The photon sphere is a spherical region in space where photons are forced to travel in an orbit at $r = \frac{3GM}{c^{2}}$.

  1. Is it possible to detect these spheres?

  2. What happens if I fall through it?

  3. What happens to a photon that tangentially hits the event horizon, wouldn't this also lead to a photon orbiting the black hole?

  4. EDIT Since neutron stars also might have a photon sphere, do we know neutron stars that should have such a sphere?

  • $\begingroup$ @FlorinAndrei: Black holes and neutron stars are astronomical objects. I'd say it would be appropriate in either Astronomy or Physics (or Astrophysics if there were such a site). $\endgroup$ Commented Mar 20, 2012 at 20:23
  • $\begingroup$ "...if it does not get any good responses here..." I feel all judged and on the spot now. :) $\endgroup$
    – Andrew
    Commented Mar 21, 2012 at 11:39
  • $\begingroup$ @Andrew: We're watching you... $\endgroup$ Commented Mar 21, 2012 at 21:23
  • $\begingroup$ It'd be difficult to determine #4 due to the uncertainities involved. $\endgroup$ Commented Jul 27, 2023 at 9:07

2 Answers 2

  1. The photon sphere is of theoretical interest only- the photon orbits are unstable [a boulder at the top of a hill, instead of the bottom of a valley], so in the real world there isn't going to be some huge herd of photons to detect or crash through or anything like that.

  2. In the same vein, if you fall through it, you fall through it. Nothing special happens to you. Other than the fact that you are almost certainly going to have a bad experience very soon, if you are not already.

  3. No. The photon is captured if its impact parameter (that is, the distance "of closest approach" to the black hole) is any less than the radius of the photon sphere. That's the definition of the photon sphere. In other words, any photon that approaches from distant space will be captured if its path goes inside the photon sphere at all. However, a photon emitted on a radial trajectory from an object between the event horizon and the photon sphere can escape. A photon emitted by an object inside the event horizon has nowhere to go but towards the singularity.

  4. Neutron stars do not have a photon sphere. The Schwarzschild radius of a neutron star is about 3 km, which means you have to compress it within that size to create a black hole. This means that the photon sphere radius would be about 4.5 km, so to have a photon sphere you would need to cram the neutron star smaller than that. [Warning: Do not try this at home. You will create a supernova, with a black hole remnant.] Neutron stars are actually more like 10-15 km in radius, so they do not have strong enough gravity to do any of that.

  • $\begingroup$ Actually, the photon sphere is the last stable circular orbit for a photon. Any photon traveling tangentially to the photon sphere will continue to orbit the black hole on a circular path. Photons within the photon sphere can escape to outside the photon sphere, if their direction is within some small angle of radially outward. Also, for really soft equations of state, which are very currently thought to be very unlikely, a neutron star could have a photon sphere as its radius would be ~5km. $\endgroup$
    – cm2
    Commented Mar 21, 2012 at 22:12
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    $\begingroup$ Doublechecked with MTW. PS is definitely last stable orbit for inward drifting orbiting rest mass particles (p. 885, IV.D); Wikipedia agrees with me that photons are unstable, but I couldn't find an explicit statement in MTW. Demorest et al. 2010 pretty much killed all possibility of that with their monster neutron star. Looks like the EOS is pretty hard. Hang on, with a really soft EOS, you get a low maximum mass anyway, so your photon sphere radius would be smaller too. I don't think you can win either way. $\endgroup$
    – Andrew
    Commented Mar 22, 2012 at 3:15
  • $\begingroup$ Right, inward drifting being the key; I mentioned photons traveling tangentially to the photon sphere have a stable orbit. RE soft EOS: yes, it is very unlikely to have such a soft EOS. If the neutron star had a soft EOS, it can still have a mass of 1.4 solar masses. Indeed, any EOS that produces a maximum mass less than the known mass of any neutron star is already ruled out. $\endgroup$
    – cm2
    Commented Mar 23, 2012 at 1:00
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    $\begingroup$ Neutron star radii could just get below 1.5 r_s. In fact the paper referenced by @cm2 has a plot which demonstrates this. These would need to be more massive than 2 solar masses and gave a hard EoS. $\endgroup$
    – ProfRob
    Commented Feb 16, 2019 at 23:49
  • $\begingroup$ The Schwarzschild radius of a neutron star is about 3 km, PER solar mass, so a 2 solar mass neutron star would have a Rs of 6km and a photon sphere raidus of 9km. In addition, "The impact parameter is the distance by which a body, if it continued on an unperturbed path, would miss the central body at its closest approach. " $\endgroup$ Commented May 22, 2022 at 10:39

I want to start by saying that the original answer to the question does a great job, but I feel a question like this deserves a more in depth (and perhaps fun) answer.

