Exercises of Plasma Physics
MEFT - Master in Engineering Physics
Vasco Guerra
February 2017
Foreword
This collection of exercises is an outcome of my teaching experience as the responsible of the Plasma Physics and Technology course from the Integrated Master in
Engineering Physics of Instituto Superior Técnico, Universidade de Lisboa, in the
period between 2013 and 2017. This is a one semester introductory course, attended
by all students of the Master. As such, it is not intended to be a comprehensive
course for students that will follow a plasma physics track. Nevertheless, the course
is designed to give a general overview of plasma physics, addressing the basic concepts and the main approaches to study the field, namely single particle motion,
fluid descriptions (both two fluid and single fluid) and kinetic theory. Each chapter
corresponds to exercises that can be used as a one-week problem sheet.
Due to the broad scope of the course, it is impossible to cover in detail all the topics
studied within the allocated time. Therefore, various exercises were conceived in
order to make students have a glance at some relatively standard phenomena not
studied in class, but that can be investigated with the knowledge already acquired,
such as the ponderomotive force, the two-stream instability and the derivation of
the fluid equations from the moments of Vlasov’s equation. Some of these problems
would be rather challenging without guidance, but with the hints included students
are expected to succeed in solving them.
Various exercises are taken or adapted from the textbook of Francis Chen (F.F. Chen,
Introduction to Plasma Physics and Controlled Fusion, Vol. 1, Plenum Press 1984).
All these exercises are duly identified along the text. Other exercises were adapted
from other books and sources available on the web. The book of Dwight Nicholson
(D.R. Nicholson, Introduction to plasma theory, John Wiley & Sons 1983) and the
very good online collection by John Howard, from the Australian National University,
deserve a special reference. To these problems I have added a significant number of
my own. Finally, several exercises used in actual written examinations are included
in this collection. They are identified with a (?) mark, which allows assessing the
intended level of the course.
Lisboa, February 2017
Vasco Guerra
4
Formulae
• Constants:
Electron mass
Electron charge
me = 9.1 × 10−31 kg;
e = 1.6 × 10−19 C;
Vacuum permittivity 0 = 8.854 × 10−12 F/m
Boltzmann constant
kB = 1.38 × 10−23 J/K
Earth’s radius
RT ' 6370 km
• Conversion factors:
1 u = 1.66 × 10−27 kg; 1 eV = 1.6 × 10−19 J;
1 atm = 760 Torr
5
1 atm = 1.013 × 10 Pa
• Mathematical relations:
~a × (~b × ~c) = (~a · ~c)~b − (~a · ~b)~c
~ × (∇
~ A)
~ =0
∇
~ × (∇
~ × A)
~ = ∇(
~ ∇
~ · A)
~ − ∇2 A
~;
∇
~ × (ψ A)
~ = ψ(∇
~ × A)
~ + (∇ψ)
~
~;
∇
×A
ˆ
∞
r
2
exp(−Ax )dx =
−∞
+∞
ˆ
2
2
π
A
x exp(−Ax )dx =
0
√
π
4A3/2
1 d2
1 d
2 dφ
In spherical coordinates and simmetry, ∇ φ = 2
r
=
(rφ)
r dr
dr
r dr2
d2 φ 1 dφ
In cylindrical coordinates and symmetry, ∇2 φ = 2 +
dr
r dr
2
6
• Basic relations and fundamental effects
Ideal gas law
Plasma frequency
P V = N kB T
q
λe = ε0nkeBe2Te
q 1
r
φ(r) = 4π
exp
−
r
λ
0
D
q
ne e2
ωpe = ε0 me
Plasma parameter
Λ = ne λ3D
Electron Debye length
Debye potential (spherical symmtry)
Electron cyclotron frequency
ωce =
Larmor radius
eB
me
v⊥
ωc
rL =
q
vt = kBmT
Thermal speed
• Drifts
ExB drift
Grad B drift
Curvature drift
Fields in vacuum
Polarization drift
Diamagnetic drift
Magnetic moment
~ ×B
~
E
2
B
2 ~
~
mv⊥
B × ∇B
~vd =
2qB
B2
2
~
mvk ~ur × B
~vd =
2
qB
Rc
~
1 2
1 ~ur × B
2
~vd = mvk + v⊥
2
2
qB
Rc
m d ~
~vd =
E⊥
qB 2 dt
~ ×B
~
1 ∇P
~vd = −
qn B 2
1
mv 2
µ= 2 ⊥
B
~vd =
• Waves
– Electrostatic electron waves
~ 0 = 0 ou ~k k B
~ 0:
∗ B
2
ω 2 = ωpe
+ 3k 2 vt2 ; vt2 = kT /m (Langmuir waves)
2
2
~ 0 : ω 2 = ωpe
∗ ~k ⊥ B
+ ωce
= ωh2 ; (upper hybrid waves)
– Ion electrostatic waves
~ 0 = 0 ou ~k k B
~ 0:
∗ B
i kB Ti
ω 2 = k 2 c2s ; c2s = γe kB Tem+γ
(ion acoustic waves)
i
γe k B T e
1
2
2 γi kB Ti
ω =k
+ mi 1+γe k2 λ2
(ion plasma waves)
mi
De
~ 0:
∗ ~k ⊥ B
ω 2 = k 2 c2s + ωl2 ;
2
ω 2 = k 2 c2s + ωci
ωl2 = ωce ωci (lower hybrid oscillations)
(ion cyclotronic waves)
7
– Electron electromagnetic waves
2
~ 0 = 0: ω 2 = ωpe
∗ B
+ k 2 c2
~ 0, E
~1 k B
~ 0:
∗ ~k ⊥ B
~ 0, E
~1 ⊥ B
~ 0:
∗ ~k ⊥ B
~ 0:
∗ ~k k B
n2 =
; (electromagnetic waves)
2 2
c k
ω2
n2 = 1 −
n2R,L = 1 −
2
ωpe
/ω 2
1∓ωce /ω ;
2
ωpe
ω2 ;
2
ω 2 −ωpe
2 ;
2
ω −ωh
=1−
(ordinary wave)
2
ωpe
ω2
(extraordinary wave)
(right and left waves)
• Transport and MHD
∂
~ · (ns~vs ) = 0
ns + ∇
∂t
h
i
∂~vs
~ v~s = qs ns E
~ + ~vs × B − ∇P
~ s − νs0 ns ms (~vs − ~v0 )
+ (~vs · ∇)
ns m s
∂t
ν = N hσvi
~Γ = nµE
~ − D∇n
~
ρm
∂~v
~ − ∇P
~
= J~ × B
∂t
~Γ = n~v
~ + ~v × B
~ = η J~
E
P = Pe + Pi
• Maxwell’s equations
~ ·B
~ =0;
∇
~ ·E
~ = ρ ;
∇
ε0
~
~ ×B
~ = µ0 J~ + 1 ∂ E
∇
2
c ∂t
~
∂
B
~ ×E
~ =−
∇
∂t
• Kinetic theory
ˆ
n(~r, t) =
f (~r, ~v , t)d3 v
;
~ = h~v i = 1
V
n
ˆ
~v f (~r, ~v , t)d3 v
∂f
~ r f + q (E
~ + ~v × B)
~ ·∇
~ v f = 0 (Vlasov eq.)
+ ~v · ∇
∂t
m
8
CHAPTER
1
Debye shielding and fundamental effects
1. Calculate the number density of an ideal gas at:
(a) p = 1 atm and T = 273 K (Loschmidth number)
(b) p = 1 Torr and T = 300 K
2. (F. F. Chen) On a log-log plot of n (cm−3 ) vs kB Te (eV) draw lines of
constant λD and Λ. On the graph place the following points:
(a) Tokamak (Te ' 1 keV, ne ' 1013 cm−3 )
(b) Solar wind near the Earth (Te ' 10 eV, ne ' 10 cm−3 )
(c) Ionosphere, ∼ 300 km above Earth’s surface (Te ' 0.1 eV, ne ' 106
cm−3 )
(d) Laser fusion (Te ' 1 keV, ne ' 1020 cm−3 )
(e) Gaseous electronics (Te ' 1 eV, ne ' 1010 cm−3 )
(f) Interstellar medium (Te ' 1 eV, ne ' 10−1 cm−3 )
(g) Flame (Te ' 0.1 eV, ne ' 108 cm−3 )
3. Consider Debye’s potential created by a punctual test charge qT that is placed
inside an homogeneous plasma.
(a) Show that the charge in the shielding cloud exactly cancels qT . Calculate
the total charge inside spheres of radii λD /2, λD and 5λD .
(b) Determine the electrostatic interaction energy between the test charge
and the particles in the plasma and the total mean energy of the plasma
particles (assume Te = Ti = T ).
4. Consider a homogeneous plasma with density n0 = 108 cm−3 occupying the
region x > 0. Outside the plasma (x < 0) there exists an uniform electric
~ = 100~ux V/cm, which penetrates the plasma. The electron and ion
field E
temperatures are equal, Te = Ti = 0.1 eV. Show that the plasma shields the
10
Chapter 1. Debye shielding and fundamental effects
field and calculate the typical shielding length. Calculate the intensity of the
electric field at x = 0.5 cm, assuming that eφ(x)/kB T 1.
5. (F. F. Chen) A spherical conductor of radius R is immersed in a plasma and
charged to a potential φ0 . The electrons remain Maxwellian and move to
form a Debye shield, but the ions are stationary during the time frame of the
experiment. Assuming eφ0 kB Te :
(a) derive an expression for the potential as a function of r;
(b) calculate the charge in the sphere;
(c) calculate the sphere capacity for R = 10 cm, Te = 1 keV and n0 = 1014
and 106 cm−3 , and show that for high electron densities the plasma
behaves as a dielectric.
6. In the deduction of the electron plasma frequency, suppose the ions are not
infinitely massive, but have a mass mi and can move. Modify the discussion
to show that the coupled oscillation of the electron and ion “slabs” is made
2
2
+ ωpi
).
with the total plasma frequency (ωp2 = ωpe
7. (∗) In this problem we want to calculate the plasma oscillation frequency for
a spherical plasma, proceeding in a similar way as it was done for the slab
configuration in the previous exercise.
Consider a spherical plasma of radius R, represented by a uniform positive
ion background of density n0 inside the sphere. Assume the ions are infinitely
massive. Initially, the electron density has the same volume distribution as
that of the ions. The “electron sphere” is then stretched to a radius R+δr and
then released. Assume at all instants that the electron density is distributed
uniformly on the spherical volume it occupies.
8. An infinite conducting plane is placed inside an homogeneous plasma and
charged to a potential φ0 . The electrons move and keep a Boltzmann distribution, with eφ/kB Te 1, while the ions can be considered stationary for
the time-scale of the experiment. Consider the xx direction perpendicular to
the plane and x = 0 coinciding with the plane.
(a) Obtain the potential as a function of x and represent φ(x)
(b) Determine the electric field as a function of x and the charge density in
the plane. Represent E(x) and compare with the solution in vacuum.
(c) Show that the plasma completely shields the charge in the conducting
plane.
9. (?) Consider an infinite line, uniformly charged with a linear charge density λ,
immersed in a homogeneous plasma. The electrons and ions follow Maxwellian
distributions, respectively with temperature Te and Ti .
(a) Show that the electrostatic potential can
written, in cylindrical co be
λ
r
ordinates, in the form φ(r) = 2πε0 K0 λD , where K0 is the modified
Bessel function of second kind of order zero, and determine λD . Derive
the expression for the electric field as a function of r.
(b) Calculate the total charge around the line, per unit length.
11
The modified Bessel equation of
order α is
x2
d2 y
dy
+ x − (x2 + α2 )y = 0 .
dx2
dx
The solution is a linear combination of modified Bessel functions of first and second kind,
Iα (x) e Kα (x), which are exponentially growing and decreasing
functions, respectively (see figure).
ˆ
K00 (x) = −K1 (x)
+∞
xK0 (x)dx = 1
0
For small x, K0 (x) ' − ln x; K1 (x) '
1
x
10. (?) Dusty plasmas.
In many plasmas there can be found large particles, besides electrons and
positive ions, known as dust particles. The aim is to study Debye shielding
around a positive test charge qT in dusty plasmas. Assume that electrons
and positive ions follow a Boltzmann distribution at temperatures Te e Ti ,
respectively, while the dust particles are infinitely massive, have a total charge
Zd and are uniformly distributed on the volume.
(a) Do you expect the charge Zd to be positive, negative, or that it can
have any of the signs? Justify.
(b) Consider a quasi-neutral plasma where the dust particles are negatively
charged. Show that Poisson’s equation can be written in the form
ne0 e2
ni0 Te
2
∇ φ=
1+
φ,
0 kB Te
ne0 Ti
where ne0 and ni0 are the non-perturbed densities of electrons and ions,
i.e., their densities at a large distance from the test charge.
(c) Determine the Debye length and tell if it is larger, smaller or equal to
the case where there are no dust particles.
(d) Obtain the expression for the electrostatic potential φ(r).
Historical note: this problem was studied by Lakhsmi, Bharuthram and
Shukla in Astrophys. Space Sci. 209 (1993) 213. Dust particles have
been observed in asteroid regions, planetary atmospheres (Earth and
Titan), comet tails and several laboratory plasmas.
11. (?) Plasma sheaths
A plasma is confined between two grounded (φ = 0) parallel plates located at
x = 0 and x = l. The ion density is ni (x) = n0 for 0 < x < l. The electron
12
Chapter 1. Debye shielding and fundamental effects
density is
ne (x) =
n0 ,
0,
if s < x < l − s
for 0 < x < s and l − s < x < l
.
The regions 0 < x < s and l − s < x < l are called the “sheath” regions.
(a) Determine the potential φ(x) and the electric field E(x) for 0 < x < l.
Find φ0 = φ(x = l/2) and plot the potential and the electric field as a
function of x.
[Hint: consider the symmetry of the problem and recall that the electric
field must be continuous]
(b) Does the electric field at the sheaths act to confine the electrons within
the bulk plasma or does it tend to destroy the confinement?
(c) Chose eφ0 = 5kB Te and find an expression for s. Discuss the plausibility
of the value given for φ0 .
CHAPTER
2
Single particle motion I
2
1. For particles with the same kinetic energy W = mv⊥
/2, compute the ratio
between the Larmor radius of a proton and an electron (mp /me = 1836).
