Planet Formation
Topic:
Viscous accretion disks
Lecture by: C.P. Dullemond
The formation of a disk
shock
shock
• Infalling matter collides with matter from the other side
• Forms a shock
• Free-fall kinetic energy is converted into heat
kT
GM*
1 2
» 2 v ff =
m mp
r
At 10 AU from 1M star:
T » 25000K
• Heat is radiated away, matter cools, sediments to midplane
• Disk is formed
The formation of a disk
3-D Radiation-Hydro simulations of disk formation
Yorke, Bodenheimer & Laughlin 1993
Keplerian rotation (a reminder)
Disk material is almost (!) 100% supported against gravity by its
rotation. Gas pressure plays only a minor role. Therefore it is a
good approximation to say that the tangential velocity of the gas
in the disk is:
Kepler frequency
The angular momentum problem
• Angular momentum of 1 M in 10 AU disk:
3x1053 cm2/s
• Angular momentum of 1 M in 1 R  star:
<<6x1051 cm2/s (=breakup-rotation-speed)
• Original angular momentum of disk = 50x higher than
maximum allowed for a star
• Angular momentum is strictly conserved!
• Two possible solutions:
– Torque against external medium (via magnetic fields?)
– Very outer disk absorbs all angular momentum by moving
outward, while rest moves inward.
Need friction
through viscosity!
Outward angular momentum transport
Ring A moves faster than ring B.
Friction between the two will try
to slow down A and speed up B.
This means: angular momentum
is transferred from A to B.
A
B
Specific angular momentum for
a Keplerian disk:
So if ring A looses angular momentum, but is forced to remain on
a Kepler orbit, it must move inward! Ring B moves outward,
unless it, too, has friction (with a ring C, which has friction with D,
etc.).
Molecular viscosity? No!
Problem: molecular viscosity is virtually zero
Reynolds number
Molecular viscosity:
Re =
uL
n
n = uT lfree
L = length scale
<u>= typical velocity
 = viscosity
lfree = m.f.p. of molecule
<uT>= velo of molecule
Typical disk (at 1 AU): N=1x1014 cm-3, T=500 K, L=0.01AU
Assume (extremely simplified) H2 (1Ang)2.
uT =
3kT
= 2.3 km/s
mm p
n = 7.3 ´106 cm2/s
1
lfree =
= 32 cm
Ns
Re = 4.7 ´109
Turbulent ‘viscosity’: Reynolds stress
The momentum equation for hydrodynamics is:
Now consider this gas to be turbulent. We want to know the motion
of average quantities. Assume that turbulence leaves  unaffected.
Split v into average and perturbation (turbulence):
v = v0 + v1
The momentum equation then becomes:
Turbulent ‘viscosity’: Reynolds stress
Now average over many eddy turnover-times, and use:
v1 = 0
and
v0 v1 = 0
but
v1v1 ¹ 0
(tensor!)
Then one obtains:
The additional term is the Reynolds stress. It has a trace
(=turbulent pressure) and off-diagonal elements (=turbulent
‘viscous’ stress).
Turbulent ‘viscosity’: Reynolds stress
Problem with turbulence as origin of viscosity in disks is: most
stability analyses of disks show that the Keplerian rotation
stabilizes the disk: no turbulence!
Debate has reopened recently:
• Non-linear instabilities
• Baroclynic instability? (Klahr et al.)
But most people believe that turbulence in disks can have only
one origin: Magneto-rotational instability (MRI)
Magneto-rotational instability (MRI)
(Also often called Balbus-Hawley instability)
Highly simplified pictographic explanation:
If a (weak) pull exists between two
gas-parcels A and B on adjacent
orbits, the effect is that A moves
inward and B moves outward: a pull
causes them to move apart!
A
B
The lower orbit of A causes an
increase in its velocity, while B
decelerates. This enhances their
velocity difference! This is positive
feedback: an instability.
A
B
Causes turbulence in the disk
Magneto-rotational instability (MRI)
Johansen & Klahr (2005); Brandenburg et al.
Shakura & Sunyaev model
• (Originally: model for X-ray binary disks)
• Assume the disk is geometrically thin: h(r)<<r
• Vertical sound-crossing time much shorter than radial drift
of gas
• Vertical structure is therefore in quasi-static equilibrium
compared to time scales of radial motion
• Split problem into:
– Vertical structure
(equilibrium
reached on short time scale)
– Radial structure
(evolves
over much longer time scale; at each time step vertical structure
assumed to be in equilibrium)
Shakura & Sunyaev model
Vertical structure
r(z)
T(z)
Equation of hydrostatic equilibrium:
Equation for temperature gradient is complex: it involves an
expression for the viscous energy dissipation (see later) radiative
transfer, convection etc.
