Do the apparent Black Hole dimensions contract as the observer falls into Black Hole?

Question

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?

Answer 1

Anixx wrote: "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."

Since the Schwarzschild/Droste $\rm dR/dr=\sqrt{g_{rr}}=1/\sqrt{1-r_s/r}=\infty$ at $\rm r=r_s$, the infinite depth-expansion and Lorentz-contraction cancel each other out, therefore in Raindrop-coordinates where the local rulers and clocks are falling with the negative escape velocity $\rm v=-v_{esc}=-c\sqrt{r_s/r}$ the gammafactor cancels the depth-expansion exactly all along the way so that $\rm \sqrt{g_{rr}}=1$ in that coordinates, and the proper distance to the singularity $\rm R$ equals the coordinate radius $\rm r$ for the raindrops, which takes them $\rm \tau=2r_s/3/c$ proper time from the horizon to the singularity.

If your velocity is faster than the escape velocity, the gammafactor will be larger than the depth expansion so that $\rm R<r$ (faster always means closer to $\rm c$, so behind the horizon a higher $\rm v$ is effectively slower and takes more proper time from the horizon to the singularity, with the maximum proper time of $\rm \tau\to \pi \ GM/c^3$ with $\rm v \to -\infty$ and a minimum $\tau\to 0$ if $\rm v\to -c$).

Answer 2

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.

Answer 3

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.