2. Consider a particle of charge q > 0 and mass m, initially at rest at (x, y, z) =
~ = B0 ~uz and E
~ = E0 ~uy ,
(0, 0, 0), in the presence of a static magnetic field B
with E0 , B0 > 0.
(a) Sketch the orbit of the particle.
(b) Derive an exact expression for the orbit of the particle.
(c) Show that the orbit can be separated into an oscillatory term and a
constant drift term. After averaging in time over the oscillatory motion,
is there any net acceleration? If not, how are the forces balanced?
(d) In a neutral plasma, with positive and negative particles and ions of
different masses, would there be any net current?
(e) Suppose the electric field were replaced by a gravitational force in the
yy direction, would there be a net current?
3. (F. F. Chen) An electron beam with density ne = 1014 m−3 and radius R = 1
~ = B0 ~uz , where B0 = 2
cm crosses a region with a uniform magnetic field B
T and the zz axis is aligned with the direction of propagation of the beam.
~ ×B
~ drift at r = R (note
Determine the direction and magnitude of the E
~
that E is the electrostatic field created by the charge of the beam).
4. (F. F. Chen) Suppose electrons obey the Boltzmann relation in a cylindrical
symmetric plasma column, ne (r) = n0 exp(eφ/kTe ). The electron density
varies with a scale length λ, i.e., ∂ne /∂r ' −ne /λ.
~ = −∇φ,
~ find the radial electric field for given λ.
(a) Using E
~ ×B
~
(b) For electrons, show that rL = 2λ when the E
p drift velocity, vE , is
equal to the thermal speed, defined here as vt = 2kTe /m (this means
14
Chapter 2. Single particle motion I
~ ×B
~ drift
that the finite Larmor radius effects are important if the E
velocity is of the order of the thermal speed).1
5. (F. F. Chen) Suppose the earth’s magnetic field is 3 × 10−5 T at the equator
and falls off as 1/r3 as in a perfect dipole. Let there be an isotropic population
of 1 eV protons and 30 keV electrons, each with density n = 107 m−3 at r = 5
earth radii in the equator plane.
~ drift velocities.
(a) Compute the ion and electron ∇B
(b) Does an electron drift eastward or westward?
(c) How long does an electron take to encircle the earth?
(d) Compute the current ring density in A/m2 .
Note: the curvature drift is non-neglible... but neglect it anyway.
6. (?) Hall thrusters are widely used in space propulsion. A stationary plasma
thruster (SPT) is schematically represented in the figure.2
The thruster has a cylindrical
shape, with an open chamber defined by inner (Ri ) and outer
(Re ) radii and height L, where
the anode is placed. In the chamber there is a magnetic field,
pointing from Ri to Re . An axial electric field points outwards
from the anode. The thruster experiences a propelling force if it is
able to eject positive ions along
the direction of the electric field.
Xenon is injected into the chamber, and electrons coming from
the cathode ionize the Xe atoms,
creating new electron/ion pairs.
Our purpose is to study the motion of electrons and ions created by ionisation
of Xe in the thruster. Consider Re = 5 cm, Ri = 3 cm and L = 30 cm. The
fields in the chamber are approximately E = 5kV /m and B = 5 mT, and the
mass of Xe is 131.3 u (1u = 1.66 × 10−24 g).
(a) Assuming that the electrons are created with a speed perpendicular to
~ ~v⊥ , describe qualitatively their motion and draw schematically their
B,
~
trajectory (neglect any possible curvature, ∇B
and centrifugal force
drifts).
(b) A simple image of the thrust operation can be obtained by calculating
the electron and ion Larmor radii, rLe and rLi . Assuming the velocity of
ions and electrons to be, respectively, v⊥,i = 100 m/s (ions are formed
from ionisation of the injected neutral Xe atoms) and v⊥,e = 1 × 104
m/s (electrons coming from the cathode are accelerated from the E
1 Notice
the definition of the thermal speed from Chen’s book, a factor of
defined in this text.
2 M. Keidar and I. I. Beilis, IEEE Trans. Plasma Sci. 34 (2006) 804
√
2 larger than
15
field), calculate rLe and rLi and compare it with the relevant thruster
dimensions. Does the thruster experience a propelling force in these
conditions?
(c) Solve the equations of motion for the electrons created in an ionising
collision, assuming they are created with zero speed. Consider a plane
geometry (i.e., solve the equations in cartesian axes), with the xx-axis
~ Determine the amplitude of oscillation in the xx direction.
along E.
(d) In the conditions of c., what would be the amplitude of oscillation of the
positive ions in the xx direction, if they were created withe zero speed?
Comment the results of c. and d..
Note: a tutorial on the physics and modelling of Hall thrusters is presented by J. P. Boeuf, J. Appl. Phys. 121 (2017) 011101.
7. (?) Consider a particle of charge q, initially at rest and placed at xi =
qE0
~
− mω
2 , that moves under the effect of a high-frequency electric field, E =
E0 cos(ωt)~ux .
(a) Solve the equation of motion and describe the trajectory.
(b) Assume now that the amplitude of the electric field slowly varies in space,
E(x, t) = E0 (x) cos(ωt), where E0 is a growing function of x.
i. Describe qualitatively the trajectory.
[Suggestion: think on the values of the field E0 on the turning
points of the trajectory]
ii. To solve for the trajectory, decompose the motion on a slowly
varying component, x0 , denoted as oscillation center, and a highfrequency component x1 , where x1 is measured from the oscillation
centre, x = x0 + x1 (note that this is analogous to the decomposition of motion in the guiding centre motion and an oscillating
component, used to describe motion on a magnetic field). Expand
E0 (x) around x0 and:
A. define condition of validity of the expansion and show that the
equation of motion can be written in the form
dE0 (x0 )
m(ẍ0 + ẍ1 ) = q E0 (x0 ) + x1
cos(ωt) ;
dx
B. start from the equation of motion to justify that, approximately,
ẍ1 =
qE0
cos(ωt)
m
and solve for x1 (t);
C. start from the equation of motion to show that the oscillation
centre feels a force
Fp = −
q2 d 2
E .
4mω 2 dx 0
[Suggestion: start by taking the temporal average of the equation of motion on the sort time 2π/ω and note that hxi = x0 ]
16
Chapter 2. Single particle motion I
Note: the ponderomotive force is very important in several applications and basic research phenomena, such as the study of the
interaction of intense lasers with dense plasmas and plasma acceleration.
CHAPTER
3
Single particle motion II
1. (F. F. Chen, Fermi acceleration of cosmic rays). A cosmic ray proton is
trapped between two moving magnetic mirrors with mirror ratio Rm = 5.
Initially its energy is W = 1 keV and v⊥ = vk at the mid-plane. Each mirror
moves toward the mid-plane with a velocity vm = 10 km/s and the initial
distance between the mirrors is L = 1010 km.
(a) Using the invariance of µ, find the energy to which the proton is accelerated before it escapes.
(b) How long does it take to reach that energy?
[Suggestions: i) suppose that the B field is approximately uniform in the
space between the mirrors and changes abruptly near the mirrors, i.e.,
treat each mirror as a flat piston and show that the velocity gained at
each bounce is 2vm ; ii) compute the number of bounces necessary; iii)
assume that the distance between the mirrors does not change appreciably during the acceleration process.]
2. (F. F. Chen) A plasma with an isotropic distribution of speeds is placed inside
a magnetic mirror with mirror ratio Rm = 4. There are no collisions, so
that the particles in the loss cone escape, while the others remain trapped.
Calculate the fraction of particles that remains trapped.
3. (F. F. Chen) The magnetic field along the axis of a magnetic mirror is B( z) =
B0 (1 + α2 z 2 ), where α is a constant. Suppose that at z = 0 an electron has
2
.
velocity v 2 = 3vk2 = 32 v⊥
(a) Describe qualitatively the electron motion.
(b) Determine the values of z where the electron is reflected.
(c) Write the equation of motion of the guiding center for the direction
~ and show that there is a sinusoidal oscillation. Calcule the
parallel to B
frequency of the motion as a function of v.
18
Chapter 3. Single particle motion II
~ = −Ėt~uy ,
4. Consider a particle moving in a time-dependent electric field E
~
where Ė is a constant, and a uniform magnetic field B = B0 ~uz .
~ ×B
~ drift.
(a) Calculate the E
(b) Relate the resulting accelerated drift with a force and verify that the
drift due to that force is the polarization drift.
5. (F. F. Chen) A plasma is created in a toroidal chamber with average radius
R = 10 cm and square cross section of size a = 1 cm. The magnetic fiel
is generated by an electrical current I along the symmetry axis. The plasma
is Maxwellian with temperature kT = 100 eV and density n0 = 1019 m−3 .
There is no applied electric field.
~ field, for both positive
(a) Sketch the typical drift orbits in the non-uniform B
ions and electrons with vk = 0.
(b) Calculate the rate of charge accumulation (Coulomb per second) due to
the curvature and gradient drifts on the upper part of the chamber. The
magnetic field in the center of the chamber is 1 T and you can use the
approximation R a if necessary.
~ = E0 exp(iωt) ~ux ,
6. (?) Consider an electron moving in an oscillating electric field, E
~
perpendicular to a constant and uniform magnetic field, B = B ~uz .
(a) Calculate the drifts existing on the particle motion and describe qualitatively the motion.
(b) Try now to confirm the results you have already obtained, starting directly from the equations of motion. In particular, show that you can
indeed recover the results from a) for low frequencies of the field, i.e.,
ω ωce
[Suggestion: i) search for solutions of the form ~v = ~vk +~vL +~vD exp(iωt),
where ~vL is the velocity of the cyclotron motion and ~vD is constant and
~ ii) verify you can obtain an equation for ~vD in the
perpendicular to B;
~ 0 − ev~D × B;
~ iii) make the cross product with B
~
form iωm~vD = −eE
~
and eliminate ~vD × B].
CHAPTER
4
Fluid drifts
1. (F. F. Chen 3.6) An isothermal plasma is confined between the planes x =
~ = B0 ~uz . The density distribution is n(x) =
±a in a magnetic
field B
2
2
n0 1 − x /a .
(a) Derive an expression for the electron diamagnetic drift velocity, as a
function of x.
(b) Draw a diagram showing the density profile and the direction of the
~ points out
electron diamagnetic drift on both sides of the midplane, if B
of the paper.
(c) Evaluate vD at x = a/2, if B = 0.2 T, kTe = 2 eV and a = 4 cm.
~ field
2. (F. F. Chen 3.7) A cylindrically
symmetric plasma
in a uniform B
column
2
has n(r) = n0 exp − rr2
0
e ni = ne = n0 exp
eϕ
kTe
.
~ ×B
~ (~vE ) and electron diamagnetic drifts (~vDe ) ãare
(a) Show that the E
equal in magnitude and have opposite directions.
(b) Show that the plasma rotates as a rigid body.
(c) In the reference frame that rotates with velocity ~vE there are drift waves
that propagate with speed vϕ = 0.5vDe . What is vϕ in the laboratory frame? Represent on a r − θ diagram the directions and relative
magnitudes of ~vE , ~vDe and ~vϕ in the lab frame.
(d) Obtain the diamagnetic current, J~D , as a function of r.
(e) Calculate JD for B = 0.4 T, n0 = 1016 m−3 , kTe = kTi = 0.25 eV and
r = r0 = 1 cm.
3. A cylindrical plasma column of an isothermal plasma of radius R = 8 mm
and equal ion and electron temperatures, kB T = 5 eV, is immersed on on
a magnetic field B = 0, 6 T, aligned with the cylinder axis (coincident
with
the zz axis). The density has a profile n(r) = n0 J0 2, 4 Rr , where J0 is
20
Chapter 4. Fluid drifts
12
−3
the Bessel function of first kind of order zero and
n0 = 10 0 cm . Assume
eϕ
you can consider ni = ne = n = n0 exp kT . Note: J0 (x) = −J1 (x),
J0 (1, 2) ' 0, 67; J1 (1, 2) ' 0, 49.
(a) Obtain the expressions for the ion and electron diamagnetic drift as a
function of r. Justify qualitatively the direction of the drifts.
(b) Calculate the diamagnetic current density at r = R/2 (value and direction).
CHAPTER
5
Waves in non-magnetized plasmas
1. (F. F. Chen 4.6) Compute the effect of collisional damping on the propagation
of Langmuir waves, by adding a term −mnν~v to the electron equation of
motion and rederiving the dispersion relation for Te = 0 (plasma oscillations).
Show that the wave is damped in time.
2. (F. F. Chen 4.13) An 8 mm microwave interferometer is used on an infinite
plane-parallel plasma slab 8 cm thick.
(a) What is the plasma density if a phase shift of 1/10 fringe is observed?
Assume a uniform density and note that one fringe corresponds to a
360o phase shift.
(b) Show that if the phase shift is small, then it is proportional to the density.
3. (Exam 2015/2016) The international space station (ISS) orbits approximately
400 kms above the surface of the Earth. The average profile of the electron
density in the ionosphere is shown in the figure. If the astronauts in the ISS
want to communicate with the Earth, in which range of frequencies shall they
tune their radios?
Note: Anyone can communicate by radio with the ISS astronauts. The details and the frequencies actually used can be found in NASA’s webpage
(http://spaceflight.nasa.gov/station/reference/radio/)
22
Chapter 5. Waves in non-magnetized plasmas
4. (F. F. Chen 4.10) Hannes Alfvén (Nobel Prize in Physics in 1970) has suggested that perhaps the primordial universe was symmetrical between matter
and antimatter. Suppose that the universe was at one time a uniform mixture
of protons, antiprotons, electrons and positrons, each species having a density
n0 .
(a) Obtain the dispersion relation for high-frequency electromagnetic waves
in this plasma, neglecting collisions, annihilations and thermal effects.
(b) Obtain the dispersion relation for ion waves. Use Poisson’s equation,
neglect Ti (but not Te ) and assume that all leptons follow the Boltzmann
relation.
5. (Exam 2013/2014) We want to study electrostatic longitudinal waves in a
non-magnetized plasma. Consider a unidimensional problem and that you
can neglect the thermal motion of the positive ions, but not of the electrons.
