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Radiation Transport Around Kerr Black Holes Jeremy David ...

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134 CHAPTER 5. STEADY-STATE α-DISKS<br />

NT use the first model, which gives the total flux integrated over frequency as<br />

( ) 1/2 ( ) 9/4 ρ<br />

F = 8.05 × 10 7 Ts<br />

erg/cm 2 /s. (5.34)<br />

g/cm 3<br />

When integrating the stellar structure equations in the diffusion limit, we find the<br />

exponential model to more accurately approximate the atmospheric density profile.<br />

In that case, the integrated flux is given by<br />

◦<br />

K<br />

) −1/3 ( ) 17/6 Ts<br />

erg/cm 2 /s. (5.35)<br />

F = 1.3 × 10 4 ( H<br />

cm<br />

◦<br />

K<br />

Combining equations (5.28) and (5.35) gives the boundary condition for the disk’s<br />

surface temperature:<br />

T s = 0.28F 2/5 R −2/15 h −2/15 . (5.36)<br />

To simultaneously satisfy all three boundary conditions, the system of differential<br />

equations is then solved using a “shooting” method, starting at z = h and integrating<br />

p, T, and q z inwards to z = 0. We iterate this approach for a series of initial values for<br />

h: the solution is given by the value of h that matches the inner boundary condition<br />

of q z (0) = 0. Repeating this entire procedure for each value of r gives the complete<br />

structure of the accretion disk outside of the ISCO.<br />

To get a good starting guess for the value of h, we derive here an approximate<br />

solution of the vertical structure equations. The result is somewhat different from<br />

that given in NT, as they ignore the (often significant) gas pressure in the inner<br />

disk, and also we use different methods of averaging the vertical structure over z.<br />

The difference in disk thickness turns out to be a factor of at least 2 − 3 for typical<br />

stellar-mass black holes. Starting with the pressure balance equation (5.19), with the<br />

pressure going to zero at the surface of the disk p(h) = 0, the disk thickness can be<br />

approximated in terms of the central pressure p c and density ρ c :<br />

dp<br />

dz ≈ p c<br />

h<br />

Taking the density profile as roughly linear with ρ(h) = 0,<br />

〈ρz〉 = ρ c<br />

h<br />

∫ h<br />

0<br />

= 〈ρz〉R. (5.37)<br />

(<br />

1 − z )<br />

zdz = ρ ch<br />

h 6 , (5.38)<br />

so our first estimate for h is<br />

h =<br />

( ) 1/2 6pc<br />

. (5.39)<br />

ρ c R

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