1. The photon sphere is the radius at which the velocity of a circular orbit is the speed of light. We can no more detect it than we can detect the Moon's radius around Earth; we can measure it, but there is really nothing about it to detect. The photon sphere has no physical substance whatsoever.

2. If you fall through the photon sphere, you're in trouble. Since your velocity cannot exceed the speed of light, you cannot go fast enough at the periapsis of a hypothetical orbit to rely on projectile motion bringing you back out. However, if you had some sort of propulsion system (and I think it'd be kind of silly to approach a black hole without being in a rocket or something), you could safely establish almost any kind of orbit around the hole at less than the speed of light because you could use the rocket to lessen the net centripetal force localized around you.
In other words, if you're planning a trip inside a photon sphere, gas up first.

3. This is a fun question. In pretty much every other situation, the term escape velocity actually means escape speed because the direction of the speed is irrelevant (as long as it isn't pointed at the gravitational body). However, for a black hole, the escape velocity at the Schwarzschild radius is a velocity and its direction vector must be radial to the black hole. The reason for this is due to the massive effects gravity has at that point (no pun intended). For a planet (say Earth), the gravity field has the well-known and measured effect of curving/warping spacetime. But for a black hole, at the event horizon the gravity is so strong that space isn't just curved, it is actually falling into the black hole as well! I'll give that a moment for it to sink in for anyone who hasn't heard that before....
Ok moment's up. Right so at the event horizon we have this, you're falling in and space is falling in. For Earth, the space is curved but relatively stationary. At any direction, as long as you reach escape speed, you're free and clear. But when the space around you is falling into the hole as well, you can just imagine that if you were to travel in any direction other than radially outward at the escape velocity, the spacetime around you would drag you in with it anyway.
So let's go back and answer the question, if a photon were to hit the event horizon tangentially, and if space weren't falling into the hole, then sure as grape soda it would orbit circularly (why grape soda? Because in my opinion, that's a drink that seems pretty sure of itself). But let's take that photon in the circular orbit and then make the fabric of space start falling into the black hole, the photon is still in that circular orbit; it will revisit the same points in space over and over, but since those points in space are falling inward, so too does the photon. This is also why the circular orbit velocity at the photon sphere is equal to the escape velocity of a lower orbit; the photons do not orbit circularly in the classical sense where they revisit the same point in space periodically, rather they are visiting always new points in space while in the process of escaping but are forced to maintain a constant distance from the black hole because of the space around them falling inward.

Aside: This also is why physicists say once your in a black hole there is no out. It's because the fabric of space inside the event horizon is falling toward the singularity so fast that no matter which way you turn, there's the singularity staring you in the face. The event horizon is nature's Vegas; what happens there, stays there.

4. I'm not much of an astronomer, but based on the debate in the comments on whether or not neutron stars can have a photon sphere, I feel I can say with greater than 50% certainty that we don't know any that should have one.

  • $\begingroup$ Do you have a source for the space falling in thing? I haven't heard that before. $\endgroup$ Commented Jul 20, 2015 at 1:33
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    $\begingroup$ @PyRulez No, that's a handy interpretation. It's a convenient way of thinking about it that illustrates the effects but isn't directly represented in the mathematics. I don't recall seeing it in print, only hearing it in conversations $\endgroup$
    – Jim
    Commented Jul 20, 2015 at 13:07
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    $\begingroup$ @PyRulez This is also known as the waterfall model. See jila.colorado.edu/~ajsh/insidebh/waterfall.html $\endgroup$
    – PM 2Ring
    Commented Jul 14, 2022 at 1:05

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