Resistive steady states
I’m preparing some lecture notes on MHD and was thinking about the induction equation when there is no flow, so that we have only resistive diffusion, which I assume here to be uniform so that \[ \frac{\partial{\bf B}}{\partial t} = \eta\Delta{\bf B}. (*) \] My question is: what are the steady states of this equation?
Since \(\Delta{\bf B} = -\nabla\times\nabla\times{\bf B},\) it is clear (and very well known) that potential fields \((\nabla\times{\bf B}={\bf 0})\) are automatically steady states of this equation. Moreover, potential fields have no Lorentz force either so they are not going to generate any motion that would break my no-flow assumption.
But what I noticed was that these are not the only steady states, because we can also have \[ \nabla\times{\bf B} = \nabla\zeta \] for any scalar function \(\zeta.\)
First observation. These non-potential steady states cannot be force-free \((\nabla\times{\bf B}=\alpha{\bf B}).\) To see this, consider the evolution of magnetic energy under equation (*): \[ \frac{\mathrm{d}}{\mathrm{d}t}\int_V \frac{B^2}{2}\,\mathrm{d}V = ... = -\eta \int_V|\nabla\times{\bf B}|^2\,\mathrm{d}V - \eta\oint_{\partial V}(\nabla\times{\bf B})\times{\bf B}\cdot\mathrm{d}{\bf S}. \] You can check that if \(\nabla\times{\bf B}=\nabla\zeta,\) then these integrals do indeed cancel out. But if we had a force-free field, then the surface integral would vanish, meaning that the only possible force-free steady states would have \(\nabla\times{\bf B}={\bf 0}.\)
Second observation. It is possible to find magnetostatic equilibria \(((\nabla\times{\bf B})\times{\bf B} - \nabla p=0)\) with \(\nabla\times{\bf B}\neq {\bf 0}\) that are steady states of (*). Probably this is well known but I’ve never come across it before.
To demonstrate the existence of such equilibria, I’m going to simplify things drastically and consider only magnetic fields of the form \({\bf B}=B_z(x,y){\bf e}_z.\) In that case, the condition \(\nabla\times{\bf B} = \nabla\zeta\) reduces to the Cauchy-Riemann equations \[ \frac{\partial\zeta}{\partial x} = \frac{\partial B_z}{\partial y}, \qquad \frac{\partial\zeta}{\partial y} = -\frac{\partial B_z}{\partial x}. \] So any function \(B_z(x,y)\) that satisfies \(\Delta B_z=0\) will be suitable. I will take the simple example \(B_z=x,\) in which case \[ \nabla\times{\bf B} = -{\bf e}_y = -\nabla y, \] which is clearly neither a potential field nor force-free. The Lorentz force is \[ (\nabla\times{\bf B})\times{\bf B} = -x{\bf e}_x = -\nabla(x^2/2), \] so we see that taking \(p=-x^2/2\) will give a magnetostatic equilibrium that does not decay under resistive diffusion! I guess this is an example of the general property that magnetostatic equilibria are much less constrained than force-free fields, at least if your pressure doesn’t have to satisfy any other constraints.
It would be interesting to investigate whether more interesting such equilibria can be found.