Structure and evolution of protoplanetary disks Part I Accretion disk theory Lecture by: C.P. Dullemond

Structure and evolution ofprotoplanetary disks

Part I

Accretion disk theory

Lecture by: C.P. Dullemond

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 ≅ ΩK r =GM*

r

€

ΩK ≡GM*

r3

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

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* r

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

€

Re =u L

ν

Typical disk (at 1 AU): N=1x1014 cm-3, T=500 K, L=0.01AU

Assume (extremely simplified) H2 (1Ang)2.

€

uT =3kT

μmp

= 2.3km/s

€

lfree =1

Nσ= 32 cm

€

ν =7.3 ×106 cm2/s

€

Re = 4.7 ×109

L = length scale<u>= typical velocityν = viscosity

Molecular viscosity:

€

ν = uT lfree

lfree = m.f.p. of molecule<uT>= velo of molecule

Turbulent ‘viscosity’: Reynolds stress

The momentum equation for hydrodynamics is:

€

∂ ρv( )∂t

+∇ ⋅ ρvv+ P( ) = 0

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:

€

∂ ρv0( )∂t

+∂ ρv1( )

∂t+∇ ⋅ ρ (v0 + v1)(v0 + v1) + P( ) = 0


€

∂ ρv0( )∂t

+∂ ρv1( )

∂t+∇ ⋅ ρ (v0 + v1)(v0 + v1) + P( ) = 0

Now average over many eddy turnover-times, and use:

€

v1 = 0

€

v1v1 ≠ 0(tensor!)

€

v0v1 = 0and but

€

∂ ρv0( )∂t

+∇ ⋅ ρv0v0 + ρ v1v1 + P( ) = 0

Then one obtains:

The additional term is the Reynolds stress. It has a trace (=turbulent pressure) and off-diagonal elements (=turbulent ‘viscous’ 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 in the last decade:• 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)


Vertical structure

€

ρ(z)

€

T(z)

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

Equation of hydrostatic equilibrium:

€

dP

dz= −ρ (z)ΩK

2 z

€

d

dz

ρ (z)k T(z)

μ mp

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= −ρ (z)ΩK

2 z


Vertical structure

€

d

dz

ρ (z)k T

μ mp

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟= −ρ(z)ΩK

2 z

Because of our assumption (!) that T=const. we can write:

€

k T

μ mp

dρ(z)

dz= −ρ(z)ΩK

2 z

This has the solution:

€

ρ(z) = ρ 0 exp −z2

2h2

⎛

⎝ ⎜

⎞

⎠ ⎟

€

h =kTr3

μmpGM*

with

A Gaussian!


Define the surface density:

€

Σ(r) ≡ ρ−∞

+∞

∫ (r,z) dz

Radial structure

€

∂Σ∂t

+1

r

∂(Σrvr)

∂r= 0

Integrate continuity equation over z:

(1)

€

∂(Σvr )

∂t+

1

r

∂(Σrvr2)

∂r+

∂(Σcs2)

∂r− Σ

vφ2

r= −Σ

GM

r2

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+

1

r

∂(Σ r vr l)

∂r=

1

r

∂

∂rΣν r3 ∂Ω

∂r

⎛

⎝ ⎜

⎞

⎠ ⎟

Integrate tangential momentum equation over z:

(3)€

( l ≡ vφr)

€

(Ω ≡ vφ /r)


(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 << v.€

∂(Σvr )

∂t+

1

r

∂(Σrvr2)

∂r+

∂(Σcs2)

∂r− Σ

vφ2

r= −Σ

GM

r2

€

vφ2 =

GM

r

€

vφ =GM

r≡ ΩK r

That means: from the radial momentum equation follows the tangential velocity

Conclusion: the disk is Keplerian


Let’s now look closer at the tangential momentum equation:

€

∂(Σl)

∂t+

1

r

∂(Σ r vr l)

∂r=

1

r

∂

∂rΣν r3 ∂Ω

∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ (3)

€

Σvr

∂lK

∂r=

1

r

∂

∂rΣν r3 ∂ΩK

∂r

⎛

⎝ ⎜

⎞

⎠ ⎟

€

lK ≡ ΩK r2

€

∂ GM*r

∂r= 1

2

GM*

r= 1

2 ΩK r

€

∂∂r

GM*

r3= − 3

2

GM*

r5= − 3

2

ΩK

r

€

vr = −3

Σ r

∂

∂rΣν r( )€

Σvr

∂(ΩK r2)

∂r=

1

r

∂

∂rΣν r3 ∂ΩK

∂r

⎛

⎝ ⎜

⎞

⎠ ⎟

Now use continuity equation

€

∂Σ∂t

+1

r

∂(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+

1

r

∂(Σ r vr lK )

∂r=

1

r

∂

∂rΣν r3 ∂ΩK

∂r

⎛

⎝ ⎜

⎞

⎠ ⎟


€

vr = −3

Σ r

∂

∂rΣν r( )

€

∂Σ∂t

+1

r

∂(Σrvr)

∂r= 0

Radial structure

Our radial structure equations have now reduced to:

with

Missing piece: what is the value of ?

