It's well-known that gravity is the weakest of the 4 fundamental forces. With this knowledge, would it be possible to cancel out the gravitational forces of a black hole with it's charge?

Say a black hole consisted (previously) of only protons and positrons. A large mass of hydrogen approaches. Because of the charge, and the strength of the Electromagnetic force, the electrons are first stripped from the hydrogen atoms. Then, because the Hydrogen atoms are now H+ ions, they repel this black hole, with more force than gravity provides, making the H+ cloud inadmissible to the mass of the black hole.

The formulae for calculating attraction is the same for both electrostatic force, as well as gravitational force, the only difference being the constants and input units:

$ F = C \frac{m_{1} * m_{2}}{r^{2}} $

In this case, C is the replaced constant, for gravity, that constant is:

$ 6.67 × 10^{-11} N·m^{2} / kg^{2}$

and for electrostatic force, the constant is:

$8.98 × 10^{9} N·m^{2} / C^{2} $

The difference is about 21 orders of magnitude. considering that the mass beyond the event horizon is gone forever, meaning the electrostatic force is irrelevant beyond the event horizon. Even if such a charge of the black hole were to blow the mass apart normally, the event horizon makes it impossible. The electrostatic force would need to accelerate the charges beyond the speed of light to escape.

Is such an object possible to create?

If such an object were to exist, would the electrostatic force make additions to the mass of the hole impossible? (ie nothing could further enter the event horizon which was positively charged)?

Is there a critical point where the above would be true? (For example, if C > m/(10^21) where C is coulombs of charge and m is mass in kg, then the effects of gravity would be canceled out by electrostatic force for an object with the same charge.)

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    $\begingroup$ Try physics.stackexchange.com $\endgroup$ Dec 19, 2021 at 23:44
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    $\begingroup$ You won't find answer there (physics se) since this extremal bh concept is both hypothetical and paradoxical. Been there, done that ;D $\endgroup$
    – user6760
    Dec 20, 2021 at 2:47
  • $\begingroup$ Could this be adapted to be on-topic here? $\endgroup$ Dec 20, 2021 at 4:02
  • $\begingroup$ The Wikipedia article on Charged black holes says they are mathematically possible but suspected to not form in nature. $\endgroup$
    – Daron
    Dec 20, 2021 at 14:57
  • $\begingroup$ This thread suggests you could build a highly charged (relative to its mass) black hole by (a) starting with a large uncharged black hole (b) shooting as many protons as it can handle before it starts repelling the protons instead and (c) waiting for the mass to decay via Hawking radiation. physics.stackexchange.com/questions/168891/… $\endgroup$
    – Daron
    Dec 20, 2021 at 15:01

3 Answers 3


The metric.

A charged blackhole follows Reissner–Nordström metric, which is: $$ c^2 d\tau^2 = \left(1 - \frac{r}{r_s} + \frac{r^2}{r_Q^2}\right) c^2 dt^2 - \left(1 - \frac{r}{r_s} + \frac{r^2}{r_Q^2}\right)^{-1}dr^2 - r^2 d\theta^2 - r^2\sin^2\theta d\phi^2 $$

Where: $$ r_s = \frac{2GM}{c^2},\quad\quad r_Q^2 = \frac{GQ^2}{4\pi\epsilon_0 c^4} $$

Notice if the charge is zero $Q=0$, then $r_Q = 0$, and we are back to a Schwarzschild metric, the usual metric around blackholes. For every discussion we gonna make, feel free to set $r_Q = 0$, in order to visualize what would happen to a 'normal' blackhole.

Just so we know what we are talking about, I better explain the concept of a metric. It measures the spacetime distance, or, the proper time $d\tau$, between two events very close to each other. An event is a spacetime point $(t,x,y,z)$, or, in spherical coordinates, $(t, r, \theta, \phi)$.

Example, if the events are separated by $dt = t' - t$ and $dr = r' - r$ and so on, then, the spacetime difference is given by above formula. Of course, $r'-r$ must be pretty small for this to work, and the formula is exact for infinitesimal distances.

The horizon.

The problem, is the $dr$ term. Look at it! The term that multiplies $dr^2$ in the metric, is something: $(\cdot)^{-1}$. There's a power -1. In other words, its in the denominator: $1/(\cdot)$. What happens if the denominator is zero? Then it means for an infinitesimal distance $dr$, we have potentially infinite spacetime distance $d\tau$. Do you get it? Yes! You do! We are talking an event horizon here. An horizon, where events are not accessible, as it takes an infinite amount of time to be seen.

So let's do it! Let's make it zero and locate the horizon. In the uncharged case, $Q = 0$, $r_Q = 0$, we have: $$ 1 - \frac{r}{r_s} = 0 \quad\implies\quad r = r_s $$

Thus, the event horizon of the blackhole is a spherical surface with radius equal the Schwarzschild radius, $r_s$.

