0
$\begingroup$

So my understanding is that in ideal magnetohydrodynamics (MHD), we assume a conductivity of infinity, and that makes the electric field in a comoving frame equal to zero. As I was watching one derivation of the equations of ideal MHD, I also noticed that the conservative energy PDE was treated as adiabatic, e.g., excluding terms for heat transfer.

So my question is this: Is the adiabatic treatment of the fluid part of the assumptions of idealization of MHD? Or was this probably just a simplification of the individual deriving the equations? Are there any other assumptions we make for ideal MHD? Commenting on how good these (or any other) assumptions are would also be appreciated.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ can you (one time is enough) write down what MHD is supposed to stand for? $\endgroup$
    – Quantumwhisp
    Commented Dec 15, 2021 at 20:07

1 Answer 1

Reset to default
3
$\begingroup$

Ideal magnetohydrodynamics (MHD) means isentropic. No entropy is produced. So dissipative heat transfer terms are excluded. The condition that the electric field vanish in the rest frame is a consequence of this since otherwise entropy would be produced due to Ohm's law.

A (relativistic) demonstration of this is given in Harris, Phys. Rev. 108, 1357 (1957). And it is also summarized in section II.B of arXiv:1412.3135.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Isn't the isentropic aspect a consequence of other assumptions, not the cause? That is, there are lots of assumptions in ideal MHD that end up resulting in it being isentropic, not one starting from the assumption of isentropic flow, right? $\endgroup$
    – honeste_vivere
    Commented Dec 20, 2021 at 14:27
  • 1
    $\begingroup$ @honeste_vivere, there are a lot of conditions that must be satisfied for a real fluid (without an electromagnetic field) to be treated as a perfect fluid too. I think you're philosophically right to say that the electric field vanishing in the rest frame is not 'caused' by the isentropic property, rather it's the reverse. As far as deriving the equations though, I think beginning with the isentropic property is conceptually cleaner. $\endgroup$
    – octonion
    Commented Dec 20, 2021 at 16:49

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.