<p>There does exist a solution for a massless point particle (&quot;photon&quot;) that asymptotes to the light ring. It has a surprisingly simple closed form.</p>
<p><span class="math-container">$$r =\frac{GM}{c^2}\left(3+\frac{9}{-2+\cosh\phi} \right)  $$</span></p>
<p>with
<span class="math-container">$\cosh^{-1}(2)&lt;\phi&lt;\infty$</span>.</p>
<p>This looks like this:</p>
<p><a href="https://i.sstatic.net/l6335.png" rel="noreferrer"><img src="https://i.sstatic.net/l6335.png" alt="enter image description here" /></a></p>
<p>Of course, as others have noted this solution is not stable. An infinitesimal perturbation of the initial conditions will cause the orbit to either spiral into the black hole or scatter back to infinity after a finite number of orbits around the black hole.</p>
<p>Moreover, this solution exists only because the idealized assumptions that went into it. In reality:</p>
<ul>
<li>A wavepacket of light (&quot;photon&quot;) is not a point particle, but has a finite size. This will cause the wavepacket to quickly lose energy as the parts of the wavepacket not on the perfect solution bend off into the black hole of back to infinity.</li>
<li>A wavepacket has some amount of energy affecting the spacetime curvature. In particular, a point particle traveling around a black hole would generate gravitational waves, further perturbing the orbit and causing the &quot;photon&quot; to lose energy.</li>
</ul>

CC BY-SA 4.0
2022-08-25T19:50:13.713Z
Nice! This should be the accepted answer!
pela
CC BY-SA 4.0
2022-08-25T13:29:42.960Z
2022-08-25T13:29:42.960Z
TimRias
<p>&quot;If approached from a certain angle&quot;? No.</p>
<p>The only circular orbit for a photon in the Schwarzschild metric is when it is emitted perpendicularly to a radial line towards the black hole and at a radial coordinate of <span class="math-container">$3r_s/2$</span>, where <span class="math-container">$r_s$</span> is the Schwarzschild radius. But even this is an unstable orbit. In practice <em>any</em> deviation from an exact tangential direction for the photon or from an emission point of exactly <span class="math-container">$3r_s/2$</span> will result in an exponentially growing deviation that will send the photon into the black hole or off to infinity.</p>
<p>It isn't possible to fire a photon towards a black hole in such a way that it looks like it could have been emitted from this position and in that direction. For all possible impact parameters the light will be headed inwards to some extent if it reaches <span class="math-container">$3r_s/2$</span>.</p>
<p>Pela's time-reversal argument (in comments) doesn't quite work. You can imagine falling into a black hole and emitting light in a tangential direction as you cross <span class="math-container">$3r_s/2$</span> and having the light loop around the black hole a couple of times and shoot off to infinity. One then imagines the light travelling along the directly opposite trajectory and you get an incoming photon that <em>nearly</em> gets captured into a circular orbit at <span class="math-container">$3r_s/2$</span>. However, you have to continue to follow this trajectory back beyond this point and you will find that the photon will either loop around the black hole and then escape again or will loop around and fall into the black hole in a finite (and very short) time. Which it does will depend on what perturbation from <span class="math-container">$r=3r_s/2$</span> and <span class="math-container">$\theta = 90^{\circ}$</span> was responsible for the trajectory.</p>
<p>Here are two screen shots from <a href="https://stuleja.org/grorbits/" rel="noreferrer">grorbits</a> for light arond a Schwarzschild black hole. In the first screenshot I get as close to possible to a tangentially emitted photon at <span class="math-container">$3r_s/2$</span> so that the light loops around the black hole, but then just a numerical instability in the nth decimal place of the computing algorithm means the instability allows it to fly off to infinity. If I then time-reverse that trajectory you see what happens if that photon was following the same trajectory but in the opposite direction. It just loops around the black hole a couple of times and falls in.</p>
<p><a href="https://i.sstatic.net/HzkUM.png" rel="noreferrer"><img src="https://i.sstatic.net/HzkUM.png" alt="Forward time" /></a></p>
<p><em>Forward time</em></p>
<p><a href="https://i.sstatic.net/tXBSG.png" rel="noreferrer"><img src="https://i.sstatic.net/tXBSG.png" alt="Backward time" /></a></p>
<p><em>Same parameters but going backwards in time</em></p>
<p>You could argue that you can get arbitrarily close to the limiting situation but I think the fact remains that any photon arriving from infinity must ultimately end up in the black hole or escape again. What you could do is refine the question to say you want the light to orbit a certain number of times before exiting towards infinity or falling in, to which there would be a reasonably clear answer in terms of constraints on the impact parameter of the light.</p>
<p><strong>NB: Somebody else feel free to develop these arguments for Kerr black holes or photons that arrive from sources at a finite distance!</strong></p>