Thankfully for us, for the charged case, it is just a quadratic formula: $$ 1 - \frac{r}{r_s} + \frac{r^2}{r_Q^2} = 0 \quad\implies\quad r_{\pm} = \frac{1}{2}\left(r_s \pm\sqrt{r_s^2 - 4r_Q^2}\right) $$

Keep in mind that, the greater the charge $Q$, the greater the length $r_Q$ is. We can analyze the discriminant $\Delta$.

  • If $r_s > 2r_Q$: Just a little bit of charge. In this case, the discriminant is positive, $\Delta > 0$. There are two horizons. The usual event horizon $r_+$, and an inner horizon, called Cauchy horizon $r_-$.

  • If $r_s = 2r_Q$: A specific amount of charge. In this case, the discriminant is zero, $\Delta = 0$, and both horizons meet: $r_+ = r_- = \frac{1}{2}r_s$. This blackhole is called an extremal blackhole.

  • If $r_s < 2r_Q$: Lots of charge. In this case, there are no more horizons, as $\Delta < 0$. There's no real solutions to this quadratic equation. This means, it isn't a blackhole anymore.

The forces.

The ratio of gravitational and electric forces are: $$ \frac{F_m}{F_q} = \frac{GMm}{r^2}\frac{4\pi\epsilon_0 r^2}{Qq} = 4\pi G\epsilon_0\frac{M}{Q}\frac{m}{q} = 4\pi G\epsilon_0\left(\frac{r_s c^2}{2G}\sqrt{\frac{G}{4\pi\epsilon_0 c^4 r_q^2}}\right)\frac{m}{q} $$

Thus: $$ \frac{F_m}{F_q} = \frac{1}{2}\frac{m}{q}\frac{r_s}{r_Q}\sqrt{4\pi\epsilon_0 G} $$

Where $\sqrt{4\pi\epsilon_0 G}\approx 8.61\cdot 10^{-11} C/Kg$. Normally, $m/q\ll 1$, for instance, for an electron, $m/e\approx 5.68\cdot 10^{-12} Kg/C$. And, assuming extremal blackhole, $r_s/r_q = 2$, we then have $F_m/F_q\ll 1$. So, electrostatic forces are much larger than gravitational forces.

Of course, this is a naive calculation. To get the real thing, I'd have to calculate the effective potential of this blackhole as to figure out exactly what are the conditions for like charge to enter the black hole and increase its charge.

Effective potential.

For that we need the equations of motion, and thus we need to calculate the extremum of the action.

$$ S = \int Ld\tau = \int g_{\mu\nu}(x) dx^\mu dx^\nu = \int\left[-A(r)c^r\dot t^2 + A(r)^{-1}\dot r^2 + r^2\left(\dot\theta^2 + \sin^2\theta\dot\phi^2\right)\right] d\tau $$

Where here it was considered $A(r) = 1 - \frac{r}{r_s} + \frac{r}{r_Q^2}$. Actually, this is not exactly the action (we missed a square root), but, it can be shown that the extremum of the real action coincides with the extremum of this action $S$ that we are working with. So, we are good, if our goal is to find quantities only dependent on the action extremum, such as, the equation of motion. The integrand can be seen as a Lagrangian, thus, we can plug it in Euler-Lagrange equations, and figure out everything we need. The calculations are not hard, but, I shall omit them. After doing all of that, we get the final answer: $$ \frac{1}{2}\dot r^2 + \frac{1}{2}A(r)\left(\frac{\ell^2}{r^2} + c^2\right) = \frac{1}{2}\frac{E}{c^2} $$

Where the conserved quantities $\ell$ and $E$ taken from the $\theta$ and $t$ Euler-lagrange equations, are exactly the angular momentum and energy of the system. We can compare the above equation with the classical equations of motion, and thus we can identify the effective potential from the middle term $\frac{1}{2}A(r)\left(\frac{\ell^2}{r^2} + c^2\right)$. Thus, plugging in for $A(r)$, we finally have it: $$ V_{eff}(r) = \frac{1}{2}\left(1 - \frac{r_s}{r} + \frac{r_Q^2}{r^2}\right)\left(\frac{\ell^2}{r^2} + c^2\right) $$

Plugging in $r_s$ and $r_Q$, and, expanding, we get: $$ V_{eff}(r) = \frac{c^2}{2} + \frac{\ell^2}{2r^2} - \frac{GM}{r} - \frac{GM\ell^2}{r^3 c^2} + \frac{GQ^2}{8\pi\epsilon_0 c^2 r^2} + \frac{GQ^2\ell^2}{8\pi\epsilon_0 r^4 c^4} $$

Looking at this, we can identify several terms that would show up in a newtonian theory, and several other new terms, which are there because this is general relativity. Furthermore, this is not exact, as we still would have to introduce the electrostatic potential in an ad hoc way, because we didn't put the electrostatic contribution directly into the action (it would further complicate the action significantly as I would have to include 4-potential terms in the action like $q A_\mu\dot x^\mu$ and, I am lazy, okay?!).

