# Coulomb's Law modified in general relativity?

It seems difficult to track down a clear explanation of this statement:

So although the Coulomb law was discovered in a supporting frame, general relativity tells us that the field of such a charge is not precisely $$1 / r^2$$.

Papers I've found seem to say either that the inverse square law for the Coulomb field of a massive point charge remains exactly true in general relativity, or that it has corrections on the order of $$1/r^4$$ and higher.

I suspect that I may be getting lost in the coordinate transformations and the in-context meaning of $$r$$. What I would like to know is if general relativity predicts any deviations from the inverse square law for the electric field surrounding a point massive charged particle, as seen by a distant observer.

Presumably the measurement method should be specified, so here's a possibility: Attach a charge $$Q$$ to an extremely massive particle, then probe the electric field of the charge by first measuring the distribution of electrons, then of positrons, shot at various energies toward the massive particle a la the Rutherford experiment. The difference between the two would be used to subtract out the purely gravitational attraction between the massive particle and the probe particles. I realize that this approach would be impractical for measuring extremely small deviations from Coulomb's law, but it should at least provide a way to dodge some of the difficulties associated with defining $$r$$ near a point mass. I also realize that quantum corrections would totally change things in real experiment. I'm just looking for a clear classical explanation of what happens to Coulomb's law due to the effects of gravity per general relativity.

EDIT 12/24/18: Specifically: in the scattering measurement proposed above, does it appear to a distant observer that there is a deviation from the inverse square law - a deviation that is a bit different from that due strictly to gravity?

• I would say it's a little misguided to look for a modification of Coulomb's law that works in GR. Coulomb's law doesn't even hold in special relativity, in the sense that SR doesn't allow us to have instantaneous action at a distance, and SR also requires that the field of a charge transform nontrivially under a boost. – user4552 Dec 24 '18 at 20:38
• Fair enough. The basic question, though, seems valid: "Does it appear to a distant observer that there is a deviation from the inverse square law - a deviation that is a bit different from that due strictly to gravity?" (given two massive, charged, point particles) Suggestions on how to phrase the question better are welcome! – S. McGrew Dec 24 '18 at 20:59

The Reissner-Nordstrom metric describes a massive point charge in General Relativity. The covariant component of the electrostatic four-potential, $$A_t$$, is exactly $$Q/r$$, but the contravariant component $$A^t$$ is more complicated. Similarly, the purely covariant component $$F_{rt}$$ of the electrostatic field, and also the purely contravariant component $$F^{rt}$$, is exactly $$Q/r^2$$ but the mixed components are more complicated.

This is in Schwarzschild-style coordinates where the area of a sphere around the charge is $$4\pi r^2$$ as $$r\rightarrow\infty$$.

There is no meaningful way to say which components are the potential and the field. The potential is a four-vector and the field is a four tensor, and they have both covariant and contravariant components which are equally valid.

Regardless of which components you use, the electric flux through a sphere at infinity is $$4\pi Q$$.

When you think about Coulomb’s Law, you should not think of it in the “high-school” form that the field is inverse-square. Instead you should think of it in the “university” form that $$\nabla\cdot\mathbf{E}=4\pi\rho$$. This equation, generalized to take curved spacetime into account, remains true when the effect of gravity is considered.

• That is helpful. According to the book, Gravitation and Spacetime , By Hans C. Ohanian, Remo Ruffini, the electric field strength goes as 1/r^2, BUT r is not equal to radial distance in curved spacetime. – S. McGrew Dec 24 '18 at 18:05
• Yes, but if your conclusion from that is that “Coulomb’s Law doesn’t hold”, most physicists wouldn’t agree. They generalize Coulomb’s Law so that it does hold. The point isn’t to quibble over words like “Coulomb’s Law” and over various coordinate systems. The point is whether we understand how to correctly combine classical electromagnetism with General Relativity, and we do, in any coordinate system! The vast majority of physicists agree on what the right equations are, because they follow from basic principles like general covariance. – G. Smith Dec 24 '18 at 18:13
• I haven't yet found an explicit relation between E and radial distance, but also am not really sure what "radial distance" means in a curved spacetime -- hence the proposed measurement method that would just deal with scattering distributions at infinity. – S. McGrew Dec 24 '18 at 18:41
• You get the radial proper distance by integrating $dr/\sqrt{1-2M/r+Q^2/r^2}$. See the $dr$ term in en.m.wikipedia.org/wiki/Reissner–Nordström_metric. It will be something complicated and probably ill-behaved at the horizons, which is why physicists don’t use it much. Why would you want to choose that as a coordinate? – G. Smith Dec 24 '18 at 18:51
• Can we move the discussion to chat.stackexchange.com/rooms/info/87463/… – S. McGrew Dec 24 '18 at 19:21

All of these answers miss the essential point.

Coulomb's law relates to electrostatics: what is the force between two charges at rest? Equivalently, what is the electric field of a point charge at rest? For the sake of this question (and said wikipedia page), we neglect the mass of the point charge: the R-N metric does NOT apply. Instead, we are trying to calculate the field in a "supported" frame in a weak (by GR standards) uniform gravitational field.

Why is such a frame of interest? Because the surface of the Earth (where Coulomb's law was discovered) is such a frame. In Newtonian gravity, the surface of the Earth is an inertial frame (neglecting Earth's rotation and motion through space); objects on the surface experience a gravitational force and a balancing "normal force" of non-gravitational origin. But recall that there is no such thing as a "gravitational force" in GR. We should be free-falling towards the center of the Earth; we are not because we (and everything on Earth's surface) are "supported" against this free-fall by other forces. Therefore our frame is in uniform acceleration (relative to the local spacetime); the applicable metric is the Rindler metric.

Back to the Coulomb field: an observer in flat space measures a perfect inverse square field. Does a supported observer (as defined above) measure that as well? The answer is no: there are slight deviations from the perfect inverse square law - even though both the observer and the charge being observed are equally supported. By dimensional analysis, the first-order correction is of order [g ⋅ r / c2], with g measuring the strength of the local gravitational field (i.e. the g00 component of the metric), r the distance from the charge where the electric field is measured, and c the speed of light. So the field in the supported frame would be E ~ q (1/r2 + O[g ⋅ r/c2]). This effect has nothing to do with the question of measuring the 'r' coordinate in a non-flat spacetime.