# Divergence of the time component of the stress-energy tensor in FLRW metric

I need some help with the divergence of the time component of the stress energy tensor for dust $$\nabla_{\mu}T^{0\mu}$$ given the stress energy tensor for dust is $$T^{\mu\nu}=(\rho+p)u^{\mu}u^{\nu} + g^{\mu\nu}p.$$ According to the internet, this is, using the FLRW metric (defined as $$ds^2=-dt^2+a(t)^2(dr^2/(1-kr^2)+r^2d\Omega^2$$ with $$d\Omega^2= d\theta^2=sin^2(\theta) d\varphi^2$$):

$$\nabla_{\mu}T^{0\mu}=\dot{\rho} +3(\rho +p)\frac{\dot{a}}{a}.$$

I have no idea how should I proceed but I'll show my progress:

Using the definition of the Stress-energy tensor for dust and knowing that the only non zero component in this case is $$T^{00}$$ we have:

$$\nabla_{\nu}T^{0\nu}=(\nabla_{\nu}\rho+\nabla_{\nu}p)u^{0}u^{\nu} + g_{0\nu}\nabla_{\nu}p=\nabla_0T^{00}.$$

Using the definition for covariant derivatives and given that the partial derivative of $$g_{00}$$ wrt time is 0, we have the Christoffel symbols

$$\Gamma^r_{rt}=\frac{\dot{a}}{a}=\Gamma^{\theta}_{\theta t}=\Gamma^{\varphi}_{\varphi t}.$$

That we can place in the definition of the covariant derivative:

$$\nabla_0T^{00}=\dot \rho +3\rho\frac{\dot{a}}{a}$$

Where or how can I find the p component?

## 1 Answer

You find this derivation in Chapter 8 of [1]. The upper/lower placements of your indices are a bit inconsistent so I take Carroll's: \begin{align} {T^\mu}_\nu&=\operatorname{diag}(-\rho,p,p,p)\,,\\[2mm] {\Gamma^1}_{10}&={\Gamma^2}_{20}={\Gamma^3}_{30}=\frac{\dot a}{a}\,,\\[2mm] \nabla_\mu {T^\mu}_0&=\partial_\mu {T^\mu}_0+{\Gamma^\mu}_{\mu\lambda}{T^\lambda}_0-{\Gamma^\lambda}_{\mu0}{T^\mu}_\lambda\\[2mm] &=\partial_0 {T^0}_0+{\Gamma^\mu}_{\mu0}{T^0}_0-{\Gamma^1}_{10}{T^1}_1-{\Gamma^2}_{20}{T^2}_2-{\Gamma^3}_{30}{T^3}_3\\[2mm] &=-\partial_0\rho\,-3\frac{\dot a}{a}\rho\,-3\frac{\dot a}{a}p\,. \end{align} [1] Sean Carroll, Spacetime and Geometry.