<p>For any massive object the gravitational potential energy is given by Newton's law:</p>

<p><span class="math-container">$$ V(r) = -\frac{GMm}{r} $$</span></p>

<p>The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy:</p>

<p><span class="math-container">$$ V_\mathrm{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$</span></p>

<p>where <span class="math-container">$L$</span> is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we plot this graph for the Earth-Moon system we get a graph like this:</p>

<p><a href="https://i.sstatic.net/CLzP3.gif" rel="noreferrer"><img src="https://i.sstatic.net/CLzP3.gif" alt="Effective potential"></a></p>

<p>(this comes from my answer to the question <a href="https://physics.stackexchange.com/questions/170962/could-we-send-a-man-safely-to-the-moon-in-a-rocket-without-knowledge-of-general">Could we send a man safely to the Moon in a rocket without knowledge of general relativity?</a> on the Physics SE)</p>

<p>Note that there is a minimum in the potential energy curve, and this minimum gives the radius of the stable orbit. Note also that because it's a minimum if we displace the object away from the minimum it will fall back towards the minimum again i.e. this is a stable orbit.</p>

<p>Now, for light we cannot simply use Newtonian mechanics because light is massless, but we can do the calculation using general relativity. The details are a bit intimidating, but we end up with an effective potential just as described above. For light the effective potential turns out to be:</p>

<p><span class="math-container">$$ V_\mathrm{eff}(r) = \sqrt{1 - \frac{2GM}{c^2r}}\frac{L}{r} \tag{2} $$</span></p>

<p>And if we graph this it looks like this:</p>

<p><a href="https://i.sstatic.net/gPzqV.gif" rel="noreferrer"><img src="https://i.sstatic.net/gPzqV.gif" alt="Light effective potential"></a></p>

<p>This looks very different from our first graph, and it's different because light is massless and only ever travels at the same speed of <span class="math-container">$c$</span>. The graph of <span class="math-container">$V_\mathrm{eff}$</span> for light has a maximum not a minimum. The maximum corresponds to the position of a circular orbit, just like the minimum in the first graph, and we find the radius of this circular orbit is given by:</p>

<p><span class="math-container">$$ \frac{r}{r_\mathrm s} = 1.5 $$</span></p>

<p>where <span class="math-container">$r_\mathrm s$</span> is the Schwarzschild radius. This radius gives the position of the notorious <a href="https://en.wikipedia.org/wiki/Photon_sphere" rel="noreferrer">photon sphere</a>.</p>

<p>But for light <span class="math-container">$V_\mathrm{eff}$</span> has a maximum not a minimum. That means if we displace the light by even the tiniest distance from this maximum it will lower its potential energy by moving either inwards or outwards. The orbit at <span class="math-container">$1.5r_\mathrm s$</span> is unstable and the tiniest perturbation will cause the light to spiral into the black hole or away from it. This means we cannot accumulate light in the photon sphere as the question asks. Any attempt to put light into this orbit is doomed to failure as even the tiniest perturbation (e.g. from other objects orbiting the black hole, or even from passing gravitational waves) will destabilise the orbit and the light will be lost.</p>

