THE DETAILS
This is the idea that we want to generalize to a 3D
plasma. I’m following Freiberg’s presentation here. A
general static plasma configuration has potential energy
¶
Z µ 2
B
+ ρe dV
(20.2)
W =
8π
The first term is the magnetic energy, the second the internal energy, and the integral is taken over the plasma
volume. Now, let each piece of the plasma be displaced
by some local δx. The fluid quantities are then disturbed as
v1 =
dδx
dt
B1 = ∇ × (δx × Bo )
(20.3)
ρ1 = −∇ · (ρo δx)
p1 =
po
γpo
ρ1 − po (δx · ∇) ln γ
ρo
ρo
(The first of these should be obvious; the last 3 of these
come from the D/Dt expressions, integrated once with
resepect to time).
Now, the (linearized) force on that piece of the
plasma due to the perturbation is
F(δx) = −∇p1 +
j1
jo
× Bo + × B1
c
c
(20.4)
112
(perturbed position)
11
00
00
11
and the change in the potential energy, due to this perturbation, is
ξ
r
11
00
00 (equilibrium position)
11
1
δW = −
2
r
o
(origin)
Figure 20.2. A small displacement of a plasma element
by ξ~ from its equilibrium position (ro ) to a new position r . .
. relative to the origin O. Our δx is ξ~ in this figure.
Following Priest Figure 17.2.
1
δW =
2
Z ·
Z
δx · FdV
(20.5)
where the explicit form (20.4) is used in the integrand
(the 1/2 is there to take a rough mean over the displacement from 0 to x.) Expanding this out, and writing
B1 = ∇ × (δx × Bo ), one explicit form for the perturbation of the “fluid+field” energy is
¸
B12
j
2
− δx⊥ · × B1 + γp|∇ · δx| + (δx⊥ · ∇p)∇ · δx⊥ dV
4π
c
In addition, many applications require the addition of a “boundary” or “surface” term:
¸
Z ·
B · B1
1
− γp∇ · δx − δx⊥ · ∇p δx · n̂dS
δWS =
2
4π
(20.6)
(20.7)
An alternative version of this which explicitly uses the field line (inverse) curvature, ~κ = b · ∇b if b is the unit
vector along B, is
Z " 2
B1⊥ B 2
1
δW =
+
|∇ · δx⊥ + 2δx⊥ · ~κ|2
2
4π
4π
#
(20.8)
j
k
+ γp|∇ · δx|2 − 2(δx⊥ · ∇p)(~κ · δx⊥ ) − (x⊥ × b) · B1 dV
c
This last form is the so-called “intuitive form”1 ; each
term allows a simple physical interpretation. The B1⊥
term represents the energy needed to bend field lines.
The second term is the energy needed to compress the
magnetic field. The γp term, the third term, is the energy needed to compress the plasma. Each of these contributions is stabilizing. The last two terms can be positive or negative, and thus can drive instabilities. The
first of these depends on ∇p ∼ j⊥ × B, while the second depends on the parallel current jk . Thus, a vacuum
field is stable (but not very interesting), while either
perpendicular or parallel currents are potential sources
of instability.
Thus: the energy method is applied starting with
1
oh yeah?
some equilibrium configuration (as in Chapter 17, say),
and asking whether some perturbation δx can be found
for which δW < 0. If this is the case, then we know
our initial configuration is unstable. (The converse isn’t
necessarily useful: if we try some perturbation and find
δW > 0, this doesn’t prove stability unless all perturbations have been considered).
B. Apply: Pinch Instabilities
1.
THETA PINCH
We first look at stability of the θ pinch: which is simply
described (cf. §17.3) by
p(r) +
B2
Bz2
= o
8π
8π
(20.9)
113
if Bo is the confining field. We recall that this needs
an outer wall, to hold everything together; it is not a
self-confined system. We will find, however, that it is
stable to the type of MHD perturbations we’re considering here.
Consider displacements
δx(r) = δx(r)ei(mφ+kz)
(20.10)
and write δx = (ξr , ξφ , ξz ). First, note that the incompressibility condition ∇ · v = 0 gives
d
(rξr )
(20.11)
dr
The terms that go into the integrand for δW are
−ikrξz − imξφ =
B1⊥ = ikBz k⊥ = ikBz (ξr , ξφ , 0)
∇ · δx⊥ =
1 d
im
(rξr ) +
ξφ
r dr
r
(20.12)
jk = 0
Putting these into (20.7), taking a as the pinch radius
and L as its length, we have
Z
πL a
δW (r)dr
(20.13)
δW =
4π 0
where the integrand can be written (with ko2 = k 2 +
m2 /r2 )
¶2
µ
δW (r)
im d
(rξr )
= ko ξφ −
Bz2
ko r2 dr
(20.14)
µ
¶2
d
k2
2 2
+ 2 2
(rξr ) + k ξr
ko r
dr
Now, the first term is the only one containing ξφ : the
perturbed δW can therefore be minimized by choosing
im
d
(rξr )
(20.15)
2
2
+ k r dr
But then, with this choice, the integrand in (20.13) for
δW is positive definite for any choice of ξr . Thus, the
minimum of δW is positive for k > 0, and approaches
zero as k → 0: this system is stable for finite wavelengths and approaches marginal stability for very long
wavelengths.
What is the underlying physics? This equilibrium
has no parallel currents, so current-driven modes (such
as the pinch) can’t be excited. In addition, as the field
lines are straight, pressure-driven modes (such as the
kink) can be excited. Any perturbation to this equilibrium either bends or compresses the field lines, and
both are stabilizing influences.
m2
Z PINCH
Now, let’s do an unstable one. The equilibrium here is
Bφ d
dp
+
(rBφ ) = 0
dr 4πr dr
(20.16)
This is self-confined; all pressures drop at large radii.
This system is particularly interesting if the source current is carried by the plasma itself. However, we will
see that it is severly unstable to large-scale perturbations which can disrupt the equilibrium.
We again consider a perturbation (20.9); but now
the m = 0 and m 6= 0 modes must be considered separately. We will work through m 6= 0 here. The expressions analogous to (20.11) are
B1⊥ =
~κ = 0
ξφ =
2.
imBφ
(ξr , 0, ξz )
r
~κ = − (1/r, 0, 0)
µ ¶
d ξr
∇ · δx⊥ + 2δx⊥ · ~κ = r
+ ikξz
dr r
(20.17)
jk = 0
Again evaluating δW (r) (to go in 20.12), one finds
m2 Bφ2 ¡ 2
¢
(20.18)
2
ξ
+
ξ
δW (r) =
r
z
r2
·
¸2
µ ¶
8π dr 2
d ξr
2
+ Bφ r
ξ
+ ikrξz +
dr r
r dr r
Now, this expression can be rearranged so that ξz appears only in quadrature. Thus, as with ξφ above, the
energy perturbation is minimized here if
µ ¶
d ξr
ikr2
(20.19)
ξz = 2
m + k 2 r2 dr r
When this is chosen, the energy perturbation can be
written
µ
¶ 2
dp
ξr
δW (r) = 8πr + m2 Bφ2
dr
r2
(20.20)
µ µ ¶¶2
m2 r2 Bφ2
d ξr
+ 2
m + k 2 r2 dr r
Still hunting for a minimum of δW , we note that k →
∞ (long wavelengths) kills the last term. Thus, our final
expression to be analyzed for the energy perturbation is
¶ 2
Z µ
dp
πL a
2 2 ξr
8πr + m Bφ
rdr (20.21)
δW =
4π 0
dr
r2
114
And now: the magnitude and sign of dp/dr control the
result. In particular, if
2r
2 2
dp m Bφ
+
<0
dr
4π
(20.22)
requires a particular mix of axial and azimuthal fields.
In particular, it can be shown that the axial magnetic
field must be made strong enough, and the plasma column fat enough, that no part of the field between the
plasma and the wall closes on itself over the length of
the plasma column. This is the Kruskal-Shafranov stability criterion, which can be written
anywhere, then one can find a δx function localized in
this region which will make the full integral negative.
Thus, this equilibrium is unstable against m 6= 0 perturbations. This is the classic kink instabilitey.
If we use the starting equilibrium relation, (20.15),
the stability condition (20.21) can be rewritten in two
ways:
if L is the length of the plasma column. Thus, this imposes a severe limit on the current than can be driven
through the plasma.
1 d ¡ 2¢
rBφ < m2 − 1
Bφ2 dr
µ ¶
2r2 d Bφ
< m2 − 4
r
Bφ2 dr
References
I’ve mostly followed a mixture of Priest & Bateman,
here.
(20.23)
Now, for typical Z pinches, Bφ /r is a decreasing function of r. For such profiles, (20.22) predicts stability for
m ≥ 2. Also, at large radii Bφ ∼ 1/r corresponding
to a vacuum field; but near the origin, in the currentcarrying region, Bφ ∼ r. Thus, a Z pinch is always
instable to the m = 1 mode. Figure 20.7 shows the
physical picture of this instability. Physically, the kink
instability may be attributed to a localized twist in the
column, pushing the field lines together on the inside
edge of the kink, and pulling them apart on the outside
edge. This sets up an internal magnetic pressure gradient which acts to enhance the kink.
A similar analysis can be done for m = 0 perturbations; it turns out that one can’t assume incompressibility in this case, so it must be treated separately. One
finds an analogous result: the system is unstable if
−
2γBφ2 /4π
r dp
>
p dr
γp + Bφ2 /4π
(20.24)
This is called the pinch instability or the sausage instability. This physical picture for this instability is given
in Figure 20.5. Physically, we can attribute the pinch
instability to a local constriction in the plasma column
causing a local increase in jk , and thus of the value of
Bφ there. This provides an extra inwards tension, which
tends to enhance the constriction.
We can stabilize a Z pinch by adding an axial magnetic field, Bz . For the pinch mode, the axial field
provides a pressure which fights back against the constricting Bphi field. For the kink mode, the axial field
provides a tension which fights back against the tendency of the azimuthal field to kink. General stability
rBz (r) > LBφ (r)
(20.25)
115
21. RESISTIVE MHD INSTABILITIES
In the previous section we considered ideal MHD instabilities – those for which resistivity is not important.
Another very important type of MHD behavior involves
the effects of resistivity. Resistivity allows plasma to
move across magnetic field lines. This violates magnetic flux conservation (what we called “frozen in field”
or “flux freezing”). Unlike the ideal case, resistive effects allow changes in the topology of the field lines.
The most important application of this is in field line
reconnection, which we treat in the next chapter.
In this chapter we consider resistive instabilities, in
which spontaneous field line readjustment occurs. Resistive instabilities are most relevant in a current sheet.
That is, a thin region, generally formed at the interface
of two different plasmas, whre B varies rapidly. Several such instabilities are known; here we consider the
tearing mode, which is fundamental to magnetic reconnection.
21.2. Now, consider a finite resistivity. This will allow
the field to diffuse through the fluid (or, vice versa, will
allow the fluid to move across the field lines). Under the
right conditions (mainly long wavelength disturbances;
see below), the new, reconnected state has lower energy
than the initial one. Thus the system is unstable.
