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I am reading about landau damping and the author states that $\omega$ is never real (due to collisions). https://cds.cern.ch/record/1982428/files/377-404%20Herr.pdf \begin{equation} 1 + \frac{\omega_e^2}{k^2} \int \frac{\partial v_x g(v_x)}{\omega/k-v_x} dv_x = 0 \end{equation}

What exactly does it mean for omega to be complex valued?

Mauricio
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jay121
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  • Related: https://physics.stackexchange.com/a/135182/59023 – honeste_vivere Jul 18 '22 at 13:06
  • The imaginary part of a frequency is conventionally taken as meaning the growth/damping rate with respect to the temporal dimension. The imaginary part of the wavenumber, correspondingly, is taken as the growth/damping rate with respect to the spatial dimension. – honeste_vivere Jul 18 '22 at 13:08

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A complex frequency $\omega$ is related to dissipation. Is the idea that if your wave is described by $\mathbf A =\mathbf A_0 \exp(\mathrm i\omega t)$ as a function of time $t$, where $\mathbf A_0$ is an amplitude (vector if needed), then you can decompose it such that $$\mathbf A=\mathbf A_0'(t) \exp[\mathrm i\Re(\omega) )t],$$ where $\Re$ is the real part, and $$\mathbf A_0'(t)=\mathbf A_0\exp[-\mathrm \Im(\omega) t] ,$$ where $\Im$ is the imaginary part. The whole idea idea is that the wave now is described by a new amplitude $\mathbf A_0'(t)$ that decays exponentially with time (note the lack of imaginary $\mathrm i$).

Note: for this to work, you have to choose a convention, I choose here that the imaginary part of $\omega$ is always positive. The opposite convention exists too but you have to change the sign of the whole exponent.

Mauricio
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