If you are referring to the fact that the connection coefficients all vanish in certain coordinate charts, the reason for that has nothing to do with a coordinate basis vs. a non-coordinate basis. It has to do with using Riemann normal coordinates (in which the metric is Minkowski and the connection coefficients all vanish) in a local patch of spacetime vs. some other chart.However, as above, in the general case they all remain. MTW does give the equation in this form, but it is still valid in curved spacetime with just partial derivatives.See exercise 22.8.

#### PeterDonis

Mentor
In the first article in this series, we looked at the Einstein Field Equations in a static, spherically symmetric spacetime. In this article, we are going to build on what we saw in the first article to show what Maxwell’s Equations in a static, spherically symmetric spacetime look like.
The electromagnetic field tensor is given in general by
$$F_{ab} = \partial_b A_a – \partial_a A_b$$
where $A_a$ is the electromagnetic 4-potential.
In covariant form, Maxwell’s Equations in general are:
$$\partial_c F_{ab} + \partial_b F_{ca} + \partial_a F_{bc} = 0$$
$$\nabla_a F^{ab} = \partial_a F^{ab} + \Gamma^a{}_{ac} F^{cb} = 4 \pi j^b$$
The first equation is an identity given the definition of ##F_{ab}##; the second...

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• elevin, vanhees71, Demystifier and 2 others
Just want to point out that the energy-momentum tensor discussed is not indeed the most general electromagnetic tensor possible. Non-linear correction of electrodynamics affects the form of the tensor.

You can have (at least theoretically, probably not in nature) a non-rotating, electrically charged black hole with big enough charge that non-linear corrections are in order.
https://arxiv.org/abs/1307.4951

• hunt_mat and professorFJA
andresB said:
Just want to point out that the energy-momentum tensor discussed is not indeed the most general electromagnetic tensor possible. Non-linear correction of electrodynamics affects the form of the tensor.

You can have (at least theoretically, probably not in nature) a non-rotating, electrically charged black hole with big enough charge that non-linear corrections are in order.
https://arxiv.org/abs/1307.4951
Is any of this relevant if you assume purely classical EM? It looks like it is all QED corrections to Maxwell EM.

• vanhees71
I just wanted to point out the possibility of generalizations. However, there are real stars with high enough magnetic fields that these corrections do need to be considered (though, in those cases spherical symmetry is lost).

• PAllen
andresB said:
I just wanted to point out the possibility of generalizations.

These are possible, but beyond the intended scope of this particular article. The article is only intended to cover the standard classical Einstein-Maxwell equations.

• vanhees71
Are you sure this form of the SET is correct? When trying to calculate it, I get a factor of 7/8 for the tt and rr components.

SWystub said:
Are you sure this form of the SET is correct?

Yes. You can find it in many references, including MTW, which is where I first saw it.

SWystub said:
When trying to calculate it, I get a factor of 7/8 for the tt and rr components.

How are you calculating it?

PeterDonis said:
Yes. You can find it in many references, including MTW, which is where I first saw it.

How are you calculating it?

I'm asking because of the Kronecker Delta - in some references (Sean Carroll. Spacetime and Geometry: An Introduction to General Relativity) it is given with an \eta_{\mu \nu} instead.

The factor of 7/8th was a simple mistake on my side.

SWystub said:
I'm asking because of the Kronecker Delta - in some references (Sean Carroll. Spacetime and Geometry: An Introduction to General Relativity) it is given with an \eta_{\mu \nu} instead.

If you write it with both indexes up or both indexes down, the metric will appear (which in a general curved spacetime will be ##g_{\mu \nu}## or ##g^{\mu \nu}##, not ##\eta_{\mu \nu}## or ##\eta^{\mu \nu}##). But if you write it with one index up and one index down, as I did, then the metric becomes ##\delta^\mu{}_\nu##.

what are the formulas that you used

BONKA THE WORM HOLE said:
what are the formulas that you used

What it says in the article. If you don't know what terms like "the Einstein Field Equation" refer to, you need to spend some time studying a basic GR textbook. Sean Carroll's online lecture notes are a good introduction to the subject.

• professorFJA and vanhees71
My textbooks (Wald and MTW) give the first of Maxwell's equations as something equivalent to:

$$\nabla_c F_{ab} + \nabla_b F_{ca} + \nabla_a F_{bc} = 0$$

While I believe this is equivalent to what Peter wrote in a coordinate basis, I don't think it's equivalent in a general basis.

I.e. in a coordinate basis we can replace the above with the simpler

$$\partial_c F_{ab} + \partial_b F_{ca} + \partial_a F_{bc}$$

but if we use a non-coordinate basis (for instance, an orthonormal basis), we need to keep the covariant derivative.

pervect said:
My textbooks (Wald and MTW) give the first of Maxwell's equations as something equivalent to:

$$\nabla_c F_{ab} + \nabla_b F_{ca} + \nabla_a F_{bc} = 0$$

While it is true that MTW does give that equation in this form, it turns out that the connection coefficients all cancel, so the equation is still valid in curved spacetime with just partial derivatives. See exercise 22.8.

pervect said:
While I believe this is equivalent to what Peter wrote in a coordinate basis, I don't think it's equivalent in a general basis.

If you are referring to the fact that the connection coefficients all vanish in certain coordinate charts, the reason for that has nothing to do with a coordinate basis vs. a non-coordinate basis. It has to do with using Riemann normal coordinates (in which the metric is Minkowski and the connection coefficients all vanish) in a local patch of spacetime vs. some other chart.

However, as above, in the particular case of Maxwell's Equations, none of that matters, since the connection coefficient terms, even if they would be present in the particular chart being used, all cancel.

• vanhees71 and (deleted member)
For the Maxwell equations (at least the free ones) you don't need a metric or a connection but you can formulate everything in terms of alternating differential forms. That's why the Maxwell equations written in terms of covariant derivatives boil down finally to equations with only partial derivatives.

vanhees71 said:
For the Maxwell equations (at least the free ones) you don't need a metric or a connection but you can formulate everything in terms of alternating differential forms. That's why the Maxwell equations written in terms of covariant derivatives boil down finally to equations with only partial derivatives.
It's true for the definition
$$F_{ab}=\partial_aA_b-\partial_bA_a \;\;\; (1)$$
and for the mathematical identity
$$\partial_{\{a}F_{bc\}}=0 \;\;\; (2)$$
but not for the physical assumption of no source
$$\nabla_bF^{ab}=0 \;\;\; (3)$$
Eq. (3) cannot be derived from (2). Eqs. (1) and (2) are pure geometry, but (3) is physics. You need metric to raise indices in getting ##F^{ab}## from ##F_{ab}##, so the covariant derivative in (3) cannot be replaced with the ordinary derivative.

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• vanhees71
It's also true, because due to the antisymmetry of the Faraday tensor
$$\nabla_{b} F^{ab}=\frac{1}{\sqrt{-g}} \partial_b (\sqrt{-g} F^{ab}),$$
i.e., the covariant "divergence" can be written in terms of partial derivatives. Admittedly here you indeed need the metric, because ##g=\mathrm{det}(g_{ab})##.

• Demystifier