In this paper, on page 3, it states that to trigger an instability, a disturbance needs to have a wave vector, $\boldsymbol{k}$, which satisfies, $$\boldsymbol{k}\cdot\boldsymbol{B}=0,$$ where $\boldsymbol{B}$ gives the magnetic induction. In astrophysical contexts, $\boldsymbol{B}$ is referred to as the magnetic field. Do you know where the equation above comes from? Im guessing this formula only applies to instabilities in a linearised system?


I am by no means an expert in plasma physics, but I think the equation of your interest comes from the linear perturbation analysis of the equations of ideal, inviscid magnetohydrodynamics, and studying the perturbations plus linearising the resulting perturbation equations, one gets1

\begin{align} \vec{k} \cdot \delta\vec{B} &= 0 \quad,\\ \\ \omega \cdot \delta \vec{B} - \vec{k} \times \left(\vec{B}_0 \times \delta\vec{v} \right) &= 0 \quad, \end{align}

as conditions for the magnetic field instability respectively.

I hope this helps somewhat.

$$ **\quad Edit \; 01 \quad** $$

Although not open access, I found2 those specific types of MHD instabilites are apparently referred to as Interchange instabilities and in the book by Boyd & Sanderson elaborates on this specific condition in more depth through the following two paragraphs,

By introducing magnetic shear we can make sure that for any given mode with propagation vector $\vec{k}$ this least stable condition, $\vec{k} \cdot \vec{B} = 0$, is restricted to specific layers and does not occur throughout the plasma. Thus, in our analysis we want to allow for arbitrary orientation of $\vec{k}$ to $\vec{B}$ and vertical variation of the direction of $\vec{B}$. Without loss of generality, we may choose coordinates such that the $y$-axis is vertical and the $z$-axis is parallel to $\vec{k}$, the direction of propagation of the mode under investigation.


Surfaces where $\vec{k} \cdot \vec{B}_0 = 0$ play a crucial role in stability analysis quite generally and are called resonant surfaces.

Footnotes & References:

1 Bartelmann, M.: Theoretical Astrophysics - An Introduction. Wiley-VCH Verlag, 2013, p. 266ff.

2 Boyd, T. J. M. & Sanderson, J. J.: The Physics of Plasmas. University of Cambridge Press, 2010, p. 124ff.

  • $\begingroup$ Thanks. So does the first equation come from $$\boldsymbol{\nabla}\cdot\boldsymbol{B}=0$$ and the second from $$\frac{\partial \boldsymbol{B}}{\partial t}=\boldsymbol{\nabla}\times(\boldsymbol{v}\times\boldsymbol{B})?$$ I get why these equations have to be satisifed. But the paper states that $\vec{k}\cdot \vec{B}=0$, where $\boldsymbol{B}$ is the background field not the perturbation? Also, I don't have access to that book do you know any other books which are open access? $\endgroup$ – Peanutlex Aug 27 '20 at 21:19
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    $\begingroup$ You are welcome @Peanutlex, I am glad to be of some help here. Yes exactly, those two ideal MHD equations you mentioned turn into the two algebraic expressions above for small, i.e. linearily approximable perturbations. Maybe this is just an issue of notation? My quick guess would be that they refer to small changes, i.e. perturbations, in the background field? $\endgroup$ – Diazenylium Aug 27 '20 at 21:26
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    $\begingroup$ As of yet, I have not found anything useful online or as open access, but I keep looking. I just like to give any references, so that you can potentially look it up there if you have access to a specific (university) library. $\endgroup$ – Diazenylium Aug 27 '20 at 21:28
  • $\begingroup$ Thanks. Maybe one way of thinking about it is that the magnetic tension force needs to equal zero i.e. $\frac{1}{\mu}(\boldsymbol{B}\cdot\boldsymbol{\nabla})\boldsymbol{B}=0$? Since the tension force acts as a restoring force for any perturbations? $\endgroup$ – Peanutlex Aug 27 '20 at 21:57

All magnetic fields satisfy $$i\mathbf{k} \cdot \mathbf{B}(\mathbf{k}) = 0 $$ at all times, since this is the Fourier transform of $$\mathbf{\nabla} \cdot \mathbf{B}(\mathbf{x}) = 0 $$

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    $\begingroup$ Thanks. Unfortunately, I still do not understand. Let $\vec{B}_0$ be a uniform background field and $\delta \vec{B}(\vec{x})$ be a small perturbation. Therefore, the total mangetic field is given by $\vec{B}(\vec{x})=\vec{B}_0 + \delta \vec{B}(\vec{x})$. Therefore, $\nabla\cdot\vec{B}=\nabla\cdot\delta\vec{B}=0$. Therefore, we only require $\vec{k}\cdot\delta\vec{B}=0$ as opposed to $\vec{k}\cdot\vec{B}=0$? $\endgroup$ – Peanutlex Aug 28 '20 at 10:09
  • $\begingroup$ @Peanutlex - Yes, when you linearize the only terms that have any temporal or spatial variation are the perturbed ones, e.g., $\delta \mathbf{B}$. $\endgroup$ – honeste_vivere Aug 28 '20 at 13:25
  • $\begingroup$ In MHD, often one is interested in instabilities with a nonuniform background field. In general, these instabilities are not plane waves. However, I just looked at the paper and in this case they are indeed considering plane waves on uniform background field, so my answer is not much use. $\endgroup$ – Daniel Aug 28 '20 at 16:22
  • $\begingroup$ For the particular case of Kruskal-Schwarzschild instabilities, the top of page 305 of Bellan and the preceding derivation answers your question. It's possible that these are the only instabilities the authors are worried about? $\endgroup$ – Daniel Aug 28 '20 at 16:24

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