# 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 $$(5)$$?

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 $$n$$.

• What about the covariant derivative of $\rho$ , why does it vanish if it can also be a scalar function of $x^{\mu}$ ? – MicrosoftBruh Oct 21 at 13:11
• yeah, sorry, didn't notice this first. The short answer would be that you just have to project your eq. (3) to $g_{\mu\nu} +u_\mu u_\nu$ to get the Euler's equation. I will edit the answer to comment on that a little more – basicquestions Oct 21 at 13:13
• I tried your method but I really can´t get equation (2) from it. By projection do you mean instead of $g_{\mu\nu}$ I should write $g_{\mu\nu}+u_{\mu}u_{\nu}$ ? By the way, if I assume that $x^{\mu}$ is a geodesic path then: $\rho_{;\nu}=0$ right? – MicrosoftBruh Oct 21 at 17:13
• By orthogonal projection I mean that instead of $\nabla_\mu T^{\mu\nu} = 0$ you should consider the equation $(g_{\mu\nu} + u_\mu u_\nu)\nabla_\rho T^{\rho\nu} = 0$, simply because it is easier. Look at your equation (3) - if you contract a free index with $(g_{\mu\rho} + u_\mu u_\rho)$ you will see that the first term on the righthand side vanishes (because it is proportional to $u^\mu$, not because $\nabla_\nu u^\nu = 0$) and the same happens with the derivative of $(\rho + p)$ in the third term - it is again proportional to $u^\mu$. The rest will give you exactly your (2). – basicquestions Oct 21 at 21:50
• I am not sure what do you mean by considering $x^\mu$ as a geodesics. In order to say that something is moving along geodesic you first need to specify some element of the fluid. When we say that $\rho$ is a function of $x^\mu$, however, it does not mean that we are considering some particular element of fluid and looking at what is happening with it (there are approaches that describe the fluid that way, though) but rather that we fix some point and see what happens there as the fluid flows. In other words, $\rho$ in a function of all four coordinates, not the proper time of some worldline. – basicquestions Oct 21 at 21:58

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.