0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks, I was quite puzzled by the minus sign. $\endgroup$
    – kstar
    Commented Aug 11, 2023 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.