<p>As already explained in <a href="https://astronomy.stackexchange.com/a/29777/16685">John Rennie's answer</a>, the effective potential of the photon has a maximum at the photon sphere (<span class="math-container">$r = 1.5 r_s$</span>). It can be shown that, in Newtonian mechanics, the effective potential of a spherical distribution of matter presents always one minimum and cannot have any maximum. This may be the reason why the orbits of the photon look so unfamiliar to someone accustomed to see classical planetary orbits.</p>
<p>While a minimum in the potential makes it possible to have stable orbits (that in Newtonian mechanics are the familiar ellipses), a maximum allows for unstable orbits, and other more interesting mechanics unseen in the Newtonian counterpart.</p>
<p>When a photon gets close to <span class="math-container">$r=1.5r_s$</span> it starts climbing the potential hill and as it does so it starts circling the black hole closer and closer.</p>
<p>If the photon has enough energy to overcome the potential hill, it will eventually start descending the opposite side and fall into the black hole.</p>
<p>If instead the energy is not sufficient, the photon will not reach the peak of the hill will invert its radial motion and start to get farther away from the black hole, always circling it, until it eventually escapes.</p>
<p>The edge case is when the photon has just the right energy to reach the top, but no more. In this case it will circle the black hole forever, getting closer and closer to <span class="math-container">$r=1.5r_s$</span> but never reaching it in a finite time.</p>
<p><strong>Remark</strong>: the behaviors I have just described are not exclusively a general relativistic effect. It happens every time the effective potential has a maximum. If Newtonian theory allowed maxima in its effective potential (but it doesn't), we would be able to observe planets or asteroids following the orbits just described.</p>
<p>A slightly more physical example comes from electromagnetism. Electromagnetic charges (unlike mass) come with two signs. As a consequence, a spherically symmetric distribution of charges can in principle generate an electrostatic potential with a local maximum in the radial profile. Charged particles in this potential would again behave as described here for the photon.</p>
<p>I would like to end with some visual treats. Gifs of a simulation I did some time ago about the orbits of a photon in the Schwarzschild metric. The yellow lines are the trajectories of some photons near a black hole. The black circle is the event horizon, the two outer circles represent respectively <span class="math-container">$r=1.5r_s$</span> and <span class="math-container">$r=3r_s$</span>. Enjoy</p>
<p><a href="https://i.sstatic.net/oFqeS.gif" rel="noreferrer"><img src="https://i.sstatic.net/oFqeS.gif" alt="lens1" /></a></p>
<p>Here some photons with different impact parameter are scattered by the black hole. Some are captured, some are just slightly deflected.</p>
<p><a href="https://i.sstatic.net/oaRb4.gif" rel="noreferrer"><img src="https://i.sstatic.net/oaRb4.gif" alt="lens2" /></a></p>
<p>Here the range of impact parameters is chosen more closely to the one needed for the photons to overcome the potential energy barrier.</p>
<p><a href="https://i.sstatic.net/uL6et.gif" rel="noreferrer"><img src="https://i.sstatic.net/uL6et.gif" alt="lens3" /></a></p>
<p>Here the range of impact parameters is still narrower. To the point that for a couple of orbits it seems that there is only one photon. But then the instability of the orbit kicks in. Some photons fall into the black hole, others are forced to get away.</p>

