We know the escape velocity from the event horizon (EH) $= c$. Anything that crosses the EH is destined to reach the singularity. Whether the escape velocity (beyond the EH) can be calculated, or not, can we say that something falling into a BH, would move faster than $c$ after crossing the event horizon? Or, does the acceleration ends at EH? Why? This is when something is free-falling into a BH, ignoring resistance. I am using the fact that the speed of (hypothetical) absolute free fall will be same as the escape velocity at any point.
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Would move faster relative to what? Things that are not causally connected with you don't have a measurable velocity. The question is ill posed. – CuriousOne Mar 06 '16 at 00:55
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Relative to its own speed (c) at horizon. It was moving at c, still accelerated, what happens? – kpv Mar 06 '16 at 00:58
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1If you want to measure a velocity, you have to have causal contact, but anything that is past the event horizon has left your light cone and with that the physics building. All you can see of it is a red-shifted echo that keeps fading away until the last photon escapes and that was that. Ghosts don't have velocity. – CuriousOne Mar 06 '16 at 01:08
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Or, Ghosts, can travel FTL :). What would an observer falling in feel being accelerated when already at c. – kpv Mar 06 '16 at 01:15
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Nothing massive moves at c, not even ghosts. The light cones are simply cut off by event horizons. – CuriousOne Mar 06 '16 at 02:23
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You do not exceed or even reach the speed of light when crossing an event horizon. The easiest way to check this is to carry a flashlight with you. When you turn it on the photons emitted from the flashlight will still move away from you. This "flashlight" analogy can be carried out in GR by calculating the trajectory of a photon with the right initial position and velocity.
To reconcile your intuition it is easiest to think in terms of curved spacetime rather than forces. Once you are inside the event horizon the reason you cannot escape is that spacetime is severely warped and all trajectories with speeds less than $c$ move further into the black hole. See wiki for an illustration.
James Rowland
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Escape velocity is a hypothetical cut off, where something with a greater velocity might escape (depending on its direction) but something with less will not.
So if you can't move at speed $c$ or above and someone says that $c$ (or above) is the escape velocity they are just saying you can't escape.
I am using the fact that speed of free fall will be same as escape velocity at any point.
This is not a true fact, don't use it. That would be like using 0=1 to deduce something. Don't do it. Your free fall speed depends on how fast you were going in the past. It could be lower than escape velocity (for example, if you are bound, like an elliptical orbit such as the earth going around the sun) it could exceed the escape velocity if you are unbound (for instance if you loaded up a rocket with lots of fuel and got up to a big speed by burning lots of fuel) or it could be equal (like above, but use less fuel).
Timaeus
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I used the free fall analogy to compare with the escape velocity. Free fall means free fall. But I get your point, there may be no absolute free fall, I will edit. – kpv Mar 06 '16 at 05:11
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@kpv I listed the case of falling faster than escape velocity (build a rocket, use lots of fuel). If someone wants to bring up a speed that involves being at rest at conformal spatial infinity they'd have to say that. And you don't always have a spatial infinity. A bounded cosmological model for instance doesn't. But that precludes an EH too, since that is a global structure that requires reference to an infinitely far away, an EH is not local physics. – Timaeus Mar 06 '16 at 05:16
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The question is about "beyond event horizon". Suppose, it is moving at c-1 as it crosses the EH. But it is still accelerating towards the singularity. What will happen a fraction of second later. Is that scenario defined to be limit of c, or the scenario is undefined? – kpv Mar 06 '16 at 05:18
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@kpv It seems like you are trying to talk about gravity without learning general relativity. Speed relative to what frame? What do you mean accelerating? Is it going on a geodesic? Is it firing rockets? Is it reacting with the black hole? Merging to form a larger one? And what do you mean a second later? The proper time along the world line, or that frame in which it had nonzero velocity? If you fire thrusters you increase your momentum but velocity is always less than $c$ because in a local SR frame $\vec v=c^2\vec p/E$ and $E^2=(mc^2)^2+(\vec p c)^2$ so $E$ is bigger than $pc$. – Timaeus Mar 06 '16 at 05:25
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Yes, the question is not about GR. You may use your GR knowledge to answer it. Before it hits the EH, it is accelerating (or following the geodesic). After it has crossed EH, is the geodesic known to be such that there is no acceleration? Or it is not known beyond EH? If it is known, it seems to be a non-accelerating geodesic path and speed stays < c. If it is not known, then obviously, the question can not be answered by GR. I know GR math will work where it works and speeds can not be > c in those regions. – kpv Mar 06 '16 at 05:32
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@kpv Oh no no no no no. You can't bring up an EH and claim this isn't about general relativity. And you can't say it is accelerating and following a geodesic that's not a mere oxymoron that's an outright contradiction in terms. If you want a Newtonian answer about a dense Newtonian star, then ask that as a separate question, tag it without using 'black hole' or 'event horizon' and explicitly say you want a Newtonian answer, and ask away. It will be utterly boring and unrelated to the real world. – Timaeus Mar 06 '16 at 05:38
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After going back and forth in comments, I guess this is what I should have asked - See if you can answer it here and then I accept the answer. Is GR known to work everywhere, always? If not, where does it break? Beyond EH, or at singularity? – kpv Mar 06 '16 at 05:49