Planet Formation
Topic:
Turbulence
Lecture by: C.P. Dullemond
The idea behind the α-formula
Viscosity is length times velocity:
n turb = LV = LeddyVeddy
Maximum height of an eddy:
Maximum velicity of an eddy:
Leddy = a L H £ H
Veddy = aV cs £ cs
n turb = a LaV cs H = a LaV
a L £1
aV £1
cs
2
WK
ºa
a £1
cs
2
WK
The idea behind the α-formula
Time scales:
teddy = Leddy / Veddy
Maximum height of an eddy:
Maximum velicity of an eddy:
Leddy = a L H £ H
Veddy = aV cs £ cs
aL H
aL
a L torbit
teddy =
=
º
aV cs aV WK aV 2p
Simulations show:
teddy
torbit
»
2p
aL = aH = a
Reynolds Number
Reynolds Number:
Re =
LeddyVeddy
n mol
n turb
=
n mol
Typically Re>>1
Turbulent eddies cannot have smaller LV than:
LeddyVeddy > n mol
i.e.
Re >1
because such eddies are quickly viscously dissipated.
Kolmogorov Theory of
Turbulence:
The Turbulent Cascade
Kolmogorov turbulence
Eddies contain eddies, which contain eddies,
which contain eddies, which contain eddies, which contain eddies, which contain
eddies, which contain eddies,
which contain eddies, which contain eddies, which contain
energy input
thermal energy dissipation
cascade
log(l)
log(k)
2p
k=
l
Kolmogorov turbulence
energy input
(turbulent driving)
Kolmogorov turbulent cascade
(must be a powerlaw!)
log(E(k))
energy dissipation
(molecular viscosity)
at the „Kolmogorov scale“
kin
kin <<< kh
kh
log(k)
Kolmogorov turbulence
Driving turbulence with energy input ε [erg/gram.s]
For scales
kin < k < kh
l >l >l
i.e.
we can
in
h
use dimensional analysis to get the powerlaw slope. The question
is: What combination of k and ε gives E? Dimensions (using
erg=gram cm2 s-2):
1
k=
length
length 2
e=
3
time
length 3
E(k) =
2
time
Only possible combination with the right dimensions:
E(k)µ e k
2/3 -5/3
Kolmogorov turbulence
Now a similar dimensional analysis for the typical velocity v
of turbulent eddies at each scale l=2π/k:
l = length
length 2
e=
3
time
length
v=
time
Only possible combination with the right dimensions:
æe ö
v » (el) » ç ÷
èkø
1/3
1/3
Kolmogorov turbulence
Eddy turn-over time scale as a function of l=2π/k:
l = length
length 2
e=
3
time
length
v=
time
Only possible combination with the right dimensions:
teddy
l
» » e -1/3l 2/3
v
So while the biggest eddies (driving scale) have turn-over time
scales ~ tkepler, the smaller eddies have shorter turn-over time
scales.
Kolmogorov turbulence
Contribution of subsubsub-eddies to the viscosity:
n turb,scale (l) = lv » l(el) µl
1/3
4/3
As you see: for ever smaller l (ever bigger k) the contribution to the
viscosity becomes smaller.
The viscosity is dominated by the biggest eddies!
However, the small eddies may play a role later, for the motion of
dust/rocky particles.
Kolmogorov turbulence
At which scale does the turbulence dissipate (i.e. what is the
value of kη)? Answer: at the scale where Re(k)=1:
1 = Re(lh ) =
lh v(lh )
n mol
e lh
»
n mol
1/3 4/3
This gives the Kolmogorov dissipation scale:
æn ö
lh » ç
÷
è e ø
1/4
3
mol
For a real Kolmogorov turbulent cascade to exist, one must have:
lh << lin º Leddy
Kolmogorov turbulence
Back to the energy input ε: Let us check if this is consistent with
the viscous heating coefficient Q+ we derived in the previous chapter.
In the cascade region we have (see few slides back):
veddy (leddy ) = (eleddy )1/3
Let us now make the bold step to assume that this also holds for
the biggest eddies (i.e. that the Kolmogorov powerlaw extents
3
to the largest eddies):
e=
Veddy = (e Leddy )1/3
V eddy
Leddy
For Veddy and Leddy we have expressions from α-turbulence theory:
a c a c 2
e=
=
WK
a L H a L WK
3
V
3
s
3
V
2
s
Kolmogorov turbulence
a c 2
e=
WK
a L WK
3
V
2
s
æ aV ö
æ aV ö
c
2
= ç ÷ a LaV
WK = ç ÷ n turbW2K
WK
è aL ø
è aL ø
2
2
2
s
We also know from viscous disk theory:
3
9
2
2
Q+ º Se =
MWK = Sn turbWK
4p
4
It follows that the two formulae can only be mutually consistent if:
aV 3
= »1
aL 2
aV » a L » a
(keep in mind, however, the approximations made!)
