How fast should the outside world seem to an observer falling into a black hole from initially at rest very far away from a Schwarzschild Black Hole? Since Special relativity cancels out the general relativistic effects, the outside world should seem to have the same speed as his, i.e. as per his clock he should observe no time dilation looking outside, before and after falling through the horizon. Am I right?

This question was previously deleted with a link to how falling clocks slow down with respect to remote observers. My question not about how fast outside observers see the falling clocks tick. I asked what a falling clock, or a falling body sees looking outside, i.e. how fast distant clocks seem to him.

  • $\begingroup$ Good question +1. I don't think the GR and SR effects cancel each other exactly. For example, in a radial fall, the falling observer looking exactly back sees the universe twice redshifted when he approaches the horizon, but blueshifted in other directions. See this and link in comments: physics.stackexchange.com/questions/436274/… $\endgroup$
    – safesphere
    Commented Sep 16, 2021 at 14:39

1 Answer 1


Things that move at the same velocity look normal.

Things that move at different velocity look Doppler-shifted.

(definition of same velocity: distance does not change)

(reason for this: gravity field does not exist for a free falling person)

  • $\begingroup$ So, with respect to the black hole, if I sit still at infinite distance, I would have a relative velocity with respect a falling observer. Should he se my clock running slow? $\endgroup$
    – Nayeem1
    Commented Sep 16, 2021 at 5:20
  • $\begingroup$ @Nayeem1 Yes, sure. That's a quite well known fact. My answer is just little bit more general. $\endgroup$
    – stuffu
    Commented Sep 16, 2021 at 5:37
  • $\begingroup$ Well, that's not supposed to happen while observing far away bookkeeper. Think like this: for a falling observer the clock local shell observer (sitting still nearest to him) will appear to tick slower because of the special relativistic effects. But to that shell observer the remote bookkeeper will appear fast because of GR. These two effects should cancel out, and the far away bookkeeper's clock should appear ticking normally for a falling observer. $\endgroup$
    – Nayeem1
    Commented Sep 16, 2021 at 5:57
  • $\begingroup$ @Nayeem1 Hmm. Falling towards a black hole and falling towards the floor of an accelerating rocket are similar experiences. Equivalence principle. There is a slowing down of descending clocks in the black hole case, and a slowing down of descending light too. In the rocket case neither of those exist ... so those two things cancel out each other in the black hole case so that the rocket case does not look different. Does this sound OK? $\endgroup$
    – stuffu
    Commented Sep 16, 2021 at 6:34
  • $\begingroup$ So, a falling observer should see the clock of a far away person sitting at rest with respect to the black hole tick at the same rate as his? $\endgroup$
    – Nayeem1
    Commented Sep 16, 2021 at 6:38

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