Tensor differentiation. Help with a step.

[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.

 

[stevendaryl]

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.

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

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

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

[itex]\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}[/itex]

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

 

[Sagar_C]

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

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

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

[itex]\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}[/itex]

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

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?

 

[stevendaryl]

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?

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

 

[Sagar_C]

Edited: Wrong post! Sorry!