(a) Show that the dispersion relation can be written in the form
2
2
ω 2 = ωpi
+ ωpe
ω2
ω 2 − γe vt2 k 2
(b) Obtain and discuss the limiting case Te → 0
(c) Obtain and discuss the limiting case M → ∞
(d) Obtain and discuss the limiting case m → 0
(e) Obtain and discuss the limiting case m → 0 e λD k 1
6. (Exam 2013/2014, two-stream instability ) Consider the one-dimensional propagation of waves in a cold (Te = Ti = 0) non magnetized plasma, where the
ions are initially stationary (i.e., the zeroth order term of the ionic velocity is
zero), but the electrons travel at speed v0 (i.e., the zeroth order term of the
electronic speed is v0 ). Neglect the effect of the collisions.
23
(a) Use the two-fluid equations and the Poisson’s equation to show that the
dielectric constant of the plasma can be written in the form
(k, ω) = 1 −
2
2
ωpi
ωpe
−
.
ω2
(ω − kv0 )2
(b) Verify that the dispersion relation is a polynomial function of fourth
order (so that for each real value of k there are four solutions for ω).
ω2
ω2
pe
Sketch approximately the function f (ω) = ωpi2 + (ω−kv
2 for a fixed k
0)
and mark on the graph where the four roots are ω [you do not need to
give the exact values, we are only interested in understanding the form
of the function].
(c) In some situations the dispersion relation has only two real roots, which
happens for small enough kv0 (convince yourself this is the case, by
looking at the graph you have just drawn). In that case, one of the
imaginary roots corresponds to an unstable wave, growing exponentially
in time. Show that, if kv0 = ωpe ω, the instability growth rate is
1/3
√
me
1
ωpe s−1 [Suggestion: start by expanding the
given by 23 21/3
mi
last term of (k, ω) to the first order in ω/ωpe ].
(d) From the general relations in a) e b), derive the dispersion relation in the
limit mi → ∞. Then obtain the limit v0 → 0 (while keeping mi → ∞).
Comment the results.
7. (Exam 2015/2016) Consider a plasma formed by electrons and two species of
positive ions, a light species (a) and a heavy species (b). We want to study
the propagation of low-frequency electrostatic waves in this plasma. As the
plasma is quasi-neutral, the non-perturbed electron and ion densities verify
the relation ne0 = na0 + nb0 .
(a) Justify why you can use the plasma approximation, neglect the electron
inertia and consider isothermal electrons.
(b) Write the relevant fluid equations and linearize them, keeping only the
terms up to first order.
(c) Show that the first order perturbations of the electron and ion-a densities
are related by
r
n ,
na1 = ω2 ma
Ta e1
k2 kB Te − γa Te
where r = na0 /ne0 .
(d) What is the relation between the first order perturbation on the electron
and ion-b densities?
(e) Show that the dispersion relation can be written as
1=
ω2
k2
r
+
ma
Ta
kB Te − γa Te
a
(1 − r) m
mb
ω2
k2
ma
kB Te
a Tb
− γb m
mb Te
(f) Verify that on the limit Ta = Tb = 0 we have ion acoustic waves,
corresponding to an ion of effective mass M given by
1
r
1−r
=
+
.
M
ma
mb
24
Chapter 5. Waves in non-magnetized plasmas
If both ion species have similar densities, which one determines the
plasma behaviour? Comment the result.
[Historical note: These waves were experimentally observed by Nakamura and Saitou, Plasma Phys. Control. Fusion (2003) 45 759; the
case Ta , Tb 6= 0 gives two solutions, a fast acoustic wave and a slow
acoustic wave and it is much more complex to analyse.]
8. (Exam 2016/2017) Consider an electromagnetic wave in a cold, non-magnetised
plasma. Neglect the ion motion but consider the effect of collisions between
electrons and neutrals, assuming a constant collision frequency ν.
(a) Write the relevant system of equations to study this system.
(b) Linearize the system of equations in the usual form and keep only the
first order terms.
(c) Show that the dispersion relation can be written in the form
2
ωpe
c2 k 2
=
1
−
.
ω2
ω(ω + iν)
(d) Assuming ν/ω 1, show that the skin depth (attenuation distance) is
given by
!1/2
2
ωpe
2c ω 2
1− 2
.
2
ν ωpe
ω
[Suggestion: use the dispersion relation and take a real frequency ω and
an imaginary wavenumber k = iα + β.]
9. (Exam 2017/2018) In this problem we study classical and quantum ion acoustic waves.
(a) Consider first classical ion acoustic waves, i.e., those studied in class.
i. The dispersion relation for ion acoustic waves was derived under
the plasma approximation. Explain what does this approximation
mean in the study of linear waves in plasmas and why the dispersion
relations obtained in this way should be valid for small values of k.
ii. Show that the dispersion relation of ion plasma waves does reduce
to that of ion acoustic waves in the limit k → 0. Explain which of
the two terms on the r.h.s. of the dispersion relation must dominate
for the expression to be valid.
(b) We turn now our attention to the quantum case. Quantum hydrodynamic models (QHM) generalize the fluid equations for plasmas with
the inclusion of a quantum correction term to the force balance equation. It is possible to show that, in the limit me /mi → 0 and for
TF i < Ti < Te TF e , where TF s denotes the Fermi temperature of
species s and the remaining quantities have their usual meaning, this
equation leads to the following relationship between the potential φ and
the electron density:
2 2
n2
eφ
1
1
cs
∂ √
= − + e2 − √ H 2
ne ,
(5.1)
2kB TF e
2 2n0
2 ne
ωpi
∂x2
25
where the quantum sound speed is given by
cs =
2kB TF e
mi
1/2
and
H=
~ωpe
2kB TF e
is a dimensionless parameter characterizing the importance of quantum
diffraction effects (it is the ratio between the plasmon energy and the
Fermi energy).
The relevant fluid equations are the ion continuity and momentum conservation equations, Poisson’s equation and equation (5.1). Neglect the
ion thermal agitation.
i. Write the four fluid equations in terms of the four unknowns ni , vi ,
ne and φ.
ii. Show that the linearizarion of equation (5.1) leads to
"
2 #
1 2 cs k
eφ1
ne1
= 1+ H
,
2kB TF e
4
ωpi
n0
where n0 is the unperturbed electron (and ion) density.
iii. Use the ion equations to derive
ni1
k 2 eφ1
= 2
.
n0
ω mi
iv. Obtain the dispersion relation for the quantum ion-acoustic waves,
2 c2 k 2
2
ωpi
1 + H4 ωs 2
pi
k2 .
ω 2 = ω2
pi
H 2 c2s k2
2
1 + 4 ω2
c2 + k
s
pi
v. Discuss the limits of small and large wave numbers in when H → 0.
[Historical note: the dynamics of a quantum electron gas is an important
issue for a variety of physical systems, such as ordinary metals, semiconductors and astrophysical systems under extreme conditions (e.g., white
dwarfs). This problem was studied by F. Haas et al., Phys. Plasmas 10
(2003) 3858.]
26
Chapter 5. Waves in non-magnetized plasmas
CHAPTER
6
Waves in magnetized plasmas
1. (F. F. Chen 4.7) For the upper hybrid oscillations, show that the elliptical
orbits are always elongated in the direction of ~k (hint: derive an expression
for vx /vy ).
2. (F. F. Chen 4.22) Faraday rotation of an 8-mm wavelength microwave beam
in a uniform plasma in a 0.1 T magnetic field is measured. The plane of
polarization is found to be rotated 90o after traversing 1 m of plasma. What
is the density?
3. (F. F. Chen 4.21) Show that in a positronium plasma, i.e., a neutral plasma
of electrons and positron, there is no Faraday rotation [suggestion: write the
system of linearized equation in matrix form, Ax = 0, and ask Mathematica
..
for help to calculate det(A) ^].
4. (F. F. Chen 4.25) A microwave interferometer employing the ordinary wave
cannot be used above the cutoff density. To measure higher densities, one
can use the extraordinary wave.
(a) Write an expression for the cutoff density for the X wave.
(b) On a vφ2 /c2 vs. ω diagram, show the branch of the X-wave dispersion
relation on which such interferometer would work.
5. (Exam 2014/2015) We want to study electrostatic waves in an electronegative
plasma, formed by electrons, positive ions and negative ions. The plasma is
quasi-neutral, i.e., the non perturbed densities are n+0 = n0 , ne0 = (1 − ε)n0
e n−0 = εn0 , respectively for the positive ions electrons and negative ions.
Assume that initially there is no electric field and all the fluid velocities are
zero. The plasma is immersed on a constant and uniform magnetic field,
~ = B0 ~uz . Consider only longitudinal perpendicular waves, with the wave
B
~ 1 , aligned along the xx axis.
electric field, E
(a) Write the fluid equations relevant to study this problem.
28
Chapter 6. Waves in magnetized plasmas
(b) Linearize the equations for the electrons, keeping only the first order
terms. Show that
ωce
vex ,
vey = i
ω
where (obviously) the speeds vex and vey are the components of the first
order correction to the electron velocity.
(c) Still using only the electron equations from c), show that
ne1 = −i
k(1 − ε)n0
eE1 ,
me Ω2e
2
where Ω2e = ω 2 − ωce
− γe k 2 c2se and c2se =
kB Te
me .
(d) Defining Ω+ e Ω− similarly toá Ωe , what are the expressions for n+1 e
n−1 ?
(e) Show that, in the plasma approximation, the dispersion relation can be
written in the form
ε
me 2 2 m−
me m− 2 2
Ω Ω +
(1 − ε)Ω2+ Ω2− +
Ω− Ωe = 0 .
m+ + e m+
m2+
(f) Obtain the dispersion relation in the limit ε = 0 and me m+ . Comment the result.
Historical note: these waves were predicted theoretically by N. D’Angelo,
IEEE Trans. Plasma Sci. 20 (1992) 658 and detected experimentally by
T. An, R. L. Merlino e N. D’angelo, Phys. Fluids 5 (1993) 1917.
6. (Exam 2015/2016) Consider an electromagnetic mode with E1 ⊥ B0 and
k k B0 .
(a) Write the system of linearised (vectorial) equations that leads to the
dispersion relation for these waves, neglecting ion motion (mi → ∞),
electron thermal motion (Te → 0) and collisions.
(b) It can be shown that the system you just wrote leads to
Ex (ω 2 − c2 k 2 − α) + Ey iαωce /ω
2
2 2
Ey (ω − c k − α) − Ex iαωce /ω
where
α=
=0
=0
ωp2
.
2 ω2
1 − ωce
Continue from here to obtain the dispersion relation for this wave (in
the form given in the formulae for the exam).
(c) Show (briefly) that the modes are right and left hand circularly polarized,
and identify which is which.
(d) Define and obtain the cutoff frequencies. Comment the results.
7. (Exam 2017/2018) We want to study with generality wave propagation in
~ 0 = B0 ~uz ).
cold, collisionless, homogenous, infinite, magnetized plasmas (B
Assume the ions to be stationary.
29
(a) Let us start by rewritting Maxwell’s equations in an appropriate form.
i. Linearize Faraday’s and Ampère’s laws and show that plane waves
can be described by the general equation
2
~k(~k · E
~ 1 ) − k2 E
~1 + ω · E
~1 = 0 ,
2
c
(6.1)
where the dielectric tensor is related to the conductivity tensor σ
by
ic2 µ0
=I+
σ,
ω

 
σxx σxy σxz
Ex
~ 1 = σyx σyy σyz  Ey 
I is the identity matrix and J~1 = σ · E
σzx σzy σzz
Ez
ii. Equation (6.1) can be written using the dispersion matrix
 D =
kxx kxy kxz
2
{~k~k − k 2 I + ωc2 }, where ~k~k is the tensor kyx kyy kyz , in the
kzx kzy kzz
form
~1 = 0 .
D·E
Assuming D is known, how would you derive the dispersion relation?
(b) To use the procedure delineated above to study waves in plasmas it is
necessary to calculate the conductivity tensor.
i. Linearize the force equation for the electrons to show that
e iωEx + ωce Ey
2
m ω 2 − ωce
e iωEy − ωce Ex
vy = −
2
m ω 2 − ωce
e iEz
vz = −
m ω
vx = −
ii. Show that the conductivity tensor is

σ⊥ σ×
−σ× σ⊥
0
0
given by

0
0 ,
σk
where
σk =
ne2 i
me ω
;
σ⊥ =
ne2
iω
2
me ω 2 − ωce
;
σ× =
ne2 ωce
.
2
me ω 2 − ωce
(c) The above formalism provides the dispersion relation for any cold, magnetized plasma (!!!) (inclusion of ion motion is straightforward). It is
~ the dispersion matrix takes the form
possible to show that when ~k ⊥ B


S
−iD
0
2 2
 ,
0
D = iD − kωc2 + S
k2 c2
0
0
− ω2 + P
where S = S(ω, ωpe , ωce ), D = D(ω, ωpe , ωce ) and P = 1 −
2
ωpe
ω2 .
30
Chapter 6. Waves in magnetized plasmas
i. Show that the system comprises two independent dispersion relations. Obtain one of them explicitly and the other one in terms of
S and D.
ii. Which wave corresponds to the latter dispersion relation?
CHAPTER
7
Diffusion and transport in weakly ionized plasmas
1. The cross section for electron-neutral momentum transfer in Ar can be approximated by the relation σ(u) = αu[eV ], with α = 1.37×10−20 m2 /eV (see
figure, σ[10−16 cm2 ] vs. u [eV]). Calculate the mean collision frequency for
momentum transfer in an argon plasma at p = 5 Torr and Tg = 20 o C, characterized by an electron temperature kTe = 1 eV, i.e, assuming a Maxwellian
distribution of velocities (pay attention to the definition and normalization of
the distribution!),
3/2
mv 2
m
2
f (v) = 4πv
exp −
.
2πkTe
2kTe
Compare with the value you would obtain if the cross section were constant,
with the value corresponding to the mean energy.
4.50E+01
4.00E+01
3.50E+01
3.00E+01
2.50E+01
2.00E+01
1.50E+01
1.00E+01
5.00E+00
0.00E+00
0
5
10
Real
15
20
25
30
35
Linear Approxima8on
2. (F. F. Chen, 5.1) The electron-neutral collision cross section for 2 eV electrons
in He is about 6πa20 , where a0 = 0.53 × 10−8 cm is the radius of the first
32
Chapter 7. Diffusion and transport in weakly ionized plasmas
Bohr orbit of the hydrogen atom. A positive column with no magnetic field
has p = 1 Torr of He (at room temperature), and kTe = 2 eV.
