Understanding Electromagnetism & General Relativity

[Anypodetos]

I'm trying to understand how the various EM tensors work in General Relativity. The only source I've found is https://en.wikipedia.org/wiki/Maxwell's_equations_in_curved_spacetime, but there are two things I don't get.

Why do they use ordinary partial derivatives instead of covariant ones? It's clear for the definition of F_μν because here the correction terms (Christoffel symbols) cancel, but I don't see how that should work for the definition of J^μ. And further down, the continuity equation is given as ##\partial_μ J^μ =0 ##. Isn't this equation coordinate system dependent? How can this be a law?

The second (possibly related) issue is their use of tensor densities for D and J. This means the time component of J has a dimension of charge per unit cube, whereas using a vector would mean charge per lengh cubed; is that right? Could that somehow fix the problem with the partial derivative? The maths is beyond me, I'm afraid.

 

[PeterDonis]

I don't see how that should work for the definition of J_μ.

Have you tried working it out?

The second (possibly related) issue is their use of tensor densities for D and J. ... Could that somehow fix the problem with the partial derivative?

Yes. (I'm not sure about your description of the units involved, but the fact that D is a tensor density is critical, yes.)

The maths is beyond me, I'm afraid.

Then you won't be able to see for yourself why it's true, I'm afraid.

 

[Anypodetos]

Have you tried working it out?

Yes. Setting μ₀=1, If ##\nabla_μ (F^{μν} \sqrt{-g}) = \partial_μ (F^{μν} \sqrt{-g}) ##, then
$$ (Γ^μ_{αμ} F^{αν} + Γ^ν_{αμ} F^{μα} - Γ^α_{αμ} F^{μν}) \sqrt{-g} =0$$
Then divide by ##\sqrt{-g}##, and then I'm stuck. I tried to express the Γ's by g's, but there doesn't seem to be anything I can eliminate.

Then you won't be able to see for yourself why it's true, I'm afraid.

I meant I can't figure out how to solve this by myself. I do unterstand tensor calculus (hopefully enough of it), so I think I'd be able to work it out given some hints...

 

[Orodruin]

And further down, the continuity equation is given as ∂μJμ=0∂μJμ=0\partial_μ J^μ =0 . Isn't this equation coordinate system dependent? How can this be a law?

Try writing ##J^\mu = \sqrt{|g|} j^\mu##. Since ##J^\mu## is a vector density of weight one, this will make ##j^\mu## an actual vector. See what you can deduce for ##\partial_\mu J^\mu## from this.

 

[Anypodetos]

See what you can deduce for ##\partial_\mu J^\mu## from this.

I assumed that the covariant derivative of ##\sqrt{|g|}## vanishes, of which I am not certain (that's the "?").
$$ \partial_μ \left( \sqrt{|g|} j^μ \right) = \sqrt{|g|} \partial_μ j^μ + j^α \partial_α \sqrt{|g|} \overset{?}=
\sqrt{|g|} \partial_μ j^μ + j^α Γ_{μα}^μ \sqrt{|g|} = \sqrt{|g|} (\partial_μ j^μ + Γ_{μα}^μ j^α) = \sqrt{|g|} \nabla_μ j^μ $$
So if ∇ of the vector is zero, so is ∂ of the vector density? Nice!

 

[TeethWhitener]

Sorry to barge in on this thread, but I'm trying to learn GR and tensor analysis and I need all the help I can get. I understand the post above, but I still have some questions about showing that ##\nabla_{\nu}\mathcal{D}^{\mu\nu} = \partial_{\nu}\mathcal{D}^{\mu\nu}##. I get to the same point as Anypodetos in post 3:
$$\Gamma^{\mu}_{\mu\alpha}F^{\alpha\nu}+\Gamma^{\nu}_{\alpha\mu}F^{\mu\alpha} = \Gamma^{\alpha}_{\alpha\mu}F^{\mu\nu} $$
I'm pretty sure the first term on LHS cancels the term on RHS because the index pattern is the same. This leaves ## \Gamma^{\nu}_{\alpha\mu}F^{\mu\alpha} =0##, and I think this is true because of index (anti-)symmetry:
$$\Gamma^{\nu}_{\alpha\mu}F^{\mu\alpha} =\Gamma^{\nu}_{\mu\alpha}F^{\mu\alpha} =-\Gamma^{\nu}_{\mu\alpha}F^{\alpha\mu}$$
Since the pattern of indices is the same, we can replace the dummy indices to get:
$$\Gamma^{\nu}_{\alpha\mu}F^{\mu\alpha}=-\Gamma^{\nu}_{\alpha\mu}F^{\mu\alpha}=0$$
Is my reasoning correct here?

 

[Anypodetos]

Is my reasoning correct here?

Yes, that's right. Thanks for pointing this out!

 

[Orodruin]

I assumed that the covariant derivative of ##\sqrt{|g|}## vanishes, of which I am not certain (that's the "?").
$$ \partial_μ \left( \sqrt{|g|} j^μ \right) = \sqrt{|g|} \partial_μ j^μ + j^α \partial_α \sqrt{|g|} \overset{?}=
\sqrt{|g|} \partial_μ j^μ + j^α Γ_{μα}^μ \sqrt{|g|} = \sqrt{|g|} (\partial_μ j^μ + Γ_{μα}^μ j^α) = \sqrt{|g|} \nabla_μ j^μ $$
So if ∇ of the vector is zero, so is ∂ of the vector density? Nice!

It is rather simple (and a good exercise) to show that ##\partial_\mu \ln(\sqrt{|g|}) = \Gamma_{\nu\mu}^\nu## for the Levi-Civita connection. This relation is often quite useful.

 

[Anypodetos]

It is rather simple (and a good exercise) to show that ##\partial_\mu \ln(\sqrt{|g|}) = \Gamma_{\nu\mu}^\nu## for the Levi-Civita connection. This relation is often quite useful.

Thanks, I'll try to figure that out as time allows...

 

[Orodruin]

Thanks, I'll try to figure that out as time allows...

Oh, and for many purposes, it is easier to write it on the form ##\partial_\mu \sqrt{g} = \sqrt{g} \,\Gamma_{\mu\nu}^\nu## ...

 

[Anypodetos]

Oh, and for many purposes, it is easier to write it on the form ##\partial_\mu \sqrt{g} = \sqrt{g} \,\Gamma_{\mu\nu}^\nu## ...

I was suspecting that