1
$\begingroup$

According the Special Relativity, all objects contract their length if moving fast.

As the observer falls into black hole, his relative speed to it increases up to the speed of light. But from his point of view, it is the Black Hole moving, not himself.

As such, the thickness of the Black Hole in the direction of the falling observer should contract.

At the moment of the impact on the event horizon, the speed of the observer will be equal to the speed of light, so the black hole should appear flat. Moreover, the point of impact on the event horizon would councide with the center of the black hole, e.g. singularity. As such, it seems the observer has no time to travel towards singularity after moving past the event horizon, but rather he will hit the horizon and singularity at the same time.

Also, the fact that the black hole shrinks may lead to the blueshift in the apparent Hawking radiation as its wavelength is proportional to the Black Hole dimentions. If this is true, the observer should experience very strong blue-shifted radiation due to dynamic Casimir effect, which may correspond to the firewall idea.

Is my picture correct?

$\endgroup$
  • $\begingroup$ You forget the radial space expansion (space curvature) near a black hole that becomes infinite at the horizon. Thus your picture is incorrect. Also a free falling observer doesn't see the Hawking radiation. $\endgroup$ – safesphere Aug 29 at 15:52
  • $\begingroup$ @Ben Crowell. Spacetime is locally flat, that is the Einstein equivalence principle. An observer crossing the horizon does not notice anything different, apart from the tidal force. However the black hole extention from singularity to the horizon, that is not a local region of spacetime, so special relativity is not applicable. $\endgroup$ – Michele Grosso Aug 29 at 16:20
  • $\begingroup$ @MicheleGrosso: Thanks for the correction, I just typed "locally" when I meant "globally." I've converted the comment into an answer, with the typo fixed. $\endgroup$ – Ben Crowell Sep 2 at 18:05
0
$\begingroup$

Your question is really three different questions.

Lorentz transformations only apply to flat spacetime, and spacetime is not globally flat. It doesn't make sense to talk about Lorentz contraction of a black hole.

The Lorentz contraction also doesn't describe what is seen optically. Re the end of your question, which is about optical observations, see What will the universe look like for anyone falling into a black hole? .

Your third question has to do with Hawking radiation. You don't need to recalculate Hawking radiation. You can just reuse the standard description and then apply the appropriate Doppler shift. So there could be either a blueshift or a redshift, depending on the observer's state of motion.

$\endgroup$
-1
$\begingroup$

According the Special Relativity, all objects contract their length if moving fast.

According the Special Relativity, an object contracts if its speed goes from a small fraction of the speed of a beam of light next to the object, to a large fraction of the speed of a beam of light next to the object.

So if a guy jumps out of a hovering platform near the event horizon of a black hole, he will consider the length of his buddy that stays on the platform to first contract and then go back to normal.

And the falling guy considers some distant black hole to never contract very much. The same is true for the black hole below him - the two black holes should behave the same way.

According to the falling guy the speed of the buddy stays below c, but the speed of a light beam next to said buddy becomes 1000 c or more, as the falling guy becomes very time dilated. That is the reason for the very small contraction.

$\endgroup$
  • $\begingroup$ This answer is kind of garbled from start to finish. Starting with the first sentence: an object contracts if its speed goes from a small fraction of the speed of a beam of light next to the object, to a large fraction of the speed of a beam of light next to the object. This description makes it sound as though Lorentz contraction only occurs for objects whose motion is noninertial, which is false. $\endgroup$ – Ben Crowell Sep 2 at 18:08
  • $\begingroup$ @BenCrowell I was talking about the falling guys answers, when we keep asking "what is the size of that black hole". If decreasing size is not a Lorentz contraction in this case, then we'll call it "decrease of size". I thought there is a Lorentz contraction type of thing occurring. $\endgroup$ – stuffu Sep 2 at 19:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.