- #1
Ancient_Nomad
- 15
- 0
Hi everyone,
I know that there is a result that corresponding to a closed p-form α, I can find a p-1 form β,
such that dβ = α.
I wanted to ask, what the tensorial analog of this would be. I mean would it be right to say that on a manifold with a lorentzian metric,
If I have a vector field A, such that
[itex]\nabla_{\mu} A^{\mu} = 0[/itex]
then there exists an anti-symmetric tensor B such that,
[itex]\nabla_{\nu} B^{\mu \nu} = A^{\mu} [/itex]
where [itex]\nabla [/itex] is the covariant derivative comaptible with the metric
(So that covariant derivative of the metric is zero)
Also, if possible, please tell me where I can find the proof of the correct statement.
Thanks in advance.
I know that there is a result that corresponding to a closed p-form α, I can find a p-1 form β,
such that dβ = α.
I wanted to ask, what the tensorial analog of this would be. I mean would it be right to say that on a manifold with a lorentzian metric,
If I have a vector field A, such that
[itex]\nabla_{\mu} A^{\mu} = 0[/itex]
then there exists an anti-symmetric tensor B such that,
[itex]\nabla_{\nu} B^{\mu \nu} = A^{\mu} [/itex]
where [itex]\nabla [/itex] is the covariant derivative comaptible with the metric
(So that covariant derivative of the metric is zero)
Also, if possible, please tell me where I can find the proof of the correct statement.
Thanks in advance.