Accretion

1  Introduction

2  Accretion disks

As mentioned, accretion disks play an extremely important role in explaining the observational features and properties of several astrophysical objects, ranging from Cataclysmic variables and centers of galaxies to protoplanetary disks. In essence, the gas in the disk accretes towards a central compact object(s), loosing angular momentum along the way. The excess angular momentum is carried outwards by an infinitesimal amount of matter which is believed to be due to the effects of viscosity. Viscosity plays a central role in the physical interpretation of the accretion disk phenomenon, but since its own origin is unclear, is generally parameterized using the famous a prescription.
We outline the basic equations that govern the evolution of an Accretion disk, and then specialize on 'steady state' disks.

2.1  Basic Physics

We assume a thin disk such that in cylindrical coordinates (R,f,z) most of the matter lies close to the z-plane []. The Keplerian value of the angular velocity of the material comprising the disk is given by:
W = WK = ( G M

R3
)1/2
(1)
We have the mass conservation equation:
R S

t
+

R
(RSvR) = 0
(2)
and the angular momentum conservation equation:
R

t
(SR2 W) +

R
(RSvR R2 W) = 1

2p
G

R
(3)
Here,
S(R,t) is the Surface Density of the disk,
vR is the radial `drift' velocity of the disk,
G(R,t) is the viscous torque exerted by the disk on the binary and is given by...
G(R,t) = 2 pR nSR2 W¢
(4)
with n being the viscosity, which needs to be specified in order to make progress. Once this is done, one can in principle solve the above equations to get analytical solutions for the disk. Using (1)2 and (3), we get:
S

t
= - 1

2 pR

R
é
ë
1

(R2 W)¢
G

R
ù
û
(5)
In case of a Keplerian disk, using (1) and (4), we get:
S

t
= 3

R

R
ì
í
î
R1/2

R
(nSR1/2) ü
ý
þ
(6)
We shall eventually try and obtain solutions for this equation using self-similar solution techniques. For now though, we concentrate on steady state solutions from which, we claim, we can build quasi-steady state solutions which faithfully mimic the complete SS solutions, except near the boundaries.

2.2  The Steady state

To obtain the steady state, we set the time derivative in (6) to zero. Then by a convenient change of variables, x = (R/R0)1/2, and setting s = nSR1/2, we get
d2 s

dx2
= 0
(7)
This has the obvious solution:
s(x) = A + Bx
(8)
with A and B being constants to be determined by boundary conditions. Now, its worth noting that the quantity s(x) introduced above can be recast as G, the torque, as can be seen from equations (4) and (5). In other words, equation (7) is a second order Differential Equation for the torque, and (8) is a solution for the disk in terms of the variable `x'. To achieve further insight into the equations, we note that the specific angular momentum is defined as:
h = (G M R)1/2 = (G M R0)1/2 x
(9)
Thus, essentially h µ x and consequently, one can write the solution (8) as:
G(h) = G0 - 3 pB

R01/2
  h = G0 + C h
(10)
where G0 = -3 p(G M)1/2 A and C = - 3 pB/R01/2 is another constant. Now, applying the boundary conditions: G(xin) = Gin and, G(xout) = Gout, and on rearranging we obtain:
G(h) = æ
è
h - hin

hin - hout
ö
ø
Gout + æ
è
h - hout

hin - hout
ö
ø
Gin
(11)
Thus we have the behavior of the torque in the accretion disk as a function of the specific angular momentum at the location in the disk. Once we have a prescription for the viscosity, we can use equation (4) to determine the surface density and using the energy equation we can obtain the temperature and so on.
Should i explicitly write all the equations down here?
We now turn to the time-dependant case, that of an evolving disk and obtain a general solution for the same using the Self-Similar solution technique. Then we extend our steady state solution to an evolving disk, by considering the time-evolution solution to be successive steady-state solutions.

