This might seem like a trivial question but it is not for me. So, I was reading on group and phase velocities from A.P. French where he calculates the phase and group velocities for a superposition of sine waves of different speed and wavelength. I will write down a brief analysis:

$$y(x,t)=A\sin(k_1x-\omega_1t) + A\sin(k_2x-\omega_2t) $$ which simplifies to


Now, what I usually find in literature is that for a general wave,the velocity of a wave is defined as $v=\omega/k$ and here we see that by simplifying the superposition, we get a slow moving and a fast moving term and

(1) For the slow moving wave which represents the group envelope, we call the velocity as group velocity $v_g=\Delta \omega/\Delta k=\partial\omega/\partial k$ (for waves with small differences in $\omega$ and $k$).

(2) For the fast moving wave which represents the ripples, we call the velocity as phase velocity $v_g=\bar \omega/\bar k=\omega/ k$ (if $\omega$ is given as a function of $k$).

What I don't understand in this analysis is

  1. Why this definition of velocity? Why did we just divide the factors of x and t and call that as velocity? For a single sine wave, I understand how we can find the displacement of a maxima or minima and see how much it moves in some time t and define that as velocity (just like mentioned here). But is there any similar treatment possible for this?

  2. How did we identify which one was the group and which one was the phase velocity? Also, it is not very intuitive at first site for a person who didn't know this before that there are actually 2 velocities embodied in such a solution?

I would be very grateful if someone could possibly have an answer to these questions.


3 Answers 3


The top two graphs from the MakeaGIF.com website are for waves of one frequency/wavelength travelling at different phase speed as shown by the motion of the red and blue dot sitting on top of a crest.
The term phase is used because you are observing the particle which make up the medium at their maximum upward excursion from the equilibrium position and the speed of that crest is measured as the distance moved by a crest divided by the time taken to move that distance.
You could have equally well chosen to follow a trough or when the particles had zero displacement or the phase $kx-\omega t = \text{constant}$.
Differentiating this expression gives the phase speed as $\left (\dfrac{dx}{dt}\right)_{\rm phase} = \dfrac \omega k$

enter image description here

The bottom graph is the addition of the top two graphs and you will note that a modulating envelope the peak of which as shown by the red dot travels at the group velocity where group refers to the motion of a number (group) of waves added together and $\left (\dfrac{dx}{dt}\right)_{\rm group} = \dfrac {\Delta\omega} {\Delta k}$.
This is derived from French's cosine term where you want the term in the bracket to be a maximum with $\omega = \omega_1-\omega_2$ and $\Delta k = k_1- k_2$ and follow the movement of that maximum.

Hopefully the gif animations below from the Institute of sound and Vibration Research (isvr) will help you to differentiate between group velocity and phase velocity.

enter image description here

  • 2
    $\begingroup$ Very nice. You could go a little bit further and specify what is the group velocity in the particular example of the OP $\endgroup$
    – lcv
    Mar 22, 2020 at 22:58
  • $\begingroup$ I like to give the OPs something to do! $\endgroup$
    – Farcher
    Feb 14, 2022 at 8:38

The way to get the velocity is exactly the same as given in the link you shared, with some re-interpretations. For the case of a single, travelling sine wave, we demanded the condition that the phase remains constant, i.e., $$kx-\omega t=\text{constant}$$ This implies that $$k\frac{dx}{dt}-\omega=0$$ which gives us the wave velocity Here, we start from the equation, $$y(x,t)=2\text{A }\text{sin}\bigg(\frac{k_1+k_2}{2}x-\frac{\omega_1+\omega_2}{2}t \bigg)\text{cos}\bigg(\frac{k_1-k_2}{2}x-\frac{\omega_1-\omega_2}{2}t \bigg)$$ We can think of this in two ways, $$y(x,t)=\text{C}_1(x,t)\text{ }\text{cos}\bigg(\frac{k_1-k_2}{2}x-\frac{\omega_1-\omega_2}{2}t \bigg)$$ which is the equation of the a travelling wave with angular frequency $(\omega_1-\omega_2)/2$ and wave number $(k_1-k_2)/2$ whose amplitude, instead of being constant, is a function of space and time (For comparison, remember that something similar happens in the case of Bloch functions). In this form, the equation describes the dynamics of the envelope, and absorbs the effects of the ripples within the envelope into the modulation of the amplitude. Now, applying the same condition as before, we get the velocity of the envelope to be, $$v_g=\frac{\omega_1-\omega_2}{k_1-k_2}$$ and we call this the group velocity. Similarly, we could consider the dynamics of the individual ripples while treating the envelope as an amplitude modulation. In this case, we work with $$y(x,t)=\text{C}_2(x,t)\text{ }\text{cos}\bigg(\frac{k_1+k_2}{2}x-\frac{\omega_1+\omega_2}{2}t \bigg)$$ and applying the condition that the phase is constant gives us the velocity of the ripples, which we call the phase velocity, $$v_p=\frac{\omega_1+\omega_2}{k_1+k_2}$$ This kind of separation of slow and fast degrees of freedom is ubiquitous throughout physics.

