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Does a black-hole of a given mass and angular momentum have the same space-time geometry regardless of its charge? I'm pretty sure that an electric field can accelerate a charged particle but doesn't curve space-time so the only way an electric field can affect space-time is by accelerating a mass which then produces a different gravitational field because it has a different position and velocity because of its acceleration. Does that mean a charged black hole will have the same gravitational field outside its event horizon because the charge in gravitational field inside the event horizon produced by accelerating charges can't escape the black hole and therefore the electric field of a charged black hole will not accelerate another charged black hole?

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So the answer is no.

An electric field has energy and energy generates a gravitational field, just like any mass.

See the charged black hole solution is the Wikipedia article https://en.m.wikipedia.org/wiki/Charged_black_hole

The charge of a black hole, if nonzero, changes the metric and solution to account for the charge and electric field. That Wikipedia article has reference to the solution, called Reissner Nordstrom, for spherically symmetric, non rotating black holes. If they rotate it is the Kerr Newman solution.

Both exist because charges have electric fields, and those have energy. And because charge is a conserved quantity charge is not radiated away in a black hole (the no hair theorem came about because conserved quantities can not be radiated away, and that is mass, angular momentum and charge).

So charged black holes have a different, but similar, gravitational field as uncharged ones. See the Wikipedia articles.

Since black holes have charge, and a static electric field that is manifest outside the black hole (just like the static gravitational field is), they definitely can interact with charged particles or bodies, including othe charged black holes.

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  • $\begingroup$ Mass is also a conserved quantity, despite that it is radiated away by Hawking radiation. $\endgroup$ – peterh Jul 24 '16 at 21:17
  • $\begingroup$ How do you know an electric field accelerates a charged black hole? Has it been proven that an electric field accelerates all charged particles including black holes? I'm not totally convinced that an electric field could accelerate a charged black hole? I don't see how the equations could predict it. $\endgroup$ – Timothy Jul 24 '16 at 22:39
  • $\begingroup$ A black hole far enough away from its horizon is just a body, with whatever mass, angular momentum it has. You can treat them with Newtonian mechanics, at worse using a first order approximation as a correction, due to GR. A semi classical netwtonian object, with charge and mass. The equations definitely predict it. Never observed because nobody has detected a charged black hole, highly unlikely, and it'll have to wait observational confirmation based on some astrophysical black hole event. As for Hawking radiation, my answer obviously did not include quantum effects, it was pure classical GR. $\endgroup$ – Bob Bee Jul 24 '16 at 23:58
  • $\begingroup$ Why does an electric field accelerate a charged black hole? Isn't a black hole just curved space time so its curvature should be unaffected by an electric field? I don't understand much about how general relativity works so can somebody give me an explanation that I can understand that figures out which assumption I'm making and breaks that assumption. $\endgroup$ – Timothy Jul 25 '16 at 1:36
  • $\begingroup$ A BH has mass, and may have charge and angular momentum. At a distance it behaves just like any body with the same mass, charge and angular momentum. The BH can also be accelerated by the gravitational field of another body; the two BH's merging in 2015 were accelerated towards each other by each gravitational field acting on the other. BH's and neutron starts are attracted to each other (in neutron starts merging with BH's) by their gravity. They get attracted by gravity because have mass. Also attracted or repulsed with any interaction with another charged object, if the first BH has charge $\endgroup$ – Bob Bee Jul 25 '16 at 1:53
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The change in the geometry is not entirely certain. One can appeal to exact solutions. With the Schwarzschild solution for a non-rotating black hole the conformal diagram

enter image description here

illustrates region I that is the exterior universe and region III for the black hole interior. The line separating them is the event horizon. There is another region II that is another exterior region, either another universe or some distant region in this universe. The two $45$ degree lines are event horizons and where they cross is where the black hole is a non-traversable wormhole connected by the event horizon. The horizon however splits and the two horizons separate at the speed of light. I illustrate this in the second diagram below enter image description here

There is a picture of a rocketship entering the wormhole. It can only traverse the bridge if it meets the horizon right at the vertex of this split. Without going into much complexity the problem with identifying this is that one would need a clock able to mark Planck units of time and this interacting with the black hole leads to further uncertainty. The region III expands, or spatial regions expand until the horizontal region at the top is reached. In this diagram this is seen by the expanding tube or bridge. This is the singularity, which is a spatial region where the curvature diverges. This rocketship in region III is unable to ever reach the region II. The singularity represents the collapse of this bridge that squeezes anything in there completely.

Let us now consider the case with electric charge, which is similar to including angular momentum. The Schwarzshild horizon occurs at $r~=~2GM/c^2$ for a black hole of mass $M$. For the Kerr-Newman solution there are two horizons that occur as $r_\pm~=~m~\pm~\sqrt{m^2~-~Q^2}$ for $m~=~GM/c^2$. This Kerr-Newman diagram is shown below enter image description here

I have changed this diagram a bit from the usual form to include Hawking radiation. This is not seen in the simple diagram for the Schwarzschild black hole. However, in that case the singularity is reached as a spatial surface for an interior observer corresponding to when the black hole finally evaporates. In this third diagram the one on the left is most important. There is the singularity $r_+$ that separates our region, say the blue region on the right, from the spacelike region interior to the black hole. However, $r_-$ the exists that leads to an additional timelike region with a timelike singularity. In addition the $r_-$ is reached by any observer at the moment all quanta that make up the black hole reaches it, and everything is blue shifted. This inner event horizon is then a Cauchy horizon, Cauchy in reference to his famous theory of accumulation points. Whether one survives this is a bit academic. Everything reaching this horizon is blue shifted arbitrarily, but so is any observer frame dragged across it. So this Cauchy horizon is a bit of a "soft singularity."

There is again this non-traversable wormhole as with the Schwarzschild case. The bridge expands, and in fact expands enormously, but is bounded by the maximal complexity permitted by this black hole, which is related to the entropy bound of the black hole and Hawking radiation. This then with the inner horizon. We might be tempted to treat $r_-$ as a singularity and deform the $r_-$ into spacelike regions with infinite curvature. This is if the Cauchy horizon is "not so soft" as a singularity. This is in part where the uncertainty in the geometry comes in. There also may be other quantum mechanical issues.

I illustrate something interesting about this in that an observer that registers qubits falling into the black hole can solve very complex algorithms. The piling up of these qubits is a asymptotic speed up of processing. If the black hole were eternal this observe could be a hyperTuring machine able to surmount the limits of Godel-Turing limits on computation. However, black holes are not eternal. However, this might have a bearing on NP-complete problems and their relationship is P-problems.

The Kerr-Newman black hole is much more of a "processing system" in terms of qubits. The inclusion of angular momentum or electric charge is definately a game changer. It also looks as if the same many be the case with gauge fields and quantum fields in general. This is particularly the case with the extremal black hole where $r_+~-~r_-~\rightarrow~0$.

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