# Same Mass But Less Gravitational Effect Due To Electric Charge?

The Schwarzschild metric describes a spherical, eternal, static black hole with no rotation and no electrical charge, so it can be used to understand gravity around objects with negligible rotations (compared to the speed of light) like the earth and the sun .

The Reissner–Nordström metric describes the same black hole, but with an electric charge.

This is the Schwarschild metric:

$$c^2d\tau^2 = \left(1-\frac{2MG}{c^2r}\right)c^2dt - \left(1-\frac{2MG}{c^2r}\right)^{-1}dr$$

This is the Reissner–Nordström metric:

$$c^2d\tau^2 = \left(1-\frac{2MG}{c^2r} + \frac{Q^2G}{4\pi\epsilon_{0}c^{4}r^{2}}\right)c^2dt - \left(1-\frac{2MG}{c^2r} + \frac{Q^2G}{4\pi\epsilon_{0}c^{4}r^{2}}\right)^{-1}dr$$

Ignoring $$\theta$$ and $$\phi$$ coordinates.

What makes me wonder is the plus sign in front of the charge-dependent term in the Reissner–Nordström metric. Unlike mass, electric charge causes geodesics around the black hole to move away from it, even if objects near the black hole are neutral (free of electrical charge).

The effect of this is that a black hole with mass M will have a greater gravitational effect on surrounding objects than a black hole with the same mass M but with electric charge. It is as if electric charge has an "anti-gravitational" effect.

In extreme, probably impossible, cases, a black hole with an electric charge much greater than its mass numerically speaking, would repulse objects instead of attracting them gravitationally.

I know this is a problem for singularities like those inside black holes, but an object doesn't have to be a black hole to be described by these metrics.

I wonder if it would be possible for an object to have more charge than mass, but not dense enough to be a black hole, to end up having an anti-gravitational effect because it has more electric charge than mass. Would it be possible? Disregarding of course the fact that such an electric charge would quickly attract opposite charges and become neutral.

One can argue that this "anti-gravitational" effect is due to the pressure of same-charged particles repeling eachother, but it affects even not charged particles outiside the event horizon, so what then? How to explain it?

• Adding lots of electric charge of the same sign should also increase net energy of the system due to Coulomb potential energy of that charge, and thus this should increase net mass as well. So it may be that it is not possible to increase $Q$ without corresponding increase in $M$. en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6m_metric Feb 10 at 0:33
• I forgot some terms, sorry. I'll correct it Feb 10 at 2:17
• I corrected the metrics, but the question is still the same. Feb 10 at 2:25
• @JánLalinský I had commented something else about irreducible mass, but I thought better of it and I think you might be right. I'll look into that. Feb 10 at 2:28