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Say a black hole is travelling at $c/2$, does the shape of the event horizon change? What about the location of the event horizon? If it is travelling at a hypothetical $c$, does the event horizon simply resemble a light cone?

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    $\begingroup$ You know... velocity is relative. Consequently, "traveling at $c/2$" doesn't mean a whole lot... Perhaps it's better to study some special relativity first :) $\endgroup$ – Danu Mar 25 '15 at 12:54
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    $\begingroup$ The event horizon is a three sub-manifold of space-time and as such is absolute. The spacial slices of the event horizon, which is what you are thinking about, will be relative. The area will be the same, but the shape will be different. Imagine a vertical cylinder if you slice it horizontally the cross sections will be circles, if you slice it at an angle the cross sections will be ellipses. The cylinder is the same the slices can have different shape. Something similar happens with black hole horizons except with one more dimension (and the geometry is not Riemanian). $\endgroup$ – MBN Mar 25 '15 at 13:19
  • $\begingroup$ @MBN: Will the area be the same? In your analogy the circumference of the ellipse increases (due to Euclidean geometry), in relativity it seems to decrease (due to Lorentz contraction, Minkowski geometry). In the extreme relativistic case, the area should just equal twice the area of a circle (area of a pancake :). $\endgroup$ – kristjan Mar 25 '15 at 13:52
  • $\begingroup$ In my analogy, the analog of area would be the circumference since we are suppressing one dimension. And you are right the circumference will not be the same, but this is why I made the comment about the Riemannian geometry. In the Lorenzian case one can show that the area of the cross sections will be the same. $\endgroup$ – MBN Mar 25 '15 at 13:55
  • $\begingroup$ Ok, so with special relativity, to get a black hole to a certain velocity from stationary it will also increase in mass/energy right? so the area of the event horizon must increase.. I assume. $\endgroup$ – Ilya Grushevskiy Mar 25 '15 at 13:57
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It is convenient to go from Schwarzschild coordinates $$ ds^2 = \left( 1 - \frac{r_s}{r}\right) c^2 dt^2 - \left( 1 - \frac{r_s}{r}\right)^{-1}dr^2 - r^2 d\Omega $$ to Isotropic coordinates $r\rightarrow r^2(1+ r_s/4r)^4$, then to quasi-Minkowski using the usual spherical-cartesian coordinate transformation to get $$ ds^2 = -\left( \frac{1 - r_s/4r}{1 + r_s/4r} \right)^2 c^2 dt^2 + \left( 1 + r_s/4r\right)d\vec{x}^2 $$ Now you can finally use the familiar Lorentz boost formula \begin{align*} ct & \rightarrow \gamma\left(ct - \vec{\beta}\cdot\vec{x}\right) \\ \vec{x} & \rightarrow \vec{x} - \gamma \vec{\beta}ct + \frac{\gamma-1}{\beta^2}\left(\vec{\beta}\cdot\vec{x}\right)\vec{\beta}\\ \Rightarrow r^2 &\rightarrow \left| \vec{x} - \vec{\beta}ct\right|^2 + \gamma^2 \left(\vec{\beta}\cdot\left( \vec{x} - \vec{\beta}ct\right) \right)^2 \end{align*}

Only now you can start asking physical questions using the new coordinate system. For example let's look at the event horizon. In the old coordinates (pre boosted isotropic quasi-minkowski) requiring $g^{00}=0$ gave us the spherical surface surface $r=r_s/4$. Now let's look at our new $g^{00}$ \begin{align*} g^{00} = \gamma^2\left( -\left( \frac{1-r_s/4r'}{1+r_s/4r'}\right)^2 + \beta^2 \left(1+r_s/4r' \right) \right) \end{align*} So using $g^{00}=0$ the horizon in the new coordinate system is found to be $$ r' = \frac{r_s}{4}\frac{1+\beta}{1-\beta} $$ $\Rightarrow$ $$ \left| \vec{x} - \vec{\beta}ct\right|^2 + \gamma^2 \left(\vec{\beta}\cdot\left( \vec{x} - \vec{\beta}ct\right) \right)^2 = \left( \frac{r_s}{4}\frac{1+\beta}{1-\beta} \right)^2 $$ Now since our new coordinates are isotropic the metric is unchanged if we had aligned the $z-$axis with $\vec{\beta}$, in which case you get an ellipsoid with moving center along the $z-$axis $(x_0,y_0,z_0) =(0,0,vt)$ and contracted along the $z$-axis.

Note: proof read for arithmetic mistakes.

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    $\begingroup$ There is no such thing as a Lorentz boost on a general spacetime. You can manipulate coordinates however you like, but that doesn't guarantee that the results have any physical significance. $\endgroup$ – Ben Crowell Mar 4 at 15:25
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The event horizon is by definition a set of points in spacetime that is independent of the observer. Therefore its location is independent of the state of motion of the black hole and the observer relative to each other.

We could ask whether the moving observer thinks the event horizon is spherical. This is not a question that is meaningful in general relativity, unless you specify what kind of observations are involved. When we define spherical symmetry in GR, we take pains to do it in a way that is independent of coordinates or observers. You might think that you could take the intersection of the event horizon with a surface of simultaneity for a particular observer, but GR doesn't define that kind of surface of simultaneity.

If you want to define how the horizon appears to an observer, one natural way to do it would be to use optical observations. In SR, the optical silhouette of a sphere is always a disk, regardless of the state of motion, so I suspect that the same would be true for the silhouette of a black hole. This does seem to be true, as far as I can tell, in figures 25-28 of Riazuelo, https://arxiv.org/abs/1511.06025 , although it's not especially clear. I've done some similar simulations myself, but I only did radial motion, so I haven't actually seen what happens for transverse motion.

But of course, even if we established that optical measurements showed the silhouette of a Schwarzschild black hole as a disk for an observer in any state of motion, that isn't an answer as to the question of whether the event horizon appears length contracted, since the result is the same as the SR result for the silhouette of a sphere. Perhaps there is some other type of observation that you could define that would give a natural sense in which this becomes a meaningful question, but it's not immediately obvious to me what type of observation that would be.

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