(a) Compute the electron diffusion coefficient in m2 /s, assuming that hσvi
is equal to the product σv for 2 eV electrons.
(b) If the current density along the column is 2 kA/m2 and the plasma
density is 1016 m−3 , what is the electric field along the column?
3. (Exam 2013/2014) Suppose that the electron distribution function in a homogeneous plasma can be approximated by a superposition of two Maxwellians
at different temperature, i.e., f (~v ) = α1 f1 (~v ) + α2 f2 (~v ), where
fj (~v ) = n
m
2πkB Tj
3/2
mv 2
exp −
,
2kB Tj
with j = 1, 2, α1 + α2 = 1 e v = |~v |.
(a) Verify that the distribution function is correctly normalized, i.e.,
n.
´´´
f (~v )d3 v =
(b) Show that the average value q
of the absolute value of the velocity of each
8k T
B j
of the maxwellians is hvj i =
πm and calculate the average value of
the absolute value of the velocity of the distribution.
(c) The cross section for electron-neutral momentum transfer can be approximated by σm (u) = βm u, where βm is constant and u is the electron
energy. Show that the mean collision frequency for momentum transfer
associated to each Maxwellian is νm = 2N βm kB Tj hvj i and calculate
the average value of the momentum collision frequency of the distribution.
(d) The ionization cross section of the same gas can be approximated by
σi (u) = 0, if u < ui , and σi (u) = βi , if u ≥ ui , where ui is the
ionization threshold. Show that the
ionization
frequency
associated with
ui
ui
each Maxwellian is νi = N βi hvj i kTj + 1 exp − kTj and calculate
the mean ionization frequency of the distribution.
(e) Calculate the values in items c. and d. for kB T1 = 1 eV, kB T2 = 16 eV,
α1 = 0.99, α2 = 0.01, βm = 10−20 m2 /eV, βi = 10−20 m2 , ui = 15
eV and N = 1023 m−3 . Comment the results.
(f) (to answer in the las problem sheet) Suppose that an electronic wave is
excited on a plasma with an initial distribution with a shape similar to
the one used in this problem. Is there a region of wavelengths where,
at least in principle, these waves are unstable. If yes, can you define an
interval of phase speeds where to search for these waves?
Useful integrals:
ˆ ∞
√
2
2
x exp −Ax
dx =
0
ˆ ∞
0
x
2
π
4A3/2
2
exp (−Ax) dx =
A3
ˆ ∞
3
2
x exp −Ax
dx =
1
2A2
2
(Ax2
3
2
i + 1) exp(−Axi )
x exp −Ax
dx =
xi
2A2
0
ˆ ∞
33
4. (Lieberman and Lichtenberg 5.2) A steady-state argon plasma is created at
high pressure between two parallel plates located at x = ±L/2 by illuminating
the region between the plates with ultraviolet radiation (which ionizes the
neutrals). The radiation creates a uniform number of electron-ion pairs per
unit volume and pe unit time, G0 (m−3 s−1 ), everywhere within the plates.
The electrons and ions are lost by ambipolar diffusion to the walls.
(a) Show that in the limit Ti Te the ambipolar diffusion coefficient is
given by Da ' µi (kTe /e).
(b) Assuming Da uniform in space and constant in time, obtain the stationary plasma profile, n(x), and the value of the density at the center, n0 ,
assuming you can impose the boundary condition n(x) ' 0 at the walls.
5. (Exam 2015/2016) Consider an axisymmetric cylindrical weakly-ionized plasma
~ = E~ur , B
~ = B~uz and ∇P
~ i,e = ∂Pi,e /∂r~ur . Neglect the convective
with E
term and consider the stationary case. Assume neutrality, the same temperature for electrons and ions and that you are on a reference frame where the
average velocity of the neutrals is zero.
(a) Write the expressions for the r and θ components of the two fluid force
equations.
(b) Solve the previous equations for vr and vθ and verify that:
i. for the r component,
ver = −µer E − Der
1 ∂n
n ∂r
where
µer =
µe
1+
,
2
ωce
νe2
Der =
De
1+
2
ωce
νe2
and
µe =
e
me νe
,
De =
kB Te
;
me νe
ii. for the θ component
veθ =
vE + vD
1+
νe2
2
ωce
where
vE = −
E
B
,
vD = −
kB Te 1 ∂n
.
eB n ∂r
(c) Find the expression of E that ensures ambipolarity along the radial direction.
(d) Obtain the expression of Der for very intense B-fields (ωce νe ) and
verify which is the length scale of the associated “random walk” motion.
Comment the result.
Note: it is interesting to compare the results of this exercise with exercise 4
from the problem sheet 8.
34
Chapter 7. Diffusion and transport in weakly ionized plasmas
6. (Exam 2016/2017) A steady-state nitrogen plasma is created between two
parallel plates at high pressure by an external electric field. The plasma
contains electrons and two main types of positive ions, N2+ and N4+ . The
electrons and ions are lost by ambipolar diffusion to the walls.
(a) Assuming a strongly collisional regime (high pressure), a constant electronneutral collision frequency and an isothermal plasma (γ = 1), obtain the
expressions for the mobility and diffusion coefficients of species s, as well
as for the ratio Ds /µs . Justify all the approximations you make.
(b) Write the quasi-neutrality and congruency hypotheses in this case.
(c) Show that the ambipolar electric field is
~
~
~
~ = D1 ∇n1 + D2 ∇n2 − De ∇ne ,
E
µ1 n1 + µ2 n2 + µe ne
where the indexes 1 and 2 represent each the two positive ions and e
the electrons.
(d) Further assuming the proportionality hypothesis,
~ 1
~ 2
~ e
∇n
∇n
∇n
'
'
,
n1
n2
ne
~ s for all species, where the ambipolar diffusion
show that ~Γs = −Das ∇n
coefficients for the positive ions (s = 1, 2) are
Das = Ds − µs
D1 n1 + D2 n2 − De ne
.
µ1 n1 + µ2 n2 + µe ne
(e) Show that, in the limit Te Ti (i = 1, 2), Dsi ' Di TTei .
(f) Within the conditions of the problem, justify that we must have ne Dse =
n1 Ds1 + n2 Ds2 .
[Historical note: the expressions for the ambipolar diffusion coefficients for a
plasma comprised of electrons and several positive ions are given, e.g., in V.
Guerra, P. A. Sá and J. Loureiro, Eur. Phys. J. Appl. Phys. 28 (2004) 125]
7. (Exam 2017/2018) Consider a weakly ionized plasma diffusing in the ambipolar regime. Assuming Da uniform in space and constant in time, obtain the
radial profile of the fundamental diffusion modes for a cylindrical plasma column of radius R and show that the characteristic decay time for this mode is
R2
τ0 ' 2.405
2D .
a
8. (Exam 2017/2018) Consider a stationary weakly ionized plasma in a strongly
~ = B~uz .
collisional regime, in the presence of an external magnetic field, B
Assume you can neglect thermal effects.
(a) Show that the equation for the perpendicular velocity of the electrons is
given by
µe
1
~
~ve⊥ = −
~v ,
2 E⊥ +
ωce
ν2 E
1 + ν2
1 + ω2e
e
where ~vE
mobility.
ce
~ ×B
~ drift velocity and µe =
is the E
e
m e νe
is the electron
35
~
(b) Show that the electron conductivity tensor, σ e , defined by J~e = σ e · E,
can be written in cartesian coordinates in the form


σ⊥ −σH 0
σ⊥
0 ,
σ e = σH
0
0
σk
and give the expressions for the longitudinal conductivity σk , perpendic2
ular conductivity σ⊥ and Hall conductivity σH in terms of σ0 = mnee νe ,
ωce and νe .
[Suggestion: start by defining J~e⊥ and J~ek in terms of ~v⊥ and ~vk ]
(c) Discuss the limiting cases ωce = 0 and νe = 0.
36
Chapter 7. Diffusion and transport in weakly ionized plasmas
CHAPTER
8
Diffusion and transport in fully ionized plasmas
1. (F. F. Chen 5.9) Suppose the plasma in a fusion reactor is in the shape of
a cylinder 1.2 m in diameter and 100 m long. The 5 T magnetic field is
uniform, except for short mirror regions at the ends, which we may neglect.
Other parameters are kTi = 20 keV, kTe = 10 keV and n(r = 0) = 1021 m−3 .
The density profile is found experimentally to be approximately as sketched
in the figure.
(a) Assuming classical diffusion, calculate D⊥ at r = 0.5 m
(b) Calculate dN/dt, the total number of electron-ion pairs leaving the central region radially per second.
(c) Estimate the confinement time, τ by τ ' −N/(dN/dt).
2. (F. F. Chen 5.11) A cylindrical plasma column has a density distribution
n = n0 1 − r2 /a2 , where a = 10 cm and n0 = 1019 m−3 . If kTe = 100 eV,
kTi = 0 and the axial magnetic field is B0 = 1 T, what is the ratio between
the Bohm and the classical diffusion coefficients perpendicular to B0 ?
3. (F. F. Chen 5.18) If a cylindrical plasma column diffuses at the Bohm rate,
calculate the steady-state radial density profile, n(r), ignoring the fact that it
38
Chapter 8. Diffusion and transport in fully ionized plasmas
may unstable. Assume the density is zero at r = ∞ and has the value n0 at
r = r0 .
~ =
4. (F. F. Chen 5.15) Consider an axisymmetric cylindrical plasma with E
~ = B~uz and ∇P
~ i = ∇P
~ e = ∂P/∂r~ur . Neglect the convective term
Ee ~ur , B
and consider the stationary case.
(a) Write the two-fluid equations.
(b) From the θ components of these equations, show that vir = ver .
(c) From the r components, show that vsθ = vE + vDs (s = i, e).
(d) Find an expression for vir and show it does not depend on Er .
5. (Exam 2014/2015)
(a) Use the MHD equations to derive the expression
ρm
∂~v
~ × B)
~ + σ0 (~v × B)
~ ×B
~ − ∇P
~ ,
= σ0 ( E
∂t
where σ0 is the plasma conductivity (σ0 = 1/η).
~ in
(b) Solve the equation for the velocity components perpendicular to B
~
the case E = 0 and P = const., to show that the characteristic time
for diffusion across the magnetic field is
τ=
ρm
,
σ0 B 2
i.e., ~v⊥ (t) = ~v⊥ (0) exp (−t/τ ).
6. (Exam 2015/2016) Consider a fully ionised plasma where the density varies
~ = B0 (x)~uz .
slowly along ~ux and where the magnetic field is given by B
(a) Use the MHD equations to show that, in stationary regime,
∂P
∂x
= Jy B0 .
(b) The MHD equations provide a macroscopic image of the plasma. Explain
the physical meaning of the expression obtained.
(c) On a more microscopic image, since the positive ions are typically heavier
and colder than the electrons, the electric current density calculated in
a) is carried essentially by the electrons. Calculate the electron velocity
associated with that current. Comment the result.
7. (Exam 2016/2017) As seen in class, the generalised Ohm’s law can take the
form
~
~ + ~v × B
~ = η J~ + 1 J~ × B
~ + 1 ∇P
~ e + me ∂ J .
E
2
en
ne
en ∂t
During a substorm in the nightside magnetotail (disturbance in the magnetosphere) the following values have been measured: E ' 0.1 mV/m;
v ' 100 km/s; B ' 1 nT; J ' 1 nA/m2 ; n ' 1 cm−3 ; Pe ' 0.1 nPa.
In these circumstances, the characteristic length scale is L ' 104 km, the
characteristic time scale is τ ' 10 s and the effective resistivity is less than
1 mS−1 .
Compare the magnitudes of the various terms in Ohm’s law in this case.
Comment the results.
39
8. (Exam 2016/2017) MHD equations can be used to investigate the origin of
the magnetic fields in stars, planets and the universe.
(a) Consider Ohm’s law in its common form (i.e., where the r.h.s contains
only the resistivity term) and assume that η is spatially uniform. Further assume the displacement current can be neglected in Ampère’s law.
Derive the following closed equation for the magnetic field,
~
∂B
~ × (~v × B)
~ + χ∇2 B
~ ,
=∇
∂t
where χ = η/µ0 is the magnetic diffusivity.
(b) The previous equation does not explain the origin of magnetic fields in
~ the
a medium initially non-magnetised [as the equation is linear in B,
~ = 0) = 0 implies B(t
~ > 0) = 0].
initial condition B(t
Repeat the previous question keeping as well the electron pressure gradient term in Ohm’s law, to show that the equation for the temporal
evolution of the magnetic field has now a source term creating a magnetic field in the direction perpendicular to the gradients of density and
electron temperature.
[Historical note: this source term is known as the Biermann battery and
~ b = kB Te ∇n.
~
can be conveniently expressed in terms of the field E
The
en
Biermann battery provides the first seeds of the magnetic field, which are
~
~
then efficiently amplified by the dynamo associated with the ∇×(~
v × B)
term. The historical reference is L. Biermann, Z. Naturforsch. 5a (1950)
65.]
(c) Knowing that in Earth’s core χ ' 2 m2 s−1 and that the Earth’s core
radius is R ' 3.5×106 m, make a rough order of magnitude estimation of
the decay time of Earth’s magnetic field due only to magnetic diffusion.
9. (Exam 2017/2018) In the deduction
of the MHD equations, it is argued the
mi me n ∂
J~
electron inertia term
can often be neglected in comparison
e
∂t n
~ Establish and discuss the conditions of
with the Hall term (mi − me )J~ × B.
validity of this approximation.
∂
10. (Exam 2017/2018) Consider a cylindrically symmetric plasma column ( ∂z
=
∂
∂
0; ∂θ = 0) of radius R, in equilibrium ( ∂t = 0), confined by a magnetic field.
(a) Show that the radial component of the MHD force equation can be
written as
∂P (r)
= Jθ (r)Bz (r) − Jz (r)Bθ (r) .
∂r
(b) [2.5 val] Use Ampere’s law neglecting the displacement current to eliminate J and show that the former equation can take the form
∂
1 2
1 2
1 Bθ2 (r)
P (r) +
Bz (r) +
Bθ (r) = −
,
∂r
2µ0
2µ0
µ0 r
which allows defining a magnetic pressure
~ force.
J~ × B
1
2
2µ0 B
associated with the
40
Chapter 8. Diffusion and transport in fully ionized plasmas
(c) Imagine a situation with Bθ = 0 and where the plasma density decreases
radially. Interpret and discuss the meaning of the previous equation in
this case.
