# What is the EM Lagrangian in curved spacetime?

• pellman
In summary: Kelley submitted a solution to the first question. He says that in this case, the Lagrange density would be a scalar, which is the form it assumes in general. He writes "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). 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}Wald also uses covariant derivatives on p.

#### pellman

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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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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bcrowell said:
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.

pellman said:
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.

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

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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## 1. What is the EM Lagrangian in curved spacetime?

The EM Lagrangian in curved spacetime is a mathematical expression used to describe the behavior of electromagnetic fields in a curved space. It is based on the principles of general relativity and quantum field theory.

## 2. How is the EM Lagrangian derived in curved spacetime?

The EM Lagrangian in curved spacetime is derived by combining the Lagrangian density for electromagnetism with the gravitational field equations from general relativity. This results in a modified Lagrangian that takes into account the effects of gravity on electromagnetic fields.

## 3. What is the significance of the EM Lagrangian in curved spacetime?

The EM Lagrangian in curved spacetime plays a crucial role in understanding the behavior of electromagnetic fields in the presence of strong gravitational fields, such as those around black holes. It also helps to bridge the gap between general relativity and quantum field theory.

## 4. Can the EM Lagrangian in curved spacetime be applied to other fields besides electromagnetism?

Yes, the Lagrangian formalism can be applied to other fields in addition to electromagnetism. For example, it has also been used to describe the behavior of other fundamental forces, such as the strong and weak nuclear forces.

## 5. How does the EM Lagrangian in curved spacetime differ from the Lagrangian in flat spacetime?

The EM Lagrangian in curved spacetime takes into account the effects of gravity, while the Lagrangian in flat spacetime does not. This results in a modified Lagrangian that includes additional terms to account for the curvature of spacetime. Additionally, the equations of motion derived from the EM Lagrangian in curved spacetime are more complex and require a different approach to solving them compared to the flat spacetime case.