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1

So I think $\partial_\mu F$ should also be 4 vectors, each being the directional derivative along a coordinate axis. It's a single covector, not a vector and not a collection of vectors. A gradient of a scalar field is a classic example of how we obtain a covector. In index notation, a free index such as your $\mu$ is interpreted as having the potential ...


0

The covariant derivative should introduce an extra index, which I presume its rho in your case. If you apply the general formula defining the covariant derivative of a tensor you will get your result.


3

On a two-index tensor, swapping the two indices is equivalent to transposing a matrix. You may not see many authors spending a lot of effort on this issue simply because an awful lot of the tensors we deal with are symmetric. This includes the metric, Ricci tensor, Einstein tensor, and stress-energy tensor. Therefore there is no special interest in ...


3

Comments to the question (v1): Indices are raised and lowered vertically by the pertinent metric tensor of the theory. The horizontal position of indices is important for a tensor that is not totally symmetric, e.g., the EM field strength $F_{\mu\nu}$ or the Riemann curvature tensor $R_{\mu\nu\lambda\kappa}$, etc, in order to properly identify which ...


3

It's almost the defition. A tensor $T_{ab}$ of rank $2$ is symmetric if, and only if, $T_{ab}=T_{ba}$, and antisymmetric if, and only if, $T_{ab}=-T_{ba}$. So from this definition you can easily check that this decomposition indeed yields a symmetric and antisymmetric part. Edit: Let $S_{bc}=\dfrac{1}{2}\left(A_{bc}+A_{cb}\right)$. Then ...



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