CC BY-SA 4.0
2021-08-02T16:06:19.543Z
Excellent GIFs. You may enjoy this recent Nature article, [Divergent reflections around the photon sphere of a black hole](https://www.nature.com/articles/s41598-021-93595-w).
PM 2Ring
CC BY-SA 4.0
2022-08-24T18:29:12.473Z
"The edge case is when the photon has just the right energy to reach the top, but no more. In this case it will circle the black hole forever, getting closer and closer to r=1.5rs but never reaching it in a finite time." is this right?? I think it is an exponentially divergent behaviour. It loops around and either dives in or dives out depending on which side of the line it is.
ProfRob
CC BY-SA 4.0
2022-08-24T21:11:19.053Z
@ProfRob Yes, it is right. Going by memory, if you write the time it takes to reach the top, with exactly the right energy, it is something like $\tau = \int_{r_0}^{r_{max}} \frac{dr}{\sqrt{E^2-V_{eff}(r)}}$ where in $r=r_{max}$ both the denominator and its first derivative vanish, so that the integral diverges. If you write a question about it, I can go look for my notes and find the formal derivation. My textbook was [Ferrari, Gualtieri, Pani](https://www.routledge.com/General-Relativity-and-its-Applications-Black-Holes-Compact-Stars-and/Ferrari-Gualtieri-Pani/p/book/9780367625320)
Prallax
CC BY-SA 4.0
2022-08-24T21:22:17.663Z
@ProfRob the same happens when you throw a ball on a _spherical_ dome, with exactly the right energy to reach the top, but no more. Formally, it takes an infinite amount of time to reach the top. If it took a finite time, it would break time inversion symmetry: under time inversion, you would see a ball that sits on the top of the dome for some time and then spontaneously choses to fall on a random direction, which is not a solution of the equation of motion (unless it is a [Norton's dome](https://en.m.wikipedia.org/wiki/Norton's_dome), not a spherical one)
Prallax
CC BY-SA 4.0
2021-08-02T15:49:51.333Z
2021-08-02T15:49:51.333Z
Prallax
<p>The <a href="https://www.ru.nl/astrophysics/black-hole/black-holes/what-does-black-hole-look-like/" rel="nofollow noreferrer">shadow</a> that is cast by a black hole was beautifully captured by the event horizon telescope. A concentration of light in the <a href="https://en.m.wikipedia.org/wiki/Photon_sphere" rel="nofollow noreferrer">photon sphere</a> around the hole (adjacent to the event horizon) gravitationally &quot;lenses&quot; the image seen by the telescope (it looks as if we look from the top at a accretion disc but this image is seen from every direction, more or less).</p>
<p>Photons involved in casting this shadow can first circle the hole and then continue their travel. This seems counterintuitive. If a mass meets another mass in space and if they have orbited one another a full 360 degrees, then they will continue to orbit. It's not that when they return where they started that they drift of in space again.</p>
<p>Is it just general relativity at work here, or do I see things wrongly somehow? Can light just make a whole hole orbit first and then continue? In the framework of GR?</p>
<p>What I mean is: the path of such a photon is drawn as an incoming line, after wich the line continues around the hole, intersects itself, and goes on into space. At the point of intersection the photon appears to have two different momenta. Incoming and outgoing. How come?</p>
<p>I mean this kind of drawings:</p>
<p><a href="https://i.sstatic.net/YDzeI.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YDzeI.png" alt="enter image description here" /></a></p>
<p>You would expect that if the photon trajectories approach the hole they would evolve from open to ones circling or spiraling down. Like when different trajectories of a mass approaching the sun <em>or</em> end up at infinity again <em>or</em> spiraling  towards the sun. They will spiral towards the sun only if the mass looses kinetic energy by means of grsvitational waves. The trajectory of such a mass is <em>or</em> closed <em>or</em> open. Not a combination. How is this different for photons approaching a black hole? Gets it impulse from the rotation of the hole? What if the hole is not rotating? This is highly unlikely though. Every object in the universe rotates. Even the universe itself maybe.</p>

CC BY-SA 4.0
2021-08-02T06:09:55.450Z
Photon orbits are unstable. See John Rennie's excellent answer: https://astronomy.stackexchange.com/a/29777/16685
PM 2Ring
CC BY-SA 4.0
2021-08-02T06:45:33.333Z
@PM2Ring I knew there is no stabe orbit (well exactly one). I dont see how this implies a motion for which the photon enters the hole, make a rotation and then ends up with a different momentum after the rotation. The orbits are drawed as incoming lines, full rotation, outgoing line ( intersecting with incoming line). Hasnt (spatial) momentum changed? There is no temporal momentum.
Deschele Schilder
CC BY-SA 4.0
2021-08-02T07:33:06.047Z
Total momentum is conserved, the BH has momentum too. It's like a hyperbolic trajectory in Newtonian gravity, where a fast body comes in at some angle and leaves at a different angle.
PM 2Ring
CC BY-SA 4.0
2021-08-02T08:22:45.070Z
@PM2Ring But the hyperbole is not closed. Is a part of the photons momentum transmitted to the hole?
Deschele Schilder
CC BY-SA 4.0
2021-08-02T08:36:53.207Z
Yes, the photon exchanges some momentum with the BH. Primarily, it gets blue-shifted (and hence increases in energy & momentum) as it falls towards the BH, and gets red-shifted as it travels away, but its direction gets changed too (unless it's heading directly towards the centre of the BH). Hyperbolic trajectories aren't closed, and neither is a photon's orbit. It either gets deflected or it falls into the BH, possibly doing 1 or more loops in either case.
PM 2Ring
CC BY-SA 4.0
2021-08-02T08:46:18.927Z
@PM2Ring So the curve *isnt* closed and self intersecting? Then what do the drawings mean? Are they just a 2d projection?
Deschele Schilder
CC BY-SA 4.0
2021-08-02T09:25:39.467Z
I can't comment on drawings that I haven't seen. ;) I presume you mean something like [this image](https://i.sstatic.net/ERoqR.gif) from here: https://physics.stackexchange.com/q/520919/123208 There's some good info on photon sphere orbits here: https://physics.stackexchange.com/q/169576/123208
PM 2Ring
CC BY-SA 4.0
2021-08-02T09:27:59.167Z
At the moment you commented I edited.
Deschele Schilder
<black-hole><general-relativity><event-horizon-telescope>
How can light circle a black hole one or more times and then escape again?
Deschele Schilder
# How can light circle a black hole one or more times and then escape again?

