Contraction of the Bianchi identity

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Homework Statement


I've been given the Bianchi identity in the form

##\nabla _{\kappa} R^{\mu}_{\nu\rho\sigma} + \nabla _{\rho} R^{\mu}_{\nu\sigma \kappa} + \nabla _{\sigma} R^{\mu}_{\nu\kappa\rho} =0##

Homework Equations

The Attempt at a Solution


In order to get from this to the Einstein tensor, I've seen online that you perform a contraction so you get the Ricci tensor and go from there. No two of my indices are the same though, can I just relabel an index? How does the contraction work? Apologies, the bottom three indices should be offset to the right but I can't make latex do it.
 
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You can put a contravariant index to the samd as a covariant one and sum over them. This is what it means to make a contraction and it results in a tensor of two ranks lower. The canonical example of this being the product of a tangent and a dual vector.

To be honest, if you are expecting to learn more advanced subjects and applications, you should make sure that you master the basics first. Based on your question, I would suggest going back to repeat the basics of tensor analysis rather than trying to understand curvature and the Einstein tensor. It will serve you better in the long run.
 
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When I say how does the contraction work, that makes it sound worse than it is, I think. I do know what a contraction is, what I wanted to double check is that I can relabel a contravariant index to be the same as the covariant. I'll definitely go back and re-read my notes though, it's always worthwhile. Thank you, I appreciate your help!
 
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To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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