What is the perceived shape of a non-spinning black hole as it is approached at high velocities?

Question

The question and answer at What is the shape of a black hole? states that an event horizon " is indeed a sphere." But if it is approached at high velocity, does the traveler calculate that the sphere is flattened due to Lorentz contraction, until it is nearly a 2D disk? The reason I ask is that Stephen Hawking said that, if the black hole was big enough that spaghettification would not be noticeable, that "you wouldn't notice anything particular as you fell into a black hole" (https://www.hawking.org.uk/in-words/lectures/into-a-black-hole). But wouldn't the traveler immediately hit the hard center?

The answers I have gotten when asked elsewhere include that Lorentz transform doesn't apply when the velocity is due to gravitational acceleration. That person referenced the "falling raindrop reference frame." Is it true that falling objects do not calculate lorentz contraction in the same way as an object would that is driven by rocket thrust?

Another answer is that one cannot define the distance between the event horizon and the center of a black hole in terms of length, but you can in terms of time. If that answer is correct, I supppose velocity must not be definable within an event horizon, else velocity*time gives a length. Also, would the shape be calculated as a flat disk immediately outside of the event horizon and then it only becomes undefinable after crossing the event horizon?

The third answer I've gotten is that no one knows a thing about what happens inside an event horizon. But then should Stephen Hawking have said "something may happen at the event horizon, but we don't know what" rather than his comment about not noticing anything in particular?

Answer 1

Event horizons don't Lorentz contract.

You could argue that the concept of Lorentz contraction doesn't even make sense in general relativity, but there is a way to generalize it. In special relativity, it refers to the fact that if you slice spacetime into parallel spacelike hyperplanes, the intersection of a hyperplane with the worldtube of some object depends on the angle between them. (This happens in Euclidean geometry too: e.g. if you slice a drinking straw at an angle, the opening is an oval, not a circle.) In general relativity you can't generally slice into parallel planes, but you still can foliate the spacetime into spacelike hypersurfaces, and a similar thing happens.

But a Schwarzschild event horizon has the metric $ds^2 = r_s^2 d\Omega^2$, in which the $t$ coordinate doesn't appear at all, and as a result, no matter how you foliate the spacetime, the slices have the metric $ds^2 = r_s^2 d\Omega^2$, which is a sphere, not an oblate spheroid. So event horizons don't Lorentz contact.

If you head into a black hole at a high speed, you will hit the singularity more quickly than if you'd fallen in slowly, but you can't explain it by saying that the distance is shorter because of Lorentz contraction. That picture really only works in special relativity (and even in SR, I personally think it's more confusing than helpful.)

Re Hawking's statement that you won't notice anything special falling in, see this answer.

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Note that Yukterez's answer, which says that event horizons do Lorentz contract, is wrong. It quotes from a PhD thesis by Sarp Akcay. Akcay actually correctly derives the metric of the diagonally sliced Schwarzschild horizon in chapter 3 (stated in three different coordinate systems in equations (3.7) through (3.9)), and it is the metric of a sphere. Both he and Yukterez seem to believe, nevertheless, that the horizon is an oblate spheroid that happens to have the same metric as a sphere—which is just not a thing in GR.

Answer 2

Ralph Berger asked: "What is the perceived shape of a non-spinning black hole as it is approached at high velocities?"

If you move towards it the shadow looks smaller, and if you move away from it bigger, due to the aberration. If you do the raytracing in the stationary frame you can apply the special relativistic doppler with the velocity relative to a local stationary observer, see here, or here where three observers at the same location, but with different velocities relative to the BH compare their views with $360°\times180°$ full panoramas:

In the upper images the observer is at rest, in the middle moving radially towards the BH and at the bottom flying past it to the right, in the transverse direction (using this color code for the red/blueshift).

All three observers are at the same Schwarzschild coordinate $\rm r=\rm 3 \ r_s=6 \ G M/c^2$. In the frame without the accretion disc the naked shadow of the Schwarzschild BH looks perfectly round for all observers.

Ralph Berger asked: "But if it is approached at high velocity, does the traveler calculate that the sphere is flattened due to Lorentz contraction, until it is nearly a 2D disk?"

