<p>Let's answer the question from the point of view of a stable neutron star.</p>
<p>The luminosity is
<span class="math-container">$$L_0 = 4\pi R_0^2 \sigma T_{0}^4$$</span>
in the inertial frame of the neutron star surface (assume a non-spinning object).</p>
<p>For a distant observer, the radius of the neutron star is (e.g. see <a href="https://www.aanda.org/articles/aa/pdf/2001/46/aah3021.pdf" rel="nofollow noreferrer">Haensel 2001</a>)
<span class="math-container">$$R_{\infty}= R_0 \left(1 - R_s/R_0\right)^{-1/2}\, ,$$</span>
the temperature is
<span class="math-container">$$T_{\infty} = T_0 \left(1 - R_s/R_0\right)^{1/2}\ ,$$</span>
and the luminosity
<span class="math-container">$$L_{\infty} = L_0\left(1 - R_s/R_0\right)\ .$$</span>
<span class="math-container">$R_s$</span> is the usual Schwarzschild radius.</p>
<p>I think the fact that <span class="math-container">$L_{\infty} &lt; L_0$</span> is the effect you are talking about.</p>

CC BY-SA 4.0
2021-01-25T14:56:35.483Z
2021-01-25T14:56:35.483Z
ProfRob
<p>I don't know if this is even a feasible question or if I have a point in asking so.</p>
<p>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 <span class="math-container">$R_\gamma = 1.5 R_s=3M$</span>), 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!</p>
<p>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 <span class="math-container">$L = 4\pi \sigma R^2T^4$</span> 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...</p>
<p>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 &amp; Teukolsky pp. 352:<a href="https://i.sstatic.net/2aSvB.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2aSvB.png" alt="CAPTURE OF PHOTONS" /></a></p>
<p>when measuring the subtended emission angles from a local static observer.</p>
<p>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.</p>

CC BY-SA 4.0
2021-01-25T08:31:23.620Z
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?
ProfRob
CC BY-SA 4.0
2021-01-25T13:59:54.763Z
Is the question what the effective emitting surface is, when taking light-bending and self-absorption into account?
Anders Sandberg
CC BY-SA 4.0
2021-01-25T14:39:05.487Z
@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
omivela17
CC BY-SA 4.0
2021-01-25T14:42:14.437Z
@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
omivela17
<black-hole><general-relativity><luminosity><photons><black-body>
Does photon capture in a black hole (or any other "very compact" object) reduce the luminosity/photon flux from its self isotropic emission?
omivela17
# Does photon capture in a black hole (or any other "very compact" object) reduce the luminosity/photon flux from its self isotropic emission?

(Asked by: username_0 on 2021-01-25)

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:[![CAPTURE OF PHOTONS](https://i.sstatic.net/2aSvB.png)](https://i.sstatic.net/2aSvB.png)

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.

**Tags:** black-hole, general-relativity, luminosity, photons, black-body

> **Comments:**
> **username_1**: 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?  
> **username_2**: Is the question what the effective emitting surface is, when taking light-bending and self-absorption into account?  
> **username_0**: @username_1 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  
> **username_0**: @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  

---

## 1 answers

### Answer by username_1 on 2021-01-25. Score: 2

Let's answer the question from the point of view of a stable neutron star.

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](https://www.aanda.org/articles/aa/pdf/2001/46/aah3021.pdf))
$$R\_{\infty}= R\_0 \left(1 - R\_s/R\_0\right)^{-1/2}\, ,$$
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.


---

