
#1
Mar612, 04:55 AM

P: 289

Following Wald I have nearly got the right answer out for time derivative for shear...what I am left with is showing that [itex]R_{cbad} V^c V^d + h_{ab} R_{cd} V^c V^d / 3[/itex] (which is obviously symmetric and tracefree) can be written as [itex]C_{cbad} V^c V^d + \tilde{R}_{ab} / 2[/itex] where [itex]\tilde{R}_{ab}[/itex] is the spatial, tracefree part of [itex]R_{ab}[/itex], i.e. [itex]h_{ac} h_{bd} R^{cd}  h_{ab} h_{cd} R^{cd} / 3[/itex].
Is there an easy way of proving this? 



#2
Mar612, 05:07 PM

P: 52

Is the Riemann tensor symmetric in ba?




#3
Mar712, 04:26 AM

P: 289

It is when contrcted by [itex]V^c V^d[/itex] cus that means you can take it to be symmetric over c and d, this plus the usual symmetries of [itex]R_{cbad}[/itex] makes [itex]R_{cbda} V^c V^d[/itex] symmetric over a and b.




#4
Mar812, 06:36 AM

P: 52

Raychaudhuri equation for shear
Right




#5
Mar812, 10:01 PM

P: 11

You have to replace the Riemann by it's decomposition into Weyl tensor ... which is given by the eq. 3.2.28 in Wald's book.



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