Estimates and numbers
Estimates for disks & turbulence @ 1 AU
Typical accretion rate:
M » 10-7 Msun / yr
Surface density powerlaw unknown, but from previous chapter
theoretical considerations (viscous heating) give a good estimate:
Tmidplane (1AU) » 400 K
With a mean molecular weight of 2.3 this leads to
The pressure scale height then becomes:
cs » 1.2 km/s
H » 0.04AU
We have no idea what the value of α is (this is one of the big
unknowns in the entire disk & planet formation theory), but
simulations suggest α=0.01, so let us take this value.
Estimates for disks & turbulence @ 1 AU
M » 10-7 Msun /yr, cs » 1200 m/s, a =0.01, H=0.04AU
We can now calculate
n turb
at 1 AU:
c
14 cm
n turb = a
» 7.2 ´10
WK
s
2
s
2
Large eddy size:
cs
Leddy = a
» 6 ´1010 cm = 0.004 AU
WK
Large eddy velocity:
Veddy = a cs » 120 m/s
Estimates for disks & turbulence @ 1 AU
M » 10-7 Msun /yr, cs » 1200 m/s, a =0.01, H=0.04AU
n turb » 7´1014 cm2s-1
For a steady-state disk the surface density follows:
S(r) =
M
3pn turb
g
» 930
2
cm
The midplane density then follows with the scale height H:
S
g
-10
r (r, z = 0) =
» 6 ´10
3
cm
2p H
n(r, z = 0) =1.6 ´1014 cm-3
(using μ=2.3)
Estimates for disks & turbulence @ 1 AU
Mass between 0.8 AU and 1.3 AU (very rough estimate):
M disk (0.8-1.3AU) » p (1.3 - 0.8 )AU S = 114 M Earth
2
2
2
Note that this is in the form of gas + 1...2 % dust. Just about enough
to form Earth. Seem thus to be ok!
Mass within 1 large turbulent eddy:
M eddy
4p 3
»
r Leddy » 10-4 M Earth
3
Estimates for disks & turbulence @ 1 AU
M » 10-7 Msun /yr, cs » 1200 m/s, a =0.01, H=0.04AU
T = 400 K, r =6 ´10-10 cmg 3 , n =1.6 ´1014 cm-3
Molecular cross section of H2 = 2x10-15 cm2
Mean free path for gas is:
lmfp =
The molecular viscosity is then:
1
ns coll
2
cm
n molec = lmfpcs » 3.7 ´10 5
s
The Reynolds number of the turbulence is thus:
n turb
Re =
» 2 ´10 9
n molec
» 3 cm
Estimates for disks & turbulence @ 1 AU
Now calculate Komogorov scale. Remember:
n turb,scale (l) = lv » l(el) µl
1/3
4/3
At the largest eddies we have
n turb = LV = Re » 2 ´10
9
At the smallest (Kolmogorov) scale (l=lη) we have Re=1. So:
æL ö
eddy
çç
÷÷
è lh ø
4/3
n turb
9
=
» 2 ´10
n molec
lh » 60 m vh » 60 cm/s
lh » 10-7 Leddy
th » 100 s
4p 3
rlh » 700 g
3
How turbulence is (presumably)
driven:
The Magnetorotational Instability
(ref: Book by Phil Armitage)
Magnetorotational 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
Kelvin-Helmholtz Instability
Now let‘s do this a bit better. We follow a discussion from the
book of Armitage.
