# Coulomb's Law in the presence of a strong gravitational field

I was under the impression that the $1/r^2$ falloff of various forces were because of the way the area of a expanding sphere scales. But that strict $1/r^2$ falloff would only be globally true in a strictly Euclidean geometry, yes? So, if you had an electron and a positron in a warped space due to gravity, wouldn't that $1/r^2$ term in Coulomb's Law end up needing to be adjusted because of the warping of space leading to an increase or decrease in effect due to gravitational lensing or scattering?

I found one related question here: Does Gravity Warp Coulomb's force? But it only has one down-voted and unclear answer.

• Seems like this Wikipedia article might be useful. – Kyle Kanos Dec 5 '14 at 21:07
• I think if I could understand that article, I wouldn't be asking this question. – Shufflepants Dec 5 '14 at 21:51
• I suppose there is that aspect that I hadn't exactly considered. However, I suppose that it could at least impart the notion that yes, there is something to GR + E&M. – Kyle Kanos Dec 5 '14 at 22:03
• @KyleKanos I wrote that article and am interested in the answer to OPs question. It's not in there. – Ryan Unger Dec 5 '14 at 22:52
• @0celo7: The statement in the second paragraph, This Lagrangian is obtained by simply replacing the Minkowski metric in the above flat Lagrangian with a more general (possibly curved) metric $g_{\mu\nu}(x)$., seems to be exactly what you're missing & what OP wants. Note also that by scrolling just a hair, you'd see the same equations that ACuriousMind presents. – Kyle Kanos Dec 6 '14 at 1:28

Of course Coulomb's law has to be adapted! And it is therefore fortunate that there exist manifestly covariant formulations of electromagnetism that do not care how spacetime is curved. However, we should first briefly observe that Coulomb's law is not one of the fundamental laws of electromagnetism, though it has played a great role in its inception:

Coulomb's law is just the solution of Maxwell's equations for a point charge and no current in flat Minkowski space. Maxwell's equations can jointly be generalized to arbitrary spacetimes:

The electric field strength is a 2-form $F = \frac{1}{2}F_{\mu\nu}\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu$ on spacetime, and electric current is a 3-form $J = \frac{1}{6}J_{\mu\nu\rho}\mathrm{d}x^\mu \wedge \mathrm{d}x^\nu \wedge \mathrm{d}x^\rho$, as the Hodge dual of the usual vector current. Maxwell's equations now simply read

$$\mathrm{d}F = 0 \; \text{and} \; \mathrm{d}\star F = J$$

where, since the Hodge star depends on the metric, the curvature of spacetime indeed influences the form of our laws.

It has to be noted that, on arbitrary spacetimes, the notion of having "separate laws" for electric on magnetic fields doesn't really make sense anymore, since they mix in (almost) arbitrary ways, depending on the metric. You can still get the electric and magnetic fields as components $F^{0i}$ and $F^{ij}$ of the field strength, but you won't be writing any nice, frame-independent laws for them. Maxwell's equations do not dissolve nicely into "Gauss' law", "Faraday's law" or such things in a general setting.

• Man, I hate overloaded notation. The only use I know for upside-down V is as a logical "and", and super scripts are exponents. But I'm sure that's not what they mean here. – Shufflepants Dec 5 '14 at 21:56
• @Shufflepants: The superscripts are spacetime indices, essentially labelling entries in a matrix/vector. The $\wedge$ is the Wedge product. – ACuriousMind Dec 5 '14 at 21:59
• @Shufflepants I'm curious as to why you decided to accept this answer. Could you explain? It (naively) seems like this answer was too technical for you to understand much of the content (no offense meant, of course)... – Danu Dec 5 '14 at 22:05
• @Danu While I don't understand the math behind how electromagnetism is handled in a more general case, I chose this answer because it did answer my question. Those equations do have to change in light of general relativity. At the outset I had the expectation that if it was the case, that I wouldn't fully understand the math involved. But this answer is fine in that I got the answer I was looking for, and there is more detail here for anyone else who might understand the math but wasn't familiar with the physics. – Shufflepants Dec 5 '14 at 22:15
• @JamalS: I believe that "overloaded" is used in the sense of programmers here, where an "overloaded" function/notation is one that has different meanings dependent on context. – ACuriousMind Dec 6 '14 at 14:29

Yes. Strictly speaking you can't apply Coulomb's law, or in general any law about the falloff of something with distance, in curved space.

Instead you have to shift to a field-based formalism. You can calculate the way the electromagnetic field propagates through a curved background—basically you take Maxwell's equations in tensor form and replace ordinary derivatives with covariant derivatives wrt the spacetime metric (for gory details, see Wikipedia). That may result in the effective falloff being weaker or stronger than $1/r^2$ depending on specific conditions, as the EM field lines will be distorted by the curvature.

• Coloumbs law holds perfectly well for say a charged back hole. r is defined though via the surface area of a spherical surface. If things are not symmetric (e.g. electron outside a non charged black hole) then the same principle applies, but there is some more math to do to account for the non symmetric curvature. So I am not sure one needs to resort to fields for the OP. – Tom Andersen Jan 2 '15 at 15:45
• @TomAndersen It may still work in a spherically symmetric case, but the OP was asking about curved spacetime in general, IIUC. In general there is no unique notion of the distance between two points, so no way to even formulate a Coulomb-like law. – Nathan Reed Jan 4 '15 at 3:48

There's no need to invoke spacetime curvature here in order to obtain some first order nontrivial results, at least in weak gravitational field, such as Earth's.

Because of the equivalence principle, homogeneous gravitational field is indistinguishable from the accelerated frame. Therefore, freely falling observer in, e.g. Earth's gravitational field, will observe that stationary charge has the same electromagnetic field as an uniformly accelerated charge outside of the gravitational field.

On order to see what will stationary observer see, one only has to make coordinate transform from freely falling reference frame to stationary frame.

The 1/r2 law holds perfectly as long as you are willing to reconsider exactly what 'r' means. 1/r2 has to hold for the same reasons that the mass of a black hole does not change no matter how far away from it you are. Its a conservation thing.

Basically the radial coordinate is defined by measuring the surface area of a sphere around the mass (or point charge or charged black hole...) and then r is defined as AreaOfSphere/sqrt(4pi).

Here is the difference - in order to walk from say r = 2km to r = 1km you will in general walk further than 1km! The result of this is the canonical drawing of a black hole as a horn.

Reference:

...in general, the Schwarzschild radial coordinate does not accurately represent radial distances... http://en.wikipedia.org/wiki/Schwarzschild_coordinates

That section on the wikipedia page has more detail.

If things are non - symmetric like in the question, then one has to adjust for the warping. Its the same concept as for a charged black hole though - Coloumbs law works perfectly as it has to to conserve charge.