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)


Further on alpha-viscosity:

€

ν=αcsh

Here the vertical structure comes back into the radial structure equations!

€

h =kTr3

μmpGM*

€

≡cs

ΩK

€

ν=α cs2

ΩK

So we obtain for the viscosity:


Summary of radial structure equations:

€

vr = −3

Σ r

∂

∂rΣν r( )

€

∂Σ∂t

+1

r

∂(Σrvr)

∂r= 0

€

ν=α cs2

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.


Suppose 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)ν

r€

ν≡α cs2

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

∂

∂rΣν r( )

€

vr = −3

2

ν

r

€

cs2

Proportionality constants are straightforward from here on...


Examples:

€

T ~ r−3 / 2

€

Σ ~ r−0


Examples:

€

T ~ r−1

€

Σ ~ r−1/ 2


Examples:

€

T ~ r−1/ 2

€

Σ ~ r−1


Examples:

€

T ~ r−0

€

Σ ~ r−3 / 2


Formulation 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 = −3

2

ν

r

€

˙ M = 3π ν Σ

€

˙ M = 3παcs

2

ΩK

Σ

€

ν≡α cs2

ΩK

We finally obtain:

€

Σ=˙ M

3πα

ΩK μ mp

k Tmid

(but see later for more correct formula with inner BC satisfied)


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!€

Σ=˙ M

3πα

ΩK μ mp

k Tmid

1−rin

r

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2 ⎡

⎣ ⎢

⎤

⎦ ⎥


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:

€

Q+ ≡ Σν r∂ΩK

∂r

⎛

⎝ ⎜

⎞

⎠ ⎟

2

€

=9

4Σν

GM*

r3

€

vr = −3

2

ν

r

For power-law solution: use equation of previous page:

€

ν =−2

3rvr

€

Q+ = −3

4π2πΣrvr

GM*

r3

The viscous heat production becomes:



€

Q+ = −3

4π2πΣrvr

GM*

r3

Define ‘accretion rate’ (amount of matter flowing through the disk per second):

€

˙ M ≡ −2π Σ r vr = constant

€

Q+ =3

4π˙ M ΩK

2

End-result for the viscous heat production:



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

Viscous heating or irradiation?

T Tauri star

Viscous heating or irradiation?

Herbig Ae star

Flat irradiated disks

α

€

α ≅0.4 r*

rIrradiation flux:

€

Firr = αL*

4π r2

Cooling flux:

€

Fcool = σ T 4

€

T =0.4 r* L*

4πσ r3

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 4

€

T ∝ r−3 / 4

Similar to active accretion disk, but flux is fixed.Similar problem with at least a large fraction of HAe and T Tauri star SEDs.

Flared disks

flaring

irradiation

heating vs cooling

verticalstructure

● Kenyon & Hartmann 1987● Calvet et al. 1991; Malbet & Bertout 1991● Bell et al. 1997; ● D'Alessio et al. 1998, 1999● Chiang & Goldreich 1997, 1999; Lachaume et al. 2003

Flared disks: Chiang & Goldreich model

The flaring angle:

€

α =r∂

∂r

hs

r

⎛

⎝ ⎜

⎞

⎠ ⎟→ ξ

hs

r

Irradiation flux:

€

Firr = αL*

4πr2

Cooling flux:

€

Fcool = σ T 4

€

T 4 =ξ

σ

hs L*

4π r3

Express surface height in terms of pressure scale height:

€

hs = χ h

€

χ =1...6


€

T 4 =ξ

σ

hs L*

4π r3

€

hs = χ h

Remember formula for pressure scale height:

€

h =k Tr3

μmpGM*

€

T 4 =ξ

σ

χ hL*

4π r3

€

h8 =k

μmpGM*

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

4

r12 T 4

We obtain

€

h8 =k

μmpGM*

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

4

r12 ξ

σ

χ hL*

4π r3

€

h8 =k

μmpGM*

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

4

r9 ξ

σ

χ hL*

4π

€

h7 =k

μmpGM*

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

4

r9 ξ

σ

χ L*

4π


€

h7 =k

μmpGM*

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

4

r9 ξ

σ

χ L*

4π

We therefore have:

€

h = C 1/ 7r9 / 7

€

C =k

μmpGM*

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

4

ξ

σ

χ L*

4πwith

Flaring geometry:

Remark: in general is not a constant (it decreases with r). The flaring is typically <9/7

Non-stationary (spreading) disksGiven a viscosity power-law function , one can solve the Shakura-Sunyaev equations analytically in a time-dependent manner. Without derivation, the resulting solution is:

€

ν ~ rχ

Lynden-Bell & Pringle (1974), Hartmann et al. (1998)

€

Σ=C

3πν 1ϖχ

θ−(5 / 2−χ ) /(2−χ ) exp −ϖ 2−χ

θ

⎡

⎣ ⎢

⎤

⎦ ⎥

where we have defined

€

ν1 ≡ ν (r1)