Kelvin-Helmholtz instability (shear instability):
Photo credit: Beverly Shannon (1999)
Kelvin-Helmholtz Instability
However, in a rotating system the rotation can stabilize the
Kelvin-Helmholtz instability. The Rayleigh criterion says:
dl d 2
= ( r W) < 0
dr dr
instability
A Keplerian disk has:
dlK d 2
1
= ( r WK ) = rWK > 0
dr dr
2
Keplerian disks
are RayleighStable
Magnetorotational Instability
Let us study the stability of a disk with a weak vertical magnetic
field. We will use perturbation theory and we will assume
ideal MHD. The equations for ideal MHD are:
¶r
+ Ñ × ( r v) = 0
¶t
2 ö
æ
Dv ¶v
1
B
1
º + v × Ñv = - ç P + ÷ - ÑF ( B× Ñ) B
Dt ¶t
r è 8p ø
4pr
¶B
= Ñ ´ ( v ´ B)
¶t
Magnetorotational Instability
Now let‘s transform this to cylindrical coordinates. This is not
trivial. But let‘s do this for the equation of motion of a single
fluid element under influence of a force-per-mass f:
Dv
=f
Dt
Since
v º x and one could write Dv/Dt º x
X = r cos f
Y = r sin f
let us write out:
X = r cos f - rf sin f
Y = r sin f + rf cos f
X = r cos f - rf sin f - rf sin f - rf sin f - rf 2 cos f
Y = r sin f + rf cos f + rf cos f + rf cos f - rf sin f
2
Magnetorotational Instability
Taking our equations at ϕ=0:
X = r - rf
2
Y = 2rf + rf
The momentum equations for the fluid parcel thus become:
dF
r - rf = + fr
dr
2rf + rf = ff
2
f are the forces coupling
the gas to the B-field.
Note that we can write out the gravity term:
dF d æ GM ö
GM
= ç
÷=- 2
dr dr è r ø
r
which is the well-known
inverse square force law
Magnetorotational Instability
Now define a local (x,y) coordinate system:
r = r0 + x
y
f = Wt +
r0
GM
GM
GM
2
»
+
2
x
+
O(x
)
2
2
3
r
r0
r0
2
æ
yö
y
2
2
f = ç W+ ÷ » W + 2W + O(y 2 )
r0 ø
r0
è
-
Inserting this into the previous equations, and discarding quadratic
terms, yields (after some calculation):
GM
x - 2Wy = 3 3 x + f x
r0
y + 2Wx = f y
Magnetorotational Instability
Now let‘s look at an (x,y) displacement varying with height z
and time t:
x(z, t) = x1e
i(w t-kz)
y(z, t) = y1ei(wt-kz)
Remember now that gas displacements carry along the B-field.
Let‘s assume a weak vertical initial B-field. Then the displacements
create x- and y- components of this B-field:
Bx (z, t) = -ikBz x1e
i(w t-kz)
By (z, t) = -ikBz y1ei(wt-kz)
Magnetorotational Instability
Bx (z, t) = -ikBz x1e
i(w t-kz)
By (z, t) = -ikBz y1ei(wt-kz)
These produce a magnetic tension force (from -
fx (z, t) = -(kv A ) x1e
2
fy (z, t) = -(kv A ) y1e
2
i(wt-kz)
i(w t-kz)
1
4pr
( B × Ñ) B ):
B
vA =
4pr
2
z
Alfven
velocity
Magnetorotational Instability
The equation of motion then becomes:
GM
2
-w x1 - 2iw Wy1 = 3 3 x1 - (kv A ) x1
r0
2
-w y1 + 2iw Wx1 = -(kv A ) y1
2
2
Combining them yields the following dispersion relation:
é GM
ù
é
GM ù
2
2
2
w - w ê 3 + 2(kv A ) ú + (kv A ) ê(kv A ) - 3 3 ú = 0
r0 û
ë r0
û
ë
4
2
Magnetorotational Instability
Unstable Stable
Most unstable
mode
Most unstable mode: kv A =
(
)
15 / 4 W
Stable for: kv A > 3W
Magnetorotational Instability
Stable for: kv A > 3W
Conclusion: If the field is too strong, the disk is stable. So
MRI works only for weak magnetic fields!
Another conclusion: MRI does not work for too small wavelengths.
There is a minimum scale that can be driven. There is also a
certain scale where the driving is the strongest.
Let‘s assume magnetic equipartition:
Then the instability occurs at:
cs
3
<
WK
k
v A = cs
2p
l>
H>H
3
Scale larger than
disk thickness:
Equipartition disk=stable
Magnetorotational Instability
Note: This instability works only if the disk is sufficiently ionized
for ideal MHD equations to be valid.
Only a tiny bit of ionization is required.
But even that can be problematic, since dust grains very
efficiently „vacuum clean“ away free electrons.
This leads to so called „dead zones“ in disks.
The debate for what causes turbulence in disks remains wide open
today.