What is the EM Lagrangian in curved spacetime?

pellman
Messages
683
Reaction score
6
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 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}
 
Last edited by a moderator:
Physics news on Phys.org
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."
 
Last edited:
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.

<br /> A_{m;n} - A_{n;m} = A_{m,n}- A_{n,m} - {\Gamma^a}_{mn} A_a + {\Gamma^a}_{nm} A_a<br />
 
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
 
Last edited:
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top