(Asked by: username_0 on 2021-08-02)

The [shadow](https://www.ru.nl/astrophysics/black-hole/black-holes/what-does-black-hole-look-like/) that is cast by a black hole was beautifully captured by the event horizon telescope. A concentration of light in the [photon sphere](https://en.m.wikipedia.org/wiki/Photon_sphere) around the hole (adjacent to the event horizon) gravitationally "lenses" the image seen by the telescope (it looks as if we look from the top at a accretion disc but this image is seen from every direction, more or less).

Photons involved in casting this shadow can first circle the hole and then continue their travel. This seems counterintuitive. If a mass meets another mass in space and if they have orbited one another a full 360 degrees, then they will continue to orbit. It's not that when they return where they started that they drift of in space again.

Is it just general relativity at work here, or do I see things wrongly somehow? Can light just make a whole hole orbit first and then continue? In the framework of GR?

What I mean is: the path of such a photon is drawn as an incoming line, after wich the line continues around the hole, intersects itself, and goes on into space. At the point of intersection the photon appears to have two different momenta. Incoming and outgoing. How come?

I mean this kind of drawings:

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

You would expect that if the photon trajectories approach the hole they would evolve from open to ones circling or spiraling down. Like when different trajectories of a mass approaching the sun *or* end up at infinity again *or* spiraling towards the sun. They will spiral towards the sun only if the mass looses kinetic energy by means of grsvitational waves. The trajectory of such a mass is *or* closed *or* open. Not a combination. How is this different for photons approaching a black hole? Gets it impulse from the rotation of the hole? What if the hole is not rotating? This is highly unlikely though. Every object in the universe rotates. Even the universe itself maybe.

**Tags:** black-hole, general-relativity, event-horizon-telescope

> **Comments:**
> **username_1**: Photon orbits are unstable. See John Rennie's excellent answer: https://astronomy.stackexchange.com/a/29777/16685  
> **username_0**: @username_1 I knew there is no stabe orbit (well exactly one). I dont see how this implies a motion for which the photon enters the hole, make a rotation and then ends up with a different momentum after the rotation. The orbits are drawed as incoming lines, full rotation, outgoing line ( intersecting with incoming line). Hasnt (spatial) momentum changed? There is no temporal momentum.  
> **username_1**: Total momentum is conserved, the BH has momentum too. It's like a hyperbolic trajectory in Newtonian gravity, where a fast body comes in at some angle and leaves at a different angle.  
> **username_0**: @username_1 But the hyperbole is not closed. Is a part of the photons momentum transmitted to the hole?  
> **username_1**: Yes, the photon exchanges some momentum with the BH. Primarily, it gets blue-shifted (and hence increases in energy & momentum) as it falls towards the BH, and gets red-shifted as it travels away, but its direction gets changed too (unless it's heading directly towards the centre of the BH). Hyperbolic trajectories aren't closed, and neither is a photon's orbit. It either gets deflected or it falls into the BH, possibly doing 1 or more loops in either case.  
> **username_0**: @username_1 So the curve *isnt* closed and self intersecting? Then what do the drawings mean? Are they just a 2d projection?  
> **username_1**: I can't comment on drawings that I haven't seen. ;) I presume you mean something like [this image](https://i.sstatic.net/ERoqR.gif) from here: https://physics.stackexchange.com/q/520919/123208 There's some good info on photon sphere orbits here: https://physics.stackexchange.com/q/169576/123208  
> **username_0**: At the moment you commented I edited.  