[Suggestion: start by drawing a typical profile of n(r), then draw Bz (r)
and mark all forces acting on the plasma.]
CHAPTER
9
Kinetic theory I
1. Derive the continuity equation from the Vlasov’s equation (integrate in d3 v).
2. Derive the force equation from Vlasov’s equation (multiply by ~v and integrate in d3 v). The most laborious term is the one involving the gradient in
configuration space, which makes appear the average value of the tensor ~v~v .
Calculate the explicitly this term when:
(a) the thermal agitation is negligible;
(b) the average velocity is zero (i.e., the fluid is at rest and there is only
thermal agitation);
(c) in the general case where the velocoty can be decomposed as ~v = ~u + w,
~
where ~u is the average velocity of the fluid and w
~ corresponds to the
thermal agitation.
3. (Exam 2015/2016) Consider a stationary plasma, without magnetic field, un~ = −∇φ.
~
der the effect of an electrostatic field E
We want to obtain the
electrostatic potential due to a test charge placed in the plasma. We shall
look for a stationary solution of Vlasov’s equation on the separable form
fs (~r, ~v , t) ≡ fs (v, ~r) = f0s (v)ψs (~r), where f0 is the Maxwellian distribution,
f0 (~r, ~v , t) ≡ f0 (v) = n0
m
2πkB Te
3/2
mv 2
exp −
2kB Te
,
where v = |~v | and s = e, i.
(a) Show that the distribution f0 is properly normalised, i.e.,
n0 .
(b) Use Vlasov’s equation to show that çã
~ s (~r)
~ r)
qs ∇φ(~
∇ψ
=−
ψs (~r)
kB Ts
´
f0 (v)d3 v =
42
Chapter 9. Kinetic theory I
(c) Solve the previous equaiton and show that
qs φ(~r)
fs (v, ~r) = f0 (v) exp −
,
kB Ts
where n0 in the Maxwellian distribution is the plasma density faraway
from the test charge, i.e., in a region where φ(~r) = 0.
(d) Write Poisson’s equation using the distributions fs and show that
eφ
eφ
en0
2
exp
− exp −
=0.
∇ φ−
0
kB Te
kB Ti
Comment the result.
4. (Exam 2016/2017) The kinetic study of the behaviour of electrons in a plasma
can be made using a general kinetic equation of the form
∂f
~ r f + q (E
~ + ~v × B)
~ ·∇
~ v f = ∂f
+ ~v · ∇
,
∂t
m
∂t c
where the r.h.s.
´ term represents the influence of collisions. Assume the
normalisation f (~r, ~v , t)d3 v = ne (~r, t).
One of the simpler expressions for the collision term is given by the relaxation
time approximation, also known as the Krook model, which takes collisions
into account using
∂f
= −νc (f − f0 ) ,
∂t c
where νc is a constant collision frequency and f0 (~r, ~v ) is the equilibrium distribution of the electrons.
(a) Show that for a homogeneous plasma in the absence of external fields
the difference between f and f0 decays exponentially with time.
(b) Consider now electrons in an unmagnetized, homogeneous, time-independent
plasma in a weak constant electric field, E~1 . Linearise the distribution
function, f (~r, ~v , t) ≡ f (~v ) = f0 (~v ) + f1 (~v ) , where f0 is the (uniform
and stationary) unperturbed distribution, assumed to be a Maxwellian,
and f1 is a first order perturbation.
i. Show that
e2
J~ = −
νc m
ˆ ~ ·∇
~ v f0 ~v d3 v .
E
ii. Show that the electrical conductivity is given by
σc =
ne e2
.
mνc
[Note: This is one of many examples of deriving familiar macroscopic
results from underlying kinetic equations.]
CHAPTER
10
Kinetic theory II
1. (F. F. Chen 7.2) An electron plasma wave with 1 cm wavelength is excited
in a 10 eV plasma with n = 1015 cm−3 . The excitation is then removed and
the wave Landau damps away. How long does it take for the amplitude to
fall by a factor of e?
2. (Exam 2014/2015) Consider a one dimensional electron distribution function
of the form
2
nb
np
u
(u − ub )2
,
f0 (u) = 1/2 exp − 2 + 1/2 exp −
vt
vb2
π vt
π vb
resulting from the injection of an electron beam of average speed
ub and
q
kB T
density nb on a Maxwellian plasma of density np , where vt =
m is the
thermal electron speed. Suppose as well that vt ∼ vb and ub vt , nb np ,
and neglect ion motion.
´ +∞
(a) Calculate −∞ f0 (u)du to verify that the distribution function is correctly normalised.
(b) Sketch f0 (u) and show where do you expect that unstable waves may
exist.
(c) In which interval of phase speeds would you search for unstable waves?
[Suggestion: determine where the two components of f0 give the same
contribution]
(d) Determine the frequency, wave number and growth rate for the fastest
growing mode.
3. (F. F. Chen 7.3) An infinite, uniform plasma with fixed ions has an electron
distribution function composed of (1) a Maxwellian distribution of “plasma
electrons” with density np and temperature Tp at rest in the laboratory frame,
and (2) a Maxwellian distribution of “beam electrons” with density nb and
temperature Tb centered at ~v = V ~ux . If nb np , plasma oscillations in
44
Chapter 10. Kinetic theory II
the x-direction are Landau damped. If nb is large, there will be a two-stream
instability. The critical density for the onset at the instability can be estimated
by setting the slope of the total distribution function to zero, as follows:
(a) write expressions for fp (v) and fb (v), using the abbreviations v = vx ,
2kB Tp
B Tb
and b2 = 2km
;
a2 = m
(b) assuming that the value of the phase velocity vϕ will be the value of v
at which fb (v) has the largest positive slope, find vϕ and fb0 (vϕ );
(c) find fp0 (vϕ ) and set fp0 (vϕ ) + fb0 (vϕ ) = 0;
(d) para V b show thatthe beam
critical density is given approximately
√ T V
nb
V2
b
by np = 2e Tp a exp − a2 .
4. (Exam 2015/2016, Gardner’s theorem) We want to study the propagation of
Langmuir waves starting from Vlasov’s equation. As it has been shown in
class, if we assume immobile positive ions (mi → ∞), the dispersion relation
can be written in the form
2 ˆ +∞
ωpe
1
∂g
du = 0
(k, ω) = 1 − 2
k
∂u
u
−
ω/k
−∞
where g(u) is the unidimensional distribution function
ˆ +∞ ˆ +∞
1
g(u) =
f0 (u, vy , vz )dvy dvz .
n0 −∞ −∞
(a) Justify that if g is Maxwellian and the wave phase speed is much larger
than the electron thermal speed we can, on a first approximation accounting only for the contribution of the electrons of the body of the
distribution, neglect the pole on the integral. Obtain the dispersion
relation in this case
´ +∞
[Suggestion: recall that for a Maxwellian and u vϕ , −∞ g(u)/(u −
vϕ )2 du ' 1/vϕ2 + 3vt2 /vϕ4 ]
(b) In fact ω can be complex. There are unstable modes if the imaginary
part of the frequency is positive. We want to show Gardner’s theorem,
establishing that a single-humped velocity distribution is always stable.
The proof can be made by contradiction. Consider ω = ωr + iγ
in the expression from a), where
ωr and γ are the real and imaginary parts of the frequency, respectively. Assuming γ > 0 the
integral in the dispersion relation
can be made along the real axis u,
since the pole is above that axis.
g(u)"
u"
v0"
i. Show that the dispersion relation can be written in the form
∂g
ωr
2 ˆ +∞
ωpe
∂u u − k
r (k, w) = 1 − 2
2
2 = 0
k
−∞
u − ωkr + γk
45
2
ωpe
γ
i (k, ω) = − 2
k k
ˆ
+∞
u−
−∞
∂g
∂u
ωr 2
k
+
γ 2
k
=0
ii. Show that
2
ωpe
1+ 2
k
ˆ
+∞
−∞
u
∂g
∂u (v0 − u)
2
2
− ωkr + γk
=0,
where v0 is the value of u corresponding to the hump in the distribution function (see figure).
[Suggestion: consider the linear combination r − i (kv0 − ωr )/γ]
iii. Show that the expression from the previous question can never be
satisfied and conclude about the stability of single-humped distributions.
5. (Exam 2016/2017) Consider longitudinal oscillations of electrons in the absence of a magnetic field. Collisions with neutrals are taken into account
by the Krook model (relaxation time approximation), so that electrons are
described by the kinetic equation
∂f
~ rf − e E
~ ·∇
~ v f = −νc (f − f0 ) ,
+ ~v · ∇
∂t
m
where νc is a constant collision frequency and f0 (v) an unperturbed velocity
distribution corresponding to n0 particles per unit volume. The dynamics of
the ion motion are neglected, the ions act merely as a uniform background of
positive charge.
(a) Linearize the equation in the usual way and show that
f1 =
1
eE1 ∂f0
,
i(kv − ω − iνc ) m ∂v
where v ≡ vx .
(b) Show that the dispersion relation can be written in the form
ˆ +∞
k2
dg
1
−
dv = 0 ,
2
ωpe
dv
v
−
ω/k
− iνc /k
−∞
´
where g(vx ) = n10 dvy dvz f0 (vx , vy , vz ).
(c) Determine the damping rate of the wave for small collision frequencies.
Comment the results, referring the conditions where Landau damping
can be observed, if any.
[Suggestion: Recall that without collisions you have the same dispersion
relation as in 5b, with νc = 0 and where ω is complex, ω = ωr + iωi ; in
that case the result is ωi '
3
πωpe
dg
2k2 du (u
=
ωr
k ).]
6. (Exam 2017/2018) The purpose of this problem is to make a kinetic study of
~ 0 = 0). A complete delinear, transverse waves in non-magnetized plasmas (B
scription of the plasma is given by the Vlasov equation and the Maxwell equations. The linearization is made about equilibrium zero order, space and timeindependent, isotropic distribution functions, fs (~r, ~v , t) = f0s (v)+f1s (~r, ~v , t),
where s ∈ {e, i}. Assume the unperturbed plasma is quasi-neutral.
46
Chapter 10. Kinetic theory II
~ r, t)
(a) Write the expressions that allow the calculation of ρ(~r, t) and J(~
from the distribution functions and show they are first order quantities.
(b) Looking for plane waves, show that the first order distributions are given
by
qs /ms ~ ~
f1s =
E1 · ∇v f0s (v) .
i(ω − ~k · ~v )
~ 0 = 0 and B~0 = 0, but that in principle E
~1 =
6 0 and
Note: recall that E
~
B1 6= 0.
(c) Use Maxwell’s equations to show that for transverse waves
2
~1 .
~ 1 = iωµ0 J~1 + ω E
k2 E
c2
(d) Show that
~ 1 (ω 2 − k 2 c2 ) = −i ω
E
0
X e2 ˆ
1
~ ·∇
~ v f0s (v)~v d3 v .
E
~k · ~v ) 1
m
s
i(ω
−
s
~ 1 = E1 ~uy and ~k = k~ux and neglect the ion motion (mi → ∞).
(e) Assume E
Simplify the expression above to derive the dispersion relation
ˆ +∞
g(vx )
2
2 2
2
ω − k c = ω ωpe
dvx ,
ω
− kvx
−∞
where
1
g(vx ) =
n0
ˆ
+∞
ˆ
+∞
f0e (v)dvy dvz .
−∞
−∞
[Suggestion: integrate by parts in dvy ]
(f) For non-relativistic plasmas the phase velocity of the wave is always much
larger than the thermal speed. Therefore, approximate ω − kvx ' ω and
obtain the dispersion relation of these waves.
(g) Would you expect these waves to be significantly damped?
SOLUTIONS TO CHAPTER
1
Debye shielding and fundamental effects
1. From P V = N kB T and n = N/V ,
n[m3 ] =
1.013 × 105 p[Torr] 1
p[Torr]
' 9.66 × 1024
.
−23
1.38 × 10
760 T [K]
T [K]
Substituting,
(a) n ' 2.69 × 1025 m−3 = 2.69 × 1019 cm−3 (Loschmidth number);
(b) n ' 3.22 × 1022 m−3 = 3.22 × 1016 cm−3 .
2. Using the definitions of λD and Λ,
λD [m] '7.44
kB T [eV]
n0 [cm−3 ]
1/2
Λ '4.12 × 108 n0 [cm−3 ]
kB T [eV]
n0 [cm−3 ]
3/2
.
The different values obtained are summarized in the table below and represented in figure 1.1
a)
b)
c)
d)
e)
f)
g)
λD (m)
7.4 × 10−5
7.4
2.4 × 10−3
2.4 × 10−8
7.4 × 10−5
24
2.4 × 10−4
Λ
4.1 × 106
4.1 × 109
1.3 × 104
1.3 × 103
4.1 × 103
1.3 × 109
1.3 × 103
48
Solutions to chapter 1. Debye shielding and fundamental effects
ne [cm-3]
10
10
10
20
Laser fusion
16
12
10
Tokamak
Gaseous electronics
Flame
8
Ionosphere
10
4
Solar wind
10
Interstellar medium
0
10-1
100
101
kTe [eV]
102
103
104
Figure 1.1: Problem 2: (—) constant Λ; (– –) constant λD
3. (a) The Debye potential created by a punctual test charge qT inside an
homogeneous plasma is given by
qT 1
r
φ(r) =
exp −
.
4π0 r
λD
Therefore, from Poisson’s equation, ∇2 φ = − ρ0 and using spherical
coordinates, ∇2 φ(r) =
1 d2
r dr 2 (rφ),
1 d2
qT
r
exp
−
r dr2 4π0
λD
1
= 2 φ(r) .
λD
∇2 φ =
Hence, for all points except the origin,
r
1 qT
ρ(r) = − 2
exp −
,
λD 4π0 r
λD
which corresponds to the charge density of the shielding cloud.
Note: it is of course possible to obtain the result following the standard
49
derivation of the Debye length,
ρ = e(ni − ne )
eφ
eφ
= n0 e exp −
− exp +
kB Ti
kB Ti
eφ
eφ
' n0 e 1 −
−1−
kB Ti
kB Te
n0 e2 1
1
1
=
+
≡ 2 φ0 ,
kB
Ti
Te
λD
where the usual assumptions of i) potential energy much smaller than
kinetic energy; ii) electrons and ions in equilibrium with the electrostatic
field (“adiabatic response” of both electrons and ions); and iii) quasineutrality [ni (r → ∞) = ne (r → ∞)] have been considered. Note
as well that the test charge can be considered to get the total charge
density with the additional term qT δ(~0).
