# Raychauduri Equation in "The Large Scale Structure Of Space-Time" book by Hawking & Ellis

Previously to Raychauduri equation, Hawking-Ellis obtain equation (4.25) (pg. 84) namely

$$\frac{d \theta_{\alpha\beta}}{ds} = - R_{\alpha 4 \beta 4} - \omega_{\alpha\gamma} \omega_{\gamma\beta} - \theta_{\alpha\gamma} \theta_{\gamma\beta} + \dot{V}_{(\alpha;\beta)} + \dot{V}_\alpha \dot{V}_\beta ,\tag{4.25}$$

and literally they say:

This equation and (4.23) can be expressed in terms of a general, non-orthonormal, non-Fermi-propagated basis by replacing the ordinary derivatives with Fermi derivatives and projecting everything into the subspace orthogonal to V.

Then, they take the trace of (4.25) and the result is Raychauduri eq. (4.26). Now, according to (4.14) (pg. 83) $$\theta = \theta_{ab} h^{ab} = V_{a;b} h^{ab} = V^a_{\ \ \ \ ;a}.\tag{4.14}$$ Therefore following the above hint and the last equation I can write

$$\frac{D_F \theta_{ab}}{\partial s} h^{ab} = ( - R_{a 4 b 4} - \omega_{ac} \omega^c_{\ b} - \theta_{ac} \theta^c_{\ b} + \dot{V}_{(a;b)} + \dot{V}_a \dot{V}_b ) h^{ab}.$$ Using property (ii) and rules (ii) y (iii) of Fermi derivative (pg. 81) we can transform the left hand side as \begin{align} \frac{D_F}{\partial s} (\theta_{ab} h^{ab} ) - \theta_{ab} \frac{D_F h^{ab}}{\partial s} &= ( - R_{ a 4 b 4} - \omega_{ac} \omega^c_{\ b} - \theta_{ac} \theta^c_{\ b} + \dot{V}_{(a;b)} + \dot{V}_a \dot{V}_b ) h^{ab}\\ \frac{d \theta}{ds} &= -h^{ab} R_{a b} - h^{ab}\omega_{ac} \omega^c_{\ b} - h^{ab}\theta_{ac} \theta^c_{\ b} + h^{ab}\dot{V}_{(a;b)} + h^{ab}\dot{V}_a \dot{V}_b . \end{align} (Evidently the second term (on lhs) vanishes by property (ii)). I didn't have problems in operating each term of these expression but the first one on the right hand side $$-h^{ab} R_{a b} = -h^{ab} R_{a b} = -(g^{ab} + V^a V^b) R_{a b} = -R - R_{a b} V^a V^b .$$ Here's my question: How to get rid of Ricci scalar $$R$$ (or scalar curvature)? I'm pretty sure that this term have to be zero to obtain Raychauduri eq. It's well known fact that according to Einstein field eqs. this term is not zero because $$R = -\frac{8\pi G}{c^4} T$$ In advanced thank you for answering my question.

Note that 4.25 is written in the Fermi propagated bases. You must be careful when going from $$\alpha, \beta...$$ to $$a,b,...$$. In particular,
$$\frac{d}{ds} \theta = \frac{d}{ds} \left(\theta_{\alpha \beta} h^{\alpha \beta}\right) = h^{\alpha \beta}\frac{d}{ds} \theta_{\alpha \beta}\;.$$ Then $$-h^{\alpha \beta} R_{\alpha 4 \beta 4} =-h^{\alpha \beta} R_{\alpha c \beta d} V^c V^d = -g^{ab} R_{a c b d} V^c V^d = -R_{c d} V^c V^d\;.$$ The first equality is true since $$E_4 = V$$, and the second equality is true since $$R_{a c b d} V^a V^c V^d = 0 = R_{a c b d} V^b V^c V^d\;.$$ Similarly, the other two terms work out. It is a simple exercise to show that $$h^{\alpha \beta} \left(\dot V_{(\alpha;\beta)} +\dot V_\alpha \dot V_\beta\right)= \dot V^a ~_{;\;a}\;.$$