Derivation Raychaudhuri equation

I'm trying to derive the Raychaudhuri equation; I found some trouble in this identity:

$$u_{d;f}u^{f;d}=2(\sigma^2-\omega^2)+\frac{1}{3}\Theta^2$$

where $\Theta=u^a\vphantom{}_{;a}$, $\sigma^2=\sigma_{ab}\sigma^{ab}$, $\sigma_{ab}=u_{(a;b)}-\frac{1}{3}\Theta^2h_{ab}+\dot{u}_{(a}u_{b)}$ and $\omega^2=\omega_{ab}\omega^{ab}$ with $\omega_{ab}=u_{[a;b]}+\dot{u}_{[a}u_{b]}$.

Computing $u_{d;f}u^{f;d}$ i find that:

$$u_{d;f}u^{f;d}=2(\sigma^2-\omega^2)+\frac{1}{3}\Theta^2 + \frac{2}{3}\Theta h^{df}\sigma_{fd}+\dot{u}_fu_d\dot{u}^du^f$$

How can I reach the equation above from the last one?

• Where are you reading this? The definitions are not the usual ones (Wald, Hawking & Ellis). May 26 '17 at 0:48
• I'm reading this arxiv.org/pdf/gr-qc/0511123.pdf just to find the right and simple way, but I follow Hawking & Ellis Books. May 26 '17 at 7:50

Assuming the terminology is standard (compare Hawking and Ellis, The large scale structure of space-time), we have $u_au^a=-1$ and $h_{ab}$ is the projector onto the subspace (of the tangent space) orthogonal to $u^a$. We then have
(i) $u_a\dot u^a=0$. This is a standard argument, and follows from applying $u^b\nabla_b$ to $u^au_a=-1$: $$\dot u^au_a+\dot u_au^a=0.$$ Fix the indices and you have the result.
(ii) $h^{ab}\sigma_{ab}=0$. Note that the first two terms in $\sigma_{ab}$, $u_{(a;b)}-\frac{1}{3}\Theta h_{ab}$, are annihilated by the $h$-trace. The remaining term, $h^{ab}\dot u_{(a}u_{b)}$, is zero since $h^{ab}u_b=0$ by definition.