The charge in the shielding cloud inside a sphere of radius R centred in
qT is (to get the total charge it is necessary to add qT )
˚
ˆ
Q(r ≤ R) =
R
r2 ρ(r) dr
ρ(r) dV = 4π
r≤R
=−
qT
λ2D
ˆ
0
0
R
r
r exp −
dr .
λD
Integrating by parts, with u = r, du = dr, v = exp − λrD and dv =
− λ1D exp − λrD ,
(
R ˆ R
)
r
r
qT
r exp −
−
exp −
dr
Q(r ≤ R) =
λD
λD 0
λD
0
R
R
R
=qT
exp −
+ exp −
−1
λD
λD
λD
Substituting values,
Q(r ≤ λD /2) = − 0.09 qT
Q(r ≤ λD ) = − 0.26 qT
Q(r ≤ 5λD ) = − 0.96 qT
lim Q(r < R) = − qT
R→∞
This example shows that the Debye length is a characteristic length and
not the distance for a perfect shielding.
Alternative solution:1
1 Thanks
due to my former student to Ricardo Barrué
50
Solutions to chapter 1. Debye shielding and fundamental effects
The electrostatic field can be readily obtained from the potential,
1
dφ
qT
r
1
~
+
E(r)
= − ~ur =
exp −
dr
4π0
λD
r2
rλD
Using Gauss’ law for the charge enclosed by a spherical surface of radius
r = R,
‹
R
R
qT
2
~
exp −
1+
,
E · ~n dS = E(R) · 4πR =
0
λD
λD
Q(r ≤ R) + qT
≡
,
0
where Q(r ≤ R) is defined as before. Hence,
R
R
R
+ exp −
−1 .
Q(r ≤ R) = qT exp −
λD
λD
λD
(b) The electrostatic energy of interaction in a system of point charges is
1 X qi qj
W =
,
4π0 i<j rij
where rij is the distance between charges i and j. We may then consider
that the interaction energy between charge i and all the other charges is
X qj
1 1
Wi =
qi
,
2 4π0
rij
j6=i
so that the total interaction energy is W =
of only two point charges).
P
i
Wi (think on a system
Note: in this view the interaction energy of a system of two equal point
charges is distributed equally among both charges [see, e.g., B. Jayaram
and A. Das, J. Mol. Struct. 543
P (2001) 123]. Different definitions
can be used, such as W = 12 i Wi , and the interaction energy of
charge i with all others is twice that defined above; this difference in the
definitions makes a difference of a factor of two in the resolution of the
exercise and is completely irrelevant for the point being made.
With the definition above, the interaction energy between the test charge
qT and the remaining charges is
˚
1 1
ρ(r)
Wq =
qT
dV
2 4π0
r
ˆ +∞
1
1 qT
r
1
2
=
qT
4πr − 2
exp −
dr
8π0
λ
4πr
λ
r
D
0
D
=−
qT2
.
8π0 λD
The mean kinetic energy of the test charge is 23 kT , so that its total
energy is
3
qT2
Etot = kT −
.
2
8π0 λD
51
Considering a typical case with qT = +e and Λ ∼ n0 λ3D 1 (many
particles inside a Debye sphere), it is immediate to conclude that the interaction potential energy of the particle is much smaller than its kinetic
energy, as it should be from the very definition of a plasma:
n0 λ3D 1
kT 0
λD 1
n0
n0 e2
e2
0 λD
qT2
3
kT .
2
8π0 λD
kT 4. The configuration of the problem is depicted in figure 1.2
plasma
E
x=0
Figure 1.2: Problem 4
Inside the plasma (x > 0), Poisson’s equation ∇2 φ = − ρ0 reads
−
d2 φ(x)
e
[ni (x) − ne (x)] .
=
2
dx
0
Since Te = Ti = T , the ion and electron densities can be written as
eφ(x)
eφ(x)
ni (x) = n0 exp −
' n0 1 −
kB T
kB T
eφ(x)
eφ(x)
ne (x) = n0 exp +
' n0 1 +
,
kB T
kB T
52
Solutions to chapter 1. Debye shielding and fundamental effects
and
1
d2 φ(x)
= 2 φ(x) ,
2
dx
λD
1/2
0 kB T
.
with the Debye length defined as λD = 2n
2
e
0
The solution is
x
φ(x) = φ0 exp −
λD
x
+ φ1 exp +
λD
,
where φ0 and φ1 are constants. The condition
lim φ(x) = 0
x→+∞
implies φ1 = 0. φ0 can be determined from the electric field outside the
plasma, assuming continuity at x = 0,
φ0
x
dφ
=
exp −
,
E(x ≥ 0) = −
dx
λD
λD
i.e., E(x = 0) = λφD0 ≡ E0 . Moreover, the expression above shows that the
plasma shields the external electric field, within a characteristic distance λD .
Substituting values, λD ' 1.66 × 10−2 cm and
0.5
E(x = 0.5 cm) ' 100 exp −
' 9 × 10−12 V/cm .
0.0166
We verify that for x = 0.5 cm∼ 30λD the electric field already decays about
13 orders of magnitude.
5. (a) In the conditions of the problem, ni = n0 and ne = n0 exp kBeφTe , so
that
eφ
ρ = e(ni − ne ) = n0 1 − exp
kB Te
eφ
n0 e2
' n0 e 1 − 1 −
=−
φ.
kB Te
kB Te
Poisson’s equation takes then the form
∇2 φ = −
with λD =
0 kB Te
no e 2
1/2
1
ρ
= 2 φ,
0
λD
. Writing the Laplacian in spherical coordinates
and assuming spherical symmetry, φ(~r) = φ(r),
1 d2
1
(rφ) = 2 φ .
r dr2
λD
Further defining ϕ = rφ,
d2
1
ϕ= 2 ϕ,
dr2
λD
53
whose solution is
r
ϕ(r) = C1 exp −
λD
r
+ C2 exp +
λD
,
where C1 and C2 are constants. The two boundary conditions are
lim ϕ(r) = 0 ⇒ C2 = 0 ,
r→+∞
φ(r = R) =
ϕ(R)
= φ0 ⇒ C1 = Rφ0 exp
R
R
λD
.
Finally, the electrostatic potential is given by
r−R
R
.
φ(r) = φ0 exp −
r
λD
Notice that when λD R we have exp λRD ' 1 and the usual
expression for the Debye potential created by a point charge is recovered,
whereas for R < r λD the potential is approximately the same as
created by a point charge in vacuum (as it should be!).
(b) The charge in the sphere, Q, can be calculated from the condition of
total shielding by the plasma (see problem 3),
˚
Q=−
ρ(r) dV .
Alternatively, as the charge in a conductor is distributed on its surface,
we may use the discontinuity condition for the electric field,
E(r = R+ ) − E(r = R− ) =
σ
,
0
where σ is the charge density on the surface of the conducting sphere
(σ = Q/4πR2 ) and E(r=R− ) is the electric field inside the conductor
and hence it is zero.
~ = −∇φ
~ = − dφ ~ur ,
In the plasma (r > R), E
dr
R
d
R
r
E(r) = −
φ0 exp
exp −
dr
r
λD
λD
R 1
1
r−R
= φ0
+
exp −
.
r r
λD
λD
Substituting,
1
1
+
R λD
1
1
σ = 0 φ0
+
R λD
R
Q = 4π0 φ0 R 1 +
.
λD
E(r = R+ ) = φ0
54
Solutions to chapter 1. Debye shielding and fundamental effects
(c) The capacity is given by
C=
Q
R
= 4π0 R 1 +
.
φ0
λD
Since the capacity of a conducting sphere immersed in a dielectric of relative permittivity r is C = 4π0 r R, the plasma behaves as a dielectric
of relative permittivity
R
.
r = 1 +
λD
In the limit λD R, when the characteristic length for the shielding
is much larger than the dimension of the sphere (i.e., poor shielding
for distances comparable to the radius of the conductor), we obtain the
result for a conducting sphere in vacuum. In the opposite limit, λD R,
corresponding to a very efficient shielding of the sphere by the plasma,
r ' R/λD .
For Te = 1 keV the Debye length is λD ' 2.35 m and 2.35 × 10−5 m,
respectively for n0 = 106 cm−3 and n0 = 101 4 cm−3 . The Debye length
is to be compared with R = 10 cm, showing the two values of the density
correspond to the two limiting cases discussed above. For completeness,
the capacity of the sphere is C0 ' 1.1 × 10−11 F in vacuum, C '
1.6 × 10−11 F∼ C0 for n0 = 106 cm−3 , and C ' 4.7 × 10−8 F C0 for
n0 = 1014 cm−3 .
6. Consider two slabs of length L and cross sectional area ∆S, one “holding”
singly charged positive ions with mass mi and number density n0 and the other
“holding” electrons of mass me and the same number density n0 (to ensure
quasi-neutrality). The slabs can move along the x-axis, perpendicular to their
cross section. The deviation of the ion and electron slabs in relation to their
equilibrium position is defined by the displacements xi and xe , respectively
(see figure 1.6).
The region of width |xi − xe | where the net charge density is different from
zero has a total charge ∆Q = n0 e(xe −xi )∆S. The configuration corresponds
to a parallel plate capacitor, for which the electric field is uniform and along
the x-axis,
~ = ∆Q/∆S ~ux = n0 e (xe − xi )~ux .
E
0
0
The total charge in the electron slab is Qe = −(n0 e∆SL), so that the force
on it is
2
~ = − (n0 e) (xe − xi )∆SL~ux .
F~e = Qe E
0
Similarly, the force on the ion slab is
2
~ = + (n0 e) (xe − xi )∆SL~ux .
F~i = Qi E
0
The equations of motion for each slab are (note that the total mass of the
electron slab is Me = me n0 ∆SL; similarly for the ion slab)
F~e = Me~ae = me n0 ∆SLẍe ~ux ,
55
Figure 1.3: Problem 6
F~i = Mi~ai = mi n0 ∆SLẍi ~ux .
Equating both sets of equations,
ẍe = −
n0 e2
2
(xe − xi ) = −ωpe
(xe − xi ) ,
me ε0
n0 e2
2
(xe − xi ) = +ωpi
(xe − xi ) ,
mi ε0
2 1/2
n0 e
and ωpi = m
.
i ε0
ẍi = +
with ωpe =
n0 e 2
me ε 0
1/2
Finally, defining χ = xe − xi , we have, successively,
χ̈ = ẍe − ẍi
2
2
χ̈ = −ωpe
(xe − xi ) − ωpi
(xe − xi )
2
2
χ̈ = −(ωpe
+ ωpi
)χ
χ̈ = −ωp2 χ
2
2
with ωp2 = ωpe
+ ωpi
.
7. The total number of electrons in the plasma, N , is N = 43 πR3 n0 . Therefore,
the electron density when the plasmas occupies a sphere or radius R + δr is
ne (δr ) =
N
R
=
n0 .
(R + δr )3
+ δr )3
4
3 π(R
56
Solutions to chapter 1. Debye shielding and fundamental effects
The electron field inside the plasma can be obtained from Gauss’s law, applied
to a spherical surface of radius r < (R + δr ),
‹
~ · ~n dS = Qint ,
E
0
where
4
Qint =e(n0 − ne ) πr3
3
4 3
R3
=en0 1 −
πr .
(R + δr )3 3
Hence,
4 3
en0
1
4πr E(r) =
1−
πr ,
3
0
(1 + δr /R) 3
1
en0 1
r 1−
.
E(r) =
0 3
(1 + δr /R)3
2
Expanding in Taylor series f (x) =
1
(1+x)3
for small x, f (x) ' 1 − 3x,
δr
en0 r
en0 1
r 1−1+3
δr .
=
E(r) '
0 3
R
0
R
The equation of motion for an electron placed at the surface of the plasma
sphere reads
d2
me 2 (R + δr ) = −eE(r=R+δr ) ,
dr
leading to
e2 n0
d2 δ r
R + δr
me 2 = −
δr
dr
0
R
2
e n0
'−
δr
0
Finally,
d2 δr
2
= −ωpe
δr ,
dr2
q
e 2 n0
where ωpe =
me 0 is the usual plasma frequency. The latter equation
corresponds to an harmonic oscillator of angular frequency ωpe , proving the
result.
8. (a) The electron and ion densities are given by
eφ
eφ
' n0 1 +
ne =n0 exp
kB Te
kB Te
ne =n0 .
57
For all points except the plane x = 0,
∇2 φ = −
ρ
,
0
2
2
where ∇2 φ ≡ ddxφ2 and ρ = e(ni −ne ) ' keBnT0e φ . Substituting, Poisson’s
equation reduces to
1
d2 φ
= 2 φ,
dx2
λD
q
with the Debye length given by λD = 0ek2Bn0Te . The general solution is
(see also problem 4)
x
x
φ(x) = C1 exp −
+ C2 exp +
,
λD
λD
where C1 and C2 are constants.
The regions x > 0 and x < 0 have to be solved separately. For x > 0,
the first boundary condition is
lim φ(x) = 0 ,
x→+∞
implying C2 = 0. The additional condition
lim φ(x) = φ0
x→0+
further determines C1 = φ0 . Hence, for x > 0 we have
x
φ(x) = φ0 exp −
.
λD
Due to the symmetry of the problem, φ(−x) = φ(x), as represented in
figure 1.4.
(b) The electrostatic field is readily obtained as
~ = − dφ ~ux = 1 φ0 exp − x ~ux ,
E
dx
λD
λD
for x > 0. In the limit x → 0+ we should recover the field created by
an infinite plane charged with surface density σ in vacuum,
lim E(x) =
x→0+
σ
.
20
This condition establishes the relation between φ0 and σ, φ0 =
Accordingly, for x > 0
φ0
x
E(x) =
exp −
.
λD
λD
σλD
20 .
For x < 0 we have E(−|x|) = −E(|x|), as represented in figure 1.5.
The discontinuity of the electric field at x = 0 is evidently σ/0 , as
required by Maxwell’s equations.
58
Solutions to chapter 1. Debye shielding and fundamental effects
ϕ/ϕ0
1
0,5
-5
-2,5
0
2,5
5
x/λD
Figure 1.4: Problem 8 - electrostatic potential.
