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$$v_g\equiv\frac{\partial\omega}{\partial k}.$$ The above equation is for Group velocity of waves, so what is the Physical Interpretation of this Equation? As we know $k$ is wave number which shows how many times the wave repeats itself and $\omega$ is the angular frequency which means the number of rotations per second. So how can my group velocity changes when I increase the wave-number?

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  • $\begingroup$ Your question is not clear. Have you tried applying it to, say, waves on deep water where $\omega=\sqrt{gk}$? Here the phase velocity is $\omega/k= \sqrt{g/k}$ while the group velicity $(1/2) \sqrt{g/k}$ is one half of this. If you throw a stone in pond you can easily see the difference as you watch the waves spread. $\endgroup$ – mike stone Oct 11 '20 at 12:27
  • $\begingroup$ More on group velocity. $\endgroup$ – Qmechanic Oct 11 '20 at 13:08
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The phase velocity, $v_{ph} = \omega/k$, relates the period and the wave length, whereas the group velocity, $v_{g} = \partial\omega/\partial k$, can be shown to describe (approximately) the speed of propagation of a wave packet, and therefore the speed with which information propagates (since communicating information can be understood as modulation of the wave.)

The dependence of the frequency on the wave number is often referred to as dispersion. In some media, such as vacuum, the group and the phase velocities are equal, and such media are called dispersion-less. However, this is by no means a general case - neither for the electromagnetic waves, nor for other types of waves.

The interpretation of the group velocity as the velocity characterizing propagation of information is especially relevant in relativistic domain, since the phase velocity can, in principle, exceed the speed of light, while the group velocity cannot.

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  • $\begingroup$ What if group velocity exceeds from the speed of light? Is there any case where group velocity exceeds and Phase velocity remains constant? $\endgroup$ – Junaid Ihsan Oct 11 '20 at 14:49
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    $\begingroup$ Should the the second $v_{ph}$ be $v_{g}$? $\endgroup$ – user45664 Oct 11 '20 at 17:19

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