Understanding Electromagnetism & General Relativity

• I
• Anypodetos
In summary: Thank you so much!In summary, The EM tensors in General Relativity are defined using ordinary partial derivatives instead of covariant ones. This is because the correction terms (Christoffel symbols) cancel for the definition of Fμν, but not for the definition of Jμ. However, this issue is resolved by using tensor densities for D and J, which allows the time component of J to have a dimension of charge per unit cube. This, in turn, fixes the problem with the partial derivative. It is also worth noting that the continuity equation, ∂μJμ=0, is coordinate system independent due to the use of tensor densities. Additionally, it is helpful to know that ∂μln(√|g|
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 Fμν because here the correction terms (Christoffel symbols) cancel, but I don't see how that should work for the definition of Jμ. 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.

Anypodetos said:
I don't see how that should work for the definition of Jμ.

Have you tried working it out?

Anypodetos said:
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.)

Anypodetos said:
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.

PeterDonis said:
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...

Anypodetos said:
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 and vanhees71
Orodruin said:
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!

Last edited:
vanhees71
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?

Last edited:
Anypodetos
TeethWhitener said:
Is my reasoning correct here?
Yes, that's right. Thanks for pointing this out!

Anypodetos said:
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.

Orodruin said:
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...

Anypodetos said:
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## ...

Orodruin said:
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

1. What is electromagnetism?

Electromagnetism is a branch of physics that studies the interactions between electrically charged particles. It is a fundamental force of nature and is responsible for many everyday phenomena, such as electricity, magnetism, and light.

2. How does an electromagnet work?

An electromagnet is created by running an electric current through a wire wrapped around a metal core. This creates a magnetic field around the wire, which can attract or repel other magnets. The strength of the electromagnet can be increased by increasing the current or the number of wire coils.

3. What is the difference between electromagnetism and general relativity?

Electromagnetism and general relativity are both theories in physics, but they explain different phenomena. Electromagnetism deals with the interactions between electrically charged particles, while general relativity explains the effects of gravity on the fabric of space and time.

4. How does general relativity explain gravity?

According to general relativity, gravity is not a force between masses, but rather the result of the curvature of space and time caused by the presence of massive objects. This curvature causes objects to follow a straight path through curved space, which we perceive as the force of gravity.

5. Can electromagnetism and general relativity be unified?

Currently, there is no complete theory that unifies electromagnetism and general relativity. However, many physicists are working on theories, such as string theory and quantum gravity, that aim to unify these two fundamental forces of nature.

• Special and General Relativity
Replies
5
Views
491
• Special and General Relativity
Replies
25
Views
1K
• Special and General Relativity
Replies
5
Views
2K
• Special and General Relativity
Replies
16
Views
2K
• Special and General Relativity
Replies
3
Views
1K
• Special and General Relativity
Replies
5
Views
2K
• Special and General Relativity
Replies
11
Views
1K
• Special and General Relativity
Replies
17
Views
824
• Special and General Relativity
Replies
8
Views
1K
• Special and General Relativity
Replies
42
Views
4K