€

ϖ ≡r /r1

€

θ ≡t / ts +1

with r1 a scaling radius and ts the viscous scaling time:

€

ts =1

3(2 − χ )2

r12

ν 1

Non-stationary (spreading) disks

Time steps of 2x105 year

Lynden-Bell & Pringle (1974), Hartmann et al. (1998)

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

Disk formation and spreading

Molecular cloudMolecular cloudcore:core:









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

€

kT

μ mp

≈ 12 vff

2 =GM*

r

• Heat is radiated away, matter cools, sediments to midplane• Disk is formed

€

T ≈ 25000KAt 10 AU from 1M star:

The formation of a disk

3-D Radiation-Hydro simulations of disk formation

Yorke, Bodenheimer & Laughlin 1993

Formation & viscous spreading of diskA numerical model





Formation & viscous spreading of disk

Hueso & Guillot (2005)

Coupling to 2D disk structure models

Dullemond et al. in prep.Dullemond et al. in prep.

t=0.1 Myr



t=0.2 Myr



t=0.4 Myr



t=0.8 Myr



t=1.6 Myr



t=3.2 Myr



t=6.4 Myr

Disk dispersal

Haisch et al. 2001

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?

Geometry: only gravity vs pressure

ρΩρΩKKzz

dP/dzdP/dz

Geometry: adding vertical B-field

Flux freezing: MHD in a nutshellStrong field: matter can only move along given field lines (beads on a string):

Weak field: field lines are forced to move along with the gas:


ρΩρΩKKzz

dP/dzdP/dz


Slingshot effect. Blandford & Payne (1982)

Gravitational potential:

€

Φ=−GM

r2 + z2

Use cylindrical coordinates r,z

€

Φ=−GM

r0

1

2

r

r0

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+r0

r2 + z2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

Effective gravitational potential along field line (incl. sling-shot effect):

(courtesy:C. Fendt)

Geometry: adding vertical B-fieldBlandford & Payne (1982)

€

Φ=−GM

r0

1

2

r

r0

⎛

⎝ ⎜

⎞

⎠ ⎟

2

+r0

r2 + z2

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

Critical angle: 60 degrees with disk plane. Beyond that: outflow of matter.

Gas will bend field lines

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.

€

vesc ≈GM

r

⎛

⎝ ⎜

⎞

⎠ ⎟

1/ 2

Evaporation proceeds for radii beyond:

€

r ≥GM

csHII

2≡ rgr

Extreme-UV Photoevaporation

Hollenbach 1994; Clarke et al. 2001Hollenbach 1994; Clarke et al. 2001Alexander, Clarke & Pringle 2006Alexander, Clarke & Pringle 2006

EUV versus FUV Photoevaporation

EUV Photoevap: EUV Photoevap: - Weak, butWeak, but- Works around 1 AUWorks around 1 AU

FUV Photoevap: FUV Photoevap: - Strong, butStrong, but- Works > 50 AUWorks > 50 AU

Hollenbach et al. 1994Hollenbach et al. 1994Gorti & Hollenbach 2007Gorti & Hollenbach 2007

1 AU1 AU

50 AU50 AU

Far-UV Photoevaporation

• Solve full 2-D (=1+1D) structure of disk on-the-fly while doing disk evolution and compute from this the mass loss at each radius.

Gorti & Hollenbach (2007)Gorti & Hollenbach (2007)

Gorti, Dullemond &Gorti, Dullemond &Hollenbach (in prep)Hollenbach (in prep)

Far-UV Photoevap: First Results




‘Dead zone’MRI can only work if the disk is sufficiently ionized.

Cold outer disk (T<900K) is too cold to have MRI

Cosmic rays can ionize disk a tiny bit, sufficient to drive MRI

Cosmic rays penetrate only down to about 100 g/cm2.

full penetration of cosmic rays

partial penetration of cosmic rays

‘Dead zone’Hot enough to ionize gas

Only surface layer is ionized by cosmic rays

Tenuous enough for cosmic rays

Above dead zone: live zone of fixed Σ = 100 g/cm2. Only this layer has viscosity and can accrete.

Accumulation of mass in ‘dead zone’

€

vr = −3

2

ν

r

Remember:

€

ν ≡rχ

€

vr = −3

2rχ −1

Stationary continuity equation (for active layer only):

€

∂(ΔΣr vr)

∂r~ −ΔΣ

∂( rχ )

∂r≠ 0

For χ>0 we have mass loss from active layer (into dead zone)

Gravitational (in)stabilityIf disk surface density exceeds a certain limit, then disk becomes gravitationally unstable.

Toomre Q-parameter:

€

Q =hΩK

2

π GΣ

€

≈h

r

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!!

Gravitational (in)stability

Spiral waves act as `viscosity’

Rice & Armitage

Documents

Structure and evolution of protoplanetary disks Part I Accretion disk theory Lecture by: C.P. Dullemond