1. The formation of stars and planets Day 3, Topic 1: Viscous accretion disks Lecture by: C.P. Dullemond

2. 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 Heat is radiated away, matter cools, sediments to midplane Disk is formed At 10 AU from 1M  star:

3. The formation of a disk 3-D Radiation-Hydro simulations of disk formation Yorke, Bodenheimer & Laughlin 1993

4. 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

5. The angular momentum problem Angular momentum of 1 M  in 10 AU disk: 3x10 53 cm 2 /s Angular momentum of 1 M  in 1 R  star: <<6x10 51 cm 2 /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!

6. Outward angular momentum transport AB 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: 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.).

7. Molecular viscosity? No! Problem: molecular viscosity is virtually zero Reynolds number Typical disk (at 1 AU): N=1x10 14 cm -3, T=500 K, L=0.01AU Assume (extremely simplified)  H 2   (1Ang) 2. L = length scale = typical velocity = viscosity Molecular viscosity: l free = m.f.p. of molecule = velo of molecule

8. 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): The momentum equation then becomes:

9. Turbulent ‘viscosity’: Reynolds stress Now average over many eddy turnover-times, and use: (tensor!) andbut Then one obtains: The additional term is the Reynolds stress. It has a trace (=turbulent pressure) and off-diagonal elements (=turbulent ‘viscous’ stress).

10. 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)

11. 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

12. Magneto-rotational instability (MRI) Johansen & Klahr (2005); Brandenburg et al.

13. 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)

14. Shakura & Sunyaev model Vertical structure 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:

15. Shakura & Sunyaev model Vertical structure Because of our assumption (!) that T=const. we can write: This has the solution: with A Gaussian!

16. Shakura & Sunyaev model Define the surface density: Radial structure Integrate continuity equation over z: (1) 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... Integrate tangential momentum equation over z: (3)

17. 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 v r << c s << v . That means: from the radial momentum equation follows the tangential velocity Conclusion: the disk is Keplerian

18. Shakura & Sunyaev model Let’s now look closer at the tangential momentum equation: (3) Now use continuity equation The derivatives of the Kepler frequency can be worked out: That means: from the tangential momentum equation follows the radial velocity

19. Shakura & Sunyaev model 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 has. This depends on the details of the source of viscosity. But from dimensional analysis it must be something like: Alpha-viscosity (Shakura & Sunyaev 1973)

20. Shakura & Sunyaev model Further on alpha-viscosity: Here the vertical structure comes back into the radial structure equations! So we obtain for the viscosity:

21. Shakura & Sunyaev model Summary of radial structure equations: If we know the temperature everywhere, we can readily solve these 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.

22. Shakura & Sunyaev model Suppose we know that is a given power-law: Ansatz: surface density is also a powerlaw: Stationary continuity equation: from which follows: The radial velocity then becomes: Proportionality constants are straightforward from here on...

23. Shakura & Sunyaev model Examples:

24. Shakura & Sunyaev model Examples:

25. Shakura & Sunyaev model Examples:

26. Shakura & Sunyaev model Examples:

27. Shakura & Sunyaev model Formulation in terms of accretion rate Accretion rate is amount of matter per second that moves radially inward throught he disk. Working this out with previous formulae: We finally obtain: (but see later for more correct formula with inner BC satisfied)

28. 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!

29. 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: For power-law solution: use equation of previous page: The viscous heat production becomes:

30. Shakura & Sunyaev model How do we determine the temperature? Define ‘accretion rate’ (amount of matter flowing through the disk per second): End-result for the viscous heat production:

31. 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: One obtains:

32. 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): The optical depth is defined as: with the Rosseland mean opacity. We finally obtain:

33. 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.

34. Shakura & Sunyaev model Example: Dust power-law opacity Remember from previous page: Remember relation between  and  : Remember from power-law solutions: