I was watching a OCW from UC Irvine on GR. It starts with some discussions about special relativistic. This one about perfect fluid caught me because is not obvious how the Euler equation is derived. So I did it here.

It is an interesting topic in astrophysics since most of the objects we study can be considered (general) relativistic perfect fluid. As a first step, perfect fluid in special relativity is studied.

Starting by write down the stress-energy tensor
T^{\alpha\beta}
at a point
x_0
. First create the reference frame at that point, and the tensor looks like

\begin{align*}
T^{00} &= \rho \\
T^{i0} &= 0 \\
T^{ij} &= p \delta^{ij}\quad. \\
\end{align*}

Then generalize it using Lorentz transformation:

\begin{align*}
T^{\alpha'\beta'}&=\Lambda^{\alpha'}{}_\alpha\Lambda^{\beta'}{}_\beta T^{\alpha\beta} \\
\end{align*}

where consider only the boost

\begin{align*}
\Lambda^{0}{}_0 &= \gamma \\
\Lambda^{0}{}_j &= \gamma v_j \\
\Lambda^{i}{}_0 &= \gamma v_i \\
\Lambda^{i}{}_j &= \delta^{i}{}_j + (\gamma-1)v_i v_j / v^2 \\
\end{align*}

where

\gamma \equiv 1/\sqrt{1-v^2}
is the Lorentz factor, then the tensor

T^{\alpha\beta}
is

\begin{align*}
\newcommand{\Lij}[2]{\Lambda^{#1}{}_{#2}}
\newcommand{\Lzz}[0]{Lij 00}
\newcommand{\Lzj}[1]{Lij 0{#1}}
\newcommand{\Liz}[1]{Lij {#1}0}
\newcommand{\LUDzz}[0]{\gamma}
\newcommand{\LUDzj}[1]{\gamma v_{#1}}
\newcommand{\LUDiz}[1]{\gamma v_{#1}}
\newcommand{\LUDij}[2]{\delta^{#1}{}_{#2}+(\gamma-1) v_{#1}v_{#2}/ v^2}
T^{0'0'} &=
{\Lij 00}{\Lij 00}T^{00}+
{\Lij 00}{\Lij 0j}T^{0j}+
{\Lij 0i}{\Lij 00}T^{i0}+
{\Lij 0i}{\Lij 0j}T^{ij} \\
&=
\gamma^2 \rho +
\gamma^2 v^2 p \\
T^{0'b'} &= \Lambda^0{}_0\Lambda^{b'}{}_\beta T^{0\beta} + \Lambda^0{}_i\Lambda^{b'}{}_\beta T^{i\beta} \\
&= \gamma\Lambda^{b'}{}_\beta T^{0\beta} + \Lambda^0{}_i \Lambda^{b'}{}_\beta p\delta^{i\beta} \\
&= \gamma\Lambda^{b'}{}_0 T^{00} + p \Lambda^0{}_i \Lambda^{b'}{}_i \\
&= \rho\gamma^2v_{b'} + p (\LUDzj{i})(\LUDij{b'}{i}) \\
&= \rho\gamma^2v_{b'} + p (\gamma v_{b'} + \gamma (\gamma -1 )v_{b'}) \\
&= \rho\gamma^2v_{b'} + p \gamma^2 v_{b'} \\
&= (\rho + p)\gamma^2v_{b'} \\
T^{a'b'} &=
{\Lij {a'}0}{\Lij {b'}0}T^{00}+
{\Lij {a'}0}{\Lij {b'}j}T^{0j}+
{\Lij {a'}i}{\Lij {b'}0}T^{i0}+
{\Lij {a'}i}{\Lij {b'}j}T^{ij}
\\
&=
\rho{\Lij {a'}0}{\Lij {b'}0}+
p{\Lij {a'}i}{\Lij {b'}j}\delta^{ij}
\\
&=
\rho\gamma^2v_{a'}v_{b'}+
(\LUDij {a'}i)(\LUDij {b'}i)p
\\
&=
\rho\gamma^2v_{a'}v_{b'}+
p(\delta^{a'b'}+2(\gamma-1)v_{a'}v_{b'}/v^2 + (\gamma-1)^2v_{a'}v_{b'}/v^2)
\\
&=
\rho\gamma^2v_{a'}v_{b'}+
p(\delta^{a'b'}+(\gamma^2-1)v_{a'}v_{b'}/v^2)
\\
&=
\rho\gamma^2v_{a'}v_{b'}+
p(\delta^{a'b'}+\gamma^2v_{a'}v_{b'})
\quad\quad \text{ since } (\gamma^2 -1) / v^2 = \gamma^2
\\
&=
\rho\gamma^2v_{a'}v_{b'}+
p\delta^{a'b'} \quad.\\
\end{align*}

Combine the equations:

T^{\alpha\beta} = p \eta ^{\alpha\beta} + (\rho + p)u^\alpha u^\beta\quad.

