Maxwell equations in curved spacetime

I am a physics student and I am working on master thesis from quantum mechanics now.

My thesis advisor told me that Maxwell equations exist only in spacetime where the scalar curvature equals zero, $$R=0$$

Is it true? And what is the problem here and the reason that we need extra dimensions?

• Did you check Wikipedia? – Qmechanic Mar 13 '17 at 20:49
• I cant find the answer. The point is, that we somehow cant find gauge field if is there is nonzero curvature. This is non trivial. And I am surprised. Schwarzschild spacetime has nonzero curvature and we live in it peacefully. With all Maxwell equations. I cant figure out what is problem – marek Mar 13 '17 at 21:19
• What does "only exists" mean? What's wrong with writing down $dF=0$ and $d\star F=J$? I can do that without any reference to curvature. Regarding your comment, $R=0$ is not the same as $\mathrm{Riem}=0$, and Schwarzschild does indeed have $R=0$. – Ryan Unger Mar 13 '17 at 21:30
• I'm afraid it is unclear what you're asking unless you can give a better reference for the claim that Maxwell's equations can only work at $R=0$ than "my advisor told me". – ACuriousMind Mar 13 '17 at 21:55
• Is your question about ME or about what happens to the gauge symmetry when $R\neq0$? (I have never thought the latter through, but, without thinking deeply, would be surprised if there were any issues there - you can "surely" add any closed one-form to $A$, no?). Also, list what you have researched - hopefully people will take this question more seriously then. You are validly seeking clarification on what you find a confusing comment (I presume you have already asked your adviser without getting an answer you can understand). – WetSavannaAnimal Mar 13 '17 at 23:39

We'll work with $\mu_0=1$. From the Lagrangian density $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-A_\nu j^\nu=-\frac{1}{2}\partial_\mu A_\nu F^{\mu\nu}-A_\nu j^\nu$ and the definition $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$, we can obtain the Maxwell equations in Minkowski space. In a curved spacetime, the Lagrangian density is multiplied by $\sqrt{\left| g\right|}$. Replacing partial derivatives with covariant ones doesn't change $F_{\mu\nu}$ (provided the connection is torsion-free), but we do need to promote $\partial_\mu A_\nu$ in the second formula with $\nabla_\mu A_\nu$.
Nothing "goes wrong" in an $R\ne 0$ spacetime, but there is one subtlety, which is the closest I can think of to redeeming the confusing comment your advisor meant. We can add a total derivative, say $\partial_\mu \left( \sqrt{\left| g\right|}V^\mu\right) =\sqrt{\left| g\right|}\nabla_\mu V^\mu$, to the Lagrangian density. (I'll denote equivalence up to such terms with $\approx$.) Let's rewrite the term not proportional to $j$ again, without the $-\frac{1}{2}$ factor: $$\nabla_\mu A_\nu \nabla^\mu A^\nu -\nabla_\mu A_\nu \nabla^\nu A^\mu\approx \nabla_\mu A_\nu \nabla^\mu A^\nu - \nabla_\mu A_\nu \nabla^\nu A^\mu-\nabla_\mu\left( A^\mu\nabla_\nu A^\nu-A^\nu\nabla_\nu A^\mu\right).$$This choice of total derivative obtains the popular result $\nabla_\mu A_\nu \nabla^\mu A^\nu-\left(\nabla_\mu A^\mu\right)^2$, plus two more terms that cancel when $R_{\mu\nu}=0$. Explicitly (and if you want to verify this with your own calculation, I've swapped two dummy indices in one term) $$A^\mu\left[ \nabla_\mu,\,\nabla_\nu\right]A^\nu=A^\mu R_{\mu\rho}A^\rho.$$
• @marek Just about any resource on general relativity will prove vector fields satisfy $\left[ \nabla_\mu,\,\nabla_\nu\right] X_\rho = R_{\mu\nu\rho\sigma}X^\sigma$. The rest is straightforward enough, as long as you're familiar with the electromagnetic Lagrangian. Deriving Euler-Lagrange equations in curved space can be tricky because of all the Christoffel symbols, but I present a simplified notation for such calculations in Sec. 1.3 of my thesis: goo.gl/q8Cshm – J.G. Mar 13 '17 at 23:24