Here we will assume that the disk is isothermal up to the very
surface layer, where the temperature will drop to the effective
temperature
Shakura & Sunyaev model
Vertical structure
Because of our assumption (!) that T=const. we can write:
This has the solution:
æ z2 ö
r(z) = r0 expç- 2 ÷
è 2h ø
A Gaussian!
with
h=
kT r 3
m m p GM*
Shakura & Sunyaev model
Difficulty: re-express
equations in cylindrical
coordinates. Complex due to
covariant derivatives of
tensors... You’ll have to
simply believe the Equations I
write here...
Radial structure
Define the surface density:
S(r) º
ò
+¥
-¥
r(r,z) dz
Integrate continuity equation over z:
¶ S 1 ¶(Srv r )
+
=0
¶ t r ¶r
(1)
Integrate radial momentum equation over z:
vf
¶ (Sv r ) 1 ¶ (Srv 2r ) ¶(Sc 2s )
GM
+
+
-S
= -S 2
¶t
r ¶r
¶r
r
r
2
Integrate tangential momentum equation over z:
(2)
( l º vf r)
(3)
Shakura & Sunyaev model
Let’s first look closer at the radial momentum equation:
vf
¶ (Sv r ) 1 ¶ (Srv 2r ) ¶(Sc 2s )
GM
+
+
-S
= -S 2
¶t
r ¶r
¶r
r
r
2
(2)
Let us take the Ansatz (which one can later verify to be true) that vr
<< cs << v.
GM
vf =
r
2
That means: from the radial momentum equation follows the
tangential velocity
Conclusion: the disk is Keplerian
Shakura & Sunyaev model
Let’s now look closer at the tangential momentum equation:
(3)
¶ S 1 ¶(Srv r )
+
=0
¶ t r ¶r
Now use continuity equation
The derivatives of the Kepler frequency can be worked out:
¶
vr = Sn r
S r ¶r
3
(
)
That means: from the tangential momentum equation follows the
radial velocity
Shakura & Sunyaev model
Radial structure
Our radial structure equations have now reduced to:
¶ S 1 ¶(Srv r )
+
=0
¶ t r ¶r
with
¶
vr = Sn r
S r ¶r
3
(
)
Missing piece: what is the value of ?
It is not really known what value  has. This depends on the
details of the source of viscosity. But from dimensional analysis it
must be something like:
n = a c sh
a = 0.001...0.1
Alpha-viscosity (Shakura & Sunyaev 1973)
Shakura & Sunyaev model
Further on alpha-viscosity:
n = a c sh
Here the vertical structure comes back into the radial structure
equations!
h=
kT r 3
m m p GM*
So we obtain for the viscosity:
Shakura & Sunyaev model
Summary of radial structure equations:
¶ S 1 ¶(Srv r )
+
=0
¶ t r ¶r
¶
vr = Sn r
S r ¶r
3
(
)
If we know the temperature everywhere,
we can readily solve these
c s2
equations (time-dependent or stationary, whatever we like).
If we don’t know the temperature a-priori, then we need to solve the above 3
equations simultaneously with energy equation.
Shakura & Sunyaev model
The “local accretion rate”
vr(r)
Radial velocity can (will) be a function of r. And so is the
surface density. Define now the local „accretion rate“ as the
amount of gas flowing through a cylinder of radius r:
M(r) = 2p r S(r)vr (r)
In a steady-state situation this must be independent of r:
M(r) = 2p r S(r)vr (r) = M
where Mis then the
accretion rate of the disk.
Shakura & Sunyaev model
2
c
Suppose we know that s is a given power-law:
Ansatz: surface density is also a powerlaw:
c s2 ~ r-x
S ~ r-h
The radial velocity then becomes:
3 ¶
n
vr º Sn r = 3(x + h - 2)
S r ¶r
r
Stationary continuity equation:
(
from which follows:
)
3n
vr = 2r
¶(Srv r )
=0
¶r
S ~ rx -3 / 2
h = -x + 3/2
Proportionality constants are straightforward from here on...
Shakura & Sunyaev model
Examples:
T ~ r-3 / 2
S ~ r-0
Shakura & Sunyaev model
Examples:
T ~ r-1
S ~ r-1/ 2
Shakura & Sunyaev model
Examples:
T ~ r-1/ 2
S ~ r-1
Shakura & Sunyaev model
Examples:
T ~ r-0
S ~ r-3 / 2
Shakura & Sunyaev model
Formulation in terms of accretion rate
Accretion rate M is amount of matter per second that moves
radially inward throught he disk.
