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 trace-free) 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, trace-free 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?
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.
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.