In this geometry, the unstable tearing mode causes
magnetic surfaces close to the tearing layer disrupt and
reconnect, forming a chain of filaments, as illustrated
in Figure 21.2. These filaments are also called “magnetic islands”, but note this really only describes a twodimensional slice of the system.
A. Tearing Mode: the Physics
The simplest illustration of this effect is a planar, twodimensional problem. Think about a current sheet (also
called a neutral sheet); this is a region over which the
field direction switches rapidly. To pick a geometry, let
B(x = 0) to lie totally in the yz plane. We’ll call this
symmetry plane the “tearing plane” or “tearing layer”.
There are two important pieces of physics here. One is
that parallel currents attract; a free current sheet with no
hinderance would tend to bunch together. The second is
that plasma has inertia, and can only move across field
lines on a slow, diffusion timescale.
B
j
B
Figure 21.1.
Schematic drawing of a neutral sheet
configuration which is subject to the tearing mode. Note
that there must be a current, j, in the region where the field
lines switch direction.
Now, imagine a small plasma displacement, perpendicular to the current sheet, and with some fixed
wavenumber k. If the plasma is unimportant (has very
low energy density compared to the field), then the current will bunch as expected. More important however,
is the case when the plasma pressure/inertia matters.
First think about the ideal case. The fluid must move
with the plasma; so the perturbation will only “wiggle”
the field lines – as shown in the top panel of Figure
Figure 21.2. Effect of a perpendicular plasma
displacement in a neutral sheet, leading to field compression
for the ideal case (η = 0; top figure) and to field disruption
and island formation for the resistive case (η 6= 0; lower
figure). Following Biskamp Figure 4.4.
Thus, the main features of tearing modes are: (1)
they are driven by field energy associated with shear;
(2) they alter B field topology and quickly relax the
shear, by field line reconnection and magnetic island
formation; (3) the relaxation of the magnetic shear occurs much more quickly than the pure resistive diffusion time would predict.
We will be interested in stability and growth rates.
If we have a current layer of thickness a, recall the two
characteristic timescales. The Alfven time is
τA = a/vA ,
(21.1)
where vA = B/(4πρ)1/2 is the Alfven speed; and the
resistive/diffusion time is
τD = a2 /η
(21.2)
(or 4πa2 /η, depending on author).. In most plasmas
of interest, τA ≪ τD (recall the high conductivity, and
thus small η, for low density plasmas). It turns out the
tearing mode goes on an intermediate timescale: τA ≪
τtear ≪ τD .
116
B. Tearing Mode: the Math
I mostly follow Biskamp in these notes; note however
that the original derivation, given by Furth, Kileen &
Rosenbluth (1963) is quite different on the surface, and
is often quoted. Both sources do agree on the answer,
however.
We start in the usual place, with the equations of
motion (10.2) and induction (10.4). This instability is
well described in the incompressible limit; to shorten
the notation we set ρ = 1, If we linearize the basic
equations, and ignore both pressure effects and fluid
viscosity, we get
∂
v1 = (∇ × B1 ) × Bo + (∇ × Bo ) × B1
∂t
(21.3)
∂
2
B1 = ∇ × (v1 × Bo ) + η∇ B1
∂t
Assuming incompressibility allows us to introduce a
velocity stream function, φ. Working in two dimensions allows us to use a magnetic flux function, ψ. In
this coordinate system, we thus have
v = ẑ × ∇φ ;
B = ẑ × ∇ψ
(21.4)
(Strictly speaking, these are the perpendicular velocity and field components; in the details of the analysis
we’re ignoring variations in the y direction). Resistive
instabilities are driven mainly by the parallel current:
j ≃ jk ≃ jz . Therefore we also neglect j⊥ , which
means Bz ≃ constant. Also, as we’re interested in stability and the growth rate, we write ∂f /∂t = γf (so
that γ real and positive corresponds to instability) and
make solving for γ the primary goal of our analysis.
With this, the equations (21.3) become
γ∇2⊥ φ1 = B · ∇j1 + B1 · ∇j
(21.5)
γψ1 = B · ∇φ1 + η∇2⊥ ψ1
Our plan now is (1) pick a geometry for the unperturbed field; (2) solve these equations for the spatial
behavior of ψ1 and φ1 , and (3) from these results determine γ. Remember that dissipation is a second-order
derivative term; it will be important only in small regions where things change rapidly. In this problem that
means close to the current sheet. The traditional way to
solve the system (21.5) is in two parts. Close to the current sheet, we keep all the physics, but use simplified
geometry to get an answer. We call the width of this
inner, resistive region δ. Far from the current sheet, we
use simplified physics (that is, we ignore the resistivity). We then require the two solutions to match where
they overlap (formally this is an asymptotic analysis),
and this matching allows us to solve for γ and δ. One
useful quantity in this analysis is the jump in the variation scale of ψ1 at the tearing layer. Its standard, if ugly,
notation is
¤
1 £ ′
∆′ = lim
ψ1 (δ) − ψ1′ (−δ)
(21.6)
δ→0 ψ1
Here and in the rest of this section, primes on ψ1 and
φ1 mean derivatives with respect to x; the prime on ∆′
is standard notation but does not mean it is a derivative
(mutter, mutter . . .). We will consider cases where
ψ1 varies only slowly with x near the tearing plane, so
that ψ1 . ≪ ψ1 /δ, but where the second derivative ψ1′′ ≃
ψ)1 ∆′ /δ, has a jump at x = 0.
To do the actual solution, we work first in the inner,
resistive region (where the dissipation terms are most
important, so we can ignore the inductive terms). We
then work in the outer, nearly-ideal region (where we
can ignore the dissipation terms). Finally, we require
that the two solutions match well, which gives us our
final answer for the growth rate (or timescale).
• Inner, Resistive Region Now: we pick
a wavenumber for the perturbation, ψ1 (x, y) =
ψ1 (x)eiky , and we assume the unperturbed ψ is constant across this region. Keeping the resistive terms,
the system (21.5) becomes
γφ′′1 = ixkB ′ ψ1′′ − ikj ′ ψ1
(21.7)
γψ1 = ixkB ′ φ1 + ηψ ′′
It turns out that j ′ term does not affect the resulst much,
and it is usually ignored in this analysis. Doing that, we
solve the system by (1) choosing parity: pick ψ1 (x)
even and φ1 (x) odd; (2) approximate derivatives: φ′′1 ≃
−φ1 /δ 2 , ψ1′′ ≃ −∆′ φ1 /δ. Putting these in and doing
some algebra gives us an intermediate solution for the
inner region:
γ ≃ η 3/5 (∆′ )4/5 (kB ′ )2/5
′
δ ≃ δ∆ /γ ≃ η
2/5
′ 1/5
(∆ )
(21.8)
′ −2/5
(kB )
Biskamp’s comments here: this argument assumed
∆′ > 0. A full stability analysis (cf. FKR 1963) finds
that instability requires ∆′ > 0, and in fact shows that
−∆′ is the perturbation energy, δW (so that δW < 0
gives instability). Thus our choice of ∆′ > 0 here is
OK.
•Outer, Ideal Region. In this region, we ignore
second derivatives, so the basic equation becomes
B · ∇j1 + B1 · ∇j ≃ 0
(21.9)
117
Putting in the geometry of Figure 21.1 explicitly, this
becomes
¶
µ
j ′ (x)
′′
2
=0
(21.10)
ψ1 − k +
B(x)
Now we must pick a form for B(x). The usual choice
for this problem is B(x) = tanh(x/a), which introduces a as the width of the current layer. We pick
boundary conditions ψ1 → 0 as |x| → ∞. The solution
of (21.10) is now analytic:
µ
¯ x ¯¶
1
¯ ¯
ψ1 (x) = e−k|x| 1 +
tanh ¯ ¯
ka
a
¶
µ
(21.11)
2 1
′
∆ =
− ka
a ka
Recalling that a full analysis shows instability only
when ∆′ > 0, we see from this that only large scales
are unstable: ka < 1, or λ > 2π/a
• Put These Together. All this collects to give our
final result, the growth rate and the physical scale of the
inner region. We can find these by putting the solution
(21.11) back into the intermediate solution (21.8). Doing the algebra, and working in terms of the two time
scales defined at the start, we find for ka ≪ 1,
1
1
γ≃
3/5
2/5
(ka) τ τ 2/5
D
µ
τA
τD
(a)
(b)
10
10
5
5
0
0
-5
-5
-10
-10
-1.0
-0.5
0.0
0.5
1.0
-1.0
-0.5
(c)
0.0
0.5
1.0
0.5
1.0
(d)
10
10
5
5
0
0
-5
-5
-10
-10
(21.12)
A
and for the layer thickness,
a
δ∼
(ka)3/5
The critical point is that this instability leads to a
topological change in the field lines; it is not at all
simple dissipative diffusion. The field lines reconnect
across the initial neutral sheet. Some authors discuss
the tearing mode as a change from one possible equilibrium state (say, an ideal one) to another (call it a resistive one). If the new, reconnected state has a lower magnetic energy, then the instability can go spontaneously;
the reduced magnetic energy will appear as heat.
¶2/5
-1.0
(21.13)
Thus, we find that such a current sheet is always unstable to tearing mode on long scales (low k), and that
the growth time (τtear ≃ 1/γ) is, indeed, inbetween the
Alfven (dynamic) and resistive (diffusion) timescales.
-0.5
0.0
0.5
1.0
-1.0
-0.5
0.0
Figure 21.3.
A 2D simulation illustrating the evolution
of a reconnecting tearing mode. Solid lines are magnetic
field lines (lines of constant flux function ψ); dotted lines
show the velocity field. The initial state (top left) and three
later times are shown; the formation of magnetic islands is
apparent. From Garasi, PhD thesis (NMSU), 2002.
C. Tearing Mode: the Consequences
The evolution of the tearing mode in a simple geometry
is illustrated by the figures on the next page, from an
early numerical simulation. Magnetic island formation
(that is flux ropes, thinking of the third dimension) does
take place as expected, and currents become strongest
at the “X” points inbetween the islands. The argument
above said that largest scales grow the fastest – in this
simulation the largest scales are those allowed by the
size of the numerical grid. Note the velocity streamlines, showing flow patterns into and through the magnetic islands and X points.
References
The classic, and nearly opaque, paper which introduced this subject is
Furth, H. P., Kileen, j. & Rosenbluth, M. N., 1963,
Phys. Fluids, 6, 459.
I have also followed Shavamoggi for most of the
math details.
118
22. TURBULENCE, PART I
Turbulence is an old topic which remains fresh today.
It was studied as long ago as the 16th century, when
Leonardo da Vinci studied turbulence generated by obstacles placed in a water flow. It is still being studied
today, when such issues as the role of small-scale vortex ropes, the effect of magnetic fields in MHD turbulence, or “quantum turbulence” in superfluids, are topics of current research. In fact, many of the basic questions are still unanswered today. Because turbulence
is fundamentally nonlinear, analytic solutions are hard
to come by; traditional work relies heavily on scaling
laws, while large numerical simulations are critical to
modern work.
To set the stage, I know of no better opening, than
to paraphrase Tennekes & Lumley.