CC BY-SA 4.0
2022-08-24T19:18:55.460Z
A photon that approaches a Schwarzschild BH with impact parameter $\left(\frac{3\sqrt3}2\right)r_s$ would orbit forever. But of course the orbit is unstable and in reality no photon can stay forever in the photon sphere. I have some details in https://physics.stackexchange.com/a/680961/123208 FWIW, I now have a better photon trajectory program which computes elliptic integals (using a very fast algorithm) with arbitrary precision arithmetic.
PM 2Ring
CC BY-SA 4.0
2022-08-24T19:41:25.330Z
The *Nature* article linked in my answer, [Divergent reflections around the photon sphere of a black hole](https://www.nature.com/articles/s41598-021-93595-w) by Albert Sneppen, has details of photon trajectories for both Schwarzschild & Kerr BHs.
PM 2Ring
CC BY-SA 4.0
2022-08-24T20:53:35.897Z
It's impossible for something to have a property that is exactly equal to an irrational number isn't it? @PM2Ring Isn't another way to look at it : the only photon that can orbit in a circle perfectly is one that has *always* been in that orbit?
ProfRob
CC BY-SA 4.0
2022-08-24T21:41:52.543Z
@ProfRob Why would it be any harder to have a property that is exactly an irrational number than it is to have a property that is a rational number?
TimRias
CC BY-SA 4.0
2022-08-24T21:52:20.310Z
@TimRias agreed. Not really the issue.
ProfRob
CC BY-SA 4.0
2022-08-24T21:59:55.857Z
Anyway, a general property of unstable circular orbits is that there will be a trajectory that asymptotically approaches that orbit. (Or in fact two, one approaching from the inside, the other from the outside.) This is a mathematical necessity. Of course, any deviation from the correct initial conditions will cause this to fail, but that is no different for the circular orbit solution.
TimRias
CC BY-SA 4.0
2022-08-25T03:37:55.480Z
@ProfRob To be honest, that critical impact parameter calculation ignores the spacetime curvature caused by the photon itself, and as I said in the linked answer, it also ignores the rest of the universe. :) Using Sneppen's notation, by making $\delta$ close to zero we can make trajectories that orbit for any number of times before escaping (or before falling in for $\delta<0$), but $|\delta|$ gets absurdly tiny for even modest numbers of loops.
PM 2Ring
CC BY-SA 4.0
2022-08-25T05:17:40.640Z
Eg, here's a [20 loop trajectory](https://i.sstatic.net/mX7lx.png) plotted using [Carlson's symmetric elliptic integral algorithm](https://doi.org/10.48550/arXiv.math/9409227) with 250 bit arithmetic. delta is 4.6321e-55 and the total deflection is ~7200.000278°. Of course, even for a SMBH, that delta is smaller than a Planck length. :) The blue points are the incoming trajectory and are calculated using the elliptic integral (of the 1st kind), the red points are calculated using symmetry. The connecting curves are just cubic Béziers.
PM 2Ring
CC BY-SA 4.0
2022-08-25T05:20:51.127Z
This is my point entirely and why the answer to the question as asked is no. @PM2Ring
ProfRob
CC BY-SA 4.0
2022-08-25T05:25:21.960Z
Oh, I fully agree that in reality an eternal orbit in the photon sphere is impossible.
PM 2Ring
CC BY-SA 4.0
2022-08-24T18:16:37.647Z
2022-08-24T21:56:34.300Z
2022-08-24T21:56:34.300Z
ProfRob
<p>In principle yes, but in practice no.</p>
<p>At the so-called <a href="https://en.wikipedia.org/wiki/Photon_sphere" rel="nofollow noreferrer">photon sphere</a>, gravity is exactly so strong that a photon on a tangential trajectory would stay in orbit. For a non-rotating black hole of mass <span class="math-container">$M$</span>, the radius of the photon sphere is <span class="math-container">$3/2$</span> times the radius of the <em>event horizon</em> — i.e. the &quot;surface&quot; of the black hole — which itself is given by <span class="math-container">$r_s \equiv 2GM/c^2$</span>, where <span class="math-container">$G$</span> and <span class="math-container">$c$</span> are the gravitational constant and the speed of light, respectively.</p>
<p>However, the orbits are unstable; any perturbation will cause a photon on the photon sphere to either escape or plunge into the black hole.</p>
<hr />
<p>Added 25.8.2022: To <em>enter</em> the photon sphere is not possible &quot;from a certain angle&quot;, unless you would have a perfect, unrealistic (albeit physically allowed) setup with a small perturbation close to the photon sphere. You can see this from time reversal: If the photon is really orbiting the BH at the photon sphere, it only leaves if it is perturbed. @ProfRob's answer discusses this better.</p>
<p>Added a few hours later: So it turns out that it <em>is</em> in fact possible, if you consider an asymptotically approaching orbit (still disregarding perturbations). TimRias answer gives the exact solution, and I think their answer should be accepted instead of mine.</p>
<p>Note also that my answer doesn't take into account the fact that BHs in general grow (or shrink on longer timescales), which affects the radius of the photons sphere.</p>

