Consider a cloud of particles falling into a black hole. How does the relative velocity between two such particles, one which is already at the event horizon and one that is still some distance away from the event horizon, depend on the distance between them, as considered in the reference frame of the particle that is still some distance away from the event horizon?
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asked Sep 25, 2016 at 4:31
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Based on the fact the nobody even tried to answer this question, I'll venture to answer it myself. I think I understand it well enough at this point to do so, but I'm not perfectly sure.
The relative velocity between a particle approaching the event horizon and one that is at the event horizon in the reference frame of the former is the speed of light. This relative velocity is valid regardless of the separation in time or space between the two particles, provided that we consider this separation in the reference frame of the former particle. In other words, even if the first particle is only a fraction of an atto-second away from crossing the event horizon, the difference in velocity between itself and the particle at the event horizon is the speed of light.
answered Oct 12, 2016 at 4:51
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$\begingroup$ I think you are right with two caveats: it's not really obvious what relative velocity means for separated objects in a strongly curved spacetime, and nothing ever quite reaches the horizon from the perspective of an observer outside it. Both of these can be resolved by suitable precision of terms. $\endgroup$– user107153Commented Oct 12, 2016 at 6:22
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$\begingroup$ Thanks for the comment. I thought I made it clear and unambiguous by repeatedly referring to the reference frame of the particle that falls toward the event horizon. In this sense, it is not an external observer that is stationary with respect to the black hole. Is there a way to make the statement even more precise? $\endgroup$ Commented Oct 12, 2016 at 6:27
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$\begingroup$ I don't think it matters whether the observer is infalling or not: they never see anything cross the horizon until they do. They do see things get asymptotically close to the horizon and, as you say, the recession velocity then goes asymptotically to $c$. $\endgroup$– user107153Commented Oct 12, 2016 at 20:35
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$\begingroup$ The difference between what is seen by an observer falling in and one that is stationary, is that the latter never sees it happen while for the former it happens in finite proper time. $\endgroup$ Commented Oct 13, 2016 at 4:03
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$\begingroup$ Exactly so. But they still don't see anyone cross the horizon before they do. $\endgroup$– user107153Commented Oct 13, 2016 at 7:58