General Relativistic treatment of Accretion disks

Abstract

The physical parameters of an Accretion disk near a black hole are derived using the complete general relativistic treatment following Novikov & Thorne[1]. It is seen that the Newtonian treatment can be extended in the general relativistic regime in a straightforward manner. The results are then compared to the Newtonian and post-Newtonian disks for a steady state disk.

1  Introduction

The centers of Active Galaxies are thought to contain Supermassive black holes which are accreting gas resulting in significant radiation in the ultraviolet, optical, infrared, radio and even X-rays. Presently, we model the accretion disk in the steady state and derive the resulting physical parameters. In the next section we briefly examine the Newtonian [2] and post-Newtonian treatments [3], following which we shall extend the relevant expressions to general relativity [1]. We then compare the results in the various treatments and discuss the significance of the results.

2  Standard Accretion disk, Paczynsky-Wiita disk

We are interested in the steady state properties of a thin accretion disk around a compact object. The potential is time independent and axisymmetric. Thus we can model disks around Schwarzschild and Kerr black holes as well as neutron stars and white dwarfs.
We now derive a general equation for the behavior of the torque in the disk. By general, we mean that the torque equation is valid for any potential (not necessarily Keplerian) as long as it is axisymmetric (only a function or R). For our purposes, we consider the Keplerian and Paczynski-Wiita potentials:
F(R) = - G M
R
    ,    FPW(R) = - GM
R-Rs
with Rs = 2  G M/c2 being the Schwarzschild radius. Now consider the equations for the conservation of mass:
R S
t
 +
R
 (RSvR) = 0
(1)
and the angular momentum conservation equation:
R
t
 (SR2 W) +
R
 (RSvR R2 W) = 1
2p
 G
R
(2)
Here,
W(R) = 1
R
 F
R
(3)
is the angular velocity,
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 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 (4), we get:
S
t
 = - 1
2 pR
R
 é
ë 		1
(R2 W)¢
 G
R
 ù
û
(5)
Notice that the term R2 W(R) = h(R), the specific angular momentum. In the steady state, we set the LHS to zero, and we immediately have the solution:
G(R) - G0 =
×
M
 
 (h(R)-h0)
(6)
where G0 = G(R0) is known at some reference radius R0, h0=h(R0) and [M\dot] is the known steady state mass transfer rate.
Thus, once the boundary conditions (essentially, [M\dot] & G0), are specified the torque profile in the disk is completely determined. Then, once the viscosity is prescribed, the entire structure of the disk can be determined [4].
In fig.(1), we plot the torque in the Keplerian and the PW potentials. We anticipate the torque to vanish at 3Rs (see below). We see that beyond 3Rs, the torque in the PW disk is smaller than the Keplerian disk.

Figure 1: Keplerian and PW-torque

3  Relativistic treatment

We now follow the standard treatment outlined above in determining the torque by taking General relativistic effects into account. We essentially follow Novikov & Thorne [1] and Page & Thorne [5].

3.1  Revisiting definition of disk Torque

Eqn.(4) above which is the torque in the disk, can be written in terms of the viscous stress as:
G(r) = ó
õ 		df  r ó
õ 		dz  r  Trf
(7)
Here, Trf is the azimuthal component of the viscous stress tensor. Integrating the angular momentum equation in terms of the stress tensor, we get:
×
M
 
 W
2 p
 + C
r2
 = - ó
õ 		dz Trf ,
×
M
 
 W
2 p
 R(r) = - ó
õ 		dz Trf
(8)
which is equivalent to Eqn. (6) above. Also,
R(r) = 1 - æ
è 		r0
r
 ö
ø 		2
 
 W(r0)
W(r)
   =   1- h0
h
(9)
Thus on determining the angular velocity (Eqn. 3) in various potentials, we can determine the torque profile in the disk.

3.2  Torque in disk in GR

It is possible ([6], Appendix A) to write the equation for the stress, and hence the torque in the same form as Eqn. (8), but by explicitly replacing W by the Keplerian angular velocity, WK.
×
M
 
 WK
2 p
 RT(x) = - ó
õ 		dz Trf
(10)
and R(r) is replaced by the relativistic reduction factor RT(x) which is given by:
RT(x) = C(x)
A(x)
(11)
x = R/Rg, Rg = G M/c2 is the gravitational radius. The functions C(x) and A(x) are given by:
A(x) = 1 - 2
x
 + a*2
x2
(12)

C(x) = 1 - yms
y
 - 3 a*
2y
 ln æ
è 		y
yms
 ö
ø 		- 3(y1 - a*)2
y y1(y1-y2)(y1-y3)
 ln æ
è 		y-y1
yms-y1
 ö
ø
- 3(y2 - a*)2
y y2 (y2-y1)(y2-y3)
 ln æ
è 		y-y2
yms-y2
 ö
ø
- 3(y3 -a*)2
y y3 (y3-y1)(y3-y2)
 ln æ
è 		y-y3
yms-y3
 ö
ø
(13)
with y = x1/2. Also, yms is the value of y at the marginally stable orbit, and y1,2,3 are the three roots of y3 - 3 y + 2 a* = 0. a is the spin of the black hole and a* is the black hole spin a normalized to G M/c.