CC BY-SA 3.0
2018-03-20T15:44:45.417Z
I think the problem is that you can't **enter** that orbit from a different trajectory.   Just as an asteroid from very far away (not part of the solar orbiting items) can only pass the sun on a parabolic or hyperbolic path, I submit that a distant photon can't be "trapped" into an elliptical orbit.
Carl Witthoft
CC BY-SA 3.0
2018-03-20T16:50:29.367Z
@CarlWitthoft You could emit the photon from an atom falling toward the BH, exactly at the photon sphere, tangentially hereto. A perturbation may cause the photon to escape. Time reverse, and you have a photon coming from far away, getting trapped in the photon sphere. Obviously, that requires a perfect setup, but it's physically possible.
pela
CC BY-SA 4.0
2022-08-24T17:45:27.850Z
But you can't stop the time reversal process there @pela. Keep followng that trajectory and the photon will fall into the black hole. That is because a photon that loops around a few times and then escapes was emitted not quite tangentially.
ProfRob
CC BY-SA 4.0
2022-08-24T17:59:41.457Z
Actually I think it escapes again.
ProfRob
CC BY-SA 4.0
2022-08-25T07:53:38.557Z
@ProfRob Yes, my comment above is wrong, I see, in that the time reverse is _not_ a photon "coming from far away". The time reversal would be a photon orbiting the photon sphere "forever" until at some point it's trapped by an atom. So, since the question is specifically about "coming from a certain angle", I suppose the answer should be "No, but theoretically you could have a photon orbiting (until it is perturbed, or the BH shrinks or grows)".
pela
CC BY-SA 4.0
2022-08-25T08:00:04.467Z
@ProfRob Ah, I only now see that you gave a more correct answer than mine!
pela
CC BY-SA 4.0
2022-08-25T13:38:37.147Z
@pela Your answer at this point is more correct than Rob's.
TimRias
CC BY-SA 4.0
2018-03-19T21:40:54.460Z
2022-08-25T19:53:34.840Z
2022-08-25T19:53:34.840Z
pela
<p>Could rays of light be ever trapped in a constant orbit around a black hole if approached from a certain angle ? Like light hitting a glass or liquid with critical angle.</p>

CC BY-SA 3.0
2018-03-19T21:08:51.410Z
Yes, this is referred to as the photon sphere.
Jack Moody
<black-hole><light><photons>
Can light be trapped in orbit?
Demis
# Can light be trapped in orbit?

(Asked by: username_0 on 2018-03-19)

Could rays of light be ever trapped in a constant orbit around a black hole if approached from a certain angle ? Like light hitting a glass or liquid with critical angle.

**Tags:** black-hole, light, photons
> **Comments:**
> **username_1**: Yes, this is referred to as the photon sphere.  

---

## 3 answers

### Answer by username_3 on 2018-03-19. Score: 15 | Accepted answer

In principle yes, but in practice no.

At the so-called [photon sphere](https://en.wikipedia.org/wiki/Photon_sphere), gravity is exactly so strong that a photon on a tangential trajectory would stay in orbit. For a non-rotating black hole of mass $M$, the radius of the photon sphere is $3/2$ times the radius of the *event horizon* — i.e. the "surface" of the black hole — which itself is given by $r\_s \equiv 2GM/c^2$, where $G$ and $c$ are the gravitational constant and the speed of light, respectively.

