On the Relativistic Euler Equations

Introduction

The relativistic form of the Euler Equations are used to describe relativistic fluids. These equations appear quite naturally from the continuity of the stress-energy tensor, and in the appropriate limit, reduce to the non-relativistic version.

Derivation

The continuity equation of the stress-energy tensor is given by:

\begin{equation}\label{continuity eq} T^{\mu\nu}_{;\nu}=0 \end{equation}

For a perfect fluid, the stress-energy tensor has the form:

\begin{equation} T_{\mu\nu}=\rho u_\mu u_\nu +P h_{\mu\nu} \end{equation}

Where \(h_{\mu\nu} = g_{\mu\nu} + u_\mu u_\nu\) and \(u_\mu\) is a 4-velocity element of the fluid.

From \eqref{continuity eq}:

\begin{equation} 0 = T^{\mu\nu}_{;\nu}h _{\gamma\mu} \end{equation}

\begin{equation} 0 = h_{\gamma\mu}\left(\left(\rho u^\mu u^\nu\right) _{;\nu}+P _{;\nu} h^{\mu\nu} + P h^{\mu\nu} _{;\nu} \right) \end{equation}

Using \(h^{\mu\nu} _{;\nu} =\left(g^{\mu\nu}+u^\mu u^\nu\right) _{;\nu}=\left(u^\mu u^\nu\right) _{;\nu}\):

\begin{equation} 0 = h_{\gamma\mu}\left(\rho_{;\nu}u^\mu u^\nu + \left(\rho+P\right)\left(u^\mu u^\nu\right) _{;\nu}+P _{;\nu} h^{\mu\nu}\right) \end{equation}

\begin{equation} h_{\gamma\mu}\rho_{;\nu}u^\mu u^\nu+\left(\rho+P\right)h_{\gamma\mu}\left(u^\mu u^\nu\right) _{;\nu} = -P _{;\nu} h _{\gamma\mu} h^{\mu\nu} \end{equation}

Because \(h_{\gamma\mu} u^\mu = u_\gamma + u_\gamma u_\mu u^\mu = u_\gamma - u_\gamma =0\):

\begin{equation} \label{rel euler eq} \left(\rho+P\right)u^\nu u^\mu_{;\nu} = -P _{;\nu} h^{\mu\nu} \end{equation}

These are the relativistic Euler Equations.

Reduction to Non-Relativistic Equations

In a flat spacetime, \(h^{\mu\nu} = \eta^{\mu\nu}+u^\mu u^\nu\), and the 4-velocity reduces to: \(u^\nu=\left(1,v^i\right)\). Furthermore, we can safely assume \(\rho \ggg P\), so the left hand side of \eqref{rel euler eq} becomes:

\begin{equation} \rho u^\nu u^\mu_{;\nu} = \rho\left(v^i_{,0} + v^j v^i_{,j}\right) \end{equation}

As for the right hand side, approximating \(v^i v^j = 0\):

\begin{equation} P _{;\nu} u^\mu u^\nu = P _{,0} + v^i P _{,0} \end{equation} So: \begin{equation} -P _{;\nu} h^{\mu\nu} = -v^i P _{,0} +\delta^{ij} P _{,j} \end{equation} Taking \(P _{,0}\) to be of the same order as \(v^i\), we can set \(v^i P _{,0}=0\) and find:

\begin{equation} v _{i,0} + v^j v _{i,j} = -\frac{1}{\rho}P _{,i} \end{equation}

Or in direct notation for a more familiar form:

\begin{equation} \partial_t\vec{v}+\vec{v}\cdot\vec{\nabla}\vec{v}=-\frac{1}{\rho}\vec{\nabla}P \end{equation}

Which are the Newtonian Euler Equations.

Sources

This derivation is from a problem set from my General Relativity class, 8.962, taught by Professor Scott Hughes in Spring 2024.