As part of a derivation in Bernstein '58 [1] a linear first-order (eqn. (9) in the image) appears:

Bernstein '58

But the general solution I would usually take (as appears in Gradshteyn and Ryzhik and checked in Mathematica) here would be

$$F=c_1G+\frac1G\int^\phi_1\frac{G(\phi^\prime)\Phi(\phi^\prime)}\Omega$$ where $\Phi$ is the RHS of eqn. (9). I suppose it depends on the requirement of periodicity and that $s$ has positive complex part (it comes from a Laplace transform) but I can't see how Bernstein's solution is obtained or equivalent, i.e. where the factor of $1/G$ goes. I'm sure I'm missing something silly, but I've wasted a bit of time on this already and can't get it to work out. The same derivation appears in slightly neater form in Montgomery and Tidman[2] but without further insight. Any help justifying this result would be greatly appreciated.

[1] Bernstein, I. (1958). Waves in a plasma in a magnetic field. Physical Review. Retrieved from http://journals.aps.org/pr/abstract/10.1103/PhysRev.109.10

[2] Montgomery D.C., Tidman, D.A. (1964). Plasma Kinetic Theory, McGraw-Hill advanced physics monograph series

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    $\begingroup$ Sean, quick question, have you looked to verify that Ira used the version of Equation 12 shown in his manuscript? I ask because I have found several typos in the equations of published papers, which is frustrating, but often not the author's fault. For instance, there is an incorrect equation for the cold plasma wavelength of a dispersively radiated whistler that is repeated in about ten publications. It is possible the equation was wrong, though it would beg the question of how one correctly predicts, what are now called Bernstein modes, from an incorrect equation. $\endgroup$ Commented Jan 1, 2015 at 14:57
  • $\begingroup$ @honeste_vivere, yes! as far as I can tell the factor of $1/G$ should appear in the expression for $Q$ in eqn. (21) of Bernstein, the same expression is also repeated, as in Bernstein, in Montgomery and Tidman. Of course I don't expect the Bernstein is wrong, but I'm still having a hard time figuring out how he's right. $\endgroup$
    – user21433
    Commented Jan 2, 2015 at 16:49
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    $\begingroup$ @SeanD - Do you have a copy of Gurnett and Bhattachjee's book Introduction to Plasma Physics (e.g., Chapter 9) or Stix's Waves in Plasmas? I have found the discussions and derivations much more easily followed in those works than Ira's original paper. $\endgroup$ Commented Jan 2, 2015 at 20:02
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    $\begingroup$ @SeanD - For instance, G&B write Equation 9 in a slightly different though more manageable (I think) form, which is expressed as: df/dx + P(x)*f = Q(x); which has a general solution given by: f = $e^{-\int \ dx' \ P(x')}$ * $[\int \ dx' \ Q(x') \ e^{\int \ dx" \ P(x")}]$. This looks more like what Bernstein finds in the end, right? So perhaps it's as simple as rearranging the form of the equation? $\endgroup$ Commented Jan 2, 2015 at 20:08
  • $\begingroup$ Hi Sean D. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Commented Jan 5, 2015 at 21:57

