Lecture 14: Viscous Accretion Disks - University of Toledoastro1.physics.utoledo.edu › ~megeath...
Transcript of Lecture 14: Viscous Accretion Disks - University of Toledoastro1.physics.utoledo.edu › ~megeath...
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Lecture 14: Viscous Accretion Disks
HST: 0.2 - 1 μm image of the A0e star AB Aurigae
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Subgroups in the Orion
OB1 Associations
OB1a (10 Myr)
OB1b (5 Myr)
OB1c (2.5 Myr)
OB1d (1 Myr)
The Orion OB1 association can be subdivided into four subgroups, each with a different age.
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Hernandez et al. 2008Orion OB1b
OB Associations as Laboratories for Disk Evolution
Text
Squares and Cricles: MembersFilled squares and cirles: sources with disks
Spitzer image
Spitzer image
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OB Associations as Laboratories for Disk Evolution
Hernandez et al. 2008 OB1b: 5 Myr
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Hernandez et al. 2008
OB Associations as Laboratories for Disk Evolution
OB1a: 10 Myr
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Calvet et al. 2005
OB Associations as Laboratories for Disk Evolution
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shock
shock
The formation of a disk
• Infalling matter collides with matter from the other side• Forms a shock• Free-fall kinetic energy is converted into heat
€
kTµmp
≈ 12 vff
2 =GM*
r
• Heat is radiated away, matter cools, sediments to midplane• Disk is formed
€
T ≈ 25000KAt 10 AU from 1M star:
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The formation of a disk
3-D Radiation-Hydro simulations of disk formation
Yorke, Bodenheimer & Laughlin 1993
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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:
€
vφ ≅ ΩKr =GM*
r
€
ΩK ≡GM*
r3
Kepler frequency
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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!
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Outward angular momentum transport
A B
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.
Specific angular momentum for a Keplerian disk:
€
l = rvφ = r2ΩK = GM* rSo 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.).
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Non-stationary (spreading) disks
Time steps of 2x105 year
Lynden-Bell & Pringle (1974), Hartmann et al. (1998)
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Formation & viscous spreading of disk
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Formation & viscous spreading of disk
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Formation & viscous spreading of disk
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Formation & viscous spreading of diskFrom the rotating collapsing cloud model we know:
€
rcentrif ~ t4
Initially the disk spreads faster than the centrifugal radius.
Later the centrifugal radius increases faster than disk spreading
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Formation & viscous spreading of diskA numerical model
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Formation & viscous spreading of diskA numerical model
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Formation & viscous spreading of diskA numerical model
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Formation & viscous spreading of diskA numerical model
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Formation & viscous spreading of diskA numerical model
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Viscosity
ux
y
A
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Viscosity
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Shakura & Sunyaev model
Define the surface density:
€
Σ(r) ≡ ρ−∞
+∞
∫ (r,z)dz
Radial structure
€
∂Σ∂ t
+1r∂(Σrvr)∂r
= 0
Integrate continuity equation over z:
(1)
€
∂(Σvr )∂ t
+1r∂(Σrvr
2)∂r
+∂(Σcs
2)∂r
− Σvφ2
r= −Σ
GMr2
Integrate radial momentum equation over z:
(2)
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...