E
σ/2ε0
-5
-2,5
0
2,5
5
x/λD
-σ/2ε0
Figure 1.5: Problem 8 - electrostatic field.
While the electric field created by an infinite conducting plane in vacuum
is constant and does not decay away from the plane, the presence of the
plasma shields the charge in the plane and the electric field (as well as
the potential),which decays with a characteristic decay length given by
the Debye length. Both the electrostatic field and potential are only
significantly different from zero in the vicinity of the conducting plane.
59
(c) From Poisson’s equation (see the beginning of the exercise),
1
ρ(x)
φ(x) = −
λ2D
0
and so, for x > 0,
ρ(x) = −
φ0
x
σ
x
exp
−
=
−
exp
−
.
λ2D
λD
2λD
λD
The total charge in the plasma, facing a section of area A of the plane
and in the positive-x region, is given by
˚
ˆ +∞
Q+ =
ρ(x) dV = A
ρ(x) dx
0
ˆ +∞
x
σ
exp −
dx
=−A
2λD
λD
0
x=+∞
σ
x
σ
=−A
−λD exp −
= −A .
2λD
λD x=0
2
Q+
σ
=− .
A
2
Similarly, the charge in the plasma in the negative-x region is
Therefore,
Q+ + Q−
= −σ .
A
9. (a) Proceeding in a similar fashion as in the previous exercise,
ρ(r)
0
ρ(r) =e[ni (r) − ne (r)]
eφ(r)
ni (r) =n0 exp −
' n0 1 −
kB Ti
eφ(r)
ne (r) =n0 exp +
' n0 1 +
kB Te
no e2 1
1
ρ(r) = −
+
φ(r) .
kB
Ti
Te
∇2 φ(r) = −
eφ(r)
kB Ti
eφ(r)
kB Te
Defining
1
n0 e2
n0 e2
1
1
= 2+ 2 =
+
,
2
λD
λe
λi
0 kB Ti
0 kB Ti
Poisson’s equation takes the usual form
∇2 φ(r) =
1
φ(r) .
λ2D
In cylindrical coordinates and with cylindrical symmetry,
∇2 φ =
d2 φ 1 dφ
+
.
dr2
r dr
Q−
A
= − σ2 .
60
Solutions to chapter 1. Debye shielding and fundamental effects
Multiplying by r2 ,
r2
dφ
d2 φ
r2
+
r
−
φ=0.
dr2
dr
λ2D
dφ dξ
dξ
and noting that dφ
dr = dξ dr and dr =
d dφ 1
dφ 1
(ξλD )2
+ (ξλD )
−ξ 2 φ = 0 ,
| {z } dr dξ λD
| {z } dξ λD
| {z }
| {z }
r
r2
Further defining ξ =
r
λD ,
1
λD ,
dφ
dr
dφ
dr
ξ 2 λD
dφ
d2 φ 1
+ξ
− ξ2φ = 0 ,
2
dξ λD
dξ
dφ
d2 φ
− ξ2φ = 0 ,
ξ2 2 + ξ
dξ
dξ
which is the modified Bessel equation of order zero.
The solution is
φ(ξ) =AI0 (ξ) + BK0 (ξ) ,
r
r
φ(r) =AI0
+ BK0
,
λD
λD
where A and B are constants.
As I0 grows exponentially, the boundary condition φ(r → +∞) → 0
implies A = 0. To find B, we impose
lim E(r) = Evac. (r) ,
r→0
where
λ
2π0 r
is the electrostatic field created by an infinite wire with linear charge
density λ in vacuum.
The electric field is
dφ
d
r
dK0 (ξ) dξ
E(r) = −
= −B K0
= −B
dr
dr
λD
dξ dr
1
1
r
= − B[−K1 (ξ)]
= BK1
.
λD
λD λD
Since for small ξ, K1 (ξ) ' 1ξ , i.e., K1 λrD ' λrD , for small r
Evac. (r) =
E(r) '
B
.
r
λ
The boundary condition as r → 0 leads to B = 2π
, so that
0
r
λ
r
λ
K1
; φ(r) =
K0
.
E(r) =
2π0 λD
λD
2π0
λD
61
(b) From ∇2 φ = − ρ0 =
1
φ,
λ2D
λ
r
ρ(r) = −
K0
.
2πλ2D
λD
The total charge in the plasma, around a length l of the wire, is
˚
ˆ
+∞
rρ(r) dr
ρ(r) dV = 2πl
0
ˆ +∞
λ
r
=2πl
r−
K0
dr
2πλ2D
λD
0
ˆ +∞
λ
λD ξ K0 (ξ) λD dξ
=−l 2
|{z}
| {z }
λD 0
Q=
ˆ
= − lλ
|0
r
dr
+∞
ξK0 (ξ) dξ = −lλ .
{z
}
=1
As
Q
= −λ ,
l
the plasma perfectly shields the charge in the wire.
Since the Bessel functions K0 (ξ) and Ki (ξ) decay with a characteristic
decay length of the order of 1 (see the figures in the formulation of the
problem), the potential and the electric field are shielded efficiently for
distances r λD , as expected. The shielding is complete only when
r → ∞, but is already nearly perfect for distances of a few Debye lengths.
Like in the previous exercise, it is possible to verify the importance and
efficiency of the Debye shielding of the plasma and the meaning of quasineutrality in the very definition of plasma. As an example, when the
problem is solved in vacuum the potential decreases very slowly and it
is not even possible to impose the condition φ(r → ∞) = 0, contrary
to what happens in the present case.
10. (a) As the electrons are lighter and more mobile than the ions, they tend
to arrive first at the dust particles, which get negatively charged. The
mechanism is somewhat analogous to the negative charging of the walls
of a reactor, at the origin of the phenomenon of ambipolar diffusion (cf.
chapter 7). Typically, Zd ∼ 100.
(b) The electron, ion and dust particle densities are given, respectively, by
eφ
eφ
' ne0 1 +
kB Te
kB Te
eφ
eφ
ni =ni0 exp −
' ni0 1 −
kB Ti
kB Ti
ne =ne0 exp
nd =nd0 .
62
Solutions to chapter 1. Debye shielding and fundamental effects
In turn, Poisson’s equation is
ρ
e
= − (ni − ne − Zd nd )
0
0
e
eφ
eφ
'−
ni0 1 −
− ne0 1 +
− Zd nd0
0
kB Te
kB Te
e
eφ
eφ
=
ni0
+ ne0
+ ne0 + Zd nd0 − ni0 .
0
kB Ti
kB Te
∇2 φ = −
In this case, the quasi-neutrality condition is written as
ne0 + Zd nd0 = ni0 .
Substituting in the previous expression,
eφ
eφ
ne0 e2
ni0 Te
e
ni0
+ ne0
=
1+
φ.
∇2 φ =
0
kB Ti
kB Te
0 kB Te
ne0 Ti
(c) The usual expression ∇2 φ =
is obtained defining
λD =
0 kB Te
ne0 e2
1
φ,
λ2D
1/2
where λD has dimensions of length,
1
1+
ni0 Te
ne0 Ti
1/2 = λDe 1
1+
ni0 Te
ne0 Ti
1/2 ,
where λDe is the electron Debye length in the absence of dust particles.
Without dust particles, ni0 /ne0 = 1; on the other hand, in the presence
of dust particles, ni0 /ne0 > 1, leading to a smaller Debye length.
(d) The electrostatic potential can be obtained as in the standard derivation
of the Debye length. It follows problem 5, with the exception of the
second boundary condition, which should be replaced by
lim φ(r) =
r→0
1 qT
,
4π0 r
corresponding to recovering the potential created by a point charge in
vacuum when we are very close to the charge (no shielding). With the
qT
notation of problem 5, this means setting C1 = 4π
and, accordingly,
0
φ(r) =
qT
r
exp −
,
4π0 r
λD
the usual expression of the Debye potential, but with λD as defined
above.
SOLUTIONS TO CHAPTER
2
Single particle motion I
1. From rLs = v⊥s /ωcs and ωcs = eB/ms , where s = {e, i} denotes electrons
and protons, respectively,
rLi
v⊥i mi
=
.
rLe
v⊥e me
Furthermore, the condition on the kinetic energy corresponds to
√
me v⊥e . Hence,
r
√
rLi
mi
=
' 1837 ' 43 .
rLe
me
√
mi v⊥i =
2. (a) The particle is initially accelerated by the electric field on the positive
y-direction. Its orbit is turned by the magnetic field, so that the resulting
~ ×B
~ drift in the
motion is a combination of cyclotron motion with an E
~
~
positive x-direction. The E × B drift is the outcome of the increase
of the Larmor radius with v⊥ (the velocity in the plane perpendicular
~ in this case v⊥ is the velocity in the x − y plane), which takes
to B;
place while vy > 0. From energy conservation it is clear that when the
particle returns to y=0 it has zero velocity. Moreover, since ~v (t=0)=0,
vk = vz is always zero, as there are no forces in the z-direction. The
orbit is schematically sketched in figure 2.1.
~ = B0 ~uz and E
~ = E0 ~uy , the compo(b) In cartesian coordinates, with B
~ + ~v × B)
~ are
nents of the Lorentz force F~ = q(E

 max = qvy B0
may = −qvx B0 + qE0

maz = 0
.
The last equation implies vz = cte. Using the initial condition vz (t=0)=0,
it comes vz (t) = 0.
64
Solutions to chapter 2. Single particle motion I
y
Larger rL
vy=0
Smaller rL
x
t=0; vy=0
y
vd
x
Figure 2.1: problem 2
The two first equations can be written as
q
v̇x = m
vy B 0
q
v̇y = − m
vx B0 + qE0
.
(2.1)
Differentiating the first equation and replacing v̇y ,
v̈x = −ωc2 vx +
where ωc =
Defining
qB
m ,
q 2
E 0 B0 ,
m
which is the equation of a driven harmonic oscillator.
ζ = ωc2 vx −
q 2
E 0 B0 ,
m
so that
ζ̈ = ωc2 v̈x ,
and replacing in the equation of the driven harmonic oscillator,
ωc−2 ζ̈ = − ζ
| {z }
v̈x
ζ̈ = −ωc2 ζ .
(2.2)
65
This is the equation of a simple harmonic oscillator, with solution
ζ(t) = ζ0 cos(ωc t + ϕ) ,
where the amplitude ζ0 and the initial phase ϕ have to be determined
from the initial conditions.
From (2.2) the expression for vx (t) is
vx (t) =
E0
1
ζ0 cos(ωc t + ϕ) +
,
ωc2
B0
while from (2.1)
vy (t) =
m
1
v̇x (t) = − 2 ζ0 sin(ωc t + ϕ) .
qB0
ωc
Determination of ζ0 and ϕ:
vy (t = 0) = 0 ⇒ ϕ = 0
ζ0
E0
vx (t = 0) = 0 ⇒ 2 +
=0
ωc
B0
q 2
ζ0 = −
E0 B0 .
m
The particle velocity is thus given by
E0
[1 − cos(ωc t)]
B0
E0
sin(ωc t) .
vy (t) =
B0
vx (t) =
Finally, the orbit is obtained as
ˆ t
x(t) =
vx (τ ) dτ
0
τ =t
E0
1
=
τ−
sin(ωc τ )
B0
ωc
τ =0
E0
1
=
t−
sin(ωc t) ;
B0
ωc
ˆ t
y(t) =
vy (τ ) dτ
0
E0
τ =t
[cos(ωc τ )]τ =0
B 0 ωc
E0
=
[1 − cos(ωc t)] ,
B0 ωc
=−
where the initial conditions x(t = 0) = y(t = 0) = 0 were used.
(c) From the previous item,
~v = vx ~ux + vy ~uy
E0
E0
E0
=−
cos(ωc t) ~ux +
sin(ωc t) ~uy +
~ux
B0
B0
B0
≡ ~vc + ~vd ,
66
Solutions to chapter 2. Single particle motion I
where
~vc = −
E0
E0
cos(ωc t) ~ux +
sin(ωc t) ~uy
B0
B0
corresponds to an oscillatory motion (cyclotron motion) and
~vd =
E0
~ux
B0
~ ×B
~ drift).
to a constant drift (E
Taking the time average over several gyroperiods, h~v i = ~vd = cte.,
since the time average of the sinusoidal functions is zero (hcos(· · · )i =
hsin(· · · )i = 0). Therefore, h~ai = 0 and there is no average acceleration.
~ cancels the electric force q E.
~
The average force q~vd × B
(d) The drift velocity ~vd is independent of the charge and the mass. Hence,
all the particles drift with the same velocity and there is no net current
in a neutral plasma. The particle’s trajectory for q < 0 is depicted in
figure 2.2
y
x
vd
Figure 2.2: trajectory for q < 0 (problem 2).
(e) In this case the drift velocity could be obtained simply by replacing
E = F/q in the previous result for ~vd , and would be
~vd =
F
mg
~ux =
~ux ,
qB
qB
which does depend on the charge. Accordingly, the drift velocity would
be different for particles of different mass and would have opposite directions for particles of different charge, resulting in a net electrical current.
3. The electrostatic field created by the electron beam is readily obtained from
Gauss’ law. Considering a cylinder of radius r > R and length l, as shown in
67
figure 2.3,
‹
~ · ~n dS = Qint
E
0
ene πR2 l
E(r)2πrl = −
0
en
R2
e
~
~ur
E(r)
= −
2r0
~ = R) = − ene R ~ur
E(r
20
Bz uθ
R
v
l
Figure 2.3: problem 3.
~ ×B
~ drift velocity is
The E
~vd =
~ ×B
~
E
E(R)B0
ene R
=
~uθ =
~uθ .
B2
B02
20 B0
Substituting values, vd ' 4.5 × 103 m/s.
ne
e
4. (a) Since ne (r) = n0 exp kBeφTe and ∂n
∂r ' − λ ,
∂ne
n0 e ∂φ
eφ
=
exp
∂r
kB Te ∂r
kTe
ne
n0
eφ
'−
= − exp
;
λ
λ
kTe
e ∂φ
1
=−
kB Te ∂r
λ
∂φ
kB Te
=−
.
∂r
eλ
Hence,
~ = −∇φ
~ = − ∂φ ~ur = kB Te ~ur .
