# Does photon capture in a black hole (or any other "very compact" object) reduce the luminosity/photon flux from its self isotropic emission?

I don't know if this is even a feasible question or if I have a point in asking so.

Let's say we can model the radiation coming from a very compact object, such as a BH with an accretion disc or any star with proper dimensions (for example, with its disc/gravitational radius within its photon sphere $$R_\gamma = 1.5 R_s=3M$$), as a black body radiating isotropically. I guess we can do so, since we cannot imagine any good reason for discarding this approximation in them!

What I try to ask is basically if we should modify the formula for blackbody emission or whatsoever bolometric magnitude/luminosity in such an object, given that we already know a radius and an effective temperature (we are using $$L = 4\pi \sigma R^2T^4$$ here). And also considering that due to the strong curvature some photons emitted in the surface of the disc/atmosphere will be returning inside and will never reach us...

Due to bending and lensing, supposedly we have only a sector from the compact object that will be sending photons outside, cf. MTW pp. 675 or Shapiro & Teukolsky pp. 352:

when measuring the subtended emission angles from a local static observer.

What I'm sure to know and explain with my own words is that the emitted photon flux reduces when we measure it as receivers on Earth, due to this capture.

• I can't work out where you are saying the emission is coming from. Accretion disks aren't present within the inner most stable orbit. Maybe if you rephrase and ask about emission from a neutron star surface? Jan 25 at 8:31
• Is the question what the effective emitting surface is, when taking light-bending and self-absorption into account? Jan 25 at 13:59
• @ProfRob how can I estimate the radius for the accretion disk to be located? And also, from the image we can see that photon capture happens even for a sufficiently big distance radius, so I thought that really wouldn't matter Jan 25 at 14:39
• @AndersSandberg that is in some way what I am trying to understand. I suppose when doing the integral for the flux through the effective surface would change the solid angle or the area we are taking into account for either flux or luminosity Jan 25 at 14:42

The luminosity is $$L_0 = 4\pi R_0^2 \sigma T_{0}^4$$ in the inertial frame of the neutron star surface (assume a non-spinning object).
For a distant observer, the radius of the neutron star is (e.g. see Haensel 2001) $$R_{\infty}= R_0 \left(1 - R_s/R_0\right)^{-1/2}\, ,$$ the temperature is $$T_{\infty} = T_0 \left(1 - R_s/R_0\right)^{1/2}\ ,$$ and the luminosity $$L_{\infty} = L_0\left(1 - R_s/R_0\right)\ .$$ $$R_s$$ is the usual Schwarzschild radius.
I think the fact that $$L_{\infty} < L_0$$ is the effect you are talking about.