To get the equation of motion, we can use the energy-momemtum conservation law:

\begin{align*}
\partial_\alpha T^{\alpha\beta}
&=
\partial_\alpha p \eta ^{\alpha\beta} + \partial_\alpha (\rho + p)u^\alpha u^\beta \\
&=\left. (-\dot p, \nabla \mathbf p)^\beta
+ (\dot\rho + \dot p, \nabla \rho + \nabla p)_\alpha u^\alpha u^\beta
+ (\rho + p) \partial_\alpha(u^\alpha u^\beta) \right. \\
&= 0 \quad.
\end{align*}

For

\beta=0
,

\begin{align*}
\partial_\alpha T^{\alpha0} &=
-\dot p + \gamma^2(\dot\rho + \dot p + (\mathbf v \cdot \nabla) (\rho + p))
+ \gamma^2(\rho + p) \nabla \cdot \mathbf v \\
&=(\gamma^2-1)\dot p + \gamma^2(\dot\rho + (\mathbf v \cdot \nabla) (\rho + p)) + \gamma^2(\rho + p) \nabla \cdot \mathbf v \\
&= 0 \\
v^2\dot p + \dot\rho
&= -(\mathbf v \cdot \nabla) (\rho + p) + (\rho + p) \nabla \cdot \mathbf v \\
&= -\nabla \cdot ((\rho + p)\mathbf v)
\end{align*}

For

\beta > 0
,

\begin{align*}
\nabla p +
\gamma^2 (\dot\rho + \dot p + (\mathbf v \cdot \nabla) (\rho + p)) \mathbf v
+ (\rho + p) \partial_\alpha(u^\alpha u^\beta) = 0
\end{align*}

Define

\begin{align*}
\partial_\sigma{}(u^\alpha u^\beta) &\equiv \partial_\sigma(u^\alpha u^\beta) \\
\partial_\sigma{}(u^0u^0) &= 0 \\
\partial_\sigma{}(u^iu^0) &= \gamma^2 \partial_\sigma(v^i)\\
\partial_\sigma{}(u^0u^j) &= \gamma^2 \partial_\sigma(v^j)\\
\partial_\sigma{}(u^iu^j) &= \partial_\sigma(\gamma^2 v^iv^j)\\
&= \gamma^2 \partial_\sigma( v^iv^j)
\quad.
\end{align*}

\begin{align*}
\partial_\alpha(u^\alpha \mathbf v)
&= \gamma^2 \dot {\mathbf v} + \gamma^2 (\nabla \cdot \mathbf v) \mathbf v + \gamma^2 (\mathbf v \cdot \nabla) \mathbf v
\end{align*}

\begin{align*}
(\rho + p) (\gamma^2 \dot {\mathbf v} + \gamma^2 (\nabla \cdot \mathbf v) \mathbf v + \gamma^2 (\mathbf v \cdot \nabla) \mathbf v) &= -\nabla p - \gamma^2 (\dot\rho + \dot p + (\mathbf v \cdot \nabla) (\rho + p) \mathbf v \\
\end{align*}

\begin{align*}
\dot {\mathbf v} + (\mathbf v \cdot \nabla) \mathbf v
&= -\frac{1}{\rho+p}\left(\nabla p/\gamma^2 + (\dot\rho + \dot p + \nabla \cdot ((\rho + p)\mathbf v)\right) \mathbf v)\\
&= -\frac{1}{\rho+p}\left(\nabla p/\gamma^2 + (\dot\rho + \dot p - v^2\dot p - \dot\rho\right) \mathbf v)\\
&= -\frac{1}{\rho+p}\left(\nabla p/\gamma^2 + (\dot p - v^2\dot p\right) \mathbf v)\\
&= -\frac{1-v^2}{\rho+p}\left(\nabla p + \dot p \mathbf v \right) \\
\end{align*}

This is called relativistic Euler Equaltion for Fluid Dynamics .