€
∂(Σl)∂ t
+1r∂(Σr vr l)
∂r=1r∂∂r
Σν r3 ∂Ω∂r
⎛
⎝ ⎜
⎞
⎠ ⎟
Integrate tangential momentum equation over z:
(3)€
( l ≡ vφr)
€
(Ω≡ vφ /r)
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Shakura & Sunyaev model
(2)
Let’s first look closer at the radial momentum equation:
Let us take the Ansatz (which one can later verify to be true) that vr << cs << vφ.€
∂(Σvr )∂ t
+1r∂(Σrvr
2)∂r
+∂(Σcs
2)∂r
− Σvφ2
r= −Σ
GMr2
€
vφ2 =
GMr
€
vφ =GMr
≡ ΩK r
That means: from the radial momentum equation follows the tangential velocity
Conclusion: the disk is Keplerian
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Shakura & Sunyaev model
Let’s now look closer at the tangential momentum equation:
€
∂(Σl)∂ t
+1r∂(Σr vr l)
∂r=1r∂∂r
Σν r3 ∂Ω∂r
⎛
⎝ ⎜
⎞
⎠ ⎟ (3)
€
Σvr∂ lK∂r
=1r∂∂r
Σν r3 ∂ΩK
∂r⎛
⎝ ⎜
⎞
⎠ ⎟
€
lK ≡ ΩKr2
€
∂ GM*r∂r
= 12GM*
r= 1
2ΩKr
€
∂∂r
GM*
r3= − 3
2GM*
r5= − 3
2ΩK
r
€
vr= −3
Σ r∂∂r
Σν r( )€
Σvr∂(ΩKr
2)∂r
=1r∂∂r
Σν r3 ∂ΩK
∂r⎛
⎝ ⎜
⎞
⎠ ⎟
Now use continuity equation
€
∂Σ∂ t
+1r∂(Σrvr)∂r
= 0
The derivatives of the Kepler frequency can be worked out:
That means: from the tangential momentum equation follows the radial velocity
€
∂(ΣlK )∂ t
+1r∂(Σr vr lK )
∂r=1r∂∂r
Σν r3 ∂ΩK
∂r⎛
⎝ ⎜
⎞
⎠ ⎟
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Shakura & Sunyaev model
€
vr= −3
Σ r∂∂r
Σν r( )
€
∂Σ∂ t
+1r∂(Σrvr)∂r
= 0
Radial structure
Our radial structure equations have now reduced to:
with
Missing piece: what is the value of ν?
It is not really known what value ν has. This depends on the details of the source of viscosity. But from dimensional analysis it must be something like:
€
ν =αcsh
€
α = 0.001...0.1
Alpha-viscosity (Shakura & Sunyaev 1973)
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Shakura & Sunyaev modelFurther on alpha-viscosity:
€
ν =αcsh
Here the vertical structure comes back into the radial structure equations!
€
h =kTr3
µmpGM*
€
≡csΩK
€
ν =αcs2
ΩK
So we obtain for the viscosity:
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Shakura & Sunyaev modelSummary of radial structure equations:
€
vr= −3
Σ r∂∂r
Σν r( )€
∂Σ∂ t
+1r∂(Σrvr)∂r
= 0
€
ν =αcs2
ΩK
If we know the temperature everywhere, we can readily solve these equations (time-dependent or stationary, whatever we like).
€
cs2
If we don’t know the temperature a-priori, then we need to solve the above 3 equations simultaneously with energy equation.
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Shakura & Sunyaev modelSuppose we know that is a given power-law:Ansatz: surface density is also a powerlaw:
€
cs2 ~ r−ξ
€
Σ ~ r−η
€
= 3(ξ +η − 2)νr€
ν ≡αcs2
ΩK
~ r−ξ +3 / 2
€
∂(Σrvr)∂r
= 0
Stationary continuity equation:
from which follows:
€
Σ ~ rξ −3 / 2
€
η = −ξ + 3/2
The radial velocity then becomes:
€
vr≡ −3
Σ r∂∂r
Σν r( )
€
vr = −32νr
€
cs2
Proportionality constants are straightforward from here on...
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Shakura & Sunyaev modelExamples:
€
T ~ r−3 / 2
€
Σ ~ r−0
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Shakura & Sunyaev modelExamples:
€
T ~ r−1
€
Σ ~ r−1/ 2
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Shakura & Sunyaev modelExamples:
€
T ~ r−1/ 2
€
Σ ~ r−1
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Shakura & Sunyaev modelExamples:
€
T ~ r−0
€
Σ ~ r−3 / 2
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Shakura & Sunyaev modelFormulation in terms of accretion rate
Accretion rate is amount of matter per second that moves radially inward throught he disk.
