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I was trying to prove that for any second order tensor:

$$A^{\mu\nu}_{;\mu\nu}=A^{\mu\nu}_{;\nu\mu}$$

considering the torsion free property and locally flat coordinates. Considering the point where all the Christoffel symbols vanish and applying the covariant derivatives one at a time we see that all the terms with Christoffel symbols vanish and the only term left is the one that only involves the partial derivatives and we know that partial derivatives commute. But I thought about the Riemann tensor definition and using that approach it would imply that the two derivatives of the Christoffel symbols cancel each other out. What am I missing?

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  • $\begingroup$ and did you mean to the tensor indices to be the same as the derivatives? $\endgroup$ – SuperCiocia Oct 18 at 23:07
  • $\begingroup$ Yes, that specific combination. $\endgroup$ – MicrosoftBruh Oct 18 at 23:11
  • $\begingroup$ Why do you think the christoffel symbols derivatives cancel each other out in definition of rieman tensor? The christoffel symbols vanish at the point, but this does not mean the derivatives vanish also. $\endgroup$ – Umaxo Oct 19 at 4:44
  • $\begingroup$ I know, that was my problem, how to prove that they vanish and I was thinking like this based on the definition of the Riemann tensor for a rank 1 tensor like this one found here: physics.stackexchange.com/questions/584068/… but I didn´t know that there were two terms for a rank two tensor like in the answer bellow $\endgroup$ – MicrosoftBruh Oct 19 at 10:12
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Wikipedia gives the commutator of covariant derivatives acting on a $(2,0)$ tensor:

$$\tau^{ab}{}_{;cd}-\tau^{ab}{}_{;dc}=-R^a{}_{ecd}\tau^{eb} -R^b{}_{ecd}\tau^{ae}.$$

(Carroll has the generalization to an $(n,m)$ tensor in eq. 3.68.)

When one sets $cd$ to $ab$, the two terms on the right side of this cancel. No locally-flat coordinates are necessary when doing it this generally-covariant way.

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