Most flows occuring in nature and in engineering applications are turbulent. These include
the boundary layer in the earth’s atmosphere; jet
streams in the upper troposphere; and cumulus
clouds which are in turbulent motion. Subsurface ocean currents are turbulent. Stellar atmospheres are turbulent, as is the gaseous interstellar medium. Boundary layers on aircraft wings
are turbulent. Most combustion processes involve
turbulence. The flow of natural gas and oil in
pipelines is turbulent, as is water flowing in rivers
and canals. THe wakes of ships, cars and aircraft
are in turbulent motion. In fluid dynamics laminar flow is the exception, not the rule: one must
have small dimensions and high viscosities to encounter laminar flow.
A. The transition to turbulence
The instabilities discussed in the last chapter can develop into full-fledged turbulence. To explore this, let’s
choose a particular setting: consider a boundary layer in
an initially laminar flow. In order to do stability analysis, we need a mathematical description of the boundary
layer. Here’s a sketch of one.
A standard representation of such a boundary layer
is the Blasius solution. (I follow Tritton, 2nd edition,
chapters 8, 11). Assume incompressible flow, and let u
b, the x-velocity, v be the y-velocity. The basic equations are
u
∂u
∂u
∂2u
+v
=ν 2 ;
∂x
∂u
∂x
∂u ∂v
+
=0
∂x ∂y
(22.1)
To solve these, we do dimensional analysis (more formally, look for a similarity solution). We want to find
δ (x)
y, v
x, u
Figure 22.1. Flow past a thin plate. An initially uniform
flow comes in from the left; when it encounters the plate,
the no-slip requirement at the plate surface causes a viscous
boundary layer to develop. The width of the layer, δ(x),
grows with x in this situation.
a solution of the form u = uo g(y/δ), if uo is the upstream/incoming velocity, g is some unknown function,
and δ can be a function of x. Because this is incompressible, we can use a stream function ψ: u = ∂ψ/∂y;
v = −∂ψ/∂x. We therefore look for a ψ solution with
the form ψ(x, y) = uo δf (y/δ); f is another ast-yetunknown function. If we carry out the algebra and put
these into the basic equations (22.1), we find (a) a new
ODE; and (b) a consistency condition on δ. The results
are
δ(x) ∝ (νx/uo )1/2 ;
f f ′′ + f ′′′ = 0
(22.2)
The first is our desired result: the shape of the boundary layer (the constant of proportionality depends on
just how δ is defined, for instance is it the point where
u(y) = 0.9uo ? 0.99uo ?). The ODE has to be solved
numerically; the primes represent differentiation with
respect to the similarity variable, η = y/δ(x).
Now, assume we have the full solution for the
boundary layer (rather than just the sketch above). Subject it to the same type of analytic instability analysis as we did in chapter 19. That is, assume the perturbed (x, y) velocities, and also the perturbed pressure, go as f (y)ei(kx+ωt) ; and determine whether the
frequency has any imaginary part. If it does, the system
is unstable. The key parameter here is the viscosity,
ν – which is folded into the boundary-layer Reynolds
number, Re = uo δ/ν. Figure 22.2 shows a typical result: any values of (k, Re) inside the curve are unstable, while values of (k, Re) outside the curve are stable.
Thus: there is a minimum Reynolds number for instability – for lower values viscosity stabilizes the flow.
Above this value, there is a finite range of scales which
are unstable: very small and very large scales remain
stable.
What happens next? From this type of analysis, as
well as from experiment, we expect that all flows become turbulent at high enough Re. Linear stability theory can tell you when it starts .. but the waves pre-
119
treatment) are still predictable. Notes, “there is every
reason to suppose that this loss of predictability occurs
as a property of the Navier-Stokes and continuity equations, although these equations contain the determinism
of classical mechanics.” It is not that the onset of turbulence represents a breakdown of these equations, but
rather that the nonlinear terms allow interesting behavior!
stable
kδ
unstable
stable
Re (crit)
Figure 22.2.
Re
Results of stability analysis on the Blasius
model of a shear layer.
dicted by such theories are only the first stage of the
process. When these waves reach a critical amplitude
- typically about 1% of the free-stream velocity – the
nonlinear terms in the driving equations start to matter, and the flow can no longer be described by analytic
means. Following Tritton, who is describing the transition to turbulence in a boundary layer:
The waves become three-dimensional, and interact
with the mean flow; growth continues until there are
transient local regions of high shear. Up to now, the
timescale of the fluctuations has remained of the same
order as the period of the initial wave; now, fluctuations with much shorter timescales appear, and develop
radpily to bursts of turbulent-like fluctuations. The
boundary layer changes now, develops hairpin-shaped
vortices, which generate more high-shear regions – so
that small areas of the boundary layer become turbulent (“turbulent spots”). These regions then grow, while
they travel downstream, and eventually merge into oneanother. Note that the transition is qualitatively similar
in other types of flows used to study turbulence – pipe
flows or free jets.
B. Turbulent flows: overview
Now that the flow has gone turbulent, what can we say
about it? It can be statistically treated as a random system (cf.§22.3, also chapter 23); but it has some characteristic properties. Many of the authors I’ve looked at
go into long qualitative descriptions at this point; in this
section I reproduce some of their discussions.
Tritton notes that turbulence is a state of continuous instability. Each time a flow changes as a result
of an instability, one’s ability to predict the details of
the motion is reduced. When successive instabilities
have reduced the level of predictability so much that we
must resort to statistics, rather than predicting the details, then we say the flow is turbulent. However, such
flow is not compoletly random. All flows involved organized structures; and the mean values (in a statistical
1.
CHARACTERISTICS
One can list characteristics of turbulent flows (I’m following Mathieu & Scott, also Tennekes & Lumley):
•It is a, irregular, random process. The flow is
random, irregular, chaotic. It is time and space dependent with a very large number of degress of freedom.
Although it’s unpredictable in detail, its statistical properties are reproducible (so that it’s deterministic in a
mean sense).
•It contains a wide range of different scales, as
can be seen from the random-looking measurements of
the flow: large scales coesits with small ones (these are
the “fur” in a time-series plot). Turbulent dynamics involve all scales, which coexist and are superimposed in
the flow, with smaller ones living inside large ones.
•It has small-scale random vorticity. It is rotational and three-dimensional, with high levels of fluctuations vorticity. The larger identifiable vortical structures are the turbulent eddies that are apparent in some
pictures. Vortex convection and stretching may be
thought of as the mechanisms by which the intense,
fine-scale vorticity fluctuations are generated and maintained. Viscous diffusion causes vortices to spread,
counteracting the amplifying and scale-reducing effects
of stretching.
•It arises at high Reynolds numbers, as we saw
above. As Re rises, the nonlinear terms on NavierStokes equation become more important, compared to
the viscous term. In addition, the tendency to instability (which is damped by viscosity) increases. Once
turbulence goes, flow instabilities keep it going; largescale eddies are themselves unstable, giving rise to
smaller ones, and so on, until viscosity comes in at
small enough scales.
•It dissipates energy. The inertial range is inviscid,
so it conserves mechanical energy. As smaller scales
form, they can be thought of as sapping some of the
KE of their parents, and transmitting it to their own
offspring. Thus, a cascade arises, which generates the
smaller scales, and allows a mean flux of energy from
large to small scales. This flux is controlled by the dynamics of the large scales, and dissipated as heat at the
120
smallest scales.
•It is intrinsically 3D. In 2D flows, there is no
vortex stretching; thus, in the absence of viscous diffusion, vorticity is passively convected, unchanged by
the flow. Therefore the high-Re energy cascade can’t
occur. 2D flows do show complicted structures, and a
degree of randomness, but they don’t have the ubiquitous fine scales associated with 3D turbulence. We will
also see that 2D turbulence admits a reverse cascade,
unlike 3D turbulence. As M&S put it, “we don’t mean
to insult the many authors who have talked of 2D turbulence....but to point out that the physical mechanisms
are very different” in the two cases.
•It is insensitive to viscosity at high enough Re.
That is, the dynamics of the large scales are essentially
inviscid (as ν → 0); and they control the system. The
size of the smallest scales adjusts to changes in ν, so
as to dissipate energy at the right rate; and that rate is
controlled by the large scales. From here, we go to the
conjecture, that ε approaches a constant, finite value as
ν → 0.
2.
BEHAVIOR
Alternatively, stepping back a bit, we can describe how
turbulent flows behave (Here I follow T&L):
•Self-Organization. Large-scale coherent structures can easily be identified in some flows. To be
specific, consider measurements at a fixed point in the
outer part of a turbulent flow. One finds periods of high
frequency fluctuations, as the fixed point encounters actuve turbulence, and quiet periods, when the fixed point
is in a quiescent region. This seems to be more important in 2D turbulence. The growth of these structures
proceeds through vortex coalesence.
•Entrainment A flow can pull the surrounding fluid
into itself and accelerate it along. This is called entrainment. We saw this process in a laminar flow (back
in roblem set 1), where viscosity accounted for the entrainment. Entrainment also occurs in a turbulent flow,
at a much higher rate. The connection here is inertial;
think of ambient fluid which is “gulped” into the turbulent flow by the large eddies at the boundary.
•Self-Preservation At large downstream distances,
the mean field in many shear flows becomes approximately self-similar. That is, quantities such as the central velocity are functions only of the local scales of
length and velocity. For instance, let δ be the width
of a turbulent jet (defined, for instance, as the halfpower width of the cross-sectional v(y) profile). Selfpreservation means that quantities such as the central
velocity can be written as functions of y/δ.
•Turbulent Mixing All transport processes can be
enhanced in a turbulent situation. Diffusion coefficients
(for transport of some trace substance), viscosity coefficients (transport of momentum, right?), and electrical
resistivity (transport of . . . ) are all enhanced. Calculating the exact transport rates involves understanding the small-scale details of the turbulence, which has
rarely been fully worked through; but dimensional scaling arguments are often used.
As an example, consider turbulent viscosity (“eddy
viscosity”). Kinetic theory says that the microscopic
viscosity coefficient ∼ vλ, where v is the mean thermal
speed and λ is the mean free path (fuller analysis puts
a numerical coefficient in front of this). A useful rule
of thumb is that turbulent viscosity can be estimated as
νt ∼ vt λt , if vt and λt are “characteristic” velocity and
length scales of the turbulence. (We will specify these
a bit more below).
•Turbulent Dissipation Viscous stresses will
quickly damp out turbulence which is not continually
driven. As with mixing, we can use scaling arguments
to estimate the timescale for viscous damping of a turbulent flow. Free turbulence will decay on several times
τt , if τu = λt /vt is the “eddy turnover time” for the
largest structures, at size λt and velocity vt as above.
•Self-Preservation At large downstream distances,
the mean field in many shear flows becomes approximately self-similar. A turbulent “free jet” (that means
not bounded by a wayy nor at a boundary layer) is one
example (which may appear in the homework). Flow
quantities such as the central velocity are functions only
of the local scales of length and velocity. For instance,
let δ be the width of a turbulent jet (defined, for instance, as the half-power width of the cross-sectional
v(y) profile). Self-preservation means that quantities
such as the central velocity can be written as functions
of y/δ.