€
˙ M
€
˙ M = −2π rvrΣ
Working this out with previous formulae:
€
vr = −32νr
€
˙ M = 3πνΣ
€
˙ M = 3πα cs2
ΩK
Σ
€
ν ≡αcs2
ΩK
We finally obtain:
€
Σ =˙ M
3παΩK µ mp
k Tmid
(but see later for more correct formula with inner BC satisfied)
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Shakura & Sunyaev modelEffect 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!€
Σ =˙ M
3παΩK µ mp
k Tmid
1− rin
r⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 2⎡
⎣ ⎢
⎤
⎦ ⎥
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Shakura & Sunyaev modelHow 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:
€
Q+ ≡ Σν r ∂ΩK
∂r⎛
⎝ ⎜
⎞
⎠ ⎟
2
€
=94Σν
GM*
r3
€
vr = −32νr
For power-law solution: use equation of previous page:
€
ν = −23rvr
€
Q+ = −34π2πΣrvr
GM*
r3
The viscous heat production becomes:
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Shakura & Sunyaev modelHow do we determine the temperature?
€
Q+ = −34π2πΣrvr
GM*
r3
Define ‘accretion rate’ (amount of matter flowing through the disk per second):
€
˙ M ≡ −2πΣr vr = constant
€
Q+ =3
4π˙ M ΩK
2
End-result for the viscous heat production:
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Shakura & Sunyaev modelHow do we determine the temperature?
Now, this heat must be radiated away.
Disk has two sides, each assumed to radiate as black body:
€
2σ Teff4 =Q+
One obtains:
€
Teff =3
8πσ˙ M ΩK
2⎛
⎝ ⎜
⎞
⎠ ⎟
1/ 4
€
=3
4π˙ M ΩK
2
€
~ r−3 / 4
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Molecular viscosity? No!Problem: molecular viscosity is virtually zero
Reynolds number
€
Re =u Lν
Typical disk (at 1 AU): N=1x1014 cm-3, T=500 K, L=0.01AUAssume (extremely simplified) σH2≈ π(1Ang)2.
€
uT =3kTµmp
= 2.3km/s
€
lfree =1Nσ
= 32 cm
€
ν = 7.3×106 cm2/s
€
Re = 4.7 ×109
L = length scale<u>= typical velocityν = viscosity
Molecular viscosity:
€
ν = uT lfreelfree = m.f.p. of molecule<uT>= velo of molecule
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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
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Magneto-rotational instability (MRI)
Johansen & Klahr (2005); Brandenburg et al.
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Gravitational (in)stabilityIf disk surface density exceeds a certain limit, then disk becomes gravitationally unstable.
Toomre Q-parameter:
€
Q =hΩK
2
πGΣ
€
≈hr
M*
Mdisk
For Q>2 the disk is stableFor Q<2 the disk is gravitationally unstable
Unstable disk: spiral waves, angular momentum transport, strong accretion!!
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Gravitational (in)stability
Spiral waves act as `viscosity’
Rice & Armitage
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FU Orionis stars
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McNeal’s Nebula: a new FU Ori?
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The Luminosity
Problem
This is a longstanding problem: the required accretion luminosities are on average higher than the observed luminosities.
Dunham et al. 2010
Dunham et al. 2010
Grey gives predicted luminosites from models
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Voroboyov & Basu
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Summary• In Keplerian orbits, the inner disk is rotating faster
than the outer disk• Viscosity acts as a friction. Mass is transported
inward, angular momentum outward.• This causes mass to accrete onto the central source
(Will Fischer’s lecture tuesday), and the disk to spread outward.
• Viscosity can be do to thermal motions in a gas (not important for disks), magneto rotational instability, or spiral waves created by gravity.
• Young stars may go through episodic accretion: gas first accumulates onto a disk until it undergoes gravitational instability, then the mass is dumped onto the disk in a major event.