# Richtmyer Meshkov instability in MHD

In magnetohydrodynamics, the Richtmyer Meshkov instability is found to get suppressed by application of longitudinal magnetic field. Exactly what happens at the interface? Why instability gets suppressed? (How one can get the physical intuition of what is happening?)

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Yes, I think that is correct. It is more of a physics question that Computational science. I was studying a paper on RM instability in which numerical simulation of the MHD system was done (Finite volume method applied to the system of equations in MHD). I could see the simulation results, but could not get any physical intuition. – Subodh Jun 4 '12 at 6:49

## 1 Answer

If you consider the case of ideal MHD (perfectly conducting fluid) we have the limiting case where the magnetic field is frozen into the fluid. Thus, manipulating the magnetic field yields manipulation of the fluid and visa-versa. The Richtmyer Meshkov (RM) instability is suppressed in this limiting case by the application of a longitudinal magnetic field (one which is parallel to the fluid interface) due to the 'control' on the fluid motion provided by the frozen-in magnetic field. To be more mathematical; in Ideal MHD the Maxwell Stress Tensor can be defined as

$T_{ij} = [(P + B^{2}/{2\mu_{0}}) \delta_{ij} - B_{i}B_{j}/\mu_{0}]$

and the momentum equation can be written

$\frac{\partial T_{ij}}{\partial r_{i}} = 0$

with a transformation to the principle axis $T_{ij}$ can be reduced to diagonal form (with i, j running from 1 to 3). The principle axis being orientated so that axis corresponding to i = 3 is parallel to $\mathbf{B}$ and the other two perpendicular. So the eigenvalues for this system may be obtained via

$|T_{ij} - \delta_{ij} \lambda| = 0$

The solution yeilds a stress tensor of the form

$T_{ij} = \mathrm{diag}(P + B^{2}/2\mu_{0}, P + B^{2}/2\mu_{0}, P - B^{2}/2\mu_{0})$

From this we see that the stress caused by the magnetic field amounts to a pressure $B^{2}/2\mu_{0}$ in directions transverse to the field and a tension $B^{2}/2\mu_{0}$ along the lines of force. In other words, the total stress amounts to an isotropic pressure which is the sum of the fluid pressure and magnetic pressures and tension $B^{2}/\mu_{0}$ along the lines of force. It is this tension that provide the suppression you are pondering about.

To form the RM instability the configuration must be perturbed, this perturbation in the case of RM instabilities is provided by MHD shocks (the particular nature of each type of MHD shock is defined by using the Rankine-Hugoniot conditions and the relevant conservation laws). [Assuming the shock is parallel to the fluid interface] Not all MHD shocks will perturb the magnetic field orientation (perpendicular/parallel shocks) but most will (oblique shocks - Alfven shocks, switch-on/switch-off shocks/ fast/slow shocks). In these cases the magnetic field is abruptly altered which provides the perturbation required for the system to enter a Rayleigh-Taylor instability phase.

I hope this helps.

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