C. Homogeneous Turbulence
Turbulence theory was initially developed for homogeneous, isotropic turbulence. In particular, it focused on
scale small compared to the overall size of the system
(so that boundaries can be ignored), and considered the
energy transfer between scales. This work is mainly
due to Kolmogorov, and dates from the 1940’s.
1.
OVERVIEW
To motivate this, start with two experimental facts,
which apply to fully developed (high Re, distant
boundary) turbulence (following Frisch here).
• Two-thirds law: the rms velocity increment
121
δv(l)2 , between two points separated by a distance l,
obeys δv(l)2 ∝ l2/3 . Put into wave-number space,1
this becomes what is now known as the Kolmogorov
law: v 2 (k) ∝ k −5/3 .
• Finite energy dissipation: in an experiment, vary
viscosity while keeping everything else the same: find
energy dissipation per unit mass behaves in a way consistent with a finite limit.. Following Frisch, think about
drag. Consider a fluid moving past a rock (or a car moving through air). The drag force F = (1/2)Cd ρAv 2 ,
(you have seen this, right?) if A = L2 is the projected
area of the rock/car, and CD is the coefficient of drag.
Experiments show that CD is only a very slow function
of Re. Thus, the work done W = F v; and the energy
dissipated per unit mass is
ε=
1
v3
W
= CD
3
ρL
2
L
EDDIES AND THE ENERGY CASCADE
One critical fact about fluid turbulence is that energy is
transferred from large scales to smaller scales. Picture,
say, initially large eddies, on the order of the driving
scale of the turbulence. This could be the width of a
boundary layer, or the diameter of a pipe. Now, consider slightly smaller eddies. Due to vortex stretching,
they are strained by the velocity field of the largest ones,
thus growing in strength; they extract energy from the
larger ones. This continues on to still smaller scales.
Thus: the turbulent kinetic energy cascades down from
large to small eddies in a series of small steps. This
process is essentially inviscid, since the vortex stretching mechanism arises from the nonlinear terms in the
equations of motion.
Following Tritton: the figure illustrates the evolution of a blob of fluid, due to the actions of the locally
turbulent velocity field. Showing (1) the process of repeated instability; each stage gives rise to motions of
greater complexity and smaller scales, than the previous
stages; (2) energy is extracted from larger scales (or the
mean flow), to smaller (refer ahead to mean-flow equations and Re stress); (3) vortex stretching is part of the
process. The random nature of turbulent motions is diffusive, as two particles that happen to be close, at a give
time, find themselves much further apart soon after. If
these two particles are on the same vortex line, we get
1
Figure 22.3.
Schematic representation of the evolution
of a “marked blob” of fluid within turbulent motion. Its
shape becomes more and more distorted by the velocity
fluctuations, with smaller and smaller scales appearing, as
time goes on. Eventually the scales become small enough
that diffusion matters, and the marked fluid mixes with its
surroundings.
(22.3)
Thus, ε does not depend (very much) on ν, and thus
will have a finite limit as ν → 0.
2.
stretching – which increases the magnitude of the vorticity (wny?), but also reduces the cross-section of the
vortex tube.
How? change variables, to k ≃ 2π/l; and by energy conservation, note that we must have v 2 (l)dl = v 2 (k)dk.
3.
THE KOLMOGOROV SCALING ARGUMENTS
We can talk about the main properties and scaling laws
for homogeneous, isotropic turbulence, following Kolmogorov’s analysis, without needing the details of the
statistics. Kolmogorov argued that properties of the
flow are determined by the scale l, and the energy rate
ε, only. (motivated by the observations, above). Then,
from this, K. argued (or “derived”), the important law:
δv(l)3 = (4/5)εl
(22.4)
What is the rate of energy transfer in the cascade?
Let vt = hvi be the mean turbulent speed, and let
λt = hλi be the largest (driving) scale.2 The time for
these largest eddies to “turn over” must be τt ∼ vt /λt .
Observations find that the large eddies transfer much
of their energy to smaller ones, in one or two τt times.
Thus the energy cascade rate, which also must become
the energy dissipation rate, is
ε≃
vt3
λt
(22.5)
Kolmogorov argued that scales in this range should
not be affected either by kt = 2π/λt (as eddies at k are
driven only by their immediate neighbors in k-space),
nor by kd = 2π/λd (basically for the same reason; energy is moving downward in k-spacd). Thus, letting
2
A formal version of this is the Taylor microscale, defined by
some authors as λ2T = [v 2 ]/[∇v]2 , and by others as λ−2
=
T
d2 C(r)/dr2 |0 , that is the curvature of the correlation function at
zero lag. Either definition has the same content: this is the scale
where most of the turbulent energy is concentrated.
122
vl be the velocity typical of scale l, the energy flow
rate ε = vl3 /l must be independent of the scale l (or
k = 2π/l). What power spectrum W (k) is consistent
with this picture? The constancy of ε tells us that
1/3
vl ≃ (εl)
;
vk ≃
µ
2πε
k
¶1/3
(22.6)
But also, the power “at k” is vk2 ≃ kW (k). Thus we
find the equilibrium spectrum,
W (k) ∝ ε2/3 k −5/3
(22.7)
This is the Kolmogorov spectrum; and the range of
wavenumbers to which it applies is called the inertial
range: kt ≪ k ≪ kd .
Where does this cascade end? It cannot to on to
infinitely small scales. In fact, Kolmorogov suggested
that the smallest scales to which the cascade reaches are
those on which the viscous dissipation rate is equal to
the energy cascade rate. But we can get this by dimensional analysis. The units of ε are cm2 /s3 , and those of
viscosity are cm2 /s. Thus, the dissipation scale must be
λd ∼
µ
ν3
ε
¶1/4
(22.8)
(Some authors call λt the outer scale, and λd the inner scale.) Note that the ratio of the inner and outer
scales just depends on the Reynolds number3 : λt /λd ∼
Re3/4 .
W(k)
k
−5/3
(direction of
energy flow
in the cascade)
(inertial
range)
k
driving
scale
damping
scale
Figure 22.4. Sketch of the energy spectrum in
Kolmogorov turbulence. Energy comes in at the driving
scale (think of the radius of a turbulent jet, or the width of
the bounday layer). It cascades forward due to wave-wave
interactions (or vortex stretching) at each intermediate scale
k; and dissipates by small-scale viscous damping at the
damping scale.
3
Check: can you show this? You need to define Re in terms of the
turbulent quantities vt and λt .
Finally, it’s worth noting that we can also find the
dissipation scale by noting that the local energy transfer rate, vl3 /l, must equal the energy dissipation rate,
νvl2 /l2 , on this scale. Equating these two recovers
(22.8) nicely.
Below kt the power falls off, due to a lack of driving, and the fact that turbulent energy only cascades forward in hydrodynamic turbulence. Above kd the power
spectrum is usually taken to fall off more steeply, ad
W (k) ∝ k −3 . The data agree with this very well (lots
of people have looked at this).
4.
WHAT IF THE FLUID IS MAGNETIZED ?
Just a note here looking ahead. Everything we’ve seen
in this chapter is well understood and well supported by
data. Isotropic turbulence in non-MHD fluids is very
well described by the Kolmogorov model. The situation changes dramatically, however, when the fluid is
magnetized. We’ll talk about this a bit in chapter 24.
References
Qualitative descriptions of turbulence can be found
in many books, as can the Kolmogorov scaling arguments. I’ve mostly followed Tennekes & Lumley, and
Tritton, who present the traditional work. Frisch’s book
is also of some use, and goes into some more modern
work; Hinze is one good source for the more mathematical (Fourier space) treatments.
D. Appendix: fun facts from Fourier transforms
To work in any detail with turbulence, we need some
statistical tools, from the theory of random variables.
Say we measure some property of the turbulence – the
velocity in some direction, say – either at one point,
as a function of time, or instantaneously (don’t ask me
how!) as a function of position. We will treat the velocity v(x) as a random function of position and assume
that the time measurement and the space measurement
give us the same information. Let the turbulent velocity
field (continue letting v be 1D for now) be v(x), and let
it have a Fourier transform
1
ṽ(k) =
(2π)3/2
Z
v(x)eik·x dx
(22.9)
123
with inversion
v(x) =
Z
ṽ(k)eik·x dk
(22.10)
This FT, ṽ(k), contains information on “where the
power is” in the turbulent signal. Note ṽ(k) can be
complex. This is generally described via the power
spectrum of the turbulent velocity:
P (k) =
(2π)3
(2π)3
ṽ(k)ṽ(k)∗ =
|ṽ(k)|2 (22.11)
V
V
(the normalization is somewhat arbitrary – this notation
is strictly interpreted to mean the limit as the source
volume, V , becomes very large).4
Now, consider also the correlation function of the
turbulence:
C(r) = hv(x)v(x + r)i
(22.12)
where the brackets denote a mean over the turbulent
volume. This also tells us “where the power is”; we
expect C(r) → 0 for scales r which are larger than any
correlation length of the turbulence.5 The correlation
function can also be Fourier transformed:
Z
C(r) = C̃(k)eik·r dk
(22.14)
so that c̃(k) is the FT of C(r). Now, a nice result from
Fourier theory is that C̃(k) and P (k) are related:
Z
1
P (k) = C̃(k) =
C(r)eik·r dr (22.15)
3/2
(2π)
A nice corrolary of this, is that the autocorrelation function at zero lag is directly related to the power spectrum:
Z
C(0) = P (k)dk
(22.16)
4
5
Note, in going from ṽ(k) to P (k) we have lost information on
the relative phase of the kth mode. This is assumed not to be important in the usual homogeneous turbulence models, but phase
information does matter in some applications, such as intermittency.
Some authors use the structure function:
D(r) = h[v(x + r) − v(x)]2 i
= hv(x + r)2 i + hv(x)2 i − 2hv(x)v(x + r)i
(22.13)
which is clearly a close cousin of the correlation function C(r)
D(r) = 2 [C(0) − C(r)] ;
C(r) +
1
1
D(r) = D(∞)
2
2
Thus: the Fourier transform of the autocorrelation function of the turbulence, gives us the power spectrum of
the turbulence. It can be interpreted nicely in that P (k)
describes how much energy “is contained in waves at
k”. Many authors
R work in the limit of isotropic turbulence: C(0) = 4πk 2 P (k). Thus, the isotropic spectrum is often quoted: W (k) = 4πk 2 P (k). This is often the quantity that is addressed in Kolomogorov-type
modelling.
124
23. TURBULENCE, PART II
2.
In the last chapter we talked about the basic nature of
turbulence: what it’s like and how it’s described statistically. In this chapter I carry on with more “classical
turbulence”, namely the effect of turbulent stresses on
the mean-field flow; and say a little about “modern turbulence”, including how 2D turbulence is different, and
turbulence on small scales (which includes vorticity &
intermittency).
