It is not clear to me if we can observe matter infalling to a black hole. When it is stated that time stretch to infinite for an external observer, then we should see a kind of accretion or jam at the very border. Like two compenetrating objects at the same place. I hope this is naive.

To be precise: think of the crash test mannequin that divulgative science often send to a black hole. Ignore spaghettification. Do we really see that mannequin "frozen" and never passes throughout the horizon?

If so, what happen to a second mannequin if it happens to it to fall at the same spot?

Or there is no the same spot as for the infalling matter has already shifted the horizon?

May be the mannequin is not the best testing particle. Feel free to replace mannequin with atoms or particles and so forth. I have been pictorial for sake of clarity.

Edit: the opening sentence refers to direct observation. Not to the observing of radiation due to the acceleration towards the black hole like that emitted by a hot accretion disk.

  • $\begingroup$ No external observer ever sees anything pass the horizon. This must be a duplicate! $\endgroup$ – tfb Dec 22 '17 at 11:41
  • $\begingroup$ @Alfred Centauri. Yes, indeed. Likely everything get stuffed at the horizon. ..Thanks for the warning as I was already cached by the explanation. But indeed it is about my curiosity $\endgroup$ – Alchimista Dec 22 '17 at 15:42

As measured from an external observer, the time dilation approaches infinity as the matter comes closer and closer to the event horizon. You will see a progressively slow motion having as limit the event horizon. Practically you will see the first mannequin freezing on the event horizon. If the spot is the same, the second mannequin will follow and apparently they will overlap. In any case, from an external observer perspective, the matter will never shift the event horizon.

  • $\begingroup$ Shouldn't in principle the horizon come closer to us after the first mannequin went inside? Just a bit ;) $\endgroup$ – Alchimista Dec 22 '17 at 13:50
  • $\begingroup$ Yes, but then the static Schwarzschild solution that we were using to do the analysis ceases to be valid! $\endgroup$ – m4r35n357 Dec 22 '17 at 17:59

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