https://journals.aps.org/prd/abstract/10.1103/PhysRevD.8.3259
Hanni and Ruffini (1973) calculated the electric lines of force for a charge close to the horizon. The result is that the closer the charge is to the horizon, the more spherically symmetric is the electric field, where the origin is the center of the black hole.
Richard H. Price's thesis (1971)
Richard H. Price (1971, 1972) calculated perturbations of a Schwarzschild black hole using a wave equation for a scalar field. He assumes that such a field gives us a clue of what happens with an electric charge or a small mass falling into the horizon.
His thesis claims that only the monopole field of an electric charge can survive. The monopole field is the spherical harmonic (0, 0), centered at the center of the black hole. All the other spherical harmonics are "reflected" back by the metric.
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/ \ black hole
\____/
• •
+Q -Q
What does this reflection mean? We use as an example a pair of charges, +Q and -Q placed close to each other near the horizon. Let us use these charges as a dipole antenna. We try to send very long radio waves, and move the charges extremely slowly.
The diagrams of Hanni and Ruffini (1973) show that the electric fields of the charges cancel each other out almost entirely at some distance from the horizon. Both fields are almost perfectly spherically symmetric.
We conclude that long radio waves cannot get very far: they are "reflected" back. If a very long wave is reflected from a potential wall with a 180 degree phase shift, then destructive interference cancels the wave close to the wall.
Note that if we move the charges very quickly, generating very short waves, our geometric argument of canceling of spherically symmetric electric fields does not work: the fields get substantially distorted.
Did we "prove" that the electric field of a black hole is the simple spherically symmetric field? No, because the interplay of gravity and electromagnetism could produce a surprise. There is no proof that the Einstein field equations have a solution at all for such a complex setup. Solutions for nonlinear equations could diverge.
However, if we assume that the charges and their electric field does not affect the Schwarzschild metric in any way, then it is obvious that the result of Hanni and Ruffini shows that the field is almost symmetric for a finite number of charges falling into the black hole.
Using the merger of two black holes as a model
In our June 26, 2023 blog post we studied the merger of two black holes of comparable sizes. Let us then assume that one of the black holes is microscopic and the other is a large Schwarzschild black hole.
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/ \ <-- • microscopic black hole
\____/
black hole
The merger process probably is qualitatively similar to one with two large black holes. In the Schwarzschild coordinates, the small black hole is very much squeezed in the radial direction. The two event horizons will remain separate until the microscopic black hole is within the Schwarzschild diameter of the combined system. Once within, a common event horizon quickly grows around the whole system.
What about a (classical) elementary particle falling in? A classical particle, actually, is a microscopic black hole.
Let us then consider a small body of "continuous" matter, with no point particles. In the Oppenheimer-Snyder (1939) collapse we have an analogous process. The surface of the dust ball never falls within the Schwarzschild radius of the whole system. Any internal point at a reasonable distance > 0 from the surface quickly is enclosed inside the forming event horizon.
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/ \ <-- ●P continuous body
\____/
black hole
We can guess that the same is true for the small body of matter. Let P be the point on the surface of the body, such that its radial distance is the largest. We expect P to remain outside the new forming horizon for ever.
Sometimes it is claimed that a falling particle can never cross the event horizon. That is true for a zero-mass idealized particle, but seems not to be true for a real particle.
Perturbation of the event horizon
Riccardo Bonetto, Adam Pound, and Zeyd Sam (2021) calculated the "tidal effect" on the event horizon of a Schwarzschild black hole M when a smaller black hole, 0.1 M is orbiting it.
<------ counterclockwise orbit
They try to explain why the bulge of the event horizon of the large black hole is ahead of the the orbiting small black hole. In the diagram (Figure 2) in the paper, the small black hole orbits counterclockwise. The figure uses the advanced time coordinate. The dotted line points to the small black hole. The dashed line points to the maximum deformation.
We saw a similar effect in the animation made by Teresita Ramirez and Geoffrey Lovelace (2018). We interpreted that the squeezing of the large black hole happens in the global pseudo-"Schwarzschild" coordinates whose origin is the center of mass of the system. The small black hole causes the radial metric to stretch (that is, g₁₁ becomes larger, if we use the (- + + +) signature).
In the coordinate time of the large Schwarzschild black hole, the trip of light from the orbiting black hole takes a very long time. Presumably, the horizon never "knows" that the small black hole has started to orbit the large black hole. If we assume that there are no superluminal signals, then the metric close to the horizon can never change. Thus, the form of the horizon cannot change for local observers at all. The horizon is perfectly rigid. There cannot be any bulge in the horizon rotating around.
However, in global coordinates the horizon can appear squeezed. We have to check if the authors have made some conceptual error when they think that a bulge exists.
Conclusions
In general relativity objects slip almost entirely behind an event horizon, and quite quickly. The prominent example is the Oppenheimer-Snyder (1939) collapse of a dust ball.
We have to check in more detail the paper by Bonetto, Pound, Sam (2021).
We will also look at the Kerr metric, the ergosphere, as well as various horizons in a rotating black hole.
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