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A well known conjecture of general relativity is that a "black hole has no hair", i.e. once matter has disappeared behind the event horizon, the information about what detailed properties this matter had (except mass, (angular)momentum and charge), before it went into the hole, is thought to be lost.

But I was asking myself if a black hole does, from the outset, have a spherical event horizon and an essential singularity in the center, or if the event horizon can experience dynamic surface waves ("cellulite" to put it pointedly) and if the interior solution could have no singularity at all (because the matter that collapsed inside the horizon is still falling to the center). In this picture, I would expect the spherically symmetric solution to be the equilibrium state of the black hole when all surface waves have been dissipated to the surroundings and all matter inside of it has fallen to the center.

Is this picture incorrect?

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    $\begingroup$ Related: physics.stackexchange.com/q/937/123208 I quite like this line from Stan Liou's answer: "rather than gravity having a special property that enables it to cross the horizon, in a certain sense gravity can't cross the horizon, and it is that very property that forces gravity outside of it to remain the same". $\endgroup$
    – PM 2Ring
    Commented Feb 2, 2021 at 16:45
  • $\begingroup$ @PM2Ring: but that property does not enforce a strictly spherically symmetric gravitational field outside the hole. The black hole exterior is subject to boundary conditions, just like any other gravitational field. So my question basically boils down to: can the event horizon strictly be homogeneous as a spherical boundary condition when the volume in its direct vicinity is filled with arbitrarily complex fields. Or, put as an electromagnetic analogy: the electric fields outside a spherical conductor determine the charge distribution on its surface and vice versa. If and why not for gravity? $\endgroup$
    – oliver
    Commented Feb 2, 2021 at 17:07
  • $\begingroup$ " that property does not enforce a strictly spherically symmetric gravitational field" That's true, but there's bound to be at least a high degree of circular symmetry, due to the angular momentum. $\endgroup$
    – PM 2Ring
    Commented Feb 2, 2021 at 17:37

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Due to gravitational time dilation, a distant observer never actually sees anything reach or cross the event horizon of a black hole. Instead, infalling objects are increasingly red-shifted as they approach the event horizon until they disappear for all practical purposes. Therefore an external observer has no way of telling whether infalling objects have reached the singularity at the centre of the black hole (if, indeed, there even is a singularity). The “interior” of the black hole is completely casually disconnected from its exterior - in one sense, for a distant observer, the black hole had no interior at all.

The closest thing I can think of to “surface waves” on a black hole is the process of “ringdown” when two black holes merge. The event horizon of the merged black hole undergoes a short period of rapid oscillation before it settles into a stable configuration. This process of ringdown generates gravitational waves, which have been observed.

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    $\begingroup$ So does the ringdown include a temporarily non-spherical shape of the event horizon? If it's not two black holes that merge, but ordinary matter "merging" with the black hole, then I guess a ringdown also happens, although it is a lot weaker, correct? Is it meaningful in any way to talk about an inside solution of the black hole when it is practically unobservable? Is it legitimate to say we just "know" what happens inside the black hole, not because we have seen it, but because we made a video of everything before, which allows us to extrapolate things from the Einstein field equations? $\endgroup$
    – oliver
    Commented Feb 2, 2021 at 16:22
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    $\begingroup$ @oliver The merging event horizons become a single sphere fairly quickly, see this brief anim: youtu.be/Tr1zDVbSjTM from an Ethan Siegel article. Bear in mind that an event horizon isn't a physical object, it's a mathematical surface, like Earth's equatorial and orbital planes. $\endgroup$
    – PM 2Ring
    Commented Feb 2, 2021 at 16:59
  • $\begingroup$ @PM2Ring: very interesting video and article. That doesn't look much like surface waves on coalescing water droplets. It looks more like honey on the ISS. But anyways, the event horizons are dynamic, that's what I wanted to know. $\endgroup$
    – oliver
    Commented Feb 2, 2021 at 17:31
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    $\begingroup$ @PM2Ring: Bear in mind that an event horizon isn't a physical object, it's a mathematical surface, like Earth's equatorial and orbital planes Generally, you cannot assign temperature, electrostatic potential, surface resistivity etc. to equatorial plane, but you can assign such properties to an event horizon unambiguosly. Event horizon is a physical object. $\endgroup$
    – A.V.S.
    Commented Feb 2, 2021 at 18:26

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