In many systems, we can treat the large-scale flow as
steady, or at least slowly varying, with the turbulence as
a rapidly fluctuating additive term. That is: let V be the
mean velocity, and v the turbulent one, so that the net
velocity field is V+v. Treat the pressure field similarly,
P + p; ditto for temperature. (As long as we stay in
the incompressible limit – which is where most turbulence analysis stays – there are no density fluctuations,
right?) If these sums are put into the basic dynamical equations, we can (borrowing terminology from
MHD turbulence), isolate the dynamical equations for
the “mean field” quantities, and find how the turbulence
affects the mean flow. We will find, for instance, that
the mean-flow momentum equation contains what are
called Reynolds stresses: non-zero terms involving second moments of the fluctuating velocity field. Thus,
the mean flow and turbulence are intimately connected,
with the one affecting the other.
THE CONTINUITY EQUATION
For instance, take the incompressible continuity equation:
∇ · (V + v) = ∇ · V + ∇ · v = 0
(23.1)
Now, take means: that means ensemble averages, or
time averages for the case of stationary flows. This
gives
∇ · hVi + ∇ · hvi = 0
(23.2)
But now: first, we have hVi = V; that is the “mean
flow” velocity. And, we assume the turbulent fluctuations have zero mean: hvi = 0. Thus, (23.2) becomes
∇·V =0
∂
(V + v)+(V + v) · ∇(V + v) =
∂t
(23.5)
1
− ∇(P + p) + ν∇2 (V + v)
ρ
∂Vi
∂Vi
1 ∂P
+Vj
=−
∂t
∂xj
ρ ∂xi
!
Ã
∂
∂ 2 Vi ∂ 2 Vi
−
+ν
hvi vj i
+
∂xj
∂x2i
∂x2j
(23.6)
Thus: the turbulent field contributes a net reaction
back on the mean field. This last term is called the
Reynolds stress. Even though hvi i = 0 for each component i, the mean of their product does not necessarily vanish: hvi vj i =
6 0. It turns out that this is so for
anisotropic turbulence – and real turbulence is commonly anisotropic.1
Note that the Reynolds stress arisews from averaging the nonlinear convective term in the full NavierStokes equation; it is not truly a stress term, more like
the mean momentum flux due to the turbulent fluctuations. Nonetheless, it is often talked about in terms of a
turbulent viscosity. Because this extra force term in the
basic momentum equation is the gradient of a tensor, it
is generally combined with the viscous stress tensor, as
¶
µ
∂Vj
∂Vi
+
− ρhvi vj i (23.7)
σij = −P δij + ρν
∂xj
∂xi
We can go further, and write the turbulent contribution
to the viscous stress tensor as
µ
¶
dVj
dVi
+
σij,turb = −ρhvi vj i = ρνt
(23.8)
∂xj
∂xi
This last step is an assumption which turns out to be
quite well justified. That is, the Reynolds stress is linearly proportional to the mean flow shears. But then,
(23.3)
(recovering incompressibility of our mean state). And,
subtracting (23.2) from (23.1) gives
∇·v =0
This analysis can be repeated for the other basic equations; I’m not going to write down all the steps, however. The momentum equation starts as
The mean of this equation becomes – written out in
Cartesian
A. Mean Field Equations
1.
THE MEAN MOMENTUM EQUATION
(23.4)
showing that our turbulent field is also incompressible
(by itself).
1
This is the case because, microscopically, viscosity mixes the relative phases of each vi component; they do not stay in phase, and
thus the mean of their product is non-zero. Alternatively: think
about how Brownian motion, on the microscopic level, transports momentum components laterally.. this gives us the usual
macro stress tensor. Here, the turbulent fluctuations provide a
mean transport of momentum...
125
we can see from this that the turbulent viscosity coefficient νt ∼ vt L, if L is the local gradient scale. This
connects back to our dimensional argument, above.
3.
EXAMPLE :
2D
CHANNEL FLOW
I follow Mathieu & Scott here. Consider a channel
in the x-direction, with y the transverse coordinate.
(Once again, compare this to the laminar flows in chapter 2). Assume the mean flow independent of z, and
that Vz = 0. Note, this does not imply that the flow
is fully 2D, indeed the turbulence will be 3D. Rather,
we’re assusming that it’s 2D in the mean, and all statistical properties are unchanged under reflection about
the z = 0 plane. As before, put the overall mean flow in
x direction, and let it be driven by some dP/dx. Also
as before, all quantites except the pressure depend only
on y; we extend this here to include Re stresses.
111111111111111111111111111111111111111111
000000000000000000000000000000000000000000
000000000000000000000000000000000000000000
111111111111111111111111111111111111111111
the constancy of τ we get τ = ydPw /dx + τw and
τw is another constant, the mean viscous shear stress
at the wall y = 0. We find its value by noting that
the flow must be symmetric about the midplane; thus
τw = −DdPw /dx. Using this in the definition of τ , we
get
³
dVx
y´
+ ρhvx vy i = ν
τw 1 −
D
dy
(23.11)
This is almost the answer for the mean flow .... as we’re
turbulent, we know Re ≫ 0, so that the RHS is small
except very close to the walls. Thus, away from the
boundary layer, −ρhvx vy i ≃ τw (1 − y/D), that is linear behavior. And, if this balance holds, we expect Vx
to be independent of y; and “measurements find Vx is
approximately constant in this region”. Near the wall,
we expect steep gradients in Vx , so that viscosity matters...and a more complicated behavior of the Reynolds
stress.
y
4.
dp/dx
THE MEAN ENERGY EQUATION
x
Vx
111111111111111111111111111111111111111111
000000000000000000000000000000000000000000
Figure 23.1.
Geometry and solution of turbulent channel
flow. The flow is symmetric about the midplane, y = 0; the
walls are at y = ±D. Following Mathieu & Scott Figure
4.3.
Now: the assumed symmetry in the z direction
means only the offdiagonal term hvx vy i in the Re stress
will be nonzero; and it depends only on y. Thus, the
mean force equation has two nonzero components:
dhvy2 i
∂P
+ρ
=0
∂y
dy
dhvx vy i
∂P
d 2 Vx
+ρ
ν 2 =
dy
∂x
dy
(23.9)
From the first of these, we get P (x, y) = Pw (x) −
ρhvy2 i: where Pw (x) is the pressure at the wall (where
vy = 0 by no-slip); and the second term is the diagonal
contribution of the Re stress to the net pressure. Putting
this into the second equation in (23.9), we get
¸
·
dPw
dτ
d
∂Vx
=
=
− ρhvx vy i
(23.10)
ν
dx
dy
dy
∂y
The total mean stress is defined as τ = −ρhvx vy i +
νdVx /dy. But now: the left hand side of (23.10) is
only a function of x, while the right hand side is only
a function of y; thus, each side is a constant. The left
hand side is just the driving pressure gradient. From
Finally, the mean-field energy equation is worth noting.
IF we multiply the mean-field momentum equation by
Vi and sum on i, we get
1 ∂
1 ∂Vi
1 dVi2 1 ∂Vi2
+ Vj
=
(Vi τij ) − τij
2 ∂t
2
∂t
ρ ∂xj
ρ ∂xj
(23.12)
Expanding the τij term back out, defining
¶
µ
∂Vj
1 ∂Vi
+
Eij =
(23.13)
2 ∂xj
∂xi
and noting that δij ∂Vi /∂xj = ∂Vi /∂xi = 0, we have a
useful form:
µ
¶
¶
µ
P Vj
∂
D 1 2
=
V
−
+ 2νVi Eij − hvi vj iVi
Dt 2 i
∂xj
ρ
(23.14)
∂Vi
− 2νEij Eij + hvi vj i
∂xj
Here, the first 3 terms (inside the brackets) are advective
energy transport; the term with ν is viscous dissipation;
and the last term is energy lost (in a decelerating flow)
to drive the turbulence.
5.
WHAT ABOUT THE TURBULENT TERMS ?
In these notes I’ve said nothing about the dynamics of
the turbulent terms (p, v, etc). That was on purpose
.... refer back to the scale separation we used in deriving the mean-field equations (23.3), (23.6), (23.14).
We could equally well have separated out the D/Dt
126
terms involving the turbulent fields. Remember, now,
that flow equations are nonlinear (the v · ∇v term, for
instance). When one isolates the dynamical equations
for the fluctuations, one finds higher-order terms are involved. For instance, the equation for Dhvi i/Dt involves ∂hvi vk i/∂xk ; the equation for Dhvi vj i/Dt involves ∂hvi vj vk i/∂xk ; and so on. Some additional assumption is always needed to close a system like this;
and pursuing that would take us too far afield.
B. Two-dimensional Turbulence
. . . differs significantly from turbulence in threedimensions. The most striking difference is that turbulent power in 2D cascades both forward (to higher
wavenumber) and backwards (to lower wavenumber).
This is still a topic of active discussion in the literature.
I mostly follow Biskamp’s discussion in these notes.
Recall 3D turbulence: we saw that turbulent power
undergoes only a forward cascade. We also recall that
three-dimensional fluid flow has two significant invariants:
Z
1
E=
v 2 dV, energy
(23.15)
2
and
HV =
1
2
Z
v · ∇ × vdV, velocity helicity (23.16)
(Invariance of the first should be obvious; invariance of
the second may be proved in the homework.2 ) Both energy and velocity helicity are conserved in the modemode interactions which set up the cascade. Such
small-scale interactions drive both E and HV forward,
to smaller wavenumbers. The second invariant in 2D
turbulence, however, is different. The two invariants in
2D are
Z
1
v 2 dV, energy
(23.17)
E=
2
and
1
Ω=
2
Z
ω 2 dV, enstrophy
(23.18)
(The enstrophy is a fancy name that turbulence types
like to use for the mean square vorticity.) In 2D turbulence, wave-wave interactions drive Ω forward, but
drives E to smaller wavenumbers – in an inverse cascade. This cascade drives power to larger and larger
2
The methods are similar to the proof that magnetic helicity is
invariant, back in chapter 15.
scales: one author described it as the buildup of coherent vortices, in which nonlinear distortions nearly vanish, and which continue to grow by vortex coalesence.
One can use arguments similar to those in §22.3 to
predict the turbulent spectrum. Let ko be the driving
wavenumber, and assume it is well between the system scale ks , and the dissipation scale kd . Energy cascades to lower wavenumbers; the arguments of §22.3
still apply, so that a power spectrum W (k) ∝ k −5/3
should obtain for kmin < k < ko , where kmin corresponds to some largest scale reached by the reverse
cascade. Above ko , the enstrophy cascade rules. Describing enstrophy and turnover time at scale l as ωl ∼
vl2 /l2 , τl ∼ l/vl , we get the enstrophy power to be
ǫl ∼ Ωl /τl ∼ vl3 /l3 . But this last must be independent of l (repeating the Kolmogorov argument); from
this we need vl ∝ l, so that kW (k) ∝ vk2 ∝ k −2 . This
should describe the cascade up to the dissipation range.
Thus, for 2D fluid turbulence driven at ko , we expect
W (k) ∝ k −5/3 ;
W (k) ∝ k
−3
;
kmin < k < ko
(23.19)
ko < k < kd
This is supported by observations. As in 3D turbulence,
the high-k part of the cascade can be steady-state. Energy input to the system at ko will cascade forward,
reaching the dissipation scale kdiss where it goes into
hear. On the contrary, however, the low-k part of the
cascade cannot be stable. There is no low-k dissipation
to balance the energy input. One would expect kmin to
decrease with time, until the system size ks is reached;
at and after this point energy will continue to accumulate at the largest scales allowed in the system.
