Your reasoning about the horizon radius not being t-dependend does not hold, you can also have a regular sphere in coordinates where the radius is independent of t but that doesn't mean that there is no contraction in a frame moving relative to the sphere

@Yukterez A sphere would have the metric $ds^2=-dt^2+r^2 dΩ^2$, in which the $t$ coordinate does appear as $dt$. In the metric for the event horizon, it really doesn't appear at all.

that's not true, a moving black hole is described by the en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultraboost and therefore contracted in direction of motion (in the limit of v→c to a 2D plane disk), see mat.univie.ac.at/~stein/dokumente/… at page 5

So if I fly towards an empty sphere of radius 10 m at 0.866c, I will calculate that it is squished to 5 m in my reference frame based on Special relativity.

@Ralph Berger - that's correct, in that frame the empty shell will be contracted to an ellipsoid with $\rm r_{xy}=10m$ and $\rm r_z=5m$ assuming the motion is along the $\rm z$ axis

@Yukterez. Thank you, I'm afraid I was going to follow up with a question for benrg, but when I hit the return for a new paragraph, it posted my comment. My follow up would be, "but your comment about Lorentz contraction doesn't make sense in GR implies that if I fall towards a black hole at .866c, I will not calculate that it has squished to 5m. Is that what you (benrg) are saying?"

If benrg's argument was right you could place a shell much larger than the black hole around the black hole and move towards it, which would make it contract. If the black hole inside the shell would not contract and you move fast enough the horizon radius in direction of motion would be larger than the shell radius, which would be paradox since the shell can not be outside the horizon in one frame and partly inside it in the other frame.

@Yukterez: The metric of a moving black hole is called boosted Schwarzschild metric (see e.g. my answer here). Aichelburg-Sexl solution is the limit of boosted Schw. when velocity approaches $c$ while simultaneously mass approaches zero, so it does not really has a horizon, just a $\delta$-like curvature singularity.

@A.V.S. - that may be, but the conclusion that the horizon gets lorentzcontracted in direction of motion still holds, and the plot in your linked answer also seems to show that since v=0.5 and the vertical contraction of the horizon is exactly r/γ (126px·109px).

@Yukteres: The plot in my answer is for specific coordinate system while the question is about “perceived shape” (whatever that means). I liked the renderings in your pages, btw, but do you have one for Schwarzschild metric and comparing views for different values of $E$ at different $r$.

@Yukterez The Aichelburg-Sexl ultraboost describes a light-speed particle, not a moving black hole. I showed that the event horizon doesn't contract; you can't show my argument is wrong by making an unrelated argument, unless GR is inconsistent. In your argument you implicitly used the Lorentz contraction formula, which is only valid in Minkowski spacetime. If you look at the geometry of concentric spheres in Schwarzschild spacetime when sliced by "diagonal" spacelike hyperplanes, you'll find it isn't as simple. Note you must look at the actual geometry, not the coordinates.

@Yukterez The coordinates are scaled by $γ$. You need to look at the actual geometry.

@benrg - they are scaled by γ not just for the LULz, but since the BH is moving with v in that frame. The "actual" geometry as you call it is the geometry in the rest frame of the BH.

@Yukterez There's no such thing as the geometry in the rest frame. Geometry doesn't depend on coordinates. For my argument, I chose coordinates in which the non-contraction of the horizon is easy to see. I'm free to do that because all coordinate systems are equivalent. You're trying to rederive the effect in a much more complicated way using formally Lorentz boosted Kerr-Schild coordinates. If you do it correctly you'll get the same answer, but it will be difficult not to make a mistake, especially if you don't understand the difference between coordinates and geometry.

@Yukterez What I think you'll find if you work it out is that the contracted size of a sphere of reduced-circumference radius $r$ is $r\sqrt{1-(1-r_s/r)β^2}$. That's equal to $r_s$ at $r=r_s$ (i.e. uncontracted), and it's a monotonically increasing function for $r\ge r_s$, so there's no contracted sphere with a smaller axis than a nested one. If you substitute $r_s=0$ then the spacetime becomes Minkowski and the formula becomes the special-relativistic one.

@benrg - then take a look at repositories.lib.utexas.edu/bitstream/handle/2152/23434/… at the text below equation (4.2) at page 68 where it says "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 γ=√1/(1−v²/c²)" (that is about Kerr black holes which are ellipsoid to begin with, but the relevant part with γ also applies to spherical symmetric Schwarzschild BHs)

@Yukterez In chapter 3 they derive the metric of a slice of the Schwarzschild horizon at $t=0$ where $t$ is a boosted coordinate, and the result is $ds^2=(2M)^2(d\bar θ^2+\sin^2 \bar θ d\bar\phi^2)$, i.e., a sphere (equation (3.9)). At the beginning of chapter 4 they do seem to make claims in English that contradict that formal derivation. I don't know what to say except that the claims in chapter 4 are wrong.

@benrg - you have to read what is written below the equation (3.9) that you quoted, which is the same as (3.8), there it reads: "Importantly, note that Eq. (3.7) describes the boosted apparent horizon; the simple form Eq. (3.8) that allows immediate evaluation of the surface area is the expression of this area in terms of coordinates appropriate first of all to the unboosted frame." so you have to use (3.7), not (3.8) or (3.9) if you want the boosted one, and that is lorentzcontracted by the gammafactor, so no contradiction there.

@Yukterez (3.7), (3.8) and (3.9) are all the same manifold in different coordinates. You can't change the shape of a manifold by putting a different coordinate chart on it. (3.9) is the metric of a sphere, and if you substitute in other coordinates—theirs or any others—it's still a sphere. You're right that even in chapter 3 they don't seem to understand that: they seem to think that this manifold is an oblate spheroid that happens to have the same metric as a sphere. That is just not a thing in GR. Until you understand the difference between geometry and coordinates, you don't grok GR.

@A.V.S. - "I liked the renderings in your pages, btw, but do you have one for Schwarzschild metric and comparing views for different values of E at different r" - at yukterez.net/f/einstein.equations/files/x you have 3 observers at the same r=6, the 1st at rest, the 2nd moving radial and the 3rd tangential relative to the BH

benrg: Your interpretation of the equations is wrong, but however, now we had multiple sources that confirm the horizon is contracted in the moving frame, so you have to find at least one reference that explicitly states it is a sphere in all frames if you still believe that.