Energy-Momentum Tensor of Perfect Fluid

[unscientific]

Homework Statement

I am given this metric: $ds^2 = - c^2dt^2 + a(t)^2 \left( dx^2 + dy^2 + dz^2 \right)$. The non-vanishing christoffel symbols are $\Gamma^t_{xx} = \Gamma^t_{yy} = \Gamma^t_{zz} = \frac{a a'}{c^2}$ and $\Gamma^x_{xt} = \Gamma^x_{tx} = \Gamma^y_{yt} = \Gamma^y_{ty} = \Gamma^z_{zt} = \Gamma^z_{tz} = \frac{a'}{a}$.

The energy momentum tensor can be written as $T^{\alpha \beta} = \left( \rho + \frac{P}{c^2} \right)u^{\alpha}u^\beta + Pg^{\alpha \beta}$. Show that $\frac{d(\rho a^3)}{da} + \frac{3Pa^2}{c^2} = 0$.

Homework Equations

The Attempt at a Solution

I shall let c=1 and u = (1,0,0,0) for simplicity.
[/B]
I know that the conservation requirement gives
\nabla_\alpha T^{\alpha \beta} = 0
Letting c=1, we have the relativistic continuity equation as
u^\alpha \left(\nabla_\alpha \rho \right) + (\rho + P)\left( \nabla_\alpha u^\alpha \right) = 0

For $\nabla_\alpha \rho = \partial_\alpha \rho$ and $\nabla_\alpha u^\alpha = \partial_\alpha + \Gamma^\alpha_{\alpha \mu} u^\mu =\partial_\alpha u^\alpha + u^t \left( \Gamma^x_{xt} + \Gamma^y_{yt} + \Gamma^z_{zt} \right) = 3c \left( \frac{a'}{a} \right)$.

Thus we have
u^\alpha \partial_\alpha \rho + 3(\rho + P)\left( \frac{a'}{a} \right) = 0
\partial_t \rho + 3(\rho + P)\left( \frac{a'}{a} \right) = 0

How do I proceed?

 

[unscientific]

Solved.