# Is it possible/correct to describe electromagnetism using curved space(-time)? [duplicate]

Comparing the simples form of the forces of both phenomena: the law of Newton for gravitation $V\propto \frac{1}{r}$, and the Coulomb law for electrostatics $V\propto \frac{1}{r}$, one might think that if one can be extended relativistically to a curvature in space-time, that the other one would lend itself to a similar description. Is this so?

Thanks for any useful thoughts and/or suggestions!

-

## marked as duplicate by Qmechanic♦Sep 28 '13 at 17:36

I don't really follow the text underneath the title, but the answer to this question is most certainly, yes. (For example, we witness electromagnetic waves traveling through curved space.) Here is how electromagnetism is described mathematically (I fear this answer is slightly beyond the level of the questioner -- sorry -- but perhaps not of other readers):

Maxwell's equations are best expressed in terms of the field strength tensor. That tensor is the curvature of a connection on a circle bundle (over spacetime). The connection is the four-vector electromagnetic potential, in physics terms. In this set up, two of Maxwell's equations are automatic (for example, saying that the magnetic field is locally the curl of a 3-vector is saying that it is divergence-free). The other two are equations for the "divergence" of this tensor. This way of phrasing the problem makes sense in any metric, i.e. on curved spacetimes.

-
Thanks! The answer is only slightly beyond my level (taking a general relativity course next year :)). Is the vector potential A a four- or three-vector in this setup? How does the general wave equation (second derivative to time of wavefunction = constant second derivative to space of wavefunction) relate to this? Is it automatically satisfied with inclusion of the fourth dimension in the derivatives? It would seem so to me, as both time and space are treated equally in that equation. –  rubenvb Nov 11 '10 at 14:50
$A$ is a 4-vector, where the time component is the scalar (electric) potential and the spatial components are the vector (magnetic) potential. It sounds like the general wave equation you're thinking of is the Klein-Gordon equation, which is actually not so general and not applicable to photons. See e.g. physics.stackexchange.com/questions/437/… –  David Z Nov 11 '10 at 15:47
Just to add a reference to Eric's answer: Maxwell's Eqs in Curved Spacetimes. –  Daniel Nov 11 '10 at 16:19
@David: no, the Klein-Gordon equation is afaik an attempt to make the schrödinger equation somewhat relativistic, the equation I had in mind applies perfectly to EM fields (it's derived from the Maxwell equations): –  rubenvb Nov 11 '10 at 23:45
@rubenvb: ah, I think I got thrown off when you said "wavefunction" - I'm just too used to hearing that term applied only to spin-0 and spin-1/2 particles. Anyway, the 4-potential $A$ does obey the wave equation $\partial_t^2 A^{\mu} = \frac{1}{c^2}\nabla^2 A^{\mu}$ in free space. So do the electric and magnetic fields. –  David Z Nov 12 '10 at 0:03

I'm not sure if this is exactly what you're getting at, but there have been attempts to derive electromagnetism as a consequence of general relativity. The most famous one is Kaluza-Klein theory, published in 1921. The theory is basically just general relativity in a 5-dimensional spacetime, where one of the dimensions is "curled up" (compact). It turns out that the additional equations of motion obtained from the compact dimension are equivalent to the Maxwell equations for electromagnetism (in the relativistic form Eric described).

-
this is the correct answer –  lurscher Mar 1 '11 at 15:59
+1 agreed. I would like to downvote the accepted answer, but this seems harsh, since it doesn't say wrong things. –  Ron Maimon Dec 8 '11 at 5:13

Everyone else here is right, but there is also an unmentioned complication to using curvature to explain electromagnetic forces:

We use curvature to explain gravity in an attempt to explain why intertial and gravitational masses are the same--why a bowling ball falls the same rate as a penny. Einstein's answer was that it's because they're just travelling in 'straight' lines in the same spacetime, and aren't feeling any forces at all--they're just trying to move in the straightest line they can.

But with E&M, not all charges feel the same force in a given electric field--in fact, chargless particles feel no force at all! So, if we're going to explain E&M using curvature, we're going to need to explain why different particles have different charges.

-
They have different charges because they have different momentum in the extra dimension. –  Ron Maimon Dec 8 '11 at 5:13