<p>Yes it is to do with general relativity. For a non-rotating Schwarzschild black hole the last <strong>stable</strong> orbit is at 3 times the Schwarzschild radius and hence the inner edge of the accretion disk.</p>

CC BY-SA 3.0
2015-11-20T20:16:27.857Z
2015-11-20T20:16:27.857Z
John Davis
<p>Three times the Schwarzschild radius corresponds to the closest stable circular orbit around a black hole. The general idea is that as matters moves in towards the black hole it gets stuck in an accretion disc where angular momentum has to be moved outwards in order to allow the matter to move inwards. The generic mechanism is some sort of viscosity, which heats the gas and hence you get radiation.</p>

<p>However, once the matter gets inside $3r_s$, that problem disappears. There are no stable orbits, no angular momentum loss or viscosity is needed and the material is able to flow (rapidly) straight into the black hole.</p>

<p>Thus when we observe black hole accretion discs we expect them to be truncated at $3r_s$. </p>

<p>So I think the argument then is along the lines of - the gravitational potential energy of unit mass falling to $3r_s$ is converted into an orbital kinetic energy of $0.5v^2 = GM/6r_s$ per unit mass and the rest is converted to radiation. Thus
$$L = \left[\frac{GM}{3r_s} - \frac{GM}{6r_s}\right] \frac{dM}{dt}$$
$$ L = \frac{GM}{6r_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$</p>

CC BY-SA 3.0
2015-11-21T21:51:16.233Z
Possibly worth mentioning that the innermost stable circular orbit (ISCO) gets smaller as the spin $a$ increases and the efficiency factor 1/12 increases.
Warrick
CC BY-SA 3.0
2016-11-02T10:10:25.447Z
@Rob Jeffries: That's interesting. Can you link some standard references to read up on this angular momentum non-conservation near black holes?
AtmosphericPrisonEscape
CC BY-SA 3.0
2016-11-02T21:36:55.623Z
@AtmosphericPrisonEscape Angular momentum is always conserved. It is accreted by the black hole along with the mass/energy. Angular momentum does not have to be lost outwards because there are no longer any stable orbits.
ProfRob
CC BY-SA 3.0
2015-11-20T20:19:46.193Z
2016-11-02T21:40:23.767Z
2016-11-02T21:40:23.767Z
ProfRob
<p>The <a href="http://www.astronomy.ohio-state.edu/~dhw/A142/notes10.pdf" rel="nofollow noreferrer">luminosity of a black hole accretion disc</a> gaining mass at a rate $\mathrm{d}M\over \mathrm{d}t$ can be estimated as</p>

<p>$${1\over 12}{\mathrm{d}M\over \mathrm{d}t}c^2$$</p>

<p>That is a substantial proportion of the rest mass of the in-falling matter. The linked document explains that the factor 1/12 is because at less than 3 times the event horizon radius the matter "spirals in without
radiating more energy" </p>

<p>What causes matter in the accretion disk not to radiate beyond this limit? It is twice the radius of the photon sphere, is the reason general relativity?</p>

<black-hole><accretion-discs>
Luminosity of black hole accretion disc
James K
# Luminosity of black hole accretion disc

(Asked by: username_0 on 2015-11-20)

The [luminosity of a black hole accretion disc](http://www.astronomy.ohio-state.edu/~dhw/A142/notes10.pdf) gaining mass at a rate $\mathrm{d}M\over \mathrm{d}t$ can be estimated as

$${1\over 12}{\mathrm{d}M\over \mathrm{d}t}c^2$$

That is a substantial proportion of the rest mass of the in-falling matter. The linked document explains that the factor 1/12 is because at less than 3 times the event horizon radius the matter "spirals in without
radiating more energy"

What causes matter in the accretion disk not to radiate beyond this limit? It is twice the radius of the photon sphere, is the reason general relativity?

**Tags:** black-hole, accretion-discs


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## 2 answers

### Answer by username_2 on 2015-11-20. Score: 5 | Accepted answer

Three times the Schwarzschild radius corresponds to the closest stable circular orbit around a black hole. The general idea is that as matters moves in towards the black hole it gets stuck in an accretion disc where angular momentum has to be moved outwards in order to allow the matter to move inwards. The generic mechanism is some sort of viscosity, which heats the gas and hence you get radiation.

However, once the matter gets inside $3r\_s$, that problem disappears. There are no stable orbits, no angular momentum loss or viscosity is needed and the material is able to flow (rapidly) straight into the black hole.

Thus when we observe black hole accretion discs we expect them to be truncated at $3r\_s$.

So I think the argument then is along the lines of - the gravitational potential energy of unit mass falling to $3r\_s$ is converted into an orbital kinetic energy of $0.5v^2 = GM/6r\_s$ per unit mass and the rest is converted to radiation. Thus
$$L = \left[\frac{GM}{3r\_s} - \frac{GM}{6r\_s}\right] \frac{dM}{dt}$$
$$ L = \frac{GM}{6r\_s} \frac{dM}{dt} = \frac{1}{12}c^2 \frac{dM}{dt} .$$

> **Comments:**
> **username_3**: Possibly worth mentioning that the innermost stable circular orbit (ISCO) gets smaller as the spin $a$ increases and the efficiency factor 1/12 increases.  
> **username_4**: @Rob Jeffries: That's interesting. Can you link some standard references to read up on this angular momentum non-conservation near black holes?  
> **username_2**: @username_4 Angular momentum is always conserved. It is accreted by the black hole along with the mass/energy. Angular momentum does not have to be lost outwards because there are no longer any stable orbits.  

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### Answer by username_1 on 2015-11-20. Score: 1

Yes it is to do with general relativity. For a non-rotating Schwarzschild black hole the last **stable** orbit is at 3 times the Schwarzschild radius and hence the inner edge of the accretion disk.


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