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For a non-rotating Black Hole, the event horizon can be described by the Schwarzschild metric as a sphere. Assuming external observer, away from the Black Hole and also assuming that there is no matter or gas clouds nearby, I would assume that the sphere would be perfectly symmetric and smooth but is that accurate? Are there very small (or large) fluctuations for the geometry of the event horizon?

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  • $\begingroup$ What are you perturbing the solution with? Whose coordinates are you using to define a horizon? $\endgroup$ – AHusain Jun 30 '16 at 20:30
  • $\begingroup$ Assuming external observer, away from the Black Hole. $\endgroup$ – Peter R Jun 30 '16 at 20:32
  • $\begingroup$ For an observer just "hanging" over the Schwarzschild radius it would have to be an extremely violent environment... it's really observer dependent. $\endgroup$ – CuriousOne Jun 30 '16 at 20:47
  • $\begingroup$ It seems to me that nothing quantum is ever smooth, but those fluctuations would be on a molecular diameter range, and very smooth from a classical perspective. The other problem is that the event horizon isn't a thing, so observing where it begins is quite problematic. My layman's guess is, very smooth but not perfectly smooth and enormously difficult to measure. $\endgroup$ – userLTK Jul 1 '16 at 4:18
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For non quantum effects an equilibrium spherical black hole horizon is smooth, perfectly symmetrical. It is defined by only one parameter (mass, if no charge and no rotation, and thus spherical) - thus it is said it has no hair. If it is perturbed it will deform, and acquire some hair. For perturbations that are not very large, and not spherically symmetric, it will acquire curvature and metric deformations that at a distance can be modeled as multipole moments, and probably some rotation. Those will settle down through gravitational radiation, back to the purely spherical state.

That it takes an infinite time for an outside observer at infinity is a negligible issue, as it settles back to enough of a symmetrical spherical black hole quickly enough. There can be arguments on this but a real and an apparent horizon don't look any different, from outside. It asymptotically approaches the event horizon.

If the perturbation is minimal it's really the same but a small effect, a bump in the horizon that settles quickly. Calculations and simulations of this are done at a perturbation level.

The calculations done to describe the merger of the two black holes discovery announced in February 2016 were all done with a combination of a post Newtonian general relativity approximation, plus numerical full general relativity during the merger itself (which were not small perturbations). The horizons of both black holes get affected and quickly, in that case in about The quarter of a second that we were able to capture, merge and settle down to a Kerr black hole (this was rotating). I mention this because it is easy enough to find online, rather than search for older simulations to show those small or bigger perturbations on one black hole.

In all those cases the hair or moments of the black hole quickly radiate away (not from the horizon but from outside), and the black hole settles down to a no-hair state

For quantum effects things are not known. In effect, we are just not sure of anything on quantum gravity. For quantum fields slightly perturbing a classical (general relativistic) black holes, such as virtual particles getting absorbed into the black hole and Hawking radiation being produced (a so-called semi-classical description), it doesn't look qualitatively different.

It is different if you assume some quantum gravity model near the horizon. Mostly either string theory or loop quantum gravity or others model the horizon as either strings or loops or quantum foam of some kind, and those have hair - basically each Planck size surface is a possibly different state of one of those, and so there is a lot of hair. Or maybe even worse and more like the firewall envisioned by the controversy over whether black holes loose quantum information. So at those scales the horizon is pretty rough. But on larger scales it is still no hair, as seen through the classical descriptions. (Note: the firewall controversy is still unresolved and claims the whole horizon is a huge firewall).

If you really want to see or work the details you have to work through the math. The classical calculations for small or limited perturbations, are now standard, and including the strong gravity cases of black hole mergers which use numerical calculations where an exact solution is not known (i.e., in most non highly symmetrical cases). But, the exact details of the dynamical changes in strong gravity, even for classical general relativity, are still a challenge, and it is hard to explore the general dynamics of it. The quantum cases are of course more unsettled.

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This horizon involves time, and to be a real horizon the time of observation must tend to infinity for observing a falling probe particle. The particle field is assumed to be much smaller than that of BH. In this case the space-time geometry is smooth.

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  • $\begingroup$ Are you saying that for an external observer it would take an infinite amount of time to become smooth if a probe was observed falling into the black hole? $\endgroup$ – Peter R Jun 30 '16 at 20:58
  • $\begingroup$ @PeterR: Yes, I am. $\endgroup$ – Vladimir Kalitvianski Jul 1 '16 at 6:40

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