# Deriving an equation from the energy-momentum tensor in general relativity

I have a mathy question regarding the Friedmann-Lemaître equations (FLE) in standard cosmology. For reference the FLE are as follows $$\frac{3}{R^2}\Big(k + \frac{\dot{R}^2}{c^2}\Big) = \frac{8\pi G}{c^2}\rho$$ $$\frac{3\ddot{R}}{c^2 R} + \frac{4\pi G}{c^2}\Big(\rho + \frac{3p}{c^2}\Big) = 0$$ You can combine these two equations into one equation, eliminating the curvature k, which yields $$\dot{\rho} + 3H\Big(\rho + \frac{p}{c^2}\Big) = 0$$ I have been told you can also derive this last equation by taking the covariant derivative of the energy-momentum tensor, which is defined as: $$T^{\mu\nu} = \Big(\rho + \frac{p}{c^2}\Big) u^\mu u^\nu - pg^{\mu\nu}$$.

I haven't been able to do this yet. How do you get that $$H$$ (the Hubble parameter) in there? I managed to rewrite it to $$\Big(\rho + \frac{p}{c^2}\Big) \Big[\frac{du^\mu}{d\tau_p} + \Gamma_{\sigma\nu}^{\mu}u^\sigma u^\nu \Big] = p_{,\nu}\Big(g^{\mu\nu} - \frac{1}{c^2}u^\mu u^\nu\Big)$$

I suppose you can put $$\nu = 0$$ on the right-hand side if the pressure is isotropic, but then there are also the 4-velocities $$u^\mu$$ that I don't know how to calculate.

The Christoffel symbols of the FLRW metric contain the scale factor, that is how the Hubble parameter comes about when you expand $$\nabla_\mu T^{\mu\nu} = 0$$.
The cosmic fluid is implicitly assumed to be at rest, since you can calculate that $$R_{0i} = 0$$ from the metric, therefore $$T_{0i} =0$$ from the Einstein equations. There is no three-momentum density, so the fluid must be stationary. Its four-velocity is then purely timelike, and you can find the value of the component from the normalization condition.
Then, you will find the desired condition from just the component $$\nabla_\mu T^{\mu 0} = 0$$.