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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    $\begingroup$ How are we to look at radius of event horizon? we cannot measure schwarzschild radius. $\endgroup$
    – Newman
    Commented Aug 17, 2011 at 20:43
  • $\begingroup$ We could probably measure it by the effects of gravitational lensing, or just simply the gravitational pull. $\endgroup$ Commented Aug 17, 2011 at 23:20
  • $\begingroup$ Related question for the gravitational field: physics.stackexchange.com/q/937/2451 $\endgroup$
    – Qmechanic
    Commented Feb 22, 2020 at 11:21

3 Answers 3


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.

  • $\begingroup$ ""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? $\endgroup$
    – Georg
    Commented Jul 12, 2011 at 9:21
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    $\begingroup$ 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.) $\endgroup$
    – Ted Bunn
    Commented Jul 12, 2011 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.


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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