  • $\begingroup$ One of things confusing me is first of all, what do you mean by phase remaining constant? What is phase? And why is it constant? $\endgroup$
    – Tachyon209
    Mar 22, 2020 at 21:25
  • 1
    $\begingroup$ Imagine you are looking at the wave at a particular point $(x_0,t_0)$. If you were to move at exactly the velocity of the wave (let us call this $v$), then the wave at the point $(x_0+v\delta t, t_0+\delta t)$ should be the same as your previous observation. You are moving at the velocity of the wave if and only if the wave is at rest in your frame of reference. $\endgroup$
    – NewUser
    Mar 22, 2020 at 21:31
  • $\begingroup$ The phase is whatever argument appears within the sine or cosine function describing a solitary travelling wave. Since the phase can be defined to be $c(x−vt)$, where c is a constant (from the general form of the solution of the wave equation), we say that the phase must be constant in time when we move at the velocity of the wave. $\endgroup$
    – NewUser
    Mar 22, 2020 at 21:48
  • $\begingroup$ @Newuser a typo in your second last equation? (I guess you meant something else) $\endgroup$
    – lcv
    Mar 22, 2020 at 23:01
  • $\begingroup$ @Icv oh, yes, that's a typo, thanks for pointing it out $\endgroup$
    – NewUser
    Mar 22, 2020 at 23:48

The phase velocity is just velocity of plane wave in the usual sense: $v_p=\omega/k=v$. The group velocity, with definition $$v_g=\left.\frac{\partial \omega}{\partial k}\right|_{k=k_0}\tag{1}$$ , is originally derived from the presumption of a traveling wave packet (see this Wikipedia for details. In what follows similar annotations and symbols to the article will be used) with mean momentum value of $k_0$. In that article the wave packet function in space and time is written as \begin{align} \alpha(x,t)=\int dkA(k)e^{i(kx-\omega t)}\tag{2} \end{align} Note the integral here means superposition of infinite number of complex-valued traveling plane waves $e^{i(kx-\omega t)}$, your question being asked here with superposition of only two wave numbers $k_1$ and $k_2$ is only a special case. You can rewrite (2) as

\begin{align} \alpha(x,t)=e^{i(k_0x-\omega_0 t)}\int dkA(k)e^{i(k-k_0)(x-\omega'_0 t)}\tag{3} \end{align} where Taylor expansion to first order $\omega\approx \omega_0+(k-k_0)\omega'_0$ is used. The main difference between the two integrands in (2) and (3) is that, all components of exponential term in the integral of (3) has the same "phase velocity" of $\frac{(k-k_0)\omega'_0}{(k-k_0)}=\omega'_0$, independent of the integration variable $k$. It is therefore identified as the group velocity of the wave packet envelope.

Back to your question of: how to identify which term correspond to phase velocity or group velocity? For the simple case of adding two plane waves with constant $k$ and $\omega$, the excellent visualization from the other answer shows that the wave which we call "envelope" must have longer wavelength than the "inner" waves. Since $k=2\pi/\lambda$, we can conclude that the cosine term with smaller wave number ($\frac{k_1-k_2}{2}$ as opposed to $\frac{k_1+k_2}{2}$) corresponds to the group velocity of the wave envelope. If we acknowledge that (1) is correct, then we can also see that the "phase velocity" of the cosine term is consistent with the definition of the group velocity of wave envelope: \begin{equation} \frac{(\omega_1-\omega_2)/2}{(k_1-k_2)/2}=\frac{(\omega_1-\omega_2)}{(k_1-k_2)}=\frac{\partial \omega}{\partial k} \end{equation} . The latter follows directly because all $k_1,k_2,\omega_1,\omega_2$ are constant in your example.


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