I am trying to understand a derivation from the textbook Radiation Processes in Plasmas by G. Bekefi (p. 14).

Start with the group velocity vector $$\mathbf{w}=\frac{\partial \omega}{\partial \mathbf{k}}=\hat{\textbf{e}}_{x} \frac{\partial \omega}{\partial k_{x}}+\hat{\textbf{e}}_{y} \frac{\partial \omega}{\partial k_{y}}+\hat{\textbf{e}}_{z} \frac{\partial \omega}{\partial k_{z}}. $$

The propagation vector k is of a wave of frequency $\omega$, where both k and $\omega$ which have spread (d$\omega, d\textbf{k}$) and can be expressed in spherical coordinates $$(k, \theta, \phi): \omega=\omega(k, \theta, \phi) \text{ or } k=k(\omega, \theta, \phi).$$

I understand that

$w_{k}=\frac{\partial \omega}{\partial k}$, $w_{\theta}=\frac{1}{k}\frac{\partial \omega}{\partial \theta}$, and $w_{\phi}=\frac{1}{k \sin \theta}\frac{\partial \omega}{\partial \phi}.$

But the text also says that

$w_{k}=\left ( \frac{\partial k}{\partial \omega} \right )^{-1},$ $w_{\theta}=-\frac{1}{k}\frac{\partial k}{\partial \theta}\left ( \frac{\partial k}{\partial \omega} \right )^{-1},$ and $w_{\phi}=-\frac{1}{k \sin \theta}\frac{\partial k}{\partial \phi}\left ( \frac{\partial k}{\partial \omega} \right )^{-1}$

and I don't understand how those expressions are derived. Could someone explain?


1 Answer 1


I didn't read the book and I am not expect in this field, but if $w_{k}=\frac{\partial \omega}{\partial k}=\left(\frac{\partial k}{\partial \omega}\right)^{-1}$, then the other equalities are obvious :

$$w_{\theta}=\frac{1}{k}\frac{\partial \omega}{\partial \theta}= \frac{1}{k}\frac{\partial \omega}{\partial k} \frac{\partial k}{\partial \theta} = \frac{1}{k}\frac{\partial k}{\partial \theta}\left ( \frac{\partial k}{\partial \omega} \right )^{-1}$$

$$w_{\phi}=\frac{1}{k sin \theta}\frac{\partial \omega}{\partial \phi} = \frac{1}{k sin \theta}\frac{\partial \omega}{\partial k}\frac{\partial k}{\partial \phi} = \frac{1}{k sin \theta}\frac{\partial k}{\partial \phi}\left ( \frac{\partial k}{\partial \omega} \right )^{-1}$$

The sign is specific to this field.

  • $\begingroup$ Thanks, I was quite puzzled by the minus sign. $\endgroup$
    – kstar
    Commented Aug 11, 2023 at 13:37

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