However, the orbits are unstable; any perturbation will cause a photon on the photon sphere to either escape or plunge into the black hole.

---

Added 25.8.2022: To *enter* the photon sphere is not possible "from a certain angle", unless you would have a perfect, unrealistic (albeit physically allowed) setup with a small perturbation close to the photon sphere. You can see this from time reversal: If the photon is really orbiting the BH at the photon sphere, it only leaves if it is perturbed. @ProfRob's answer discusses this better.

Added a few hours later: So it turns out that it *is* in fact possible, if you consider an asymptotically approaching orbit (still disregarding perturbations). TimRias answer gives the exact solution, and I think their answer should be accepted instead of mine.

Note also that my answer doesn't take into account the fact that BHs in general grow (or shrink on longer timescales), which affects the radius of the photons sphere.

> **Comments:**
> **username_6**: I think the problem is that you can't **enter** that orbit from a different trajectory.   Just as an asteroid from very far away (not part of the solar orbiting items) can only pass the sun on a parabolic or hyperbolic path, I submit that a distant photon can't be "trapped" into an elliptical orbit.  
> **username_3**: @CarlWitthoft You could emit the photon from an atom falling toward the BH, exactly at the photon sphere, tangentially hereto. A perturbation may cause the photon to escape. Time reverse, and you have a photon coming from far away, getting trapped in the photon sphere. Obviously, that requires a perfect setup, but it's physically possible.  
> **username_4**: But you can't stop the time reversal process there @pela. Keep followng that trajectory and the photon will fall into the black hole. That is because a photon that loops around a few times and then escapes was emitted not quite tangentially.  
> **username_4**: Actually I think it escapes again.  
> **username_3**: @ProfRob Yes, my comment above is wrong, I see, in that the time reverse is _not_ a photon "coming from far away". The time reversal would be a photon orbiting the photon sphere "forever" until at some point it's trapped by an atom. So, since the question is specifically about "coming from a certain angle", I suppose the answer should be "No, but theoretically you could have a photon orbiting (until it is perturbed, or the BH shrinks or grows)".  
> **username_3**: @ProfRob Ah, I only now see that you gave a more correct answer than mine!  
> **username_2**: @pela Your answer at this point is more correct than Rob's.  

---

### Answer by username_2 on 2022-08-25. Score: 12

There does exist a solution for a massless point particle ("photon") that asymptotes to the light ring. It has a surprisingly simple closed form.

$$r =\frac{GM}{c^2}\left(3+\frac{9}{-2+\cosh\phi} \right) $$

with
$\cosh^{-1}(2)<\phi<\infty$.

This looks like this:

[![enter image description here](https://i.sstatic.net/l6335.png)](https://i.sstatic.net/l6335.png)

Of course, as others have noted this solution is not stable. An infinitesimal perturbation of the initial conditions will cause the orbit to either spiral into the black hole or scatter back to infinity after a finite number of orbits around the black hole.

Moreover, this solution exists only because the idealized assumptions that went into it. In reality:

* A wavepacket of light ("photon") is not a point particle, but has a finite size. This will cause the wavepacket to quickly lose energy as the parts of the wavepacket not on the perfect solution bend off into the black hole of back to infinity.
* A wavepacket has some amount of energy affecting the spacetime curvature. In particular, a point particle traveling around a black hole would generate gravitational waves, further perturbing the orbit and causing the "photon" to lose energy.

> **Comments:**
> **username_3**: Nice! This should be the accepted answer!  

---

### Answer by username_4 on 2022-08-24. Score: 6

"If approached from a certain angle"? No.

The only circular orbit for a photon in the Schwarzschild metric is when it is emitted perpendicularly to a radial line towards the black hole and at a radial coordinate of $3r\_s/2$, where $r\_s$ is the Schwarzschild radius. But even this is an unstable orbit. In practice *any* deviation from an exact tangential direction for the photon or from an emission point of exactly $3r\_s/2$ will result in an exponentially growing deviation that will send the photon into the black hole or off to infinity.

It isn't possible to fire a photon towards a black hole in such a way that it looks like it could have been emitted from this position and in that direction. For all possible impact parameters the light will be headed inwards to some extent if it reaches $3r\_s/2$.

