4
$\begingroup$

How can I understand the charge of a black hole?

We can understand the charge of elementary particles, like the charge of a proton or neutron. But what does the charge of a big object like that of a black hole mean? A black hole will have so much stuff inside it. So will we add the charge of all the particles inside of a black hole to calculate the charge of a black hole as a whole? I know this sounds inappropriate, but I am very confused by the statement "Charge of a black hole". It would be very helpful if someone can clarify.

Note: There is a similar question, How can a black hole have a charge, or be charged?. I have gone through it, but I am not satisfied.

$\endgroup$
4
  • 2
    $\begingroup$ I don't see how this is different. "Take a neutral black hole. Drop in a proton. Now what? " A charged black hole is black hole with more particles of one charge than the other. What is unclear about the answers to the other question? $\endgroup$
    – James K
    Jul 28, 2022 at 5:53
  • 2
    $\begingroup$ @JamesK My question is - inside a black hole, there will be so many particles, right? It means there will be so many protons, neutrons, electrons, etc. So when we say the black hole is charged, does it mean we are adding the charge of all those protons and electrons? How is it possible to measure the charge of a black hole? $\endgroup$
    – apk
    Jul 28, 2022 at 8:47
  • $\begingroup$ @apk just like for any other object, yes. $\endgroup$
    – hobbs
    Jul 28, 2022 at 14:22
  • 1
    $\begingroup$ You may be looking for the No Hair Theorem which states what quantities can be measured for a black hole. $\endgroup$ Jul 28, 2022 at 16:59

2 Answers 2

8
$\begingroup$

You cannot dig into the black hole, count "all the quantised charges it contains", and report the result to the observer at infinity. The same applies to the BH mass -- you do not count masses of whatever particles "inside"/within the event horizon.

Instead, you can measure the electric field the BH produces far enough -- the usual way, by measuring the force upon the charged particle, subtracting the gravitational force from that (e.g. measured by the electrically neutral device/test particle of the same mass). That electric field asymptotically (at large distance $r$) scales as ~$q/r^2$. From that you infer the electric charge $q$. The force exerted upon the test particle gets somewhat more involved if the BH has non-zero angular momentum -- but the general idea is as outlined above: from observations "far enough" you find how the metric tensor and the EM vector potential depend on $r$, the coefficients at the leading terms (e.g. $\propto r^{-2}$ for the Newtonian approximation of the gravitational force and electric field) give you the mass and the charge of the BH.

$\endgroup$
6
  • 2
    $\begingroup$ Wait... but how does the electric field escape the black hole? (Layman here, explain it to me like I'm a 10 year old. 😄) $\endgroup$
    – Vilx-
    Jul 29, 2022 at 10:30
  • $\begingroup$ @Vilx: Yes, that is probably the implied question. If light can't escape, how can the electric field? If a type 7 civilization dumped gazillions solar masses worth of only electrons into a black hole, how would it behave? $\endgroup$ Jul 29, 2022 at 12:30
  • $\begingroup$ @Vilx- The answer is the same as how gravity "escapes" the black hole (it can't travel faster than light either)... and to my understanding has to do with the funny relativistic effects that happen near/at the event horizon. $\endgroup$
    – Michael
    Jul 29, 2022 at 19:26
  • 2
    $\begingroup$ Electric field does not escape the BH -- it just exists outside the BH. And if the flux of that field across a sphere around the BH (in a constant $t$ slice of spacetime) is nonzero, and there are no other charges within that sphere apart from the hypothetically charged BH -- that flux is due to the charge of that BH, and defines its charge. Source equation for EM field in GR is $\mathrm {d} \star \mathbf {F} =\mathbf {J}$, $\mathbf {J}$ being the current 3-form; like in school flat-geometry Coulomb's law "field lines end on charges"; the charges can be inside the event horizon. $\endgroup$ Jul 29, 2022 at 20:51
  • 2
    $\begingroup$ Now, if you consider the wave equation for the perturbation of EM field, you will find that perturbing the current distribution within the event horizon you do not get the perturbation of the EM field outside the event horizon. All the perturbations "stay in Vegas", invisible to the outside observer. Unlike the non-GR intuition, in which perturbations propagate through all space. EM waves, gravitational waves cannot "escape" the BH, deliver the info about what happened inside to the outside observer. Static electric field can exist around BH, it carries no info about the dynamics inside. $\endgroup$ Jul 29, 2022 at 21:06
7
$\begingroup$

The charge of something is defined as the sum of the charge it contains, $\Sigma q_i$ where $q_i$ are the individual, quantised, charges and which may be positive or negative. The same applies to a black hole.

The charge could be inferred by looking at the electric field around a black hole.

No astrophysical (stellar or supermassive) black holes are expected to be significantly charged, but the possibility exists for primordial black holes of much lower mass (e.g., Kritos & Silk 2022).

$\endgroup$
2
  • $\begingroup$ Shouldn't primordial black holes be the ones to lose their charge most easily, since they evaporate much faster? $\endgroup$ Jul 29, 2022 at 10:11
  • $\begingroup$ @TomášZato-ReinstateMonica I refer you to the paper I have cited. $\endgroup$
    – ProfRob
    Jul 29, 2022 at 10:27

Not the answer you're looking for? Browse other questions tagged .