---

## 1 answers

### Answer by username_2 on 2021-08-02. Score: 9

As already explained in [John Rennie's answer](https://astronomy.stackexchange.com/a/29777/16685), the effective potential of the photon has a maximum at the photon sphere ($r = 1.5 r\_s$). It can be shown that, in Newtonian mechanics, the effective potential of a spherical distribution of matter presents always one minimum and cannot have any maximum. This may be the reason why the orbits of the photon look so unfamiliar to someone accustomed to see classical planetary orbits.

While a minimum in the potential makes it possible to have stable orbits (that in Newtonian mechanics are the familiar ellipses), a maximum allows for unstable orbits, and other more interesting mechanics unseen in the Newtonian counterpart.

When a photon gets close to $r=1.5r\_s$ it starts climbing the potential hill and as it does so it starts circling the black hole closer and closer.

If the photon has enough energy to overcome the potential hill, it will eventually start descending the opposite side and fall into the black hole.

If instead the energy is not sufficient, the photon will not reach the peak of the hill will invert its radial motion and start to get farther away from the black hole, always circling it, until it eventually escapes.

The edge case is when the photon has just the right energy to reach the top, but no more. In this case it will circle the black hole forever, getting closer and closer to $r=1.5r\_s$ but never reaching it in a finite time.

**Remark**: the behaviors I have just described are not exclusively a general relativistic effect. It happens every time the effective potential has a maximum. If Newtonian theory allowed maxima in its effective potential (but it doesn't), we would be able to observe planets or asteroids following the orbits just described.

A slightly more physical example comes from electromagnetism. Electromagnetic charges (unlike mass) come with two signs. As a consequence, a spherically symmetric distribution of charges can in principle generate an electrostatic potential with a local maximum in the radial profile. Charged particles in this potential would again behave as described here for the photon.

I would like to end with some visual treats. Gifs of a simulation I did some time ago about the orbits of a photon in the Schwarzschild metric. The yellow lines are the trajectories of some photons near a black hole. The black circle is the event horizon, the two outer circles represent respectively $r=1.5r\_s$ and $r=3r\_s$. Enjoy

[![lens1](https://i.sstatic.net/oFqeS.gif)](https://i.sstatic.net/oFqeS.gif)

Here some photons with different impact parameter are scattered by the black hole. Some are captured, some are just slightly deflected.

[![lens2](https://i.sstatic.net/oaRb4.gif)](https://i.sstatic.net/oaRb4.gif)

Here the range of impact parameters is chosen more closely to the one needed for the photons to overcome the potential energy barrier.

[![lens3](https://i.sstatic.net/uL6et.gif)](https://i.sstatic.net/uL6et.gif)

Here the range of impact parameters is still narrower. To the point that for a couple of orbits it seems that there is only one photon. But then the instability of the orbit kicks in. Some photons fall into the black hole, others are forced to get away.

> **Comments:**
> **username_1**: Excellent GIFs. You may enjoy this recent Nature article, [Divergent reflections around the photon sphere of a black hole](https://www.nature.com/articles/s41598-021-93595-w).  
> **username_3**: "The edge case is when the photon has just the right energy to reach the top, but no more. In this case it will circle the black hole forever, getting closer and closer to r=1.5rs but never reaching it in a finite time." is this right?? I think it is an exponentially divergent behaviour. It loops around and either dives in or dives out depending on which side of the line it is.  
> **username_2**: @username_3 Yes, it is right. Going by memory, if you write the time it takes to reach the top, with exactly the right energy, it is something like $\tau = \int_{r_0}^{r_{max}} \frac{dr}{\sqrt{E^2-V_{eff}(r)}}$ where in $r=r_{max}$ both the denominator and its first derivative vanish, so that the integral diverges. If you write a question about it, I can go look for my notes and find the formal derivation. My textbook was [Ferrari, Gualtieri, Pani](https://www.routledge.com/General-Relativity-and-its-Applications-Black-Holes-Compact-Stars-and/Ferrari-Gualtieri-Pani/p/book/9780367625320)  
> **username_2**: @username_3 the same happens when you throw a ball on a _spherical_ dome, with exactly the right energy to reach the top, but no more. Formally, it takes an infinite amount of time to reach the top. If it took a finite time, it would break time inversion symmetry: under time inversion, you would see a ball that sits on the top of the dome for some time and then spontaneously choses to fall on a random direction, which is not a solution of the equation of motion (unless it is a [Norton's dome](https://en.m.wikipedia.org/wiki/Norton's_dome), not a spherical one)  

---

