Let $M$ be a Riemannian manifold with metric given by:
$ds^2=-f(r)dt^2+h(r)dr^2+r^2(d\theta^2+sin^2\theta d\phi^2)$
Let $T^{\mu \nu}$ be the tensor defined by:
$T^{\mu \nu}=(\rho+p)u^{\mu}u^{\nu}+pg^{\mu \nu}$
$u=(\frac{1}{\sqrt{f(r)}},0,0,0)$
I have to find conditions in $\rho,p$ such that $\nabla_{\mu} T^{\mu \nu}=0$
What I did:
$\nabla_{\mu} T^{\mu \nu}=\nabla_\mu (\rho+ p)u^\mu u^\nu +(\rho+p)(\nabla_\mu u^\mu u^\nu +u^\mu \nabla_\mu u^\nu)+\nabla_u p g^{\mu \nu}$
I think this must be equal to:
$\nabla_{\mu} T^{\mu \nu}=\frac{1}{f(r)}\nabla_\mu (\rho + p)\delta^\mu_0 \delta^\nu_0 +(\rho + p)\frac{f'(r)}{2fh}\delta^\mu_0 \delta^\nu_r +g^{\mu \nu}\nabla_u p$
Because $\nabla_\mu u^\mu=0$
$u^{\mu}\nabla_\mu u^\nu=u^\mu(\partial_\mu u^\nu +\Gamma^{\nu}_{\mu \sigma}u^\sigma)$
Is what I did correct? It doesn't coincide with the solution I was given.
Solution:
$\frac{1}{f}\partial_0(\rho + p)\delta^{\nu}_0+(\rho + p)\delta^\nu_r\frac{f'}{2fh}+g^{\mu \nu}\partial_\mu p=0$
I think my confusion comes from here, $u^{\mu}\nabla_\mu u^\nu=u^\mu(\partial_\mu u^\nu +\Gamma^{\nu}_{\mu \sigma}u^\sigma)$, I see repeated indices but I'm just looking at a component of a tensor I shouldn't sum over them right?
If you use Einstein notation, then you always sum over repeated indices (except if you explicitly say otherwise). So indeed, in the term $u^\mu(\partial_\mu u^\nu + \Gamma^\nu_{\mu \sigma} u^\sigma)$ you sum over $\mu$ and $\sigma$. The remaining free index $\nu$ suggests that this expression is the component of a 1-tensor. You can also see that your guess can't be correct since in the second term your $\mu$ index is not contracted, so this expression actually is the component of a 2-tensor, but the LHS is that of a 1-tensor.
However, you are on the right way. The last step is to evaluate the covariant derivatives in the second term: $$ (\nabla_\mu u^\mu)u^\nu = (\partial_\mu u^\mu + \Gamma^\mu_{\mu\sigma}u^\sigma)u^\nu = \delta^\nu_0 \Gamma^\mu_{\mu 0} u^0 u^0 = \delta^\nu_0 \frac{\Gamma^\mu_{\mu 0}}{f}. $$ Since the only non-vanishing component $u^0$ does not depend on $t$. Similarily, $$ (\nabla_\mu u^\nu)u^\mu = (\partial_\mu u^\nu + \Gamma^\nu_{\mu\sigma}u^\sigma)\delta^\mu_0 u^0 = \frac{\Gamma^\nu_{0 0}}{f}. $$ Finally, we have to calculate a few Christoffel symbols (no sums in these equations): $$ \Gamma^\mu_{\mu 0}=\frac{1}{2}g^{\mu \mu}\partial_0 g_{\mu \mu} = 0, $$ $$ \Gamma^\nu_{0 0} = -\frac{1}{2} g^{\nu \nu} \partial_{\nu} g_{00} = -\frac{1}{2} \delta^\nu_r g^{r r} \partial_{r} g_{00} = \delta^\nu_r \frac{f'}{2h}. $$ This gives you the desired result.