Why is the group velocity of the wave given as $\frac {\partial w }{\partial k}$?

I understand what the group velocity means but its always given as a definition and I would like to have some intuition behind it. The $k$ derivative confuses me.

  • 1
    $\begingroup$ Perhaps you could explain what you think the group velocity means. If you have the correct understanding, then the $k$ derivative should not be so confusing. $\endgroup$
    – Jon Custer
    Aug 30, 2018 at 13:17
  • $\begingroup$ It's the velocity of the signal. Like, the velocity of an envelope. $\endgroup$
    – user148787
    Aug 30, 2018 at 13:56
  • 3
    $\begingroup$ Correct. What would cause the envelope of a signal travel at a different speed than the phase velocity? Now, go to en.wikipedia.org/wiki/Group_velocity - they have some nice graphics showing the difference of group and phase velocity. $\endgroup$
    – Jon Custer
    Aug 30, 2018 at 14:03
  • $\begingroup$ Tip: Check the right margin for related questions. $\endgroup$
    – Qmechanic
    Aug 30, 2018 at 14:54

1 Answer 1


Let there be 2 waves of unit amplitude with similar angular frequencies $\omega_1$ and $\omega_2$, and angular wavenumbers $k_1$ and $k_2$ respectively. Therefore, adding these 2 waves, we have $$y = \sin(k_1x-\omega_1 t) + \sin(k_2x-\omega_2 t)$$ Using the factor formula, we can express this as: $$y = 2 \sin \left( \frac{(k_1x-\omega_1 t) + (k_2x-\omega_2 t)}{2}\right) \cos \left( \frac{(k_1x-\omega_1 t) - (k_2x-\omega_2 t)}{2}\right)$$ $$= 2 \sin \left( \frac{(k_1 + k_2)x-(\omega_1 + \omega_2)t }{2}\right) \cos \left( \frac{(k_1 - k_2)x-(\omega_1 - \omega_2)t }{2}\right)$$ $$= 2 \cos \left( \frac{(k_1 - k_2)x-(\omega_1 - \omega_2)t }{2}\right) \sin \left( \frac{(k_1 + k_2)x-(\omega_1 + \omega_2)t }{2}\right)$$ $$= y' \sin \left( \frac{(k_1 + k_2)x-(\omega_1 + \omega_2)t }{2}\right)$$

The important thing here is that $\omega_1$ and $\omega_2$, and $k_1$ and $k_2$ are very close to each other. Since the frequency of the sine term is the sum of $\omega_1$ and $\omega_2$, while the frequency of the cosine term is the difference of $\omega_1$ and $\omega_2$. The frequencies of the sine and cosine terms will therefore be very different. Therefore, we see that the resultant wave has a frequency of $(\omega_1 + \omega_2)$, but its amplitude, $y'$, will be perceived to be modulated by the cosine term, forming an envelope, which look like this.

The group velocity is the velocity of the envelopes, not the velocity of the individual wavefronts. We know that $$v = f \lambda = \left(\frac{\omega}{2\pi} \right) \left(\frac{2\pi}{k} \right) = \frac{\omega}{k}$$ Therefore, the velocity of the envelope, given by the cosine term, will be equal to $$\frac{\omega_1 - \omega_2}{k_1 - k_2}$$ Since $\omega_1$ and $\omega_2$, and $k_1$ and $k_2$ are very close to each other, this reduces to $$\frac{d\omega}{dk}$$ as claimed.

An intuitive way to think about it is this. The two waves interfere constructively when the cosine term is equal to 1, and destructively if it is equal to 0. Let two successive points of constructive interference be a wavelength by itself, since you are finding their velocity (which is the velocity of the envelope). Ignoring the sine term, this can be a wave on its own, completely described by the cosine term. Thus its velocity, which is the group velocity, will be exclusively calculated from the cosine term by $\frac{\omega_1 - \omega_2}{k_1 - k_2}$.


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