CC BY-SA 4.0
2019-02-27T15:17:46.540Z
@PM 2Ring left a comment on the stable photon orbit  in the Kerr black hole. What do you think?
Muze
CC BY-SA 4.0
2019-02-27T15:28:07.267Z
@Muze which comment? Do you mean *In case it isn't obvious, we can't see the photons that are in the photon sphere, only those that happen to spiral out of it & head our way.*
John Rennie
CC BY-SA 4.0
2019-02-27T15:32:58.717Z
https://duetosymmetry.com/tool/kerr-circular-photon-orbits/
Muze
CC BY-SA 4.0
2019-02-27T15:35:59.777Z
@Muze that describes a rotating black hole, i.e. the Kerr geometry. That gets mind bendingly complicated because the rotation of the hole sweeps the light sideways as it orbits. However the photon sphere is still unstable so you'd never be able to build up any accumulation of light there.
John Rennie
CC BY-SA 4.0
2019-02-27T15:38:14.753Z
The post suggests a non circular orbit and the photons released may create a thicker photon sphere by the observer concealing the black hole.horizon.
Muze
CC BY-SA 4.0
2019-02-27T15:41:27.387Z
@Muze the photon sphere in a Kerr black hole is actually an oblate ellipsoid not a sphere, but it is still a 2D surface and orbits in it are still unstable. In fact [this was discussed on the physcs SE](https://physics.stackexchange.com/questions/181288/why-is-the-photon-sphere-around-a-kerr-black-hole-spherical-and-not-ellipsoid) though the answer is rather mathematical.
John Rennie
CC BY-SA 4.0
2019-02-27T15:42:29.393Z
I understand. It would be even more unstable. Thanks
Muze
CC BY-SA 4.0
2019-02-27T23:35:45.233Z
@JohnRennie great answer!
Max0815
CC BY-SA 4.0
2021-04-10T20:20:47.087Z
Do the photons on such orbits move in the same direction as black hole's spin? If a photon shifts inward from such orbit, how many revolution it still can do before being swallowed by the event horizon?
Anixx
CC BY-SA 4.0
2019-02-27T09:46:08.047Z
2019-02-27T10:42:42.380Z
2019-02-27T10:42:42.380Z
John Rennie
<p>Can a black hole free from an accretion disk collect photons in the photon sphere in a prolonged stable orbit?</p>

<p>Could enough photons be in orbit to shade the actual black hole with light?</p>

<p>Would a rogue black hole appear to have an atmosphere made of photons from the light emitted from other distant stars entrapped and released?</p>

CC BY-SA 4.0
2019-02-16T23:05:48.373Z
See https://en.wikipedia.org/wiki/Photon_sphere
PM 2Ring
CC BY-SA 4.0
2019-02-26T07:23:44.117Z
@PM2Ring The question is more tricky as it seems. The orbit of such photons are unstable. To calculate their mean or typical stability requires the combined usage of quantum mechanics and general relativity.
peterh
CC BY-SA 4.0
2019-02-26T07:51:07.523Z
@peterh Yes, it's tricky. FWIW, I linked that article before the question was edited to mention the photon sphere.
PM 2Ring
CC BY-SA 4.0
2019-02-26T08:23:56.040Z
I very badly want to ignore GR and QM and just guess the answers to OP's query are: no, no, and no.
Florin Andrei
CC BY-SA 4.0
2019-02-26T08:43:07.953Z
@FlorinAndrei Afaik "no, no and no" would deserve a NAA flag for a question asking for "how long". I would suggest to dig the https://physics.stackexchange.com for related questions, I think you essentially misunderstood the problem.
peterh
CC BY-SA 4.0
2019-02-26T09:11:18.190Z
I very wildly guess @FlorinAndrei is right simply because we haven't seen anything like this before. If it's possible at all, it's probably very unlikely.
Everyday Astronaut
CC BY-SA 4.0
2019-02-26T09:24:07.580Z
@rehctawrats Science works on a way, that we have models. QM and GR are particularly well-working models. Questions asking about what follows from our models, are considered objective answerable even if we didn't see them experimentally.
peterh
CC BY-SA 4.0
2019-02-26T15:14:24.310Z
It should be possible to do the math based on how close the photon (defining it as a point particle) would be to a theoretical perfect orbit around the photon sphere and work it out from there.   I'd guess that just 1 atomic radius off and the photon would spiral away or in fairly quickly. A few dozen orbits perhaps.  There's also no such thing as a perfectly lined up orbit and presumably other objects would create gravitational ripples in space time too.    So stability is impossible for two reasons.   Number of orbits could be estimated based defining the parameters and lots of math.
userLTK
CC BY-SA 4.0
2019-02-26T17:56:16.217Z
@userLTK Yes. I think a more exact calculation would use quantum field theory-based heuristics, calculating the uncertainity of the position of the photon. Here doesn't the atom size matter, but the wavelength of the photon, so probably also the result would depend on its wavelength. I suspect, photons with bigger energy, thus with smaller wavelength, would remain longer on the equilibrical orbit.
peterh
CC BY-SA 4.0
2019-02-27T03:25:05.930Z
In case it isn't obvious, we can't see the photons that are *in* the photon sphere, only those that happen to spiral out of it & head our way.
PM 2Ring
CC BY-SA 4.0
2019-02-27T07:22:06.000Z
@peterh no doubt this is a good, scientific question. Just gave my best guess.
Everyday Astronaut
CC BY-SA 4.0
2019-02-27T08:07:56.973Z
[Here](https://duetosymmetry.com/tool/kerr-circular-photon-orbits/) is an app that shows photon orbits around a Kerr black hole.
PM 2Ring
<orbit><black-hole><photons>
Can a photon have a stable orbit around a black hole?
Muze
# Can a photon have a stable orbit around a black hole?

