The vector field $V^{a}$ is the normalised $\left(V_{a} V^{a}=-c^{2}\right)$ tangent to a congruence of timelike geodesics, and $B_{a b}=\nabla_{b} V_{a}$.

Show that:

(i) $V^{a} B_{a b}=V^{b} B_{a b}=0$;

(ii) $V^{c} \nabla_{c} B_{a b}=-B_{b}^{c} B_{a c}-R_{a c b}^{d} V^{c} V_{d}$.

[You may use the Ricci identity $\nabla_{c} \nabla_{b} X_{a}=\nabla_{b} \nabla_{c} X_{a}-R_{a c b}^{d} X_{d}$.]

Now assume that $B_{a b}$ is symmetric and let $\theta=B_{a}{ }^{a}$. By writing $B_{a b}=\widetilde{B}_{a b}+\frac{1}{4} \theta g_{a b}$, or otherwise, show that

$$\frac{d \theta}{d \tau} \leqslant-\frac{1}{4} \theta^{2}-R_{00}$$

where $R_{00}=R_{a b} V^{a} V^{b}$ and $\frac{d \theta}{d \tau} \equiv V^{a} \nabla_{a} \theta$. [You may use without proof the result that \left.\widetilde{B}_{a b} \widetilde{B}^{a b} \geqslant 0 .\right]

Assume, in addition, that the stress-energy tensor $T_{a b}$ takes the perfect-fluid form $\left(\rho+p / c^{2}\right) V_{a} V_{b}+p g_{a b}$ and that $\rho c^{2}+3 p>0$. Show that

$$\frac{d \theta^{-1}}{d \tau}>\frac{1}{4}$$

and deduce that, if $\theta(0)<0$, then $|\theta(\tau)|$ will become unbounded for some value of $\tau$ less than $4 /|\theta(0)|$.