What happens to the fabric of space in the wake of a moving black hole? Is space permanently deformed by a moving black hole or does it rebound as the black hole passes?
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1$\begingroup$ A black hole is only special and interesting when you are at or inside the event horizon. From far outside of the event horizon a black hole has the same effects gravitationally, as a normal star or planet of the same mass. $\endgroup$– vszCommented Dec 4, 2012 at 7:09
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6$\begingroup$ Moving relative to what? $\endgroup$– JohannesCommented Dec 4, 2012 at 7:42
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3 Answers
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A black hole creates curvature in spacetime just like any other massive object, such as a star, which makes it attract other matter gravitationally. A black hole isn't any more exotic than a star in that regard.
The black hole's gravitational field is just a consequence of its mass existing in that location. If a black hole is in motion relative to you, then the curvature of spacetime that it creates will follow it.
You could, of course, also argue that the black hole is at rest, and you're the one who's in motion. That's why it wouldn't make sense for a black hole to "permanently" deform spacetime, since that would mean that there's a preferred frame of reference in which the black hole is in motion.
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$\begingroup$ Excellent answer. I'd actually be interested to know what the "blackhole wormhole" theories say about this - can an Einstein-Rosen bridge move, and (if so) does it leave distorted spacetime in its wake? $\endgroup$ Commented Dec 4, 2012 at 10:51
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$\begingroup$ @Polynomial: As Dmitry says, you can't have a "permanent" distortion because that would mean there's a preferred frame of reference. $\endgroup$ Commented Dec 4, 2012 at 10:56
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1$\begingroup$ @Polynomial In order to leave a "wake", an object needs to travel through a medium in which a wake could be made (e.g. a meteor falling in the atmosphere). Space itself is not a "medium" of that kind. It is Lorentz-invariant, meaning that there are no preferred reference frames (no "absolute" motion through space, only relative motion). $\endgroup$ Commented Dec 4, 2012 at 16:42
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$\begingroup$ @DmitryBrant That makes sense. Nice one :) $\endgroup$ Commented Dec 4, 2012 at 16:45
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A constant velocity black hole leaves no wake whatsoever. To show this, one can rewrite the Schwarzschild black hole metric, after a change of coordinates, as:
$$g_{ab} = \eta_{ab} + \frac{2M}{r}k_{a}k_{b}$$
where $\eta$ is the Minkowski metric, and $k_{a} = (-1,\pm 1, 0, 0)$ is a null vector with respect to both the Schwarzschild and Minkowski metrics.${}^{1}$ Now, you have removed the Black hole geometry from the background geometry, and you are free to apply a Lorentz transformation to the whole thing, and you will not transform $\eta$, but you will now have a Lorentz-transformed $k$. This gives you the metric of a moving black hole. But this metric just differs from your original one by a coordinate transformation, and therefore, has the exact same physics.
Thus, in the presence of a constant-velocity black hole, you will see no wake effects with gravitational radiation.
${}^{1}$the choice of plus/minus sets whether you are in the black hole or white hole patch of the extended Kruskal solution
answered Feb 20, 2016 at 0:25
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$\begingroup$ Surely this $k$ doesn't give the correct spacelike part, as we want a factor of $\left(1-\frac{2M}{r}\right)^{-1}$ to show up. (edit) or I guess you're imagining a large r series expansion $\endgroup$– user12029Commented Jun 12, 2017 at 17:03
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1$\begingroup$ @NeuroFuzzy: this is not Schwarzschild coordinates. This is the Schwarzschild metric in Kerr-Schild coordinates, which does not have the factor you cite. It is a coordinate system that eliminates the coordinate singularity on the horizon. $\endgroup$ Commented Jun 12, 2017 at 18:24
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1$\begingroup$ And it is exact. You can show that it is a vacuum solution yourself (or go and find the coordinate functions that transform to the usual Schwarzschild coordinates) $\endgroup$ Commented Jun 12, 2017 at 18:26
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$\begingroup$ Do you mean $k_a=(-1,\pm x/r,\pm y/r,\pm z/r)$, or are you using $(t,r,θ,\phi)$ coordinates? Also probably worth mentioning that you're assuming $-{+}{+}+$ signature, otherwise it would be $η_{ab} - \ldots$. $\endgroup$– benrgCommented Nov 4, 2021 at 5:55
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$\begingroup$ @benrg: yes, I am using $(t,r,\theta, \phi)$ coordinates. $\endgroup$ Commented Nov 4, 2021 at 13:45
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There's nothing precisely analogous to a Wake in water, because the particular form of water wakes depends crucially on the fact that a Gravity wave (as opposed to a Gravitational wave) has a distinctive Dispersion relationship. For deep water it's ω=gk−−√ω=gk , where ω=2πfω=2πf is angular frequency (frequency in units of radians per second instead of hertz) and k=2π/λk=2π/λ is wavenumber in radians per metre. (The funny units are so you can write the amplitude of the wave as Acos(ωt−kx)Acos(ωt−kx) and save a lot of occurrences of 2π2π .) Then the Group velocity is ∂ω/∂k=g/k−−−√/2∂ω/∂k=g/k/2 and the Phase velocity is ω/k=g/k−−−√ω/k=g/k which are different and kk -dependent so it's all very complicated.
In particular, no matter how slow the boat goes there will be some large- kk (short wavelength) components of the disturbance it creates that go slower than it, so there'll always be at least some tiny sonic-boom-type shockwave, i.e., a V-shaped wake.
By contrast, EM and gravitational waves in vacuum are non-dispersive, which means they have ω=ckω=ck for a common frequency-independent phase/group velocity which happens to be cc for both. And since the object itself can't go faster than c, there can't be a proper wake. In particular, there's no wake of the sort that needs energy to create, so the source doesn't need to be continuously supplied with energy to keep moving.
Of course a charged particle in a refractive medium can go faster than c/nc/n ( nn being the refractive index) and create a wake in the form of Cherenkov radiation. There's very likely a similar effect for a mass particle moving through a dense dust, but I'm not up to calculating it.
