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NASA recently published a video simulation of what it would look like flying into a black hole (video can be seen here ).

What confuses me is that it appears to show the photon ring around the black hole as being visible from a distance. How is that possible?

"Seeing" a star or any visible light from space means that the corresponding star photon has travelled through space and reached the eye retina. However, I thought the photon ring around the black hole consists of photons that have been caught by the massive gravitational pull of the black hole and therefore cannot escape: rather, they keep rotating around the hole. Therefore, they cannot travel "away" and reach the eye retina of a distant observer.

Surely, one would only notice the photon ring when flying through it (i.e. reaching the ring and intersecting it: then the rotating photons could penetrate the eye retina).

Am I wrong in my assumptions?

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It's not the photons caught in the ring that you are seeing in that simulation, but think about all the photons that don't quite get caught by the ring.

They will be significantly diverted around the ring, especially those that come really close to it. This will evidence itself the ring outlined in light, which is what you see. You are not seeing the trapped photons, but those almost trapped.

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    $\begingroup$ Thank you: if I may ask an additional question (since I am still a bit confused): this seems a bit similar to the Saturn ring: those are particles caught by Saturn's gravity. If the sun reflects off these particles and the reflected sun photons cross my eye retina, I see the Saturn ring. But if those diverted photons circle around along those deflected paths, how do they reach my eye retina for me to see them? I thought the shape is the path of those photons, so they never leave? $\endgroup$
    – Jan Stuller
    Commented May 10 at 12:43
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    $\begingroup$ you aren't seeing the trapped photons. This is nothing like Saturn's rings where you see the sun reflected by the ring itself. This is more like the circle of stars you see round any dense stellar body - lensed into a ring $\endgroup$
    – Rory Alsop
    Commented May 10 at 12:46
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    $\begingroup$ Photons can't get "caught in a ring". This is the essence of an answer. $\endgroup$
    – ProfRob
    Commented May 10 at 19:41
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    $\begingroup$ @ProfRob: could you please elaborate in a full answer? Your insight sounds super interesting! $\endgroup$
    – Jan Stuller
    Commented May 10 at 19:59
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I think you may be confusing two distinct concepts (though I can fully understand why).

Photon rings on a black hole are a visual effect, visible from a distance, caused by inbound photons making (approximately) N+1/2 full circles around the black hole before escaping back the way they came. The ring structure is made up of 'subrings' named for the value of N -- thus, the N=0 ring is light that made a U-turn and came right back, the N=1 ring made one full loop and then escaped, N=2 made two full loops, and so on. This would create what is essentially a retroreflective band around the black hole, called the photon ring.

The photon shell (or photon sphere) is the lowest theoretically possible orbit around the black hole, where the orbital velocity exactly equals the speed of light. In theory, a photon that falls precisely on that line will enter an orbit and remain there eternally. Realistically, the gravity field won't be precisely spherical, so there would be bulges that disrupt the shell and allow the photons to escape it.

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  • $\begingroup$ Those rings are pretty close to the photon sphere, though. Even the 180° U-turn trajectory gets fairly close to the photon sphere, and to the distant observer, its impact parameter is very close to the BH "shadow" at the critical impact parameter (the dotted circle in this diagram of the 180° trajectory) i.sstatic.net/LsG0C.png $\endgroup$
    – PM 2Ring
    Commented May 12 at 15:32
  • $\begingroup$ +1 See also ...there is no limit on how many "laps" a hyperbolic orbit can make before it returns to infinity. and references therein. $\endgroup$
    – uhoh
    Commented May 12 at 16:28
  • $\begingroup$ Thank you so much for this answer. If the photons do N + 0.5 circles and then escape back the way they came, what would the observes see though? Would he see an image of the original photon-emitting source, or would they see an image of the photons glowing around the black hole as they make N + 0.5 circles? $\endgroup$
    – Jan Stuller
    Commented May 12 at 19:47
  • $\begingroup$ @Jan have you seen en.wikipedia.org/wiki/File:BlackHole_Lensing.gif & en.wikipedia.org/wiki/File:EnsteinRingZoomOptimised.gif from en.wikipedia.org/wiki/Einstein_ring ? $\endgroup$
    – PM 2Ring
    Commented May 13 at 4:46
  • $\begingroup$ @JanStuller The observer would see a bright ring that varies in brightness depending on how much light is coming from the observer's direction, much like a retroreflective sign on the highway, but like that sign, it's not a mirror. You'd see brightness apparently originating from a ring around the black hole, not an image. $\endgroup$
    – Darth Pseudonym
    Commented May 13 at 12:21
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Here I delve into the involved mechanics to define what is meant by a photon that's almost trapped.

Imagine a photon approaching a black hole from distant space, but not directly in line with the (apparent) center of the hole so that the photon has some angular momentum relative to the hole. If the photon is too closely aligned (not enough angular momentum), the light gets curved into the event horizon and thence oblivion. With just the right amount of misalignment/angular momentum, the light beam does not get pulled into the hole but is continually curved around it, trapped around the photosphere. This beam does not reach the photosphere in finite time (so it also avoids the event horizon), but it approaches the photosphere asymptotically.

The light you are seeing is from when the incoming beam has just a little more than that minimal misalignment/angular momentum to avoid capture into the hole. Such a beam gets curved around the hole several times, producing rings/subrings, but the hole does not have quite enough pull to trap it forever given the incoming angular momentum. So the beam spirals out again to tell the tale to our eyes/telescopes.

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  • $\begingroup$ To drive-by downvote always sucks. Care to stick around and elaborate? $\endgroup$
    – Oscar Lanzi
    Commented May 14 at 9:38

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