# Derivation of the relativistic fluid stress tensor

On Wikipedia, the definition of the relativistic fluid stress tensor is given by:

$$T_{\mu \nu} := \rho u_{\mu} u_{\nu} + p h_{\mu \nu} + (q_{\mu} u_{\nu} + q_{\nu} u_{\mu} ) + \pi_{\mu \nu}$$

Where $h_{\mu\nu}$ is the projection tensor, $\rho$ is the density, $p$ is the pressure, $q$ is the heat flux vector and $\pi$ is the viscous shear tensor. But how is this derived?

• Can you link relevant Wikipedia page? The few I see (e.g., see this one and this one) don't include the $\pi$ or the $q$ terms. – Kyle Kanos Oct 30 '17 at 15:17
• This is basically exercise 22.7 of MTW! – user154997 Oct 30 '17 at 15:45
• @Kyle Kanos here’s the link en.wikipedia.org/wiki/Fluid_solution – Horus Nov 1 '17 at 2:56

This is really just a covariant decomposition of the stress-energy tensor, with aptly given names. In particular, given a normalized timelike vector field $u^\mu$ (with convention $u^\mu u_\mu = 1$), any tensor $S_{\mu\nu}$ can be decomposed as: $$S_{\mu\nu} = S_{\sigma\tau}u^\sigma u^\tau u_\mu u_\nu \pm h_\mu{}^\sigma S_{\sigma\tau}u^\tau u_\nu \pm h_\nu{}^\tau S_{\sigma\tau}u^\sigma u_\mu + \frac{1}{3}\mathrm{Tr}(S)h_{\mu\nu} + \Sigma(S)_{\mu\nu} + \Omega(S)_{\mu\nu},$$ where the sign on the second and third terms depends on how you define the projection tensor ($h_{\mu\nu} = g_{\mu\nu} - u_\mu u_\nu$ yields $+$ while $h_{\mu\nu} = u_\mu u_\nu - g_{\mu\nu}$ yields $-$), and where \begin{align} \mathrm{Tr}(S) &= h^{\mu\nu}S_{\mu\nu}, \\ \Sigma(S)_{\mu\nu} &= h_\mu{}^{\sigma}h_\nu{}^\tau S_{(\sigma\tau)} - \frac{1}{3}\mathrm{Tr}(S)h_{\mu\nu}, \\ \Omega(S)_{\mu\nu} &= h_{\mu}{}^\sigma h_\nu{}^\tau S_{[\sigma\tau]}. \end{align} Above $S_{(\mu\nu)} = \frac{1}{2}(S_{\mu\nu} + S_{\nu\mu})$ and $S_{[\mu\nu]} = \frac{1}{2}(S_{\mu\nu} - S_{\nu\mu})$. Since the stress-energy tensor is symmetric, $T_{\mu\nu} = T_{(\mu\nu)}$, we have $\Omega(T) = 0$, and $$h_\mu{}^\sigma T_{\sigma\tau}u^\tau = h_{\mu}{}^{\tau}T_{\sigma\tau}u^{\sigma} \equiv \pm q_\mu.$$ By letting $\pi_{\mu\nu} \equiv \Sigma(T)_{\mu\nu}$, $p \equiv \mp\frac{1}{3}\mathrm{Tr}(T)$, and $\rho \equiv T_{\sigma\tau}u^\sigma u^\tau$, we get the desired expression (there must be an error in your terms involving $q$).
The interpretation of $\rho$ as energy-density, $p$ as pressure, $q_\mu$ as heat vector (or equivalently as momentum density), and $\pi_{\mu\nu}$ as viscous shear (anisotropic stress) tensor follows from the definition of the stress-energy tensor, i.e. by defining \begin{align} T^{\mu\nu} &= \text{flux of the $\mathbf{e}_ \mu$-component of 4-momentum along $\mathbf{e}_\nu$}. \end{align}
Edit: If you instead use the so called spacelike sign convention, $u^\mu u_\mu = -1$, then $h_{\mu\nu} = g_{\mu\nu} + u_\mu u_\nu$ and the sign on the second and third terms become $-$, while $p \equiv \frac{1}{3}\mathrm{Tr}(T)$, and we define $$h_\mu{}^\sigma T_{\sigma\tau}u^\tau = h_{\mu}{}^{\tau}T_{\sigma\tau}u^{\sigma} \equiv -q_\mu.$$
• Yes indeed there is an error in q thanks. However the wiki page gives the projection tensor as $h_{\mu \nu} = g_{\mu \nu} + u_{\mu} u_{\nu}$ – Horus Nov 1 '17 at 2:54
• @Horus That would be because they use a different sign convention, taking $u^\mu u_\mu = -1$. – Erik Jörgenfelt Nov 1 '17 at 7:00