# What is the EM Lagrangian in curved spacetime?

## Main Question or Discussion Point

In flat space time the Lagrangian for the EM potential is (neglecting the source term)

$$\mathcal{L}_{flat}=-\frac{1}{16\pi}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})$$

which is a scalar for flat spacetime. I would have expected the generalization to curved manifolds to be

$$\mathcal{L}_{curved}=-\frac{1}{16\pi}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu})\sqrt{-g}$$

However, the Wikipedia article for [URL [Broken] equations in curved spacetime[/url] gives the Lagrangian still in terms of the regular derivatives, not covariant derivatives. But $$\mathcal{L}_{flat}$$ is not a scalar in general, is it? I was just reading an article about covariance and Noether's theorem that likewise uses an EM Lagrangian in terms of partial derivatives instead of covariant derivatives.

So which is correct?

Edit: I just checked Gravitation (Misner, Thorne, Wheeler) and eq 22.19a gives the EM tensor as $$\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}$$

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bcrowell
Staff Emeritus
Gold Member
Wald also uses covariant derivatives on p. 455.

But did you see the following text in the WP article? "Despite the use of partial derivatives, these equations are invariant under arbitrary curvilinear coordinate transformations. Thus if one replaced the partial derivatives with covariant derivatives, the extra terms thereby introduced would cancel out."

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But did you see the following text in the WP article? "Despite the use of partial derivatives, these equations are invariant under arbitrary curvilinear coordinate transformations. Thus if one replaced the partial derivatives with covariant derivatives, the extra terms thereby introduced would cancel out."
No. I didn't read that . Thanks. I thought that might be it though and attempted to work it out myself earlier. I must have made a mistake. They didn't cancel for me. I'll give it another shot.

No. I didn't read that . Thanks. I thought that might be it though and attempted to work it out myself earlier. I must have made a mistake. They didn't cancel for me. I'll give it another shot.
They must cancel because of the symmetry of the Christoffel symbols.

$$A_{m;n} - A_{n;m} = A_{m,n}- A_{n,m} - {\Gamma^a}_{mn} A_a + {\Gamma^a}_{nm} A_a$$

dextercioby
Homework Helper
The title of the thread contains a good question. In the absence of torsion on the manifold (hypothesis which is of course not assumed in the question), the Lagrange density can contain partial derivatives, i/o covariant ones, because the necessary antisymmetry of the E-m field 2-form, coupled with the symmetry of the (probably metric compatible) connection would allow that.

A more interesting question arises though in the following case: Assuming the manifold has non-vanishing curvature and torsion and the affine connection is not metric compatible, what form would the field equations for the abelian gauge field have and moreover, assuming the existence of an action integral and a variational principle leading to the field equations, what form would the non-integrated Lagrangian have ??

I' m not aware of a full solution to my 2 queries, though I believe the first one has been already addressed (there's an article on arxiv.org as far as I recall).

Daniel

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