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I was reading about the derivation for group velocity on the wiki page here. As far as I understood, what's done here is based on these steps

  1. Get a basic solution for the wave equation, that is $exp(i(kx-\omega t))$
  2. Write some general solution using the assumption that its Fourier transform at $t=0$ is localized
  3. Consider that $\omega=\omega(k)$ and using some assumptions one can get the envelope and therefor its velocity.

What I don't understand here is related to step 3. where on the wiki site the author writes $\omega(k)\approx\omega_0+(k-k_0)\frac{d \omega}{dk}(k_0)$ as if we don't know how $\omega(k)$ looks like. However, this comes after solving the wave equation for step 1. where we already established that $\omega=ck$ as is mentioned here, where $c$ is the velocity from the differential equation.

So, my question is why was $\omega(k)\approx\omega_0+(k-k_0)\frac{d \omega}{dk}(k_0)$ needed in this scenario and if I'm missing something, what is it. In other words, did the author actually assume that he does not know something that he knows in order to get a more 'general' result?

If I'm not missing something, then where could I find a better and also general way (if it exists) of defining the group velocity?

Edit:

I felt that a clarification is needed regarding the question.

I am not implying that there are no other relation other than $\omega=ck$. My problem arose from the derivation of the solution of wave equation using separation of variables. By using that procedure, one can derive the basic solution $exp(i(kx-\omega t))$, where $k$ is introduced as the constant needed for separation of variables. Solving the ODE for the space coordinate gives the $exp(ikx)$ part, while solving for the temporal coordinate gives the other part $exp(i(ck)x)$. Only now, one can make a notation $\omega=ck$ such that the temporal part becomes something like $exp(i\omega t)$. This reasoning is the one used in the second link I used.

I will try to reformulate the question. Why am I allowed to consider the same solution $exp(i(kx-\omega t))$ that depends strictly on the above mentioned notations, even if I change the parameters $\omega$, $k$?

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  • $\begingroup$ Where in the derivation on the wiki page was $\omega=c k$ assumed? I don't see this assumption anywhere in the derivation. $\endgroup$ – Virft Aug 15 '17 at 16:12
  • $\begingroup$ When solving the wave equation by separation of variables, at the second link. The constant is taken as $-k^2$ so for spatial coordinate we get $kx$, while for the temporal part we get $ckt$. After that, the notation $\omega = ck$ is used and that's why I said that by solving the PDE we imply the dependency between $\omega$ and $k$ $\endgroup$ – Victor Palea Aug 15 '17 at 16:34
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For general waves (such as gravity or capillary waves on the surface of water) it is not true that $\omega=ck$. In other words they do not satisfy the same wave equation as for sound or for waves on a string. For example waves on the surfae of deep water have $\omega(k)=\sqrt{gk}$. Group velocity is different from the phase velocity $v_{\rm phase} = \omega/k$ when this is so. So you first need to find $\omega(k)$ for your particlular species of wave and then find $$ v_{\rm group} = \frac{\partial \omega}{\partial k}. $$ Again, for deep water waves, $$ v_{\rm group} = \frac 1 2 \frac \omega k $$ so the group velocity is half that of the wave crests, which move at $v_{\rm phase}= \omega/k= \sqrt {g/k}$.

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  • $\begingroup$ I'm not denying that there is a physical motivation for having different dependencies for $\omega(k)$. My issue is related to the way $\omega$ is introduced then solving the wave equation in the second link. There, $\omega$ is just a notation for the constant resulting from separation of variables and the velocity parameter from the PDE. How it is ok to use this 'notation' to define $\omega$ and then move to a different case without considering the changes that might appear in the solution of the PDE. $\endgroup$ – Victor Palea Aug 15 '17 at 16:41
  • $\begingroup$ @Victor: The second link applies only to the particular "wave equation" mentioned there. That wave equation only applies in very simple cases. For ant system of waves you can look for a solution of the form $e^{i(k-\omega t)}$. You plug that into the (usually complicated) equation of motion for the given system and find that it requires some (usually complicated) relation between $\omega$ and $k$. This has nothing to do with separation of variables. $\endgroup$ – mike stone Aug 15 '17 at 17:07
  • $\begingroup$ Ok, so in the best case scenario, I'm missing terms in the wave equation. Worst case scenario it's a different equation altogether. I think I get the difference. Thank you. $\endgroup$ – Victor Palea Aug 15 '17 at 17:26
  • $\begingroup$ @victor: Have a look at the derivation of the water wave equation starting on page 217 of these notes: courses.physics.illinois.edu/phys508/fa2017/amaster.pdf $\endgroup$ – mike stone Aug 15 '17 at 17:45
  • $\begingroup$ Thank you for the material. I was looking for something along the lines of page 257, that is hamiltons theory of rays. Also, regarding the page you pointed out, as I understand it, the overall idea is that dispersion is dependent on the PDE so if i take classic wave eq, i'll get a dispersion, for Schrodinger a different one, for KdV another, and so on. Is that the thing? $\endgroup$ – Victor Palea Aug 15 '17 at 21:57

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