Figure 2: Keplerian, PW and Relativistic-torque
Thus, the torque is given by:
G(r) = ó
õ 		df  r ó
õ 		dz  r  Trf
= 2 pr2 æ
è
×
M
 
 W+ G0
r2
 ö
ø 		RT(r)
= æ
è
×
M
 
 h(r) + G0 ö
ø 		RT(r)
(14)
From fig. 2 we see that the relativistic and PW torques match almost exactly. We have set the torque G0 at the inner boundary, which is pegged at the marginally stable orbit (Rms = 6 Rg in the Schwarzschild case), to zero. This has to be the case because the specific angular momentum has a minimum at Rms. Moreover, as pointed out by Paczynsky [7], the `no-torque inner boundary condition' must hold since no information can propagate upstream in the supersonic region inwards of Rin.

Figure 3: Angular momentum profiles in Keplerian and PW & Relativistic-potentials.

4  Appendix A: Conservation of Mass, Energy and Angular momentum.

4.1  Properties of Kerr Metric.

The form of the Kerr metric in the equatorial plane is given by:
ds2 = - r2D
A
 dt2 + A
r2
 (df-wdt)2+ r2
D
 dr2 + dz2
(15)
where,
D = r2 - 2 M r + a2  =  r2 D ,
A = r4 + R2 a2 2 M R a2 =  r4 A ,
w = 2Mar/A  =  (2Mar/r3) A-1
Also, we define the following functions of r and a in order to express formulae in Newtonian+correction terms.

A = 1 + a*2
x2
 + 2a*2
x3
  ,
B = 1 + a*
x3/2
  ,
C = 1 - 3
x
 + 2 a*
x3/2
  ,
D = 1 - 2
x
 + a*2
x2
  ,
E = 1 - 4 a*2
x2
 - 4a*2
x3
 + 3 a*4
x4
  ,
F = 1 - 2a*
x3/2
 + a*2
x2
Q = 1+a*y-3
(1-3y-2+2a*y-3)1/2
 1
y
 é
ë 		y-y0- 3
2
 a*ln( y
y0
 )
- 3(y1-a*)2
y1(y1-y2)(y1-y3)
 ln( y-y1
y0-y1
 )
- 3(y1-a*)2
y1(y1-y2)(y1-y3)
 ln( y-y1
y0-y1
 )
- 3(y1-a*)2
y1(y1-y2)(y1-y3)
 ln( y-y1
y0-y1
 ) ù
û
(16)
where, y=x1/2.
Thus, the angular velocity is given by:
W = df
dr
 = M1/2
r3/2 + a M1/2
 = M1/2
r3/2
 1
B
(17)
and, the specific angular momentum is:
h = uf = M1/2 r1/2 F
C1/2
(18)

4.2  Mass Conservation

As in the Newtonian case, the conservation of mass leads to a constant accretion rate in the steady state. The mass transfer flux vector, r0[u\vec] has vanishing divergence:
Ñ·(r0
®
u
 
 ) = 0
Using Gauss's theorem, we write this as
0 = ó
õ

V 
 Ñ·(r0
®
u
 
 )(-g)1/2dt dr dz df = ó
õ

V 
 r0
®
u
 
 ·d3 S
×
M0
 
 Dt = ó
õ

S 
 ár0
®
u
 
 ñ·d3 S = (-2 pr D1/2Dt) ó
õ 		+h

-h 
 ár0 v[^r] ñdz
(19)
Thus,
×
M0
 
 = - 2 pr S
®
v
 
 [^r]
 
 D1/2 = constant
(20)
where, [v\vec][^r] = 1/Sò-h+hár0v[^r] ñ  dz is the radial velocity of the gas.

4.3  Angular momentum Conservation

In this case, the divergence of the complete stress-energy tensor vanishes.
(xa Tab);b = 0
x is the Killing vector satisfying Killing's equation. Following Novikov & Thorne [1], it can be shown that
(
-
×
M0
 
 h(R)
2 p
 + r2BC-1/2DW),r + 2 rh(R) F = 0
(21)
The first term is the rate at which angular momentum of the gas increases, whilst the second term gives the rate at which shear stresses carry off angular momentum. The last term is the rate at which photons carry away angular momentum. Here, W is the integrated stress. Eqn (21) is the GR analog of Eqns. (2) &(5), the Newtonian angular momentum and angular momentum rate conservation equations. Using Eqns (20) and (21),
W =
×
M0
2p
 æ
è 		M
r3
 ö
ø 		1/2
 
 C1/2Q
BD
 = - ó
õ 		dz Trf
(22)
Thus, on comparing Eqns (22) & (10), we conclude that
RT(x) = C(x)
A(x)
 = C1/2Q
BD
On substitution, this is indeed seen to be true, i.e., the Novikov-Thorne & Krolik expressions are completely equivalent.

References

[1]
Novikov, I.D. & Thorne, K.S. 1973, in Black Holes, ed. C. DeWitt and B. DeWitt (New York: Garden & Breach).
[2]
Pringle.
[3]
Paczynsky & Wiita.
[4]
Frank, King, Raine.
[5]
Page & Thorne.
[6]
Krolik, J.H.: Active Galactic Nuclei.
[7]
Paczynsky, B.: astro-ph/0004129v1
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