- #1
marschmellow
- 49
- 0
The vanishing divergence(s) of the stress-energy tensor, which proves/demands (not sure which) the conservation laws for mass-energy and momentum, would seem to suggest to a naive person (me) that there might be some sort of "vector potential" associated with the stress-energy tensor, similar to how divergenceless vector fields can be written as the curl of another vector field. I'm not sure how the curl generalizes to tensors. In the context of vector calculus, the curl operator preserves the tensor rank of 1, and I want to write
[itex]\epsilon[/itex][itex]^{a}_{ik}[/itex][itex]\delta[/itex][itex]^{ij}[/itex][itex]\partial[/itex][itex]_{j}[/itex]A[itex]^{k}[/itex]
as the components in some coordinate system of
curl([itex]\vec{A}[/itex])
But I don't see how this generalizes. Furthermore, I'm not sure whether the "vector potential" would be a type-(1,0) tensor with a generalized curl that raises its contravariant rank or a type-(2,0) tensor with a generalized curl that preserves its contravariant rank. Of course, it's also possible that this little aspect of vector calculus simply doesn't apply to higher-order tensors. Any ideas? Thanks.
[itex]\epsilon[/itex][itex]^{a}_{ik}[/itex][itex]\delta[/itex][itex]^{ij}[/itex][itex]\partial[/itex][itex]_{j}[/itex]A[itex]^{k}[/itex]
as the components in some coordinate system of
curl([itex]\vec{A}[/itex])
But I don't see how this generalizes. Furthermore, I'm not sure whether the "vector potential" would be a type-(1,0) tensor with a generalized curl that raises its contravariant rank or a type-(2,0) tensor with a generalized curl that preserves its contravariant rank. Of course, it's also possible that this little aspect of vector calculus simply doesn't apply to higher-order tensors. Any ideas? Thanks.