# Maxwell's Equations in curved spacetime

I know that we can write Maxwell's equations in the covariant form, and this covariant form can be considered as a generalization of these equations in curved spacetime if we replace ordinary derivatives with covariant derivatives. But I have read somewhere that this generalization is not unique and it is just the simplest one. Can anyone introduce some resources about this subject and how electromagnetism and Maxwell's equations are generalized to curved spacetime?

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Can you please provide references where you've read "that this generalization is not unique and it is just the simplest one"? –  Cristi Stoica Jul 11 at 17:52
in a book by Narlikar ("introduction to cosmology"), in a section about the principle of equivalence: "this generalization of (2.65) to (2.66) is called the minimal coupling of the field with gravitation, since it is the simplest one possible." –  Mostafa Jul 11 at 18:09
Hi @Mostafa: Concerning retagging: If you haven't done this already please read the wiki tags for the ref.req. and books tags. I believe the books tag is appropriate here, while the ref.req. tag is not. –  Qmechanic Jul 11 at 18:43
Maybe he's thinking along the lines of the options with the KG equation in curved space: $(\Box+m^2+\xi R)\phi = 0$ where $\xi=0$ is minimal coupling and $\xi=\frac{1}{6}$ is conformal coupling. (Birrell and Davies eq 3.26)? –  twistor59 Jul 11 at 19:17
Maxwell's equations characterize the critical points of the Yang-Mills functional of a $U(1)$ principal connection; arguing from a subjective point of view of mathematical 'beauty', any generalization that doesn't come with an equally nice geometric story would smell iffy to me... –  Christoph Jul 12 at 0:22

Here is why I doubt there are other ways to generalize Maxwell's equations to curved spacetime.

Special relativity was obtained from the invariance of the speed of light. In special relativity, the electric field is not a vector field, and the magnetic field is not a pseudovector, but that they transform as the components of a two-form $F_{ab} = \partial_a A_b -\partial_b A_a$, where the four-vector $A_a$ contains the scalar and vector potentials.

Maxwell's equations become $$d F=0$$ $$d\ast F=J$$

When moving to curved spacetimes, they remain the same, since the Hodge dual $\ast$ is defined at each point $p$ of the manifold, on $\wedge^\bullet T^\ast_p$. When expressed in this form, the covariant derivative is not involved, although the metric is involved in the $\ast$ operator.

While I think the generalization of Maxwell's equations to curved spacetime is very rigid and I see no choice based on simplicity here, it is known that there are modified (nonlinear) versions, like the Born-Infeld theory. But they did not originate because of some freedom of generalizing Maxwell's equations to curved spacetimes.

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OK, good, so I'm not crazy. I saw this, thought the same thing (but wasn't sure), and thought I'd wait for someone else to answer this. –  Jerry Schirmer Jul 11 at 18:30
I haven't thought about this much, so forgive me if this is naive, but couldn't one conceivably for example add any terms to either of these equations that vanishes on Minkowski space and still preserve the property that the equations reduce to the ones your wrote down when doing EM in Minkowski? Of course these equations would not be as "simple" in some sense, but still I'm just trying to explore precisely how rigid the generalization is. –  joshphysics Jul 11 at 18:48
@joshphysics The easiest thing to do would be to suggest some different EM Lagrangian that is the usual, accepted one plus some higher-order deviation. This is precisely what's done for alternate theories of gravity, which are meant to replicate GR in some approximation but have differing behavior on some length scale. –  Muphrid Jul 11 at 19:27
I see Mostafa's comment giving the reference, and I agree with Narlikar and with joshphysics. The choice of the generalization is due to the minimal coupling principle, which is a consequence of the principle of equivalence. My rigidity argument contains them implicitly, because the $\ast$ operator depends only on the metric, and not on the covariant derivatives. We can write Maxwells equations using covariant derivative, but the terms containing the derivatives of the metric actually cancel each other. –  Cristi Stoica Jul 11 at 20:42