From here, all we do, is to figure out the maximum and minima of the potential, so as to figure out what is the energy barrier a particle with angular momentum $\ell$ would need to overcome, in order to get inside the charged black hole. And thus, if we set the particle with greater energy, it gets through. In order words, we should compute the derivative and set to zero, that is, $V'(r) = 0$. I shall cleverly stop here because I am lazy, because finding the potential maximum/minimum here would involve solving a third degree polynomial.

However, you could, if you want, plot $V_{eff}$, to see how it behaves. Or, if you want to solve it, go ahead.

Let's now answer your questions.

Is such an object possible to create?

Oh well. I don't see any GR objections of creating such an object. So, if you are asking it is possible to create one? Yes. Probably. Maybe.

However, if you are asking: an object like that exists out there? Unlikely. Very unlikely. Very very unlikely. No, it doesn't! Without external interference, nature tends to neutralize charge pretty quickly.

If such an object were to exist, would the electrostatic force make additions to the mass of the hole impossible?

Probably not. It would be difficult though. Like we've done, $F_m/F_q\ll 1$, so electrostatic forces dominate. Like I said before, without external interference, nature tends to neutralize charge pretty quickly.

Even though I didn't prove this, I suspect a particle with sufficient energy would be able to overcome the electrostatic potential barrier and get across. Thus, not impossible.

[...] considering that the mass beyond the event horizon is gone forever, meaning the electrostatic force is irrelevant beyond the event horizon.

You can assume that, yes. But, in reality, it does not happen. Black-holes do evaporate due to Hawking radiation.

Even if such a charge of the black hole were to blow the mass apart normally, the event horizon makes it impossible.

Even if we assume black holes do not evaporate....

What horizon? Its gone! Like we've shown, if there's too much charge such that $2r_Q > r_s$, such an object would no longer have an horizon. So, charge is free to fleeeeeeeee.

  • $\begingroup$ You mention evaporation a few times. Does the power from Hawking radiation change with the charge of the black hole? $\endgroup$
    – tuskiomi
    Dec 24, 2021 at 10:51
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    $\begingroup$ @tuskiomi Yes. You can check this post for more information. $\endgroup$ Dec 27, 2021 at 0:10

Physicist137's answer is the most mathematically-correct way to handle this, but I'd like to sidestep the GR, because it's not necessary to answer your main question. Specifically this part:

Is such an object possible to create?

The answer is no, not out of any particles we know of, for a really simple reason.

As you state in your question, you need to have more charge than mass, so let's build our black hole out of the particle that has the best charge-to-mass ratio in the universe: the electron. Let's shoot a bunch of electrons into a tiny space until they collapse into a black hole. Nothing in fundamental physics (that I know of) prevents us from doing this; we will indeed create a black hole this way.

However, even this optimal charge-to-mass black hole will still not have enough charge to cancel out its gravitational pull (it will not be extremal). The problem is, the electrons' repulsion creates potential energy, and in relativity, energy acts like mass ($E = mc^2$). That is, energy density curves space-time and causes gravitational attraction, just like mass density. And as this simple calculation shows, this energy is well enough to outweigh the charge. Even this entirely-electron negatively-charged black hole will still attract negatively-charged particles, and its charge-to-mass ratio will go down with every added particle!

So you can't get an extremal or super-extremal black hole from building it with any particles we know of. That said, if the universe already had one of these (either it existed since the beginning, or it formed through some physical process we can't explain yet), then that's where you'll need general relativity to describe how it works. I'll defer to Physicist137's answer for that.


Then, because the Hydrogen atoms are now H+ ions, they repel this black hole, with more force than gravity provides, making the H+ cloud inadmissible to the mass of the black hole.

You are missing a crucial point here: the cloud of H+ will be flying away from itself, because similarly charged particles repel each other, and they will be much closer to each other than they are to the black hole and there is nothing to keep them together.

And once they have been divided, the black hole will conquer them.

  • $\begingroup$ But once the black hole has absorbed half the ions it will become charged and exert a pushing force on the incoming ions. $\endgroup$
    – Daron
    Dec 20, 2021 at 15:03
  • $\begingroup$ I'm going to -1 this because it appears to be a comment, and doesn't answer any of the questions posed. $\endgroup$
    – tuskiomi
    Dec 20, 2021 at 21:58

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