Contraction in the Riemann Tensor

[Fraser]

Hi all,

I'm trying to follow through some of my notes of a GR course. The notes are working towards a specific expression and the following line appears:

R^{\alpha \beta}_{\gamma \delta ; \mu} + R^{\alpha \beta}_{\delta \mu ; \gamma} + R^{\alpha \beta}_{\mu \gamma ; \delta}=0

Which by contraction over \alpha and \gamma becomes

R^{\alpha \beta}_{\alpha\delta ; \mu} + R^{\alpha \beta}_{\delta \mu ; \alpha} + R^{\alpha \beta}_{\mu \alpha; \delta}=0

I'm afraid I don't understand this, it seems to relabel \gamma with\alpha. But how can we do this?

I do understand contraction in general, such that for a general tensor

T^{\alpha}_{\beta}=T^{\rho \alpha }_{\beta \rho}

But I don't see how this has been applied here?

Thanks in advance

p.s If this is more of a general maths question then please move to the appropriate forum

 

[George Jones]

From your first equation,

<br /> \delta^\gamma_\alpha \left( R^{\alpha \beta}_{\gamma \delta ; \mu} + R^{\alpha \beta}_{\delta \mu ; \gamma} + R^{\alpha \beta}_{\mu \gamma ; \delta} \right)=0<br />

Now do the sum over \gamma.

 

[Fraser]

Wow, thank you! 4 of us working together didn't think of that :(