Yes, if your velocity relative to a raindrop (or in other terms relative to the local infall velocity of space, or generally any local timelike observer) goes to $\rm c$ the distance shrinks to $0$ and your proper time $\tau$ to reach the singularity goes to $0$.

But that has nothing to do with the perceived shape, a sphere always looks like a sphere (or in the lightlike limit a point) even when it is lorentzcontracted to an ellipsoid, see here.

If "perceived" is not about what you see with your eyes, but the shape in your frame see this paper and the graphics therein:

Sarp Akcay (p. 68, eq 4.2) wrote: "When boosted, the black hole will not retain its ellipsoidal shape or axisymmetry except for boosts in the z-direction, which simply give more ‘compressed’ ellipsoids whose ‘height’ (length along the z-direction) get Lorentz contracted by a factor of 1/γ=√(1−v²/c²)."

That is about Kerr BHs which are ellipsoid to begin with, but the relevant part with $\gamma$ also applies to spherical symmetric Schwarzschild BHs.

Ralph Berger asked: "But wouldn't the traveler immediately hit the hard center?"

No, the radial depth expansion of Schwarzschild/Droste coordinates cancels exactly with the inverse gamma factor of a freefaller with the negative escape velocity, so in Gullstrand/Painlevé raindrop coordinates $g_{\rm rr}=1$, which means for the falling observer the distance from $\rm r=0$ to $\rm r=r_s$ is simply $\rm r_s$.

Only if your path is almost lightlike you hit the singularity almost immediately, see the comparison of $\rm v$ and $\rm \dot{r}=dr/d\tau$. The "slowest" observer's $\rm v$ (the one who started with $\rm v=0$ close above the horizon) behind the horizon is almost $\infty \rm c$. Nevertheless it takes him the longest proper time of $\tau = \pi \ \rm G M/c^3$ from the horizon to the singularity since the distance in the limit then lorentzexpands to $\pi \infty \ \rm G M/c^2$.

The "fastest" observer's $\rm v$ behind the horizon is just slightly above $\rm c$ (that is done by starting slightly below $\rm c$ outside), with almost $0$ proper time to reach the central singularity, since the closer to $\rm c$, from above or below, the shorter the contracted lengths in direction of motion are.

Photons with $\rm 1 c$ and $\tau=0$ are still the fastest inside and outside, and can overtake every physical timelike observer inside and outside of the horizon. For them the direction of motion is contracted to absolute $0$; thefore in ingoing Eddington/Finkelstein Null coordinates where such radially ingoing photons are the local eventmanagers $g_{\rm r r}=0$ everywhere (for them the radial distance is contracted to $0$ all the way).

Ralph Berger asked: "one cannot define the distance between the event horizon and the center of a black hole in terms of length, but you can in terms of time."

That is a common misconception. In the frame of the external observer the distances and times inside the horizon are imaginary, but in the frame of a free falling observer/raindrop with the negative escape velocity $\rm v=-c \sqrt{r_s/r}$ the distance from the horizon to the singularity is exactly ${\rm R}=\int_0^{\rm r_s} \sqrt{g_{\rm r r}} \ \rm dr = 2 \ G M/c^2$ and the proper time to reach it $\rm \tau=4/3 \ G M/c^3$.

Ralph Berger asked: "The third answer I've gotten is that no one knows a thing about what happens inside an event horizon. But then should Stephen Hawking have said "something may happen at the event horizon, but we don't know what" rather than his comment about not noticing anything in particular?"

We don't know if there's any quantum woohoo going on behind the horizon and in particular close to the singularity, but from the General Relativity perspective in which category your question and the quoted answer are the story about noticing nothing special at the horizon is correct.

benrg claims: "Note that Yukterez's answer, which says that event horizons do Lorentz contract, is wrong."

Since the guy who wrote the other answer and in my opinion confuses the constant surface area with a constant shape is of the opinion that me and my references are wrong, I have to refer to the chat where all the comments regarding this story were moved and suggest that everybody forms his own opinion, see here.