Pela's time-reversal argument (in comments) doesn't quite work. You can imagine falling into a black hole and emitting light in a tangential direction as you cross $3r\_s/2$ and having the light loop around the black hole a couple of times and shoot off to infinity. One then imagines the light travelling along the directly opposite trajectory and you get an incoming photon that *nearly* gets captured into a circular orbit at $3r\_s/2$. However, you have to continue to follow this trajectory back beyond this point and you will find that the photon will either loop around the black hole and then escape again or will loop around and fall into the black hole in a finite (and very short) time. Which it does will depend on what perturbation from $r=3r\_s/2$ and $\theta = 90^{\circ}$ was responsible for the trajectory.

Here are two screen shots from [grorbits](https://stuleja.org/grorbits/) for light arond a Schwarzschild black hole. In the first screenshot I get as close to possible to a tangentially emitted photon at $3r\_s/2$ so that the light loops around the black hole, but then just a numerical instability in the nth decimal place of the computing algorithm means the instability allows it to fly off to infinity. If I then time-reverse that trajectory you see what happens if that photon was following the same trajectory but in the opposite direction. It just loops around the black hole a couple of times and falls in.

[![Forward time](https://i.sstatic.net/HzkUM.png)](https://i.sstatic.net/HzkUM.png)

*Forward time*

[![Backward time](https://i.sstatic.net/tXBSG.png)](https://i.sstatic.net/tXBSG.png)

*Same parameters but going backwards in time*

You could argue that you can get arbitrarily close to the limiting situation but I think the fact remains that any photon arriving from infinity must ultimately end up in the black hole or escape again. What you could do is refine the question to say you want the light to orbit a certain number of times before exiting towards infinity or falling in, to which there would be a reasonably clear answer in terms of constraints on the impact parameter of the light.

**NB: Somebody else feel free to develop these arguments for Kerr black holes or photons that arrive from sources at a finite distance!**

> **Comments:**
> **username_5**: A photon that approaches a Schwarzschild BH with impact parameter $\left(\frac{3\sqrt3}2\right)r_s$ would orbit forever. But of course the orbit is unstable and in reality no photon can stay forever in the photon sphere. I have some details in https://physics.stackexchange.com/a/680961/123208 FWIW, I now have a better photon trajectory program which computes elliptic integals (using a very fast algorithm) with arbitrary precision arithmetic.  
> **username_5**: The *Nature* article linked in my answer, [Divergent reflections around the photon sphere of a black hole](https://www.nature.com/articles/s41598-021-93595-w) by Albert Sneppen, has details of photon trajectories for both Schwarzschild & Kerr BHs.  
> **username_4**: It's impossible for something to have a property that is exactly equal to an irrational number isn't it? @PM2Ring Isn't another way to look at it : the only photon that can orbit in a circle perfectly is one that has *always* been in that orbit?  
> **username_2**: @ProfRob Why would it be any harder to have a property that is exactly an irrational number than it is to have a property that is a rational number?  
> **username_4**: @TimRias agreed. Not really the issue.  
> **username_2**: Anyway, a general property of unstable circular orbits is that there will be a trajectory that asymptotically approaches that orbit. (Or in fact two, one approaching from the inside, the other from the outside.) This is a mathematical necessity. Of course, any deviation from the correct initial conditions will cause this to fail, but that is no different for the circular orbit solution.  
> **username_5**: @ProfRob To be honest, that critical impact parameter calculation ignores the spacetime curvature caused by the photon itself, and as I said in the linked answer, it also ignores the rest of the universe. :) Using Sneppen's notation, by making $\delta$ close to zero we can make trajectories that orbit for any number of times before escaping (or before falling in for $\delta<0$), but $|\delta|$ gets absurdly tiny for even modest numbers of loops.  
> **username_5**: Eg, here's a [20 loop trajectory](https://i.sstatic.net/mX7lx.png) plotted using [Carlson's symmetric elliptic integral algorithm](https://doi.org/10.48550/arXiv.math/9409227) with 250 bit arithmetic. delta is 4.6321e-55 and the total deflection is ~7200.000278°. Of course, even for a SMBH, that delta is smaller than a Planck length. :) The blue points are the incoming trajectory and are calculated using the elliptic integral (of the 1st kind), the red points are calculated using symmetry. The connecting curves are just cubic Béziers.  
> **username_4**: This is my point entirely and why the answer to the question as asked is no. @PM2Ring  
> **username_5**: Oh, I fully agree that in reality an eternal orbit in the photon sphere is impossible.  

---

