# Tensor differentiation. Help with a step.

1. Jul 8, 2012

### Sagar_C

I am not very used to jugglery of tensors...I am learning it all now-a-days....I am trying to read a paper...and stuck at a point..:( ...It will be of great help if someone could help me get at eqn (34) from eqn (32) (cf. attached.) d/d\tau=u^\alpha\partial_\alpha (I think) and semi-colon is for covariant derivative and "[]" are for cyclic permutation of indices just like in Bianci relation.

P.S.: Hope I am not breaking any forum rules as this probably doesn't count as homework.

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2. Jul 8, 2012

### stevendaryl

Staff Emeritus
Hmm. It doesn't seem to be a very big gap.

If $Q$ is a function of coordinates, then $\dfrac{d}{d\tau} Q = \dfrac{dx^{\gamma}}{d\tau} \nabla_{\gamma} Q$. So in the particular case $Q = w_{\alpha \beta} \xi^{\alpha}\eta^{\beta}$, we use the product rule to get
$\dfrac{d}{d\tau}(w_{\alpha \beta} \xi^{\alpha}\eta^{\beta}) = ((\nabla_{\gamma} w_{\alpha \beta}) \xi^{\alpha}\eta^{\beta} + w_{\alpha \beta} (\nabla_{\gamma} \xi^{\alpha}) \eta^{\beta} + w_{\alpha \beta} \xi^{\alpha} (\nabla_{\gamma} \eta^{\beta})) \dfrac{dx^{\gamma}}{d\tau}$

Using the semicolon notation, and using $\dfrac{dx^{\gamma}}{d\tau}= u^{\gamma}$, this becomes:

$\dfrac{d}{d\tau}(w_{\alpha \beta} \xi^{\alpha}\eta^{\beta}) = (w_{\alpha \beta ; \gamma}\ \xi^{\alpha}\ \eta^{\beta} + w_{\alpha \beta}\ \xi^{\alpha}_{; \gamma}\ \eta^{\beta} + w_{\alpha \beta}\ \xi^{\alpha}\ \eta^{\beta}_{; \gamma}) u^{\gamma}$

The last step doesn't have anything to do with differentiation; it's just a fact about tensors: If $w_{\alpha \beta ; \gamma}$ is anti-symmetric in the first two indices, then $w_{\alpha \beta ; \gamma} = 3 w_{[\alpha \beta ; \gamma]} - w_{\gamma \alpha ; \beta} - w_{\beta \gamma ; \alpha}$

3. Jul 9, 2012

### Sagar_C

Many thanks. Actually, I should have been more specific. What I am really confused with is how covariant derivative appeared. The source of my confusion is that earlier in the text it seemed to be defined that "d/d\tau" is for simple derivative (see attachment 1) and "D/d\tau" is for the covariant one (attachment 2)! And all the time they are in general relativistic formalism...

P.S.: Is it because the term which is being differentiated is actually a scalar and so ordinary or covariant derivatives are just the same?

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Last edited: Jul 9, 2012
4. Jul 9, 2012

### stevendaryl

Staff Emeritus
Right. If you are taking the derivative of a scalar quantity, then there is no difference between an ordinary derivative and a covariant derivative.

5. Jul 10, 2012

### Sagar_C

Edited: Wrong post! Sorry!

Last edited: Jul 10, 2012