(Asked by: username_0 on 2019-02-16)

Can a black hole free from an accretion disk collect photons in the photon sphere in a prolonged stable orbit?

Could enough photons be in orbit to shade the actual black hole with light?

Would a rogue black hole appear to have an atmosphere made of photons from the light emitted from other distant stars entrapped and released?

**Tags:** orbit, black-hole, photons
> **Comments:**
> **username_1**: See https://en.wikipedia.org/wiki/Photon_sphere  
> **username_2**: @PM2Ring The question is more tricky as it seems. The orbit of such photons are unstable. To calculate their mean or typical stability requires the combined usage of quantum mechanics and general relativity.  
> **username_1**: @peterh Yes, it's tricky. FWIW, I linked that article before the question was edited to mention the photon sphere.  
> **username_3**: I very badly want to ignore GR and QM and just guess the answers to OP's query are: no, no, and no.  
> **username_2**: @FlorinAndrei Afaik "no, no and no" would deserve a NAA flag for a question asking for "how long". I would suggest to dig the https://physics.stackexchange.com for related questions, I think you essentially misunderstood the problem.  
> **username_4**: I very wildly guess @FlorinAndrei is right simply because we haven't seen anything like this before. If it's possible at all, it's probably very unlikely.  
> **username_2**: @rehctawrats Science works on a way, that we have models. QM and GR are particularly well-working models. Questions asking about what follows from our models, are considered objective answerable even if we didn't see them experimentally.  
> **username_5**: It should be possible to do the math based on how close the photon (defining it as a point particle) would be to a theoretical perfect orbit around the photon sphere and work it out from there.   I'd guess that just 1 atomic radius off and the photon would spiral away or in fairly quickly. A few dozen orbits perhaps.  There's also no such thing as a perfectly lined up orbit and presumably other objects would create gravitational ripples in space time too.    So stability is impossible for two reasons.   Number of orbits could be estimated based defining the parameters and lots of math.  
> **username_2**: @userLTK Yes. I think a more exact calculation would use quantum field theory-based heuristics, calculating the uncertainity of the position of the photon. Here doesn't the atom size matter, but the wavelength of the photon, so probably also the result would depend on its wavelength. I suspect, photons with bigger energy, thus with smaller wavelength, would remain longer on the equilibrical orbit.  
> **username_1**: In case it isn't obvious, we can't see the photons that are *in* the photon sphere, only those that happen to spiral out of it & head our way.  
> **username_4**: @peterh no doubt this is a good, scientific question. Just gave my best guess.  
> **username_1**: [Here](https://duetosymmetry.com/tool/kerr-circular-photon-orbits/) is an app that shows photon orbits around a Kerr black hole.  