3  Self-Similar solutions and Quasi Steady-State behavior

We shall treat this case in all generality, relaxing even the Keplerian nature of the orbit, and adopting a power law prescription for the viscosity. In general even in the a prescription, the parameter a is some function of R, and so it makes sense to circumvent the a-prescription totally, by simply assuming a power law for the viscosity.
We consequently specialize to specific cases (self-gravity, a-prescription and so on) and compare the results to published results.

3.1  Self-Similar solution for an evolving disk

Using equations (2), (3) and (4) we have a general evolution equation for the surface density:

S

t
= - 1

2 pR

R
é
ë
1

(R2 W)¢
G

R
ù
û
(12)
= - 1

R

R
ì
í
î
1

(R2W)¢

R
(nSR3 W¢) ü
ý
þ
(13)
We now assume the angular velocity to be a power-law in R: W = W0 (R/R0)a, and change the variable R via the substitution, x ® (R/R0)-(a+1) 1. On substitution and some algebra, one obtains:
S

t
= - (a+1)2 a

(a+2)R02
  x[(a+3)/(a+1)] 

x
ì
í
î
x[(2a+3)/(a+1)]

x
(nS  x-[(a+2)/(a+1)]) ü
ý
þ
(14)
Now, we also assume a power law dependance for the viscosity:
n = n0 (R/R0)b (S/S0)c = n0  (S/S0)c   x-[b/(a+1)]
(15)
On substituting in equation (14), we obtain:
S

t
= - (a+1)2 a

(a+2)R02
  x[(a+3)/(a+1)] 

x
ì
í
î
x[(2a+3)/(a+1)]

x
(n0S  (S/S0)c   x-[(a+b+2)/(a+1)]) ü
ý
þ
(16)

3.1.1  General Self-Similar Solution

I'am working on trying to get a general solution by appropriate substitutions. I think it is possible. For now, i'll specialize in the simplest cases to illustrate our simple model.

3.1.2  Special Cases

Lets set a=-3/2 to represent the Keplerian case, and b=1, c=0 implying the linear viscosity regime. Then Equation (16) reduces to
S

t
= 3 n0

4 x3 R02
2

x2
(x3 S)
(17)
Rearranging and defining the dimensionless time t = (3 n0 t/4 R02), we obtain the familiar diffusion equation
s(x,t)

t
= 2s(x,t)

x2
(18)
The solutions to this equation have been dealt with by Pringle and Lybarskii. We only note that the evolution of the disk can be thought to occur in 3 stages:
  1. The disk evolves and spreads inwards till it reaches the inner boundary. This inner edge is determined by the torque being applied by the central object at that location.
  2. A quasi-stationary accretion stage near the inner edge, gradually moving outwards. Here the mass accretion rate is practically constant or increases slowly.
  3. After a long time," long" being defined as the local time-scale at the outer edge of the disk, the disk enters a new stage of equilibrium. The accretion may cease at this stage and the disk gradually spreads outward (especially in the case of Circumbinary or black hole disks) with the surface density falling continuously.
The first stage occurs on relatively short timescales, and is the transient stage in the entire evolution of the disk. The second and third stages occur on long time-scales and are independent of the initial conditions. These stages can be modelled by Quasi-stationary solutions. In fact if accretion ceases, i.e. if Macc® 0, as in the third stage, the solution is a Steady Stage solution mentioned above.

3.2  Quasi-Steady State solutions

The primary motivation for using quasi-steady state solutions in mathematical simplicity and physical insight into the disk evolution. As mentioned, this model should be relevant for stages 2  &  3 where we can appeal to dimensional and time-scale arguments to achieve mathematical simplicity and physical insight.
Firstly we briefly examine the various relevant time-scales and the regime in which we expect this treatment to mimic the complete time-dependant solution. We then explicitly calculate various physical quantities of the disk and compare it with the published complete solutions.

3.2.1  Time-Scales

4  Applications

4.1  Cataclysmic Variables

4.2  Active Galactic Nuclei

4.3  Protoplanetary Disks

5  Discussion

References


Footnotes:

1This substitution is not very useful. I should've chosen x ® (R/R0)a+2. This is more consistent with the specific angular momentum variable h.
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    On 20 May 2004, 15:57.