I have one question. I'm new in tensor calculus, so question may seem like stupid. Covariant derivative of a tensor $T^\alpha$: $$\nabla_\beta T^\alpha=\frac{\partial T^\alpha}{\partial x^\beta}+\Gamma^\alpha_{\beta\mu}T^\mu$$ But if I have a tensor as a matrix (lets say tensor with diagonal values -1;1;1;1, other equal to zero) how can I understand what is $T^\mu$? In this case I don't have any coordinates for a matrix like column and row number. Please, if It won't be hard, provide an example for a better understanding.


In addition to what the other questions state, if you're trying to work out what a tensor derivative shoudl look like, consider a tensor made of vectors:

$$T^{ab} = v^{a}v^{b}$$

Then, since covariant derivatives obey the Liebniz rule

$$\begin{align} \nabla_{c}T^{ab} &= \nabla_{c}\left(v^{a}v^{b}\right)\\ &= v^{b}\nabla_{c}v^{a} + v^{a}\nabla_{c}v^{b}\\ &= v^{b}\partial_{c}v^{a} + v^{b}\Gamma_{cd}{}^{a}v^{d} + v^{a}\partial_{c}v^{b} + v^{a}\Gamma_{cd}{}^{b}v^{d}\\ &= \partial_{c}T^{ab} + \Gamma_{cd}{}^{a}T^{bd} +\Gamma_{cd}{}^{b}T^{ad} \end{align}$$

And there you go, there's your rule for the derivative of rank-2 tensors with two "up" indices. If you have "down" indices, they follow the one-form rule (by a similar pseudo-proof), and then you're done.

  • $\begingroup$ Thank you. Now I understand. But, please, can you explain what do you mean by "one-form rule"? Just change $$\nabla_\beta T^\alpha=\frac{\partial T^\alpha}{\partial x^\beta}+\Gamma^\alpha_{\beta\mu}T^\mu$$ on$$\nabla_\beta T_\alpha=\frac{\partial T_\alpha}{\partial x^\beta}-\Gamma^\mu_{\alpha\beta}T_\mu$$ $\endgroup$
    – AlexSok
    Feb 24 at 9:30
  • $\begingroup$ @AlexSok Consider the covariant derivative of the scalar $T_\alpha T^\alpha$ and use the Leibniz rule. $\endgroup$ Feb 24 at 9:45
  • $\begingroup$ Or I just can use uppering of the index ?$T^{\mu\nu}=g^{\mu\nu}g^{\mu\nu}T_{\mu\nu}$ $\endgroup$
    – AlexSok
    Feb 24 at 9:46
  • $\begingroup$ @AlexSok : you can pull out factors of $g$ if you want, but you'll end up with factors like $g_{ab}\Gamma_{cd}{}^{b}$, which you can show will end up being the same thing as just using that other rule for one-forms. $\endgroup$ Feb 24 at 15:27
  • 1
    $\begingroup$ @JerrySchirmer Pulling out the metric is only possible for metric/Levi-Civita connections, while using the Leibniz rule is more general. So it might be a good exercise in GR, but in general it's better not to use it if not required. $\endgroup$
    – NDewolf
    Feb 25 at 9:16

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