What sets the cascade direction? I have not found
any clear answer in the literature. On small scales,
the nature of energy, enstrophy and helicity transfer in
mode-mode interactions (say, k1 +k2 → k3 ) must have
a preference for forward or reverse transfer. The details
of these processes seem not to be obvious, however.
Biskamp reports on work in the literature addressing
the (thermodynamic) equilibrium distribution of E(k),
HV (k), and Ω(k). It appears that the equilibrium distributions of both E(k) and HV (k) are both weighted
towards high k’s in 3D; while in 2D the distribution of
Ω(k) is weighted to high k, but that of E(k) is weighted
to low k. The inference, then, is that the cascade direction is set by the tendency of the system to move
towards a statistical equibrium state.
127
C. Small scales and intermittency
Much of the action these days seems to be “what’s happening on small scales”, which means close to the Kolmogorov dissipation scale. This is potentially a big
topic; I’m storing only a brief overview here.
1.
INTERMITTENCY
The term intermittency is used in two different ways —
which drove me crazy when I was trying to learn this
field.
Older: macroscopic intermittency. Turbulence
observed on large scales (comparable to the system
size, well above the dissipation range) is intermittent:
only a fraction of the volume is filled with turbulent
spots or eddies at any instant. Sit at one point in a turbulent region .. you will find periods of high frequency
fluctuations, and also quiet periods, as the turbulent
patches evolve into/out of your region. This seems to
be how the term is used in the older literature.
Newer: microscopic intermittency. Turbulence
shows quite interesting behavior when observed on
scales comparable to the dissipation scale. On these
small scales, you find intermittent random “bursts” of
energy (think of measuring the velocity as a function of
time); this which translates into non-Gaussian statistics,
which many authors use as a definition of intermittency.
Physically, careful observations and numerical simulation reveal the existence of a tangle of intense, slender
vortex ropes on these small scales. This is the way the
term is used in the current literature.
2.
TURBULENCE ON SMALL SCALES
The current discussion on small-scale structure is quite
interesting. Problems with the Kolmogorov picture
(called K41) on small scales seem to have been realized early on. It was noted that velocity derivatives did
vary with the Reynolds number (which is inconsistent
with K41 assumptions). Batchelor and Townsend interpreted this as “a tendency to form isolated regions
of concentrated vorticity”. Moffatt reports a comment
from Landau to Kolmogorov, that in local regions of
higher ε, the cascade will proceed more vigorously.
Thus an intermittent distribution of ε(x, t) is expected;
and this should affect the 5/3 exponent slightly, but
should have stronger effect on higher-order statistical
quantities (such as the velocity derivative).
This can also be interpreted statistically. A central assumption of the Kolmogorov theory is the selfsimilarity of the random (velocity) field at all scales.
Therefore, you should find the same statistics on all
time/space scales. This turns out not to be so: on
small enough scales (high enough frequencies, think of
a high-pass filter), you find intermittent “bursts” of energy. Connect this to statistics: a self-similar random
signal, which is the same on different scales, is a white
noise signal, and has Gaussian statistics. When you go
intermittent, you have a greater chance of getting large
bursts, and less chance of getting low-amplitude signal
.. so the probability distribution function (PDF) flattens. Several authors show PDF plots with tails much
flatter than Gaussian. This becomes conspicuous only
on scales comparable to, or smaller than, the dissipation scale. Thus, it is characteristic of the dissipation
range, and does not imply breakdown of the entire K41
analysis.
3.
THE ROLE OF VORTEX FILAMENTS
One more commenet on the topic of statistics: the statistical approach we introduced in §22.C was based on
the power spectrum of two-point correlations. It threw
away phase information about the Fourier components
of the flow, as well as any information about higherorder correlations. In particular, that means that the
power-spectrum approach can’t say anything about locally anisotropic structures, such as small-scale vortex
ropes. But we are learning, more and more, that vorticity is important on small scales. So, we can’t just
rely on analytic methods, let alone scaling arguments.
We need to turn to numerical simulations – which have
made great strides in recent years, as computers get
faster and sophisticated numerical codes are developed
to take advantage of them. We can also turn to careful
experiments, designed to probe small scales.
The picture emerging from simulations and experiment is as follows. Essentially all simulations show
persistent and extended tubes, sheets and blobs of
small-scale vorticity. The filaments are tubes with approximately circular cross section, and diameter on the
order of the dissipation scale. Their length is somewhere between the Taylor scale and the driving (outer)
scale. The internal structure and dynamics of these little
vortices is still unclear, it seems – attempts to use analytic models (such as the Hill’s vortex we saw in homework, or a linear vortex model called Burgers) have not
been all that successful. There is definitely a sense in
the literature that these dissipation-scale vortes tubes
play an important role in the overall structure and dynamics of turbulence ... but, again, just how that works
is far from clear. So this field is still evolving. Meanwhile, I like the quote from Moffatt, Kida and Ohkitani
(1994):
128
Just as sinews serve to connect a muscle with a
bone or other structure, so the concentrated vortices of turbulence serve to connect large eddies
of much weaker vorticity; and just as sinews can
take the stress and strain of muscular effort, so the
concentrated vortices can accomodate the stress
associated with the low pressure in their cores and
the stress imposed by relative motion of the eddies
into which they must merge at their ends.
References
The mean-field material is “traditional”, and can be
found in various books which treat turbulence mathematically. Tennekes & Lumley, or Hinze, are good
sources.
The newer turbulence material, especially on intermittency, I’ve taken from “here and there”. The Annual Review of Fluid Mech has several useful reviews
(including the Moffat homage-to-Batchelor article from
2003).
129
24. MHD TURBULENCE AND DYNAMOS
MHD turbulence can be quite different from the fluid
turbulence we considered in chapters 22 and 23. The
study of MHD turbulence is a much newer field, with
must less experimental support or motivation. Many of
the arguments and conclusions appear to be less “engraved in stone” than is the case for fluid turbulence.
This doesn’t mean the physics is unimportant, though.
A critical application is to magnetic dynamos, which
are clearly important (what maintains the earth’s B
field? The sun’s? What makes each one reverse every
so often?), but not yet understood. We do know, however, that naturally occuring (plasma/MHD) dynamos
could not exist without turbulence. So we need to understand the one, in order to understand the other.
A. Magnetic fields in Isotropic turbulence
To start, let’s think about the effects of isotropic (mainly
HD) turbulence on a passive, “isotropic” B field (one
that’s initially very small, with no imposed order, and
thus dynamically unimportant). The most basic fact is
that the turbulence will amplify the field: as the plasma
flows stretch and twist the field lines, the magnitude of
B will increase with time. We might guess that the field
will grow to “energy equipartion” with the turbulence,
< ρv 2 /2, before it can “fight back” against
i.e. B 2 /8π ∼
t
the flow and make the whole problem more complicated. Simulations show that this energy balance is
more or less true. But what is the structure of the B
field? Does it have power on large or small scales?
To approach this, you may remember that we noted
the similarity between the dynamical equation for vorticity (4.8), and that for the B field (13.7). We know
that HD turbulence contains – or is held together – by
small-scale vortex ropes, which turn up around the dissipation scale. We therefore expect MHD turbulence to
show concentrated magnetic flux ropes on small scales.
This is somewhat of a new area, but there is suggestive evidence from numerical simulations that this is
the case. Both analytic estimates and simulations –
mostly the latter – now show that the the magnetic field
in isotropic, fluid turbulence tends to be concentrated
into small-scale filaments. We can, in fact, guesstimate the size of the flux ropes. Remember the magnetic Reynolds number, Rm = vt λt /η (defined here in
terms of turbulent velocity and length scales). These filaments have length ∼ λt , and thickness ∼ λt /Rm1/2 .
Some simulations also find that most of the magnetic
energy exists around these scales: there isn’t much
large-scale B field in this situation.1
On magnetic flux ropes ... we know from astrophysics that they are common (perhaps “B fields like to
clump into flux ropes”?). One example is the B field at
the surface of the sun is concentrated in intense, narrow
flux ropes – the locations where they emerge through
the surface are sunspots. There are hints elsewhere
in astrophysical data – observations of the interstellar
medium or the plasma in galaxy clusters – which also
point to the existence of magnetic filaments on these
scales.
B. The Inertial Range in MHD Turbulence
How does this change if there is an external, ordered
B imposed on the system to start – even if it’s quite
weak? How, if at all, does a magnetic field change Kolmogorov’s arguments about the energy casacade and
the existence or spectrum of the inertial range? That’s
not yet clear. Everyone agrees that energy still cascades
forward, to smaller scales, in MHD turbulence. But just
how that happens is a matter of ongoing debate.
To be specific, assume the plasma is threaded by a
straight, uniform B field. Let’s also assume the driving
is fairly weak, so that the “turbulence” itself is fairly
weak. In this case, several authors have argued that
the turbulence can be thought of as a field of smallamplitude wavelike structures (think of Alfven waves,
or wave packets) propagating along B. This changes
the physics that governs energy transfer in the cascade,
and thus impacts the nature of the internal range (compared to simple HD turbulence). Can we come up with
scaling arguments to describe the intertial range, for turbulence in this limit? To proceed, we need to note that
there are two critical differences betwen this case and
the simpler HD turbulence described by Kolmogorov.
The first critical difference is the timescale for energy exchange between scales. Recall for HD turbulence the timescale is just the eddy turnover time at l,
namely τl ∼ l/vl . For MHD turbulence, there is a second important timescale, namely the time it takes an
Alfven wave to move through the “eddy”. Which of
these, if either, should we take as the energy transfer
time? The literature contains two important alternative
possibilities, as follows.
1
We’ll see below that if the turbulence has a particular type of
anisotropy – helicity – then the B field it generates has substantial large-scale energy. This is one form of a dynamo.
130
1.
THE KRAICHNAN MODEL
The first approach comes from the 1960’s. It is generally attributed to Kraichnan, but Iroshnikov published the same general idea in the Russian literature
a few years earlier. So this is now called IroshnikovKraichnan (IK) turbulence.