˙ = -2p rv S
M
r
Working this out with previous formulae:
˙ = 3pn S
M
vr = -
3n
2r
We finally obtain:
(but see later for more
correct formula with inner
BC satisfied)
Shakura & Sunyaev model
Effect of inner boundary condition:
Powerlaw does not go all the way to the star. At inner edge (for
instance the stellar surface) there is an abrupt deviation from
Keplerian rotation. This affects the structure of the disk out to
many stellar radii:
Keep this in mind when
applying the theory!
Shakura & Sunyaev model
How do we determine the temperature?
We must go back to the vertical structure......and the energy
equation.....
First the energy equation: heat production through friction:
9
GM*
= Sn 3
4
r
For power-law solution: use equation of previous page:
3n
2
vr = n = - rv r
2r
3
The viscous heat production becomes:
3
GM*
Q+ = 2pSrvr 3
4p
r
Shakura & Sunyaev model
How do we determine the temperature?
3
GM*
Q+ = 2pSrvr 3
4p
r
Define ‘accretion rate’ (amount of matter flowing through the
disk per second):
˙ º -2p Sr v = constant
M
r
End-result for the viscous heat production:
Shakura & Sunyaev model
How do we determine the temperature?
Now, this heat must be radiated away.
Disk has two sides, each assumed to radiate as black body:
2s Teff4 = Q+
One obtains:
~ r-3 / 4
Shakura & Sunyaev model
How do we determine the temperature?
Are we there now? Almost.... This is just the surface
temperature. The midplane temperature depends also on the
optical depth (which is assumed to be >>1):
4
Tmid
= t Ross Teff4
The optical depth
t Ross
is defined as:
t Ross º 12 S k Ross
with k Rossthe Rosseland mean opacity.
We finally obtain:
Shakura & Sunyaev model
To obtain full solutions, we must first have an expression for the
Rosseland mean opacity.
We can then express the midplane temperature as a function of
the surface density.
In the radial structure equations we can then eliminate the
temperature in exchange for the surface density.
We then have an equation entirely in terms of the surface density.
We solve this equation.
Shakura & Sunyaev model
Example:
Dust power-law opacity
g
k Ross ~ Tmid
Remember from previous page:
S ~ r-h
Remember from power-law solutions:
Tmid ~ r
-x
Remember relation between  and :
3(2 - g )
h=
2(5 - g )
Tmid ~ r-h / 4 Tmid
g /4
h+ 3
x=
4 -g
h = -x + 3/2
r-3 / 4
Non-stationary (spreading) disks
• So far we assumed an infinitely large disk
• In reality: disk has certain size
• As most matter moves inward, some matter must
absorb all the angular momentum
• This leads to disk spreading: a small amount of outer
disk matter moves outward
Non-stationary (spreading) disks
n ~, one
r c can solve the
Given a viscosity power-law function
Shakura-Sunyaev equations analytically in a time-dependent
manner. Without derivation, the resulting solution is:
2- c ù
é
v
-( 5 / 2- c )/(2- c )
S=
q
expêú
c
3pn1v
q
ë
û
C
where we have defined
v º r /r1
n1 º n (r1)
q º t /ts +1
with r1 a scaling radius and ts the viscous scaling time:
1
r12
ts =
3(2 - c ) 2 n1
Lynden-Bell & Pringle (1974), Hartmann et al. (1998)
Non-stationary (spreading) disks
Time steps of 2x105
year
Lynden-Bell & Pringle (1974), Hartmann et al. (1998)
Formation & viscous spreading of disk
Formation & viscous spreading of disk
Formation & viscous spreading of disk
Formation & viscous spreading of disk
From the rotating collapsing cloud model we know:
rcentrif ~ t 4
Initially the disk spreads faster than the centrifugal radius.
Later the centrifugal radius increases faster than disk spreading
Formation & viscous spreading of disk
A numerical model
Formation & viscous spreading of disk
A numerical model
Formation & viscous spreading of disk
A numerical model
Formation & viscous spreading of disk
A numerical model
Formation & viscous spreading of disk
A numerical model
Formation & viscous spreading of disk
Hueso & Guillot (2005)
Disk dispersal
It is known that disks
vanish on a few Myr
time scale.
But it is not yet
established by which
mechanism. Just
viscous accretion is too
slow.
- Photoevaporation?
- Gas capture
.
by planet?
Haisch et al. 2001
Photoevaporation of disks
(Very brief)
Ionization of disk surface creates surface layer of hot gas. If this
temperature exceeds escape velocity, then surface layer
evaporates.
æ GM ö1/ 2
v esc » ç
÷
è r ø
Evaporation proceeds for radii beyond:
GM
r ³ 2 º rgr
c sHII