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Navier-Stokes equation is non-relativistic, what is relativistic Navier-Stokes equation through Einstein notation?

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For starters, check Wikipedia. –  Qmechanic May 1 '13 at 19:44
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Are you referring to the exact relativistic equivalent to Navier-Stokes equation or a more general Dissipative Relativistic Hydrodynamics Equation?

The "relativistic equivalent to Navier-Stokes equation" would be something like this:

There would be an energy momentum tensor with the following form:

$T_{\mu\nu} = (e+p)u_\mu u_\nu - p g_{\mu\nu} + \tau_{\mu\nu}$

with the viscous stress tensor given by:

$\tau_{\mu\nu}= -\eta \left[(\partial_\mu u_\nu+\partial_\nu u_\mu) - u_\mu u^\alpha \partial_\alpha u_\nu - u_\nu u^\alpha\partial_\alpha u_\mu\right] - (\zeta - \frac{2}{3}\eta)\left[\partial_\alpha u^\alpha\right](g_{\mu\nu}-u_\mu u_\nu) $

And the Conservation of Energy and Momentum would lead to:

$\partial^\mu T_{\mu\nu}=0$

The idea is to try to transpose the expressions you use in usual non-relativistic NS equation and fix the terms via positivity of entropy production. The relativistic Euler equation can be obtained setting $\tau_{\mu\nu}=0$, although some effort must be made to recover the Euler equation, it can be done.

The reference that I used was L. D. Landau and E.M. Lifshitz, Fluid Mechanics, chap. XV

I'm asking this because this is not the only formulation of dissipative relativistic hydrodynamics, and it`s not necessarily the best one.

P.S.:

I'm using $g_{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$, $u^\mu$ refers to the 4-velocity of the fluid, $e$ and $p$ refer to the proper energy density and the pressure, $\zeta$ and $\eta$ would refer to the bulk and shear viscosity.

Edit: Sorry, more typo corrections

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protected by Qmechanic Feb 20 at 13:58

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