Following them, assume the main turbulent structures – call them “turbulent eddies” or “Alfven wave
packets” are isotropic, with scale l. The Alfven time is
then τA ∼ l/vA . Now, if the turbulence is weak, vl is
small, so that τA ≪ τl = l/vl . In addition, the distortion a wave packet suffers2 in one crossing is small,
δv/v ∼ τA /τl . The number of “hits” required to distort
the packet significantly, and thus to transfer energy to
another packet (as needed for the cascade, is therefore
∼ (v/δv)2 . IK therefore argued that the energy transfer
time in MHD turbulence is more like
τE ∼
τl2
τA
(24.1)
This has consequences for the MHD turbulent spectrum. The energy transfer rate is now
ε∼
vl4
vl4 τA
∼
l2
lvA
(24.2)
From here, repeating the arguments in §22.3, we find
the Kraichnan spectrum for MHD turbulence:
W (k) ≃ (εvA )1/2 k −3/2
ANISOTROPY; GOLDREICH - SRIDHAR MODEL
But there’s another critical difference between HD and
MHD turbulence. MHD turbulence is very likely to
be anisotropic: the turbulent wave packets have two
dimensions, lk and l⊥ (along and across B). Now,
Alfven waves will travel along the field, so we can define τA ∼ lk /vA . But turbulent “strains” should act
across the packet; so we expect τl ∼ l⊥ /vl . If we
retain the weak-eddy-interaction picture, from (24.1),
we come up with a more compicated “energy transfer
time:”
τE ∼
2
l 2 vA
τl2
∼ ⊥ 2
τA
l k vl
¢1/2 −2
¡
k⊥
W (k) ∼ ǫkk vA
(24.4)
The distortion can be measured by the fluctuation in the (transverse) plasma velocity, δv
(24.5)
But that’s not the whole story. GS suggested that
this argument only works for the larger scales in the
system. The “transition” spectrum, (24.5), says that
less and less energy resides in the perpendicular component (of a turbulent “eddy”), while the parallel remains unchanged. Thus, the turbulent structures become more and more elongated as one goes to smaller
and smaller scales. So, GS argue that at high enough
wavenumber (k⊥ ), the system will reach a state in
which v⊥ (l⊥ )/l⊥ ∼ kk vA . This is a state in which
the parallel Alfven wave transit time through the eddy
is equal to the perpendicular straining time (or turnover
time) of the eddy: τA ∼ τl will arise naturally, according to GS. This state is called critical balance. But look
back at (24.1): when critical balance holds, we have
τE ∼ l/vl ∼ τl ! That is, we now have only one time
scale – and so we recover the original IK turbulent spectrum, with an additional prediction on the aspect ratio
of the turbulent eddies:
(24.3)
which is flatter than the Kolmogorov spectrum. This
result should hold for 2D and 3D MHD turbulence (as
both show a forward energy cascade).
2.
What now? How do we deal with this?
One possibility is to argue that the A wave packets
are very weak, so that lk is not affected at all by interactions between them – thus we can treat it as a constant
(governed by the driving). If this is the case, then only
the l⊥ ∼ 1/k⊥ term in (24.4) is important to the cascade, and we expect
W (k⊥ ) ∼
3.
−5/3
ǫ2/3 k⊥
2/3
;
k ǫ1/3
kk ∼ ⊥
vA
(24.6)
WHAT NOW ?
Where do we stand today? Rather in the middle of a
muddle ... there is still a lot of discussion, or argument,
about which of these models works, if so where (under
what conditions), or do they work at all? Maybe IK
works here, GS works there? From what I see in the literature, the question is far from settled. Simulations are
very valuable tools, but different simulations can give
different answers (and it may be that the different physical/boundary conditions used in different simulations
truly give different turbulence physics).
What about observations? Those are also hard to
come by. Turbulence in the solar wind (which we can
measure directly, with satellite probes, at least close to
earth) was initially reported to have an IK spectrum,
W (k) ∝ k −3/2 (over a few decades – 2 or 3 I think
– in k). This was taken as strong support for the IK
model. More recently, however, more data have been
131
acquired. It is now established that solar wind turbulence is anisotropic; that’s good, because the wind definitely has an ordered background B. Unfortunately,
however, newer data don’t clearly support the IK spectrum – they suggest a steeper spectrum, maybe closer
to k −5/3 . I personally suspect that we just don’t know,
yet.
Moving to a larger playing field, there is a muchquoted paper (Armstrong etal ApJ 1995) which attempted to pull together several different types of measurement – all indirect, of necessity – of electron density fluctuations in the interstellar medium (ISM). I
have to point out that the ISM is very definitely a magnetized plasma, but also very definitely highly inhomogeneous – different experiments probe different regions and scales of the galaxy. This paper patched all
the measurements together and concluded that the ISM
obeys a Kolmogorov-type scaling, W (k) ∝ k −5/3 ,
over 12(!!) orders of magnitude in k. I remain personally skeptical of this result, because of the difficulties
and uncertainties in combining these disparate types of
measurement.
C. Cascades and Related Things
In addition to MHD effects on the inertial range, there
are several other important differences.
1.
IDEAL INVARIANTS
How to HD and MHD turbulence compare, in 2D and
3D? What are the important invariants?
We saw in Chapter 23 that magnetic helicity is invariant in MHD flows; as long as dissipation is small
(for scales above the dissipation range), total energy is
as well. This analysis can be carried out for HD and
MHD turbulence. One finds the following invariants.
R
• 3D HD: total
energy E = 12 v 2 dV ; velocity heR
licity HV = 12 v · ∇ × vdV
• 2D HD: total energy E; and total vorticity, also
~ = 1 ω 2 dV (recall ω
called enstrophy, Ω
~ = ∇ × v).
2
R • 3D MHD: total energy E; magnetic1 Rhelicity HB =
1
A · BdV ; and cross helicity, K = 2 v · BdV
2
• 2D MHD: total energy E; cross helicity
R K; and the
mean square magnetic potential, A = 21 ψ 2 dV
2.
CASCADE DIRECTIONS
In 3D HD turbulence we encountered a forward cascade (also called a direct cascade). That is, the turbulent shear stresses transfer energy to smaller scales
(higher wavenumbers). The original Kolmogorov picture involved a forward cascade, of course, as does its
modifications in the Kraichnan and GS models. Note
that this allows a steady state spectrum W (k): the forward cascade transfers energy from the input/driving
scale, to the dissipation scale, at which scale it goes
to heat. Because dissipation is more important at small
scales (high k’s), we expect always to be able to find a
steady state.
For 3D-HD turbulence, we only have a forward cascade. But in 2D HD turbulence, and in 2D and 3D
MHD turbulence, ond also finds an inverse cascade:
power flows to larger scales (smaller wavenumbers).
Inverse cascades move power to large scales, where dissipation is always small. Thus, they do not reach steady
states. They tend to evolve to lower and lower k’s, until
they reach the scale of the system, and then accumulate
power on those scales.
To summarize:
• 3D HD: all cascades are direct. This is the basic
Kolmogorov picture.
• 2D HD: enstrophy has a forward cascade, but total
energy has an inverse cascade. This seems to connect
to the coherence and large-scale eddies seen in 2D turbulent flows.
• 3D MHD: energy and cross helicity have forward
cascades; while helicity has an inverse cascade. This
has dramatic consequences for maintaining large-scale
ordered magnetic fields; this is the turbulent dynamo
which we see in the next section.
• 2D MHD: energy and cross helicity have forward
cascades; while the mean vector potential has an inverse
cascade.
3.
SELF - ORGANIZATION IN MHD
Some interesting properties of MHD turbulence are
connected to these invariants.
• Force-free fields. We have already seen that
turbulent relaxation evolves a plasma towards a selforganized state, in which j k B. This is due to the
invariance of the magnetic helicity (that is, to its much
slower rate of decay than the magnetic energy shows).
• Dynamic alignment. There is evidence (in
the solar wind, backed up by dynamical models) that
strong MHD turbulence shows a second form of selforganization. It tends to evolve to a state in which
±v k B. That is, it shows dynamic alignment of the
velocity and magnetic fields. This is due to the invariance of the cross-helicity; a formal minimization of E,
subject to K being constant, derives this condition.
• Energy equipartition. We should also note that
there is evidence that MHD turbulence reaches a state
132
of approximate equipartition between kinetic and magnetic energies: hvi2 ∼ hB 2 i/4π. This is a common
result in numerical simulations
D. MHD Dynamos
Now, a larger question: where do magnetic fields come
from? In the lab, the answer is easy: “currents”. In
magnetic solids, the currents are those of well-ordered
electrons spins in ferromagnetism. More typically, currents in the lab — and their consequent B fields —
come from obvious things like batteries and wires. The
issue is then, what drives the currents? My dictionary
defines a dynamo as
“a device for converting mechanical energy into electrical energy, usually by expending the mechanical energy in producing a periodic motion of a conductor and
a surrounding magnetic field”.
sun/earth/galaxy formed — because we know the resistivity of the plasmas in question, and thus we know
how long it would take a primordial current to dissipate.
Such calculations predict that primordial fields would
long ago have died away; but we know that stars, planets, and galaxies are still magnetized.3
Thus, we still must ask, “what supports astrophysical B fields? The answer is still, “currents”; but what
drives astrophysical currents? We can’t expect a device
such as in Figure 24.1 exists inside a planet, or star, or
whatnot ... so we need to find a way to drive fluid motions which can maintain the B fields we observe. This
question gets us into what’s called dynamo theory.
As with vorticity, the question of fluid dynamos
could occupy a full course on its own. All we can do,
once again, is a brief introduction. To set the stage,
I paraphrase Roberts & Soward (1994, Ann Rev Astr
Ap):
We know dynamos exist; one can buy them in
shops.4 What, one may then ask, is all the fuss
about? To answer, we must make an important distinction between commercial dynamos and
naturally occuring ones. In order to ensure that
the induced currents don’t short-circuit, an engineer makes sure that the geometry of the machine
is an asymmetric and multiply-connected region.
By contrast, the conducting fluid in as astrophysica body usually occupies a symemtric, simplyconnected domain – such as a sphere (the earth
or sun) or a simple disk (the galaxy). It is by no
means obvious that a dynamo can operate in such
a simple system. Than, then, is the point of this
field [by which they mean dynamo theory].
A simple lab version of this is called the unipolar
dynamo, in Figure 24.1. This involves a conducting
disk, threaded by a B field, which rotates about its axis.
This induces a radial E field, v×B/c, and thus a potential drop between the axis and the edge of the disk. If
you hook up wires in the right way you’ll have a current
– and this current will create its own magnetic field.
1.
Figure 24.1. A simple unipolar dynamo (in a less than
simple figure from www.stardrivedevice.com, the best
figure I could find). The conducting disk moves through an
(externally supported) B field as it rotates about its axis.
The resultant EMF supports a potential drop between the
axis and edge of the disk — which can drive a current.
What about astrophysical magnetic fields? To be
specific, what is the origin of the earth’s field, or the
sun’s field? It’s easy to think of what doesn’t work.
One, even solid planets like the earth can’t be ferromagnetic (because the core temperature is well above
the Curie temperature at which permanent magnetism
disappears); and clearly stars and galaxies can’t be ferromagnetic at all. Two, we can’t assume the fields
are primordial — were somehow created when the
COWLING ’ S THEOREM
We can start by seeing what won’t work. That is, most
astrophysical models assumem simple, symmetric geometries; but these can’t support a dynamo.
We can support the statement from Roberts &
Soward by proving Cowling’s theorem, namely, that it
is not possible to maintain a steady dynamo in an axisymmetric system. To prove this, we follow Cowling’s
original (1934) argument, as presented by Choudhuri.
Start by assuming we do have an axisymmetric dynamo: one with ∂/∂t = ∂/∂φ = 0. Consider a plane
through the symmetry axis: the projections of the field
3
In addition, we know that the sun’s field reverses pretty regularly,
every 11 years or so; and the earth’s field reverses less regularly,
every 104 − 105 years. This clearly requires some internal, selfgoverning mechanism.
