Raychaudhuri equation for shear

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

Is there an easy way of proving this?
 
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Is the Riemann tensor symmetric in ba?
 
It is when contrcted by V^c V^d cus that means you can take it to be symmetric over c and d, this plus the usual symmetries of R_{cbad} makes R_{cbda} V^c V^d symmetric over a and b.
 
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Right
 
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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