Deriving continuity equation from time component of $T_{\nu ; \mu}^{\mu}=0$ I know that $T_{\nu ; \mu}^{\mu}=0$.
I want to derive the continuity equation from examining the time component in the Newtonian limit.
I have 
$$T_{; \nu}^{t \nu}=\rho_{0} U^{\nu} U_{, \nu}^{t}+\rho_{0} U^{\nu} \Gamma_{\gamma \nu}^{t} U^{\gamma}+p_{, \nu}\left(g^{t \nu}+U^{t} U^{\nu}\right)$$
but I am not sure how to get $$\frac{\partial \rho}{\partial t}+\vec{\nabla} \cdot(\rho \vec{v})=0$$
from the above terms.
 A: I will be using natural units during this discussion, if you don't mind.
The energy-momentum tensor for a perfect fluid has the form $$T^{\mu\nu} = (\rho+p)U^\mu U^\nu +pg^{\mu\nu}$$
We know that $\nabla_\mu T^{\mu\nu}=0$. Treating this as essentially flat spacetime, we get the continuity equation as $\partial_{\mu}T^{\mu\nu}=0$ and the energy-momentum tensor as $$T^{\mu\nu} = (\rho+p)U^\mu U^\nu +p\eta^{\mu\nu}$$ Thus, we now have $$\partial_{\mu}T^{\mu\nu}=\partial_{\mu}(\rho+p)U^{\mu}U^{\nu}+(\rho+p)[U^{\nu}\partial_{\mu}U^{\mu}+U^{\mu}\partial_{\mu}U^{\nu}]+\partial^{\mu}p=0$$
Now, we note from the identity $U_\nu U^\nu=-1$ that $$U_\nu \partial_\mu U^\nu=\frac{1}{2}\partial_\mu U_\nu U^\nu = 0$$
Thus, we now project our equation along the four-velocity, getting $$U_\nu \partial_\mu T^{\mu\nu}=-\rho\partial_\mu U^\mu-U^\mu \partial_\mu\rho-p \partial_\mu U^\mu=-\partial_\mu(\rho U^\mu )-p \partial_\mu U^\mu=0$$
Now, taking the nonrelativistic limit, with $U^\mu=(1,v^i), |v_i| <<1, p<<\rho$, we get $$-\partial_0\rho-\partial_i\rho v^i=\frac{\partial\rho}{\partial t}+\mathbb{\nabla}\cdot(\rho\mathbb{v})=0$$
which is the continuity equation.
