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?

  • 3
    $\begingroup$ Did you check Wikipedia? $\endgroup$ – Qmechanic Mar 13 '17 at 20:49
  • $\begingroup$ 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 $\endgroup$ – marek Mar 13 '17 at 21:19
  • 3
    $\begingroup$ 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 $R=0$. $\endgroup$ – Ryan Unger Mar 13 '17 at 21:30
  • 2
    $\begingroup$ 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". $\endgroup$ – ACuriousMind Mar 13 '17 at 21:55
  • 1
    $\begingroup$ 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). $\endgroup$ – 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) $$A^\mu\left[ \nabla_\mu,\,\nabla_\nu\right]A^\nu=A^\mu R_{\mu\rho}A^\rho.$$

  • $\begingroup$ Here we go. Its not so stupid question, I asked. This should work. Could you please add any resources where I can find something more about it? $\endgroup$ – marek Mar 13 '17 at 22:43
  • $\begingroup$ @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 $\endgroup$ – J.G. Mar 13 '17 at 23:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.