B and A in Curved Space Time: Does \nabla \times A =B?

[yourWitten]

By definition of the vector potential we may write

\nabla \times A =B

at least in flat space. Does this relation hold in curved space? I am particularly interested if we can still write this in a spatially flat Friedmann-Robertson-Walker background with metric ds^2=dt^2-a^2(dx^2+dy^2+dz^2) and if not, how it should be modified.

I know this question is extremely simple but I'm still developing intuition on GR.

 

[Orodruin]

Curl as such only exists in 3-dimensional Euclidean space. You need to look at a generalisation in the 4d case of both flat and curved spacetimes, namely the exterior derivative. In relativity, both the electric and magnetic fields together form a second rank tensor F called the electromagnetic field tensor. This in turn can be written in terms of the exterior derivative as F=dA where A is the 4-potential containing both the electric scalar and the magnetic vector porential. Of course, what is what (electric/magnetic) depends on the reference frame.

 

[yourWitten]

@Orodruin I am familiar with the covariant expressions of Maxwell's equations but having trouble working this out. In particular http://sedici.unlp.edu.ar/bitstream/handle/10915/125010/Documento_completo.pdf-PDFA.pdf?sequence=1 Eq. 2.24 suggests that the formula remains the same, but I don't see why.

 

[haushofer]

@Orodruin I am familiar with the covariant expressions of Maxwell's equations but having trouble working this out. In particular http://sedici.unlp.edu.ar/bitstream/handle/10915/125010/Documento_completo.pdf-PDFA.pdf?sequence=1 Eq. 2.24 suggests that the formula remains the same, but I don't see why.

Maybe I'm wrong, but isn't this simply because the Maxwell equations in general covariant form are the same as the equations in flat spacetime because the torsion vanishes? If they retain the same form, the expressions for B and E in terms of the gauge field don't change either.