---

## 1 answers

### Answer by username_6 on 2019-02-27. Score: 13 | Accepted answer

For any massive object the gravitational potential energy is given by Newton's law:

$$ V(r) = -\frac{GMm}{r} $$

The gravitational potential energy is due to the attractive gravitational force, but for an orbiting object there is also a (fictitious) centrifugal force pushing it outwards. If we calculate the potential energy due to the centrifugal force and add it to the gravitational potential energy we get an effective potential energy:

$$ V\_\mathrm{eff}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2} \tag{1} $$

where $L$ is the angular momentum, which is a constant for an orbiting object (because angular momentum is conserved in a central field). If we plot this graph for the Earth-Moon system we get a graph like this:

[![Effective potential](https://i.sstatic.net/CLzP3.gif)](https://i.sstatic.net/CLzP3.gif)

(this comes from my answer to the question [Could we send a man safely to the Moon in a rocket without knowledge of general relativity?](https://physics.stackexchange.com/questions/170962/could-we-send-a-man-safely-to-the-moon-in-a-rocket-without-knowledge-of-general) on the Physics SE)

Note that there is a minimum in the potential energy curve, and this minimum gives the radius of the stable orbit. Note also that because it's a minimum if we displace the object away from the minimum it will fall back towards the minimum again i.e. this is a stable orbit.

Now, for light we cannot simply use Newtonian mechanics because light is massless, but we can do the calculation using general relativity. The details are a bit intimidating, but we end up with an effective potential just as described above. For light the effective potential turns out to be:

$$ V\_\mathrm{eff}(r) = \sqrt{1 - \frac{2GM}{c^2r}}\frac{L}{r} \tag{2} $$

And if we graph this it looks like this:

[![Light effective potential](https://i.sstatic.net/gPzqV.gif)](https://i.sstatic.net/gPzqV.gif)

This looks very different from our first graph, and it's different because light is massless and only ever travels at the same speed of $c$. The graph of $V\_\mathrm{eff}$ for light has a maximum not a minimum. The maximum corresponds to the position of a circular orbit, just like the minimum in the first graph, and we find the radius of this circular orbit is given by:

$$ \frac{r}{r\_\mathrm s} = 1.5 $$

where $r\_\mathrm s$ is the Schwarzschild radius. This radius gives the position of the notorious [photon sphere](https://en.wikipedia.org/wiki/Photon_sphere).

But for light $V\_\mathrm{eff}$ has a maximum not a minimum. That means if we displace the light by even the tiniest distance from this maximum it will lower its potential energy by moving either inwards or outwards. The orbit at $1.5r\_\mathrm s$ is unstable and the tiniest perturbation will cause the light to spiral into the black hole or away from it. This means we cannot accumulate light in the photon sphere as the question asks. Any attempt to put light into this orbit is doomed to failure as even the tiniest perturbation (e.g. from other objects orbiting the black hole, or even from passing gravitational waves) will destabilise the orbit and the light will be lost.

> **Comments:**
> **username_0**: @PM 2Ring left a comment on the stable photon orbit  in the Kerr black hole. What do you think?  
> **username_6**: @Muze which comment? Do you mean *In case it isn't obvious, we can't see the photons that are in the photon sphere, only those that happen to spiral out of it & head our way.*  
> **username_0**: https://duetosymmetry.com/tool/kerr-circular-photon-orbits/  
> **username_6**: @Muze that describes a rotating black hole, i.e. the Kerr geometry. That gets mind bendingly complicated because the rotation of the hole sweeps the light sideways as it orbits. However the photon sphere is still unstable so you'd never be able to build up any accumulation of light there.  
> **username_0**: The post suggests a non circular orbit and the photons released may create a thicker photon sphere by the observer concealing the black hole.horizon.  
> **username_6**: @Muze the photon sphere in a Kerr black hole is actually an oblate ellipsoid not a sphere, but it is still a 2D surface and orbits in it are still unstable. In fact [this was discussed on the physcs SE](https://physics.stackexchange.com/questions/181288/why-is-the-photon-sphere-around-a-kerr-black-hole-spherical-and-not-ellipsoid) though the answer is rather mathematical.  
> **username_0**: I understand. It would be even more unstable. Thanks  
> **username_7**: @JohnRennie great answer!  
> **username_8**: Do the photons on such orbits move in the same direction as black hole's spin? If a photon shifts inward from such orbit, how many revolution it still can do before being swallowed by the event horizon?  

---

