I'm learning through the basics of ideal MHD instabilities and the textbook (Friedberg, Plasma Physics and Fusion Energy, 12.4-12.6) writes the linearized momentum equation in terms of the perturbation ($\boldsymbol{\xi}$) as $$ -\omega^2 \rho \boldsymbol{\xi} = F(\boldsymbol{\xi}). $$ When solving for the instability in the linear $\theta$-pinch, for example, it says: since the equilibrium is symmetric with respect to $\theta$ and z, one can analyze the perturbation as follows: $$ \boldsymbol{\xi}(\boldsymbol{r}) = \boldsymbol{\xi}(r)exp[i(m\theta + kz)]. $$ I think this is a normal math question, so I'll ask it as one. Why does the solution take that form? I thought Fourier solutions involved integrals or sums, and I also don't see where it being symmetric w.r.t. $\theta$ and z comes into play. This question may be very specific, so I hope it reaches someone familiar with ideal MHD.


You actually are doing a sum and integral. What he's written is a shorthand for the inverse Fourier transform in this coordinate system, $$ \boldsymbol{\xi}(\mathbf{r}) = \sum_{m=0}^\infty \int_{-\infty}^{\infty} \boldsymbol{\xi}(r, k, m)\exp\left[i(m\theta + kz)\right]dk. $$ Since the equations are linear, everything can be moved inside the sum and integral (work this out for yourself if you don't see why), and you can look at each mode individually.

Uniformity in $\theta$ and $z$ is important because otherwise after doing this, there would be leftover functions of $\theta$ and $z$ that can't be eliminated algebraically. As a general rule, you only use this sort of mode analysis in directions that can be assumed to be uniform. That's why we aren't also transforming $r$.

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  • $\begingroup$ Is the integral over $dm$ as well? Maybe my Fourier knowledge is rustier than I thought. If you know of a reference that spells this out this process of solving differential equations like this, that would probably help me a ton. $\endgroup$ – Shawn Zamperini Oct 11 '18 at 12:35
  • $\begingroup$ @ShawnZamperini It's a sum over $m$, as I wrote. $m$ is a discrete index, not a continuous one like $k$. I'm not sure of a reference that goes into the specific math of this. It's equivalent to the Fourier transform under the stated conditions, and it's just something that shows up all over plasma physics. Have you not done, say, cold plasma waves? The same idea shows up there. $\endgroup$ – eyeballfrog Oct 11 '18 at 13:57
  • $\begingroup$ It's still difficult for me to grasp. It feels like there is a crucial section in my undergraduate math I missed or something. Also, I noticed you say the analysis can only be used in directions that are uniform, but in the book it is justified by them being symmetric. I have done some waves in plasmas, but again I just kind of believed the author in the Fourier analysis part. $\endgroup$ – Shawn Zamperini Oct 30 '18 at 17:30

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