# Detection of the Electric Charge of a Black Hole

By the "No Hair Theorem", three quantities "define" a black hole; Mass, Angular Momentum, and Charge. The first is easy enough to determine, look at the radius of the event horizon and you can use the Schwarzschild formula to compute the mass. Angular Momentum can be found using the cool little ergosphere Penrose "discovered". However, I don't know how to determine the charge of the black hole.

How can an electromagnetic field escape the event horizon of a Reissner-Nordström black hole? Is there any experiment we could theoretically do to a black hole to determine its charge?

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How are we to look at radius of event horizon? we cannot measure schwarzschild radius. – Newman Aug 17 '11 at 20:43
We could probably measure it by the effects of gravitational lensing, or just simply the gravitational pull. – Benjamin Horowitz Aug 17 '11 at 23:20

A charged black hole does produce an electric field. In fact, at great distances (much larger than the horizon), the field strength is $Q/(4\pi\epsilon_0 r^2)$, just like any other point charge. So measuring the charge is easy.

As for how the electric field gets out of the horizon, the best answer is that it doesn't: it was never in the horizon to begin with! A charged black hole formed out of charged matter. Before the black hole formed, the matter that would eventually form it had its own electric field lines. Even after the material collapses to form a black hole, the field lines are still there, a relic of the material that formed the black hole.

A long time ago, back when the American Journal of Physics had a question-and-answer section, someone posed the question of how the electric field gets out of a charged black hole. Matt McIrvin and I wrote an answer, which appeared in the journal. It pretty much says the same thing as the above, but a bit more formally and carefully.

Actually, I just noticed a mistake in what Matt and I wrote. We say that the Green function has support only on the past light cone. That's actually not true in curved spacetime: the Green function has support in the interior of the light cone as well. But fortunately that doesn't affect the main point, which is that there's no support outside the light cone.

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""Even after the material collapses to form a black hole, the field lines are still there, a relic of the material that formed the black hole."" So, the field lines end on the event horizon? Where is the charge? Inside or on the horizon? – Georg Jul 12 '11 at 9:21
Well, when we picture electric field lines, we picture them at a moment in time, so the answer to this question depends on a choice of a time coordinate (or at least of a particular foliation of spacetime into constant-time slices). If you use Schwarzschild coordinates for this, then yes, the field lines end on (or just barely outside of) the horizon. There's a good reason for this: in Schwarzschild coordinates, infalling matter appears to get "stuck" at the horizon, not crossing it until $t=\infty$. (I say "appears to" because this is just an artifact of a coordinate singularity.) – Ted Bunn Jul 12 '11 at 14:36

The Aharanov-Bohm effect (where an electrically charged particle is effected by the electric and magnetic fields even though it travels only in a region where these fields are zero, i.e. outside of a solenoid) demonstrates that from the point of view of quantum mechanics, the underlying fields are not the electromagnetic field, but instead the electromagnetic 4-potential. This says that forces are not enough to define physics, one must also use potentials (energies). So maybe the question should be not how the electric field gets out of the black hole but instead how the electric / electromagnetic potential gets out.

The electromagnetic potential $A^\mu$ is not uniquely defined. There is a gauge freedom; one can always add the gradient of a function of space and time to get a different electromagnetic potential $A'^{\mu}$ that has the same electric and magnetic field: $A'^{\mu} = A^\mu + \partial^\mu \Gamma$. For a charged black hole, this means that there can be a non-zero magnetic potential $A^j$, (which still gives a zero magnetic field).

Anyway, the point is that the question of "how does the electric field get out of a black hole" has an analog in the quantum mechanics of flat space-time; "how does the Aharanov-Bohm effect work?" In both cases, there seems to be a global requirement that things be consistent even though it appears that there might logically be no relationship.

To detect an electric field or an electromagnetic potential we use a small test charge. In the two cases, we consider interactions between "restricted" electrons and "test" electrons. The restricted electrons are stuck inside the black hole, or are on paths inside a solenoid that generate essentially no electric or magnetic fields external to the solenoid. The test electrons detect the electric field outside the black hole, and the electromagnetic potential outside the solenoid.

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I will say that this question has a bit of a misconception--you don't need to interact with the horizon or extreme-limit GR effects like the ergosphere in order to measure the parameters of the black hole. The metric tensor is different for black holes with different masses and charges, and have different $\frac{1}{r^{n}}$ falloffs for different values of the parameters. In principle, you can measure $M$, $Q$ and $a$ solely based on observations of the orbital parameters of three different planets in orbit around the black hole.

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I agree with Ted's answer that there is an electric field present. However I think his answer is insofar incomplete, as it doesn't say how to measure the charge of the black hole. (I didn't understand the answer of Carl fully. How come the "restricted" electrons are still valid as GR test particles? Or if they are inside the BH, how does the information about the measurement come out?)

The thing is, (at least in first orders) the dynamics of an uncharged test particle orbiting a black hole stays the same, regardless if the black hole is charged or not. All that one would infer from observing the test particle is a different BH mass.

Aside, I don't know if we have observable charged test particles at hand, which would change the game. I would reason not, as if they should be observable, they either have to consist of gas, which would spread out due to mutual repulsion, or being a macroscopic object like a star or another BH, in conflict with being a test particle. A midsized/smaller star would probably be fine, but I don't think there is a charged one out there, as it would attract lots of opposite charge to be neutral.

In general, the test particle in the Schwarzschild spacetime has a Hamiltonian that can be transformed into an spherical harmonic oszillator, with decreased centrifugal force (see D.D'Orazio, P.Saha 2010, http://arxiv.org/abs/1003.5659). In Reissner-Nordström spacetime it is fairly similar: Increased mass decreases the centrifugal force, but interestingly, increased net charge decreases it (own calculations, not in the paper).

A hypothetical experiment could run as follows:

1. Observe a test particle (e.g. galactic center star S2).
2. Have a process at hand that drops (lots of) charge with know mass into the BH.
3. The changed BH net charge will affect the orbit of the test particle as follows: If it increases the net charge of the BH, the (from test particle observation) inferred BH mass grows less than expected from the mass of the material dropped into the BH. If it decreases the net charge of the BH, we will have the opposite, the BH seems to be heavier than expected.

Let me know what you think.

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All you need is two test particles, one charged and one uncharged. Drop them in the vicinity of the black hole, and you'll see them follow different geodesics. – Jerry Schirmer Aug 21 '13 at 22:20
This is much more complicated and impractical than it needs to be. I agree with Ted's answer that there is an electric field present. However I think his answer is insofar incomplete, as it doesn't say how to measure the charge of the black hole. It does say. The first paragraph tells us to measure the field at some distance and infer the charge from that -- the same way we'd do it for any other charged. – Ben Crowell Aug 21 '13 at 22:31
@Jerry: The charged particle won't follow a geodesic, as we included an non-gravitational force. – test particle Aug 21 '13 at 22:36
@testparticle. Fine, Mr. Pedantry. substitute "paths" for "geodesics". You get my point. – Jerry Schirmer Aug 21 '13 at 22:42
@testparticle: You measure it the same way you would measure any other electric field. – Ben Crowell Aug 21 '13 at 23:06