8
$\begingroup$

Just as the title says, is there a relativistic version of Navier-Stokes equations?

In electromagnetic hydrodynamics it would be very useful to have relativistic version of Navier-Stokes equations, although I couldn't find one

$\endgroup$
4
  • $\begingroup$ The relativistic version of Navier-Stokes equations is a quite different theory called Israel-Stewart hydrodynamics ("relativistic Navier-Stokes" are known as "Eckart" or "Landau" relativistic hydrodynamics, but this naive relativistic generalization of Navier-Stokes does not work, it is highly unstable and acausal, so the more advanced and complex formulation of Israel-Stewart is needed). $\endgroup$
    – Quillo
    Oct 4, 2022 at 16:55
  • 2
    $\begingroup$ Possibly interesting: "Relativistic theories of dissipative fluids" J. Math. Phys. 36, 4226 (1995); doi.org/10.1063/1.530958 Robert Geroch $\endgroup$
    – robphy
    Oct 5, 2022 at 23:53
  • $\begingroup$ @robphy yes, this paper of Geroch is a seminal work! $\endgroup$
    – Quillo
    Oct 6, 2022 at 0:06
  • $\begingroup$ MHD is typically derived using Eulerian flows, which neglect viscosity, rather than NS equations. Including viscosity in such a scenario is generally called "Extended MHD" or something akin to that (with the inviscid case generally called "Ideal MHD"). $\endgroup$
    – Kyle Kanos
    Oct 27, 2022 at 20:48

1 Answer 1

17
$\begingroup$

I expand my previous comment. Yes, there are relativistic versions of the Navier-Stokes equation: They extend to special (or general) relativity the usual Navier-Stokes equations coupled to heat conduction (i.e., energy diffusion, see this answer). You can find them in several famous books, in particular:

Landau, Fluid Mechanics (volume 6 of the theoretical physics course). In the chapter dedicated to relativistic hydrodynamics, you find the famous relativistic version of the Navier-Stokes equation in the so-called "Landau frame". See, e.g., this answer.

Weinberg, Gravitation and Cosmology: here you can find the relativistic generalization of Navier-Stokes in the so-called "Eckart frame" (chapter 11).

The problem is that both these "naive" relativistic generalizations of Navier-Stokes do not work: the partial differential equations, despite being written in a covariant fashion, display instabilities and lead to non-causal propagation of signals (they display instabilities both at the computer simulation level, i.e. once discretized, as well as at the exact mathematical level!). This is a theorem based on the analysis of Hiscock and Lindblom (1985).

Beyond relativistic Navier-Stokes: To fix the stability and causality problems of relativistic versions of Navier-Stokes hydrodynamics, we have to look for a more general framework for dissipative relativistic hydrodynamics. The various possibilities, their underlying assumptions and motivations are summarized in this review. A possible (and widely used) alternative to Eckart or Landau versions of the relativistic Navier-Stokes is the so-called "Israel-Stewart hydrodynamics" (later revised and upgraded in Derivation of fluid dynamics from kinetic theory with the 14-moment approximation). You can find an introduction to Israel-Stewart hydrodynamics in the recent book by Rezzolla and Zanotti: this formulation overcomes the instability problem and is causal (signals, like sound waves, propagate subliminally, namely with a speed less than the one of light), as shown in a seminal work by Hiscock and Lindblom (1983).

Why Eckart & Landau's approaches fail: A simple explanation of why Navier-Stokes does not work in special and/or general relativity is given in "When the entropy has no maximum: A new perspective on the instability of the first-order theories of dissipation": in the formulations of Landau and Eckart, the entropy function turns out not to have a maximum (the homogeneous equilibrium state is an unstable equilibrium point). Therefore, since the fluid wants to maximise entropy, the fluid explodes because entropy "wants" to grow indefinitely.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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