Raychaudhuri and Friedmann Is it possible to derive the Friedmann equations from Raychaudhuri's equation, assuming only the Weak Energy Condition? The standard method for the derivation of Friedmann's equations is, if I am not mistaken, by assuming the FLRW metric and plugging it in Einstein's field equations. 
Now, the Raychaudhuri equation for the expansion can be derived in a geometrical manner without assumptions for the symmetry of the metric, as is done in Wald or Carroll. For example, one can replace in $$ \frac{dθ}{dτ}=-\frac{1}{3}θ^{2}-σ_{μν}σ^{μν}+ω_{μν}ω^{μν}-R_{μν}U^{μ}U^{ν} $$ the evolution of a representative length scale as $ \theta=\frac{3\dot{l}}{l} $, getting a 2nd order differential equation for the length scale l. Under what assumptions would this be enough to reproduce the Friedmann equation for $ \ddot{a} $ ?
 A: You can! Here's how I did it. We're going to need the following:
The Einstein field equations:
\begin{equation}
R_{ab} - \Lambda g_{ab} = 8 \pi G \Big(T_{ab} - \frac{1}{2} T g_{ab} \Big)
\end{equation}
The energy-momentum tensor for a perfect fluid with four-velocity $u^a$:
\begin{equation}
T_{ab} = (\rho + P) u_a u_b + P g_{ab}
\end{equation}
The geodesic congruence to the FLRW metric which is $\xi^a = \partial_t x^a = \delta^{a}_{\ t}$ (note that $\xi^a \xi_a = -1$). The expansion is $\theta := h^{ab} \nabla_b \xi_a = 3 \frac{\dot{a}}{a}$, where $h_{ab} := g_{ab} + \xi_a \xi_b$, and $\nabla_b \xi_a = \frac{\dot{a}}{a} h_{ab}$. Lastly, both $\sigma_{ab} = \omega_{ab} = 0$. This is because
\begin{align}
\sigma_{ab} &:= \nabla_{(b} \xi_{a)} - \frac{1}{3} \theta h_{ab}\\
\omega_{ab} &:= \nabla_{[b} \xi_{a]}
\end{align}
Thus, given the former, we can now deduce the second Friedmann equation. First we contract the Einstein field equations with $\xi^a \xi^b$ which gives
\begin{equation}
R_{tt} + \Lambda = 8 \pi G \Big(T_{tt} + \frac{1}{2} T \Big)
\end{equation}
We compute $\xi^a \nabla_a \theta = \partial_t \theta = 3 \Big(\frac{\ddot{a}}{a} - \Big(\frac{\dot{a}}{a} \Big)^2 \Big)$. Since $\sigma_{ab} = \omega_{ab} = 0$, using Raychaudhuri's equation we obtain
\begin{equation}
R_{tt} = -\partial_t \theta - \frac{1}{3} \theta^2 = -3\frac{\ddot{a}}{a}
\end{equation}
Plugging into the Einstein field equations, we have:
\begin{equation}
-3 \frac{\ddot{a}}{a} + \Lambda = 8 \pi G \Big(\rho + \frac{1}{2} (-\rho + 3P) \Big)
\end{equation}
\begin{equation}
\therefore \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3} (\rho + 3P) + \frac{\Lambda}{3}
\end{equation}
Which is the second Friedmann equation. Note that $T_{ab} \xi^a \xi^b = \rho \geq 0$, thus the WEC is satisfied. In fact
\begin{equation}
T^{ab} \xi_b T_{a}^{\ c} \xi_c = T_{ab} T^{b}_{\ c} \xi^a \xi^c = -\rho^2 \leq 0
\end{equation}
Which means that the DEC is also satisfied.
