# Maxwell's equations in curved space-time

We have that $$\nabla \times \vec{A}=\vec{B}$$, at least in Minkowski space.

Does this relation still hold in curved space-time, for example on a spatially flat FRW background?

• I'm not 100% percent sure, but it probably should still hold. Remember that in GR curved-spacetime has to be a manifold, and therefore it's locally Minkowski (affine). On the other hand, vector fields are defined on the tangent space $T_pM$ which is also locally defined. Therefore I would expect for the relation to remain true in a local sense, but this shouldn't be any problem because vector operations only make sense locally too. Dec 9, 2021 at 4:13
• @Chaotic Yes according to sedici.unlp.edu.ar/bitstream/handle/10915/125010/… Eq 2.24, it should remain true without any modifications. I'm just looking for a rigorous demonstration. Dec 9, 2021 at 6:53
• Possible duplicates: physics.stackexchange.com/q/70739/2451 , physics.stackexchange.com/q/175047/2451 , physics.stackexchange.com/q/531779/2451 and links therein. Dec 9, 2021 at 9:47

Actually, the things are easier than you might think.

The electromagnetic field tensor defined with covariant derivatives or with standard partial derivatives is the same simply due to the anti-symmetry of this tensor:

$$F_{\mu\nu} = \nabla_\mu A_\nu- \nabla_\nu A_\mu = \partial_\mu A_\nu - A_\tau \Gamma^\tau_{\mu\nu} - \partial_\nu A_\mu + A_\tau \Gamma^\tau_{\nu\mu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

since the Christoffel-symbols in a space without torsion are symmetric in the last two indices $$\Gamma^\tau_{\mu\nu} = \Gamma^\tau_{\nu\mu}$$.

Actually, in curved space the electromagnetic field tensor can be seen a 2-form $$F$$ which results form the exterior derivative of the 1-form $$A$$,

$$F=dA$$

the 4-potential. An exterior derivative does not need any further structure, i.e. in particular it does not need a connection like the Christoffel-symbols. So I even guess, this equation $$F=dA$$ even holds on a space with torsion.

When it comes to $$\vec{B}$$ and $$\vec{A}$$, they are just a subset of $$F$$ and $$A$$, so this should also apply for the subset. Well $$\vec{B}$$ might not transform as standard vector, but the relationship between $$\vec{B}$$ and $$\vec{A}$$ should be the same (I put a vector array on both, but I did it only in order to identify them, not because of their transformation behaviour under coordinate change).

To work out the curl in general coordinates or in FRW is a bit laborious, refer to tensor and vector transformation under general coordinates (Well, there might be tricks for exterior derivatives which make the work easier).

• the last sentence you wrote is sort of my question, the precise form of the equation should change depending on the metric, and I'm asking what the form should be for the FRW metric (we probably some scale factor raised to some power)? Dec 11, 2021 at 3:14
• FRW coordinates are orthogonal coordinates, so you can use the formulas for the curl given on the wikipedia page given on "orthogonal coordinates". Dec 11, 2021 at 13:30

On curved spacetimes, things can get a bit more complicated for two reasons: firstly, the electromagnetic field can hardly now be seen as a pair of vectors, it instead is combined in a single electromagnetic tensor $$F_{\mu\nu}$$. On flat spacetime, one has $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$, which is the relativistic version of the formula you wrote (combined also with $$\mathbf{E} = - \mathbf{\nabla}\phi - \frac{\partial \mathbf{A}}{\partial t}$$). In curved spacetime, this expression generalizes to $$F_{\mu\nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$$, where $$\nabla_\mu$$ is the covariant derivative. On an inertial coordinate system with one time component and three spatial components, the Christoffel symbols vanish and this equation reduces to the usual expression, so your formula is valid, but only locally. Nevertheless, the existence of the four-potential $$A_\mu$$ is also only ensured locally.

In short, the expression you provided holds locally for an inertial coordinate system with a timelike coordinate and three space-like coordinates.

• Alves are you saying there is no modification for an FRW background (i.e no scale factors which should be inserted anywhere?). Dec 9, 2021 at 5:28
• @JoshuaBenabou on an inertial coordinate system, the metric reads $\textrm{d}s^2 = -\textrm{d}t^2 + \textrm{d}x^2 + \textrm{d}y^2 + \textrm{d}z^2$, just like Minkowski, so there is no need for scale factors. On the standard coordinate systems of FRW cosmology, one would need to work with the covariant derivatives instead, and hence the formula will probably have some modifications due to the Christoffel symbols Dec 9, 2021 at 5:37
• @Nicolas Alves, yes this is precisely my question, how would the formula be modified? I'm not sure that it's simply a matter of replacing the nabla by a covariant derivative. Dec 9, 2021 at 6:26
• @JoshuaBenabou it is not just replacing the nabla with a covariant derivative, that would even make sense (the covariant derivative acts on spacetime tensors, not $\mathbb{R}^3$ vectors. You have to use $F_{\mu\nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu$ instead, which is the general relativistic generalization of your expression for any coordinate system Dec 9, 2021 at 7:49
• Alves, sure I understand that the general procedure is to replace any derivatives by covariant derivatives, but it's not clear if the covariant form of nabla x A=B looks different. Indeed sedici.unlp.edu.ar/bitstream/handle/10915/125010/… Eq 2.24 suggests that it's the same, but I'm having trouble justifying it rigorously. Dec 9, 2021 at 8:43

Maxwell equations when he first wrote them down was a system of twenty equations. The Maxwell's equations that we know now was written down by Heaviside using Gibb's vector calculus. This is a system of four equations.

Using tensor calculus and after Einstein, the realisation that we should think of time and space together as spacetime means we can write these four equations as simply two. Since these are tensor equations they work on any curved manifold of any dimension.

This is usually written in Cartan's formalism of differential forms as:

$$dF = 0$$ and $$\delta F = j$$

Because the Hodge star is used this means the manifold must be semi-Riemannian.