1 Answer 1


In general, one can write the linearized and Fourier transformed Vlasov equation, assuming $F_{s}(\mathbf{v})$ $\rightarrow$ $F_{so}(\mathbf{v})$ + $\delta F_{s}(\mathbf{v})$ , as:
$$ \partial_{\phi} F_{s} - i \ (\alpha_{s} + \beta_{s} \ cos \ \phi) \ F_{s} = \frac{q_{s}}{m_{s} \Omega_{cs}} \left[ \mathbf{E} + \mathbf{v} \ \times \ \left( \frac{\mathbf{k} \ \times \ \mathbf{E}}{\omega} \right) \right] \cdot \nabla_{\mathbf{v}} \ F_{so} $$ where subscript $s$ defines the particle species, $\mathbf{k}$ came from the linearized assumption that $\nabla$ $\rightarrow$ $i \ \mathbf{k}$, and $\alpha_{s}$ and $\beta_{s}$ are given by:
$$ \alpha_{s} = \frac{k_{\parallel} \ v_{\parallel} - \omega}{\Omega_{cs}} \\ \beta_{s} = \frac{k_{\perp} \ v_{\ perp}}{\Omega_{cs}} $$ If we limit ourselves to electrostatic waves, then $\mathbf{k} \ \times \ \mathbf{E}$ $\rightarrow$ 0 and $\mathbf{E}$ can be approximated by $-i \ \mathbf{k} \ \Phi$. Then the Vlasov equation above goes to:
$$ \partial_{\phi} F_{s} - i \ (\alpha_{s} + \beta_{s} \ cos \ \phi) \ F_{s} = \frac{-i \ q_{s} \ \Phi}{m_{s} \ \Omega_{cs}} \left( \mathbf{k} \ \cdot \ \nabla_{\mathbf{v}} \right) \ F_{so} $$ which can be rewritten in a general form as:
$$ \frac{d F}{dx} + P(x) \ F = Q(x) $$ This has a general solution given by:
$$ F = e^{-\int \ dx' \ P(x')} \left[ \int \ dx' \ Q(x') \ e^{ \int dx'' \ P(x'') } \right] $$ where the term $\int \ dx \ P(x)$ goes to:
$$ \int \ d\phi \ P(\phi) = -i \ \int \ d\phi \ \left( \alpha_{s} + \beta_{s} \ cos \ \phi \right) \\ = -i \ \left( \alpha_{s} \phi + \beta_{s} \ sin \ \phi \right) $$ which lets us rewrite the solution for F as:
$$ F_{s} = \frac{-i \ q_{s} \ n_{s} \ \Phi}{m_{s} \ \Omega_{cs}} \ e^{ i \ \left( \alpha_{s} \phi + \beta_{s} \ sin \ \phi \right) } \ \int \ d\phi' \ \left( \mathbf{k} \ \cdot \ \nabla_{\mathbf{v}} \right) \ F_{so} \ e^{ -i \ \left( \alpha_{s} \phi' + \beta_{s} \ sin \ \phi' \right) } $$

This is the same form as found in Bernstein's work.

After discussing with a colleague, I realized that the principle difference between the above derivation and Bernstein's work is that I did not use a Laplace transform. Physically, the Laplace transform gives you causality in the equation, it will give you a formal initial condition. The derivation I showed is only using a Fourier transform and imposing a priori the zero on the right-hand side of the Vlasov equation.

In many cases, the method used above is okay because $\Im [\omega]$ $\ll$ $\Re [\omega]$, in which case one can get away with assuming zero initial amplitudes.

The isolated $F^{*}(\mathbf{k}, \mathbf{v}, 0)$ on the right-hand side comes from a Laplace transform and integrating by parts. The the only surviving term is the Laplace transform at t = 0.

This is the reason why my result differs from Bernstein's work.

  • $\begingroup$ Thanks for the great answer and the book recommendations. Gurnett and Bhattacharjee looks good and my librarian is digging up Stix at the moment. Forgive me for being dense though but I must say I can't see how your solution (which makes sense) is equivalent to Bernstein's, in particular the presence of the $e^{i\left(\alpha_s\phi+\beta_s\sin\phi\right)}=1/G$ outside the integral which seems to be absent in eqn. (12). $\endgroup$
    – user21433
    Commented Jan 6, 2015 at 15:31
  • $\begingroup$ @SeanD - You make a valid point. Now that I look at it, I see what you mean. Though I am a bit confused about the definition of G since Bernstein's equation 11 uses a $\pm$. How does the single-signed exponent relate in that case? Did he make a deliberate choice (sorry, stream of thought here...). I will look into this more. $\endgroup$ Commented Jan 6, 2015 at 15:47
  • $\begingroup$ @SeanD - Oh, okay... Bernstein used the $\pm$ for the different particle species charge states. $\endgroup$ Commented Jan 6, 2015 at 16:07
  • $\begingroup$ @SeanD - Now I am not sure how Bernstein got his equation 6. At no point in any derivation I have seen is the reduced zeroth order VDF alone... I also think there is a typo in his equation 4, now that I look at it... $\endgroup$ Commented Jan 6, 2015 at 16:12
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    $\begingroup$ @SeanD - I will ask one of my colleagues who was friends and worked with Ira and get back to you. $\endgroup$ Commented Jan 6, 2015 at 16:14

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