E
∂r
eλ
68
Solutions to chapter 2. Single particle motion I
(b) Assuming the electrons have a Maxwellian velocity distribution, their
average kinetic energy is
K=
1
3
mhv 2 i = kB Te ,
2
2
where each component of the velocity contributes with
2
(equipartition theorem). Since v⊥
= vx2 + vy2 ,
2
hv⊥
i=
1
2 kB Te
to K
2kB Te
≡ vt2 .
m
Hence, the average Larmor radius is
hrL i =
mhv⊥ i
mvt
=
,
eB
eB
or, equivalently,
vt =
eBhrL i
.
m
On the other hand,
E
kB Te
=
B
eλB
2kB Te m
m
=
= vt2
m 2eλB
2eλB
vt
hrL ieB m
=
hrL i .
= vt
m } 2eλB
2λ
| {z
vE =
vt
The last expression can be re-written as
vE
hrL i = 2λ
.
vt
When vE = vt the Larmor radius is hrL i = 2λ.
Notice that when the magnetic field is strong enough to impose vE vt ,
the Larmor radius is much smaller than the characteristic distance for
the variation of the electron density ne , rL λ. In this case, within
a gyroperiod the electron sees a nearly homogeneous plasma and one
need not bother with the fast Larmor motion, which can be averaged
out to leave the slower guiding center drift. This procedure corresponds
to the study of phenomena that occur on time scales τ = 1/ω such
~ ×B
~
that ω ωce (τ τce ). The former picture is modified as the E
drift increases, since the Larmor radius becomes comparable with the
typical length λ and the perturbation approach may become questionable
(in particular the very use of the orbit theory in constant electric and
magentic fields, as the sources of the fields are changing significantly
within a Larmor orbit).
5. (a) At the equatorial plane the magnetic field points along the z-direction
(see figure 2.4) and, as it falls as 1/r3 ,
3
RT
~
B = B(r)~uz = B0
~uz ,
r
69
with B0 = 3 × 10−5 T and where RT ' 6370 km is the radius of the
Earth.
uz
B
Earth N
B
uθ
Figure 2.4: problem 5.
Since the gradient drift is given by
~vd =
2 ~
~
1 mv⊥
B × ∇B
,
2
q 2B
B
it is enough to calculate the gradient in the direction perpendicular to
the magnetic field,
~ ⊥=
(∇B)
B0
3
∂B
~ur = −3 4 RT3 ~ur = − B(r)~ur .
∂r
r
r
2
~ × ∇B/B
~
Accordingly, |B
| = 3/r and
vd =
2
1 mv⊥
3
.
|q| 2B r
2
For an isotropic velocity distribution, hv 2 i = 32 hv⊥
i (see exercise 4) and
3
2
so the kinetic energy is K = 4 mhv⊥ i or, alternatively,
r
v⊥ =
4K
.
3m
Substituting values, and noting that K/e is the kinetic energy in eV and
B(r = 5RT ) = B0 /25,
vd =
K
2
K 50
=
,
e B(r = 5RT )5RT
e B0 RT
1
it comes, vde ' 7.85 × 103 m/s for the electrons and vdi = ved 30×10
3 '
2.61 × 10−1 m/s for the positive ions.
~ points in the direction −~ur , so that B
~ × ∇B
~ points
(b) The gradient ∇B
in the direction −~uθ (see figure 2.4). Taking into account the charge of
the particles in the expression for ~vd , the electrons drift eastwards (i.e.,
along ~uθ ) while the positive ions drift westwards (along −~uθ ).
70
Solutions to chapter 2. Single particle motion I
(c) At the distance r = 5RT an electron has to cover a distance 2π(5RT ).
T
The time it takes to orbit the Earth is therefore T = 10πR
' 7 h.
vd
(d) J = ne(vdi − vde ) ' nevde = 1.25 × 10−8 A/m2 .
6. (a) If there would be no electric field, the electrons would simply describe
cyclotron motion (uniform circular motion around the magnetic field
~ ×B
~ drift is
lines). Since there is an electric field, to this motion a E
added. The resulting motion is depicted in figure 2.5.
ExB drift
L
Cyclotron motion
B
Re
E
E
B
Ri
2Ri
Re
E
Re
B
Cyclotron motion
ExB drift
ExB drift
Figure 2.5: simplified electron trajectory in the thruster (problem 6).
(b) The electron and ion Larmor radii are given by, respectively,
v⊥e me
' 1.14 × 10−5 m
eB
v⊥i mi
=
' 2.72 × 10−2 m
eB
RLe =
RLi
Since rLe L, the electrons are magnetized (i.e.,trapped in the magnetic field lines). For the positive ions, on the contrary, 2rLi . L, so
that a part of the positive ions can espace the confinement, by moving
to regions where the magnetic field vanishes (see figure 2.6). These ions
are ejected and communicate an acceleration to the thruster. In this
rough estimation the thruster operation does not seem very efficient,
since 2rLi /L ' 0.2. Notice, however, that the ion acceleration by the
electric field was not taken into account. Moreover, there is another
design for the thrusters, the thruster with anode layer (TAL), with a
short acceleration channel, L ∼ 3 cm, for which the current estimation
gives 2rLi ' L.
71
L
B≃0
B≠0
E
2rL
Figure 2.6: ciclotron motion in the thruster (problem 6).
(c) This exercise is very similar to problem 2, so that just an outline is given
~ and defining the y−axis along B
~ (cf.
here. Taking the x−axis along E
figure 2.6), the Lorentz force equation reads


evz B
 v̇x = − eE
 mv̇x = −eE + evz B
m + m
mv̇y = 0
v̇y = 0
−→
.


mv̇z = −evx B
v̈z = − eB
v̇
m x
The equation for vy implies vy (t) = cte. Since vy (0) = 0, vy (t) = 0.
E
2
, the
vz − ωc2 B
Substituting v̇x in the last equation and defining ζ = ωce
2
simple harmonic oscillator equation ζ̈ = −ωce ζ is recovered. Therefore,
ζ(t) = ζ0 cos(ωce t + ϕ), where ζ0 and ϕ are constants to be determined
from the initial conditions. Moving back to the original variables and
imposing vx (0) = 0 and vz (0) = 0, the solution for the velocities is
E
sin(ωce t)
B
E
vz (t) = − [cos(ωce t) − 1] .
B
vx (t) =
The trajectory is obtained by integration the velocities on time,
E 1
cos(ωce t)
B ωce
E
E 1
sin(ωce t) + t ,
z(t) = −
B ωce
B
x(t) = −
E
eE
where we identify a cyclotron motion with radius rLe = Bω
= m
eB 2
ce
and a constant velocity drift along zz. Substituting values, rLe = 1.14×
10−3 m L = 30 cm.
E
2
iE
(d) Similarly, rLi = Bω
=m
eB 2 = 2.7 × 10 m L.
ci
On the one hand, the results confirm that the electrons are magnetized,
confined the to chamber, because rLe L. On the other hand, since
rLi L, the positive ions can be accelerated and leave the chamber,
producing a thrust and pushing system, as the qualitative analysis of
6b somehow anticipated. However, the effect is significantly larger than
72
Solutions to chapter 2. Single particle motion I
~
estimated in that simple analysis, due to the ion acceleration on the E
field.
~
7. (a) From the force equation, F~ = q E,

 mẍ = qE0 cos(ωt)
ÿ = 0

z̈ = 0
;



qE0
dvx
dt = m cos(ωt)
vy (t) = vy (t = 0) =
0
vz (t) = vz (t = 0) = 0
.
Hence
ˆ
t
qE0
qE0
cos(ωτ ) dτ =
sin(ωt) ,
m
mω
0
ˆ t
qE0
qE0
−qE0 qE0
x(t) = xi +
−
cos(ωt) +
,
sin(ωτ ) dτ =
2
2
2
mω
mω
mω
mω
0
| {z }
vx (t) =
xi
qE0
x(t) = −
cos(ωt) .
mω 2
The charge as an harmonic oscillatory motion around x = 0, with fre0
quency ω and amplitude |q|E
mω 2 , as shown in figure 2.7.
x
qE0/mω2
t
-qE0/mω2
Figure 2.7: oscillatory trajectory in problem 7a.
(b)
~ is stronger in the positive direction of the x-axis.
i. The electric field E
This means the “restoring force” is stronger on the upper part of
the trajectory than on the lower part and, accordingly, the center of
mass drifts to the regions of smaller electric field. The trajectory is
schematically represented in figure 2.8.
73
x
t
Figure 2.8: qualitative sketch of the trajectory in problem 7b.
A. Following the suggestions, we have, successively,
x = x0 + x1
dE0
E0 (x) ' E0 (x0 ) + x1
(x0 )
dx
dE0
(x0 ) cos(ωt) .
E(x, t) ' E0 (x0 ) + x1
dx
The expansion is valid if
x1
dE0
(x0 ) |E0 (x0 )
dx
;
dE0
E0 (x0 )
(x0 ) ,
dx
x1
which means that the amplitude of the high-frequency oscillatory component is much smaller than the characteristic distance
of the variation of the electric field.
The result follows directly from the force equation, mẍ = qE,
which reads
dE0
m(ẍ0 + ẍ1 ) = q E0 (x0 ) + x1
(x0 ) cos(ωt) .
dx
B. Since x1 is a quickly oscillating function, ẍ1 ẍ0 and |E0 (x0 )| 0
x1 dE
dx (x0 ) , the equation of motion in the previous item reduces to
mẍ1 ' qE0 cos(ωt) .
This equation was solved in 7a, with solution
x1 (t) = −
qE0
cos(ωt) .
mω 2
74
Solutions to chapter 2. Single particle motion I
C. Taking the temporal average of the equation of motion†on the
short time interval 2π/ω (which is analogous to taking the tem~ ×B
~
poral average along a gyroperiod in the calculation of the E
~ drifts),
or the ∇B
dE
cos(ωt) .
mhẍ0 i + mhẍ1 i = qhE0 cos(ωt)i + q x1
dx
Furthermore, from the previous step, mhẍ1 i = qhE0 cos(ωt)i,
so these two terms cancel in the previous equation, which reduces to
dE
dE
qE0
2
cos (ωt)
mẍ0 ' q
hx1 cos(ωt)i = q
−
dx
dx
mω 2
q2
dE
dE0
q2
2
= −
E
E0
hcos
(ωt)i
;
=
−
0
2
2
|
{z
}
mω
dx
2mω
dx
1/2
2
ẍ0 = −
q
dE0
q2
d
E0
=− 2 2
E02 ,
2
2
2m ω
dx
4m ω dx
d
0
where the equality dx
(E02 ) = 2E0 dE
dx was used.
Finally, the charge feels a ponderomotive force
q2 d
E02 ,
2
4mω dx
which acts in the direction of decreasing E-field, as described
qualitatively in the beginning of the exercise.
Fp = mẍ0 = −
SOLUTIONS TO CHAPTER
3
Single particle motion II
1. (a) Let the indexes 0 and R denote the mid-plane of the mirrors and the
reflection points, respectively. Since
µ=
1
2
2 mv⊥
B
,
the conservation of µ reads
1
2
2 mv⊥0
B0
=
1
2
2 mv⊥R
BR
,
where the magnetic field takes the minimum value at the center, B0 =
bmin .
The conservation of energy implies
1
1
1
2
2
2
mv⊥0
+ mvk0
= mv⊥R
,
m
m
m
where the condition of zero parallel velocity at the reflection point was
used (vkR = 0).
The limit when the proton escapes corresponds to BR = Bmax .
(b) How long does it take to reach that energy? Suggestions: i) suppose that
the B field is approximately uniform in the space between the mirrors
and changes abruptly near the mirrors, i.e., treat each mirror as a flat
piston and show that the velocity gained at each bounce is 2vm ; ii)
compute the number of bounces necessary; iii) assume that the distance
between the mirrors does not change appreciably during the acceleration
process.
2. (F. F. Chen) A plasma with an isotropic distribution of speeds is placed inside
a magnetic mirror with mirror ratio Rm = 4. There are no collisions, so
that the particles in the loss cone escape, while the others remain trapped.
Calculate the fraction of particles that remains trapped.
76
Solutions to chapter 3. Single particle motion II
3. (F. F. Chen) The magnetic field along the axis of a magnetic mirror is B( z) =
B0 (1 + α2 z 2 ), where α is a constant. Suppose that at z = 0 an electron has
2
.
velocity v 2 = 3vk2 = 32 v⊥
(a) Describe qualitatively the electron motion.
(b) Determine the values of z where the electron is reflected.
(c) Write the equation of motion of the guiding center for the direction
~ and show that there is a sinusoidal oscillation. Calcule the
parallel to B
frequency of the motion as a function of v.
~ = −Ėt~uy ,
4. Consider a particle moving in a time-dependent electric field E
~
where Ė is a constant, and a uniform magnetic field B = B0 ~uz .
~ ×B
~ drift.
(a) Calculate the E
(b) Relate the resulting accelerated drift with a force and verify that the
drift due to that force is the polarization drift.
5. (F. F. Chen) A plasma is created in a toroidal chamber with average radius
R = 10 cm and square cross section of size a = 1 cm. The magnetic fiel
is generated by an electrical current I along the symmetry axis. The plasma
is Maxwellian with temperature kT = 100 eV and density n0 = 1019 m−3 .
There is no applied electric field.
~ field, for both positive
(a) Sketch the typical drift orbits in the non-uniform B
ions and electrons with vk = 0.
(b) Calculate the rate of charge accumulation (Coulomb per second) due to
the curvature and gradient drifts on the upper part of the chamber. The
magnetic field in the center of the chamber is 1 T and you can use the
approximation R a if necessary.
~ = E0 exp(iωt) ~ux ,
6. (?) Consider an electron moving in an oscillating electric field, E
~ = B ~uz .
perpendicular to a constant and uniform magnetic field, B
(a) Calculate the drifts existing on the particle motion and describe qualitatively the motion.
(b) Try now to confirm the results you have already obtained, starting
directly from the equations of motion. In particular, show that you
can indeed recover the results from a) for low frequencies of the field,
i.e., ω ωce [Suggestion: i) search for solutions of the form ~v =
~vk + ~vL + ~vD exp(iωt), where ~vL is the velocity of the cyclotron motion
~ ii) verify you can obtain an
and ~vD is constant and perpendicular to B;
~ 0 − ev~D × B;
~ iii) make the
equation for ~vD in the form iωm~vD = −eE
~ and eliminate ~vD × B].
~
cross product with B