Euler's equation from energy-momentum tensor conservation Given the energy-momentum tensor for a perfect fluid:
$$T^{\mu\nu}=(\rho+p)u^{\mu}u^{\nu}+pg^{\mu\nu}\space\space\space\space\space(1)$$
I was trying to obtain Euler´s equation:
$$ (\rho+p)u_{\lambda;\nu}u^{\nu} + p_{,\lambda}+p_{,\nu}u^{\nu}u_{\lambda} = 0\space\space\space\space(2)$$
and here is my attempt:
$$T^{\mu\nu}_{\space\space\space\space\space;\nu}=(\rho+p)u^{\mu}(u^{\nu})_{;\nu}+(pg^{\mu\nu})_{;\nu}+((\rho+p)u^{\mu})_{;\nu}u^{\nu} \space\space\space\space(3)$$
the first term in the RHS is zero if we parametrize by arc-length and on the second term I used the metric compatibility condition to bring the metric tensor out and contract the index on the covariant derivative. Then we´re left with:
$$ ((\rho+p)u^{\mu})_{;\nu}u^{\nu} + p^{;\mu}=0\space\space\space\space(4)$$
Distributing the covariant derivative on the first term:
$$ (\rho+p)u^{\mu}_{\space\space;\nu}u^{\nu} + p_{;\nu}u^{\mu}u^{\nu} + p^{;\mu}=0\space\space\space\space(5)$$
Finally using the metric tensor $g_{\mu\lambda}$ we get:
$$ (\rho+p)u_{\lambda;\nu}u^{\nu} + p_{;\nu}u_{\lambda}u^{\nu} + p_{;\lambda}=0 \space\space\space\space(6)$$
which is equation $(2)$ only with covariant derivatives. My attempt is correct so far right? From here, can I simply convert $p_{;\lambda}=p_{,\lambda}$? Also, in general, $\rho=\rho(x^{\mu})$ so why does it´s covariant derivative vanish when I go from $(4)$ to $(5)$?
 A: In Special Relativity (so a bit of a reduction from GR), you use:
$$\partial_{\mu}T^{\mu\nu} + u_{\mu}u^{\nu}\partial_{\alpha}T^{\alpha\mu} = 0$$
to obtain the SR Euler Equation.
Similarly, try using
$$\nabla_{\mu}T^{\mu\nu} + u_{\mu}u^{\nu}\nabla_{\alpha}T^{\alpha\mu} = 0$$
This equality follows from the fact that $u_{\mu}u^{\mu} = -1$ (just multiply each side by $u_{\nu}$).
You may find it to be worth it to write $T^{\mu\nu} = w u^{\mu}u^{\nu} + pg^{\mu\nu}$ so you don't have to write out the $\rho + p$ term everywhere.
So, using this and expanding, we have
$$\nabla_{\mu}T^{\mu\nu} = u^{\mu}u^{\nu}\nabla_{\mu}w + w(\nabla_{\mu}u^{\mu}) u^{\nu} + wu^{\mu}\nabla_{\mu}u^{\nu} + g^{\mu\nu}\nabla_{\mu}p + p\nabla_{\mu}g^{\mu\nu}$$
$$u_{\mu}u^{\nu}\nabla_{\alpha}T^{\alpha \mu} = u_{\mu}u^{\nu}(\nabla_{\alpha}w)u^{\alpha}u^{\mu} + u_{\mu}u^{\nu}w \nabla_{\alpha}u^{\alpha}u^{\mu} + u_{\mu}u^{\nu}w u^{\alpha}\nabla_{\alpha}u^{\mu} + u_{\mu}u^{\nu}g^{\alpha\mu}\nabla_{\alpha}p + u_{\mu}u^{\nu}p\nabla_{\alpha}g^{\alpha\mu}$$
Now, we can use the following facts:
$$\nabla_{\mu}u^{\mu} = 0$$
$$u^{\mu}u_{\mu} = -1$$
$$\nabla_{\alpha}g^{\alpha \beta} = 0$$
$$u_{\mu}\nabla_{\alpha}u^{\mu} = 0$$
Thus things simplify to
$$\nabla_{\mu}T^{\mu\nu} = u^{\mu}u^{\nu}\nabla_{\mu}w + wu^{\mu}\nabla_{\mu}u^{\nu} + g^{\mu\nu}\nabla_{\mu}p$$
$$u_{\mu}u^{\nu}\nabla_{\alpha}T^{\alpha \mu} = -u^{\nu}(\nabla_{\alpha}w)u^{\alpha} + u^{\alpha}u^{\nu}\nabla_{\alpha}p$$
So, we have
$$u^{\mu}u^{\nu}\nabla_{\mu}w + wu^{\mu}\nabla_{\mu}u^{\nu} + g^{\mu\nu}\nabla_{\mu}p -u^{\nu}(\nabla_{\alpha}w)u^{\alpha} + u^{\alpha}u^{\nu}\nabla_{\alpha}p = 0$$
Then, the first and fourth term cancel by just relabeling indices, and thus we have
$$wu^{\mu}\nabla_{\mu}u^{\nu} + g^{\mu\nu}\nabla_{\mu}p + u^{\alpha}u^{\nu}\nabla_{\alpha}p = 0$$
We can write this as
$$(\rho + p)u^{\mu}\nabla_{\mu}u^{\nu} + (g^{\mu\nu} + u^{\mu}u^{\nu})\nabla_{\mu}p = 0$$
Note that the term $u^{\mu}\nabla_{\mu}$ is just the total derivative in our metric. So it retains the familiar form of the Euler Equation.
A: The answer to your first question is yes, you just simply convert $p_{;\mu}$ to $p_{,\mu}$. The reason for that is that $p$ is a scalar function: see equation (1) in your question - you can take it as a definition of pressure and energy density:
$$p = \frac{1}{3}(g_{\mu\nu} + u_\mu u_\nu) T^{\mu\nu}.$$
Here we used the signature of the metric in which $u^2 = -1$.
As for the second question: first of all, I am not sure if I understand your comment about "arc-length parameterization" after equation (3). The Euler equation really is given by an orthogonal projection of the energy-momentum conservation on $u_\mu$. If you do this to your equation (3) then the terms with derivatives of $\rho(x)$ vanish and you should get
$$ (g^{\nu\rho} + u^\nu u^\rho)( (\rho+p)u^\mu \nabla_\mu u_\rho + \partial_\rho p) = 0 .$$
Since $u^\rho \nabla_\mu u_\rho = 0$ this is the Euler's equation. You can also project equation (3) on $u_\mu$ to obtain the entropy current conservation. Then you will have to use the derivative of $\rho+p$ to get the entropy density using that $\rho + p = sT +\mu n$ ($\mu$ is a chemical potential and $n$ is a charge density) and $dp = s dT + n d\mu$. The first term in (3) does not vanish in general in this case, rather you should use the current conservation equation $\nabla_\mu nu^\mu = 0$ to rewrite it in terms of $n$.
