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Imagine we have three observers, A, B and C, all standing on a platform hovering just outside the event horizon of an ideal Schwarzschild black hole$^1$. At time $t=0$, as measured from the platform, A jumps off the platform radially in toward the black hole. At time $t=1$, as measured from the platform, B jumps in after A.

My general question is, How does B observe A's trajectory and how does A observe B's, specifically thinking about when each observes themself and the other crossing the event horizon? I've only taken a half semester of GR, years ago so I'm not sure how to even approach this mathematically. However, I am doing my PhD in condensed matter physics so a mathematically rigorous answer should be fine for me to parse.

My (likely silly and uninformed) thoughts on the matter:

I remember learning that C will never see A or B cross the event horizon. I believe C will see A and B slowly converge on each other approaching the event horizon while being red-shifted to oblivion. Please correct me if this is wrong.

Extending the intuition about what C observes, I would guess that B cannot observe A crossing the horizon until B also crosses. Otherwise an outside observer would observe an infalling object cross the event horizon in finite time. I am very suspicious about this conclusion though since 1) B is accelerating w.r.t. A and 2) this suggests B catches up to A. In A's own reference frame, A should cross the event horizon in finite time (perhaps fast enough that in A's frame, A hasn't even observed B to have jumped yet) and so B should not be able to catch up to A. So what is it that A and B actually observe? After A crosses the horizon (in A's frame), when if ever will A observe B also enter the interior of the black hole?

Any intuition, math or explanation about this would be much appreciated!

$^1$ Assume whatever is necessary for this to be reasonable e.g. super powerful rockets to keep the platform stationary, a massive enough black hole where tidal forces / Hawking radiation is survivable, ...etc.

bRost03
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    In GR, there is no global reference frame associated to an observer like there is in SR. So rather than asking "where is B in A's reference frame" (a question that is only meaningful in flat spacetime / Galilean coordinates), you should be asking "what does B see, that is, what part of A's trajectory does the light that B registers correspond to". That problem is essentially relativistic ray tracing. In your situation, it is not too hard to do the corresponding math rigorously. You have the Schwarzschild metric, so all you need to do is to write down the geodesics for A, B and the light ray. – Prof. Legolasov May 14 '21 at 14:55
  • @Prof.Legolasov thanks for the comment. My understanding is the metric is different inside and outside the event horizon, does that not complicate things? Might you be able to suggest a reference that works though a simple problem which I could build from? I haven't done any GR in almost 10 years and even then it was very introductory. Thanks! – bRost03 May 14 '21 at 14:58
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    @safesphere I assume you're thinking of the fact that in the usual Schwarzschild coordinates, the $t$-direction is timelike outside the horizon and spacelike inside the horizon (and conversely for the $r$-direction). But A and B are both described by worldlines, so what do you mean by "They are xxx separated"? Can you clarify which pairs of events along their worldlines you are calling "They"? (Even if A and B are intertial and at relative rest in flat spacetime, some events on A's and B's worldlines are timelike separated, some are lightlike separated, and some are spacelike separated.) – Chiral Anomaly May 14 '21 at 15:28
  • @safesphere thank you for your responses. I am trying to process what you have written. "Since they are lightlike separated at the horizon, A sees B crossing at the same time" - that's very interesting. So your saying it doesn't matter how long B waits to jump (as measured in the platform's reference frame), A will always observe himself and B to cross the event horizon at the same time? Since they both jumped radially, wouldn't this mean they actually hit each other upon crossing? – bRost03 May 14 '21 at 15:34
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    For a qualitative intuition in problems like this it helps to think in terms of Kruskal-Szekeres coordinates (see my answer here), where all light rays are at 45 degree angles and all timelike worldlines are closer to vertical than 45 degrees, so the light cone structure is easy to see. In this case the event horizon behaves like a future light cone, so think about what would happen if two observers A & B crossed a future light cone in sequence in flat spacetime--B can see A the whole time, but B won't see A cross until the moment B crosses. – Hypnosifl May 14 '21 at 15:38
  • No, the horizon is not a “place” where things meet. - I think this is getting at my confusion. Is the location of the horizon reference frame dependent? Or are you saying something stronger, like "location of the horizon" is nonsense itself? I guess in mind mind I've been picturing the horizon as a spherical shell in space which is constant in time. What's the right model for thinking about the horizon? – bRost03 May 14 '21 at 15:58
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    To add to @Hypnosifi's comment: the horizon is a lightlike hypersurface. That's why B won't see A cross until the moment B crosses. It's just a special case of the more general statement that B won't see something that A did until the light from that event reaches B's eyes. That's true whether spacetime is flat or curved, and whether or not an event horizon is involved. B never catches up to A, but B does catch up to the light that A emitted earlier. That's what B encounters upon crossing the horizon: some of the light that A emitted when A crossed the horizon. – Chiral Anomaly May 14 '21 at 16:15
  • @ChiralAnomaly that's very helpful! Thank you. I think I understand what B sees. The only remaining question is what does A see? – bRost03 May 14 '21 at 16:22
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    @bRost03 Here's how you can figure it out. A and B's worldlines are radially ingoing timelike geodesics. Use outgoing and ingoing radial lightlike geodesics to model light emitted by A and B, respectively. For any ingoing lightlike radial geodesic that connects an event $b$ on B's worldline to an event $a$ on A's worldline, that tells you that event $a$ is when A sees the light that was emitted at $b$. But don't use Schwarzschild coordinates! Use a coordinate system that is nonsingular on the horizon. One nice choice is Kerr-Schild coordinates. – Chiral Anomaly May 14 '21 at 16:53
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    @bRost03 The linked answer is for an extremal charged black hole, but if you set $V(r)=R/r$ then you get the uncharged Schwarzschild black hole (in Kerr-Schild coordinates) with Schwarzschild radius $R$. – Chiral Anomaly May 14 '21 at 17:00
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    @safesphere - I didn't say that either of them cross the future light cone of an event on their own worldline, just that they both cross "a future light cone"--it would have to be the future light cone of an event E that did not lie on the worldline of A or B. – Hypnosifl May 14 '21 at 19:54
  • @safesphere - Sure, I wasn't suggesting otherwise. The point is just that from a local perspective, two observers crossing the event horizon in succession is no more exotic than this type of situation. – Hypnosifl May 15 '21 at 15:49

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