3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Kyle Kanos Oct 30 '17 at 15:17
  • 1
    $\begingroup$ This is basically exercise 22.7 of MTW! $\endgroup$ – user154997 Oct 30 '17 at 15:45
  • $\begingroup$ @Kyle Kanos here’s the link en.wikipedia.org/wiki/Fluid_solution $\endgroup$ – Horus Nov 1 '17 at 2:56
2
$\begingroup$

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 $$ h_\mu{}^\sigma T_{\sigma\tau}u^\tau = h_{\mu}{}^{\tau}T_{\sigma\tau}u^{\sigma} \equiv -q_\mu. $$

$\endgroup$
  • $\begingroup$ Yes indeed there is an error in q thanks. However the wiki page gives the projection tensor as $h_{\mu \nu} = g_{\mu \nu} + u_{\mu} u_{\nu}$ $\endgroup$ – Horus Nov 1 '17 at 2:54
  • $\begingroup$ @Horus That would be because they use a different sign convention, taking $u^\mu u_\mu = -1$. $\endgroup$ – Erik Jörgenfelt Nov 1 '17 at 7:00
  • $\begingroup$ Btw could you provide a link for further information on this decomposition? I’ve tried to find it myself but I can’t seem to find it. $\endgroup$ – Horus Nov 4 '17 at 13:21
  • $\begingroup$ @Horus I'm afraid I don't. The places I've seen it used don't tend to explain it. Is there anything you find unclear? $\endgroup$ – Erik Jörgenfelt Nov 8 '17 at 13:39
  • $\begingroup$ There’s nothing that is unclear except that I don’t know why the decomposition was done that way. For example some would do a decomposition based on symmetric and antisymmetric properties and traceless parts like the Ricci decomposition. I just don’t understand why this decomposition works. My understanding is that it splits the stress tensor into orthogonal and perpendicular components along the four vector but I’m still trying to find the orthogonal and perpendicular components in the decomposition and why it’s like that. $\endgroup$ – Horus Nov 9 '17 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.