133
(a)
Figure 24.2.
Illustration of geometry for Cowling’s
theorem. Following Choudhuri Figure 16.3.
lines on this plane must be closed curves (think of a
simple magnetic dipole). There will be at least one neutral point in this plane (a point where the closed field
lines center) – and jφ must be non-zero here, while B
has only a φ component at this point. Take a line integral of Ohm’s law (13.4) along a closed loop through
these neutral points, enclosing the symmetry axis:
I
I
I
1
jφ dl = E · dl + v × B · dl
(24.7)
σ
But now: the second term vanishes, because B k dl if
this loop goes entirely through neutral points. The first
term vanishes, because
I
Z
Z
∂B
· dS
E · dl = ∇ × E · dS = −
∂t
But this last is zero by our steady-state assumption.
However, the LHS of (24.7) is non-zero, as jφ is finite.
Thus, we have a contradiction; and Cowling’s theorem
is proved.
2.
PARKER ’ S SOLAR DYNAMO
It follows, then, that we must relax the assumption of
axisymmetry. The classic example of this is dynamo
model Parker’s (1955) model of the solar dynamo. His
model is meant to describe the solar magnetic field, in
particular explain how “turbulence” (in this case convection) offsets field stretching due to differential rotation, and support the quasi-dipolar field we observe.
His model is best presented qualitatively — refer to
Figure 24.3 for the cartoon. Say the solar field starts
mainly dipolar (this is roughly consistent with observations of the global field, just above the solar surface).
The sun does not rotate as a solid body; near the surface, the equator rotates faster than the polar regions.
This will stretch our dipolar field, generating toroidal
components. Thus, it is no problem to generate toroidal
(b)
(c)
Figure 24.3.
Parker’s model of the solar dynamo, at the
cartoon level. (a) Differential rotation (the sun’s equator
rotates faster than the poles) stretches initially dipolar field
in the toroidal direction. (b) Coriolis forces acting on
surface convective cells generates local poloidal fields. (c)
The opposite sense of the Coriolis force in the north and
south hemispheres, combined with the opposite sense of the
initially toroidal field, results in a strong net poloidal field,
rather than a randomly directed set of field loops. (These
loops are shown projected in the meridional plane.)
Following Choudhuri, Figures 16.4.
field if the body has differential rotation. But this cannot be all of the story. Such a stretched toroidal field
will have many local field-line reversals, and if nothing
else happens it will simply decay away due to resistive
dissipation.
However, the upper layers of the sun are convectively unstable. In this region, plasma blobs rise and
fall.5 Now, these vertically moving blobs are subject to
a Coriolis force, due to the sun’s overall rotation. The
blobs therefore rotate as they rise; they act like little
cyclones, and formally we say that their their motion
has a net helicity (that means the small-scale motions
do not have mirror symmetry: for instance a flow with
v · (∇ × v) 6= 0 is helical). Look at (b) of Figure
9.5: this cyclonic motion twists the magnetic field back
into poloidal loops. Remember that both the direction
of Bφ and of the Coriolis rotation are opposite in the
north and south hemispheres: this means the direction
of the poloidal field component generated is the same in
the two hemispheres. We therefore have a fully working dynamo: poloidal fields are generated by the helical
convective (turbulent) motions, while toroidal fields are
generated by differential rotation. The whole system
must be stabilized by dissipation – that is resistivity will
keep each field component from getting too large.
Thus: we have argued “by cartoon” that helical,
convective motions on the sun (or the earth) can maintain the large-scale B field. That is, we’re arguing that
small-scale turbulent motions can add up to a net largescale dynamo, as long as the turbulence is helical. Now
we need to quantify this idea.
5
In chapter 8 we talked about (in)stability to buoyancy – an unstable atmosphere will develop strong convection.
134
E. Kinematic Dynamos
F. Mean-Field Dynamos
One way to approach the problem is to assume you
know the velocity field, and solve the induction equation in order to see what happens to B. This type of
problem is usually “illustrative” rather than “a solution”; I present one such example here (taken straight
from Davidson).
We assume a large-scale (uniform, slowly varying
in space) Bo field exists to start; and pick a useful
small-scale velocity field,
v(x, t) = vo ei(k·x−ωt)
(24.8)
where vo = vo (−i, 1, 0) in Cartesian (i.e., the vx and
vy components of v are out of phase), and k k ẑ
(so the wave travels along the z-axis). This is a helical wave: ∇ × v = kv. What effect does this
have on the magnetic field? We can linearize: assume
B = Bo + b, where Bo is some background field. Because we have two very different spatial scales, we can
split the linearized induction equation into “large-scale”
and “small-scale” parts. The large-scale part is
∂B
= ∇ × (hv × bi) + η∇2 B
∂t
(24.9)
(24.10)
Note, in (24.9) we’ve made a critical assumption: that
the mean EMF from the small-scale stuff, hv × bi, has
large-scale order.
To solve this, we also assume b ∝ ei(k·x−ωt) (it
echoes the wave behavior). If we carry out the algebra,
we can solve for the amplitude and direction of b:
b=b
(Bo · k)(−ω + 1ηk 2 )
ẑ
(η 2 k 4 + ω 2 )
(24.11)
From this, we can (eventually) show that the helical v
field, and the b field that responds to it, do have a finite,
time-averaged EMF:
hv × bi = (Bo · k)
ηk 2 vo2
ẑ
(η 2 k 4 + ω 2 )
B=B+b;
v =V+v
(24.13)
and assume that b and v have zero mean. Once again,
we split the induction equation into large-scale (mean)
and small-scale (fluctuating) parts. For the mean field,
we get
∂B
= ∇ × (V × Bo ) − ∇ × ε + η∇2 B
∂t
(24.14)
where
where the first term on the RHS – the inductive term –
derives from the mean (over space & time) of the smallscale EMF. The small-scale part is
∂b
= ∇ × (v × Bo ) + η∇2 b
∂t
The above is one specific example of a “turbulent helical dynamo”. The problem is usually treated much
more formally, using statistical measures of MHD turbulence.
This analysis – backed up by numerical work –
shows that MHD turbulence can, indeed, maintain magnetic fields with structure on scales large compared to
the turbulent scales. This result derives from the fact
that magnetic helicity cascades to larger, rather than
smaller, scales.
We return to mean-field theory to describe this. Split
the velocity and magnetic fields again into mean and
fluctuating parts:
ε = −hv × bi
is the net EMF due to the fluctuating v and B̃ fields.
Once again, we’re assuming that ε belongs in the largescale equation – but we need to verify that.
So, we must ask whether ε has any interesting largescale effect. Because we’re being general (not specific
as in the previous example), we need to be formal here.
If v and b are rapidly varying, have zero mean, and
uncorrelated, we’d expect the mean of their product to
be zero. It turns out (“can be shown”) that things are
interesting (i.e., ε 6= 0) if the turbulence satisfies two
conditions: (i) it must be helical, satisfying v · ∇ ×
b 6= 0; and (ii) it must be resistive; η 6= 0. If both
of these conditions are met, one can show formally that
the turbulent EMF, ε , will be non-zero on large scales.
But we also want to explore the small-scale behavior. The small-scale field satisfies
∂b
=∇ × (V × bv × B)
∂t
(24.12)
This verifies our assumption, that hv×bi belongs in the
large-scale induction equation, (24.9). Thus, this EMF
acts back on the large-scale field, in fact acts as a source
(driving) term on that field.
(24.15)
(24.16)
2
+ ∇ × G + η∇ b
where
G = v × b − hv × bi
(24.17)
135
Now: the standard analysis argues that the ∇ × G
term can be dropped from (24.16) (due to being secondorder small), as can the dissipative term η∇2 b (assuming we are still far from the disspation lengths). This
means that we can formally integrate (24.16) to get
Z
b ≃ ∇ × [v × B] dt′
(24.18)
– which connects back to Taylor relaxation, which we
saw in Chapter 15: a plasma will spontaneously relax
to a mininum-energy state, which is force-free. The relaxation proceeds through self-generated turbulence –
thus the plasma finds its own solution of ∇ × (αB) +
η∇2 B = 0.
(where v′ is evaluated at t′ , and unprimed means evaluated at t. From this,
Z
£
¡
¢¤
ε ≃ − dt′ hv × ∇ × v′ × B i
(24.19)
Finally, a few words about trying to do this in the lab.
Everything above is still pure theory — it would be
good to verify directly that an αω dynamo (rotation plus
turbulence, as in the sun), or an α2 dynamo (pure turbulence) can really make a large-scale ordered B field.
Several groups are working on this, including NMT’s
very own Stirling Colgate. The experiments use liquid
metal — usually liquid sodium — in some sort of rotating system (the ω in an αω dynamo), and try various
ways to induce turbulence (the α) in the flow. The last I
heard, no one had sucessfully made their dynamo work
— but I think the field’s progressing. Check the Feb
2006 issue of Physics Today if you’d like more details.
But now, we can note that ε is linear in B and its curl:
ε ≃ αB + β∇ × B
Working out the integrals in detail, one finds
Z
τ
1
dt′ hv · ∇ × v′ i = HV
α=
3
3
Z
τ
1
dt′ hv · v′ i = hv2 i
β=
3
3
(24.20)
G. Astrophysical dynamos in the lab
(24.21)
if τ is the velocity correlation time. Thus: the mean
field equation now becomes
∂B
= ∇ × (V × Bo ) + ∇ × (αB) + (η + β)∇2 B
∂t
(24.22)
Thus: the β term simply adds to the dissipation – another meeting with turbulent dissipation. The α term,
however, acts as a source term: one can show that it
leads to a growth of magnetic flux.
Thus, we have a big result: helical turbulence in a
plasma can amplify the large-scale field. That’s a turbulent dynamo.
If this works – if ε = αB — then we can see two
useful astrophysical consequences. One is balancing
ohmic losses, as in the sun or the earth — and (in principle) accounting for the occasional field reversals in
each body. The second is “growing” the B field in the
first place. To see this, note that (9.14) allows solutions
B ∝ eαt , if α is constant in time. That’s a growing B
field, with growth time ∼ L/α (some large-scale length
scale L). We might expect that a small seed field would
grow exponentially until some other physics (dissipation? back reaction on the driving fluid?) comes into
play.
In addition, we note that when V = 0 (no largescale flows), (24.14) has particularly simple steadystate solutions: B k j. That is just a force-free field
References
For the IK/GS turbulent cascade discussion, I’ve
followed two excellent recent review papers: Diamond
etal, A Tutorial on basic concepts in MHD turbulence and turbulent processes (available through the
physics.ucsd web site, I can’t find the publication reference); and Schekochihin & Cowley, Turbulence and
magnetic fields in astrophysical plasmas, to appear
in MHD: historical evolution and trends (Molokov,
Moreau & Moffatt, eds, 2006). These papers have references to the original papers.
I’ve also used Biskamp’s book for discussions on
intermittency, inverse cascades & invariants, etc.
The general dynamo material is in lots of books;
I like Priest’s Solar MHD, and also Moffett’s original
Mean Field Electrodynamics.