Group velocity as a concept in Classical Waves confuse me. It's very easy to point out visually, like in this really helpful graphic here. Okay, it's the speed of the moving bulge, which, notably moves opposite to the phase velocity.

I see what it looks like quite clearly, but there are key things about it I still don't understand.

  1. What are situations where the graphic shown can describe a physical thing? What kind of waves have this property and why is it useful?

  2. Mathematically, group velocity is described as $$v_{g} = \frac{d \omega}{dk}$$

Or perhaps more loosely, the rate of change of angular frequency as a function of wavenumber. However, there is no correlation in my mind between the graphic and this equation. How can I relate my intuition to the mathematics?

  • 1
    $\begingroup$ It is the speed at which a modulation moves. A modulation can be seen as the beat of two frequencies with a small interval. An example of a physical situation is water waves: commons.wikimedia.org/wiki/File:Wave_group.gif $\endgroup$
    – user137289
    Commented Feb 4, 2018 at 18:17
  • 1
    $\begingroup$ The graphic shows a situation in which the group velocity and phase velocity have opposite signs. This can occur in metamaterials having a negative index of refraction $\endgroup$
    – garyp
    Commented Feb 4, 2018 at 18:52

6 Answers 6


With a continuous wave you cannot transmit a signal. For a signal to be transmitted, you need a modulation of the wave, e.g. amplitude modulation. For example, to transmit acoustic frequencies (speech), you modulate the high frequency electromagnetic carrier wave (on the order of MHz for medium wave transmitters) with the acoustic frequencies(up to 20kHz). This modulation produces small variations called side-bands (plus and minus 20kHz) in the transmitted waves. The group velocity of a wave describes the velocity with which such modulation of the carrier amplitude, which transmits the signal, propagates. In free space, the group velocity of an EM wave is identical to the phase velocity $c$ because the dispersion is linear $\omega=c k$. Thus also a pulse shaped modulation propagates with unchanged form. On transmission lines, there can be significant nonlinear dispersion, i.e. the phase velocity $v_{ph}= \frac {\omega}{k}$ for different frequencies is not constant and, in general, different from the group velocity $v_{gr}=\frac {\partial \omega}{\partial k}$. This leads to a loss of shape of a pulse-like modulation of the carrier wave. However, the propagation speed of such a pulse modulation can still be obtained from the group velocity.

That the group velocity is opposite to the phase velocity happens only in systems with special nonlinear dispersion relations.

  • $\begingroup$ I wonder if a tangible analogy is that you can't send smoke signals, with a only constant stream of smoke - you have to create an alternation between clear air and smoke, in order to signal (even if the lighting of the fire in the first place, is the only alternation that occurs - that is enough to send a simple signal to anyone who can see that the state of the system has changed from clear air to smoked air). $\endgroup$
    – Steve
    Commented Feb 4, 2018 at 18:51
  • $\begingroup$ I'm afraid I'm having trouble understanding this. First, I'll try and wrap my head around the first sentence. Why is a continuous wave unable to serve as a signal? With @Steve 's analogy, are you saying that a constant stream of smoke cannot serve as a signal because it has no variation of form? You need to periodically turn off and on the signal to produce a message? Kind of like morse code? $\endgroup$
    – sangstar
    Commented Feb 4, 2018 at 18:56
  • $\begingroup$ @sangstar, yes. You can't send morse code, if all you hear is either constant silence or constant tone - it's the alternation between them that sends a signal. A fire alarm would send no signal, if it sounded constantly at all times (or did not sound at all under any circumstances). $\endgroup$
    – Steve
    Commented Feb 4, 2018 at 18:59
  • 1
    $\begingroup$ It was also the fate that befell the boy who cried wolf - his cries ceased to signal that a wolf was present, because he cried whether or not the wolf was present. $\endgroup$
    – Steve
    Commented Feb 4, 2018 at 19:01
  • $\begingroup$ @Steve with you there then. Now, with sound then, for instance, it's clear that a continuous wave could be heard but wouldn't be able to send any signal for the same logic as we've just agreed upon. How then, would this modulation then, solve things? I'm afraid the answer hasn't helped me immediately grasp what this does to the continuous wave to send a signal. $\endgroup$
    – sangstar
    Commented Feb 4, 2018 at 19:03

The following is taken from the intro to this question:

Let us define some relevant parameters:

  • Wave Number $\equiv$ $\mathbf{k} = \mathbf{k}\left( \omega, \mathbf{x}, t \right)$ is effectively the number of wave crests per unit length, which is akin to a density of waves;
  • Wave Frequency $\equiv$ $\omega = \omega\left( \mathbf{k}, \mathbf{x}, t \right)$ is effectively the number of wave crests crossing position $\mathbf{x}$ per unit time, which is akin to a flux of waves;
  • Wave Phase $\equiv$ $\phi = \phi\left( \mathbf{x}, t \right) = \mathbf{k}\left( \omega, \mathbf{x}, t \right) \cdot \mathbf{x} - \omega\left( \mathbf{k}, \mathbf{x}, t \right) \ t + \phi_{o}$ is the position on a wave cycle between a crest and a trough;
  • Wave Amplitude $\equiv$ $A = A\left( \mathbf{k}, \omega, \mathbf{x}, t \right)$ is one-half the distance between the crest and trough for a symmetric, linear wave (though in most cases, $A$ is a constant).

From these definitions we can see that the wave number and frequency are defined as: $$ \begin{align} \mathbf{k} & = \frac{ \partial \phi\left( \mathbf{x}, t \right) }{ \partial \mathbf{x} } \tag{0a} \\ \omega & = \frac{ \partial \phi\left( \mathbf{x}, t \right) }{ \partial t } \tag{0b} \end{align} $$

The phase speed, $V_{ph} \hat{\mathbf{k}}$, is not just $\omega/k$, it is actually the real part of this ratio, or $\Re\left[\omega/k\right]$, since both the frequency and wavenumber can be, in general, complex. Note, this speed is not a true velocity vector, since the vector actually derives from $\mathbf{k}$.

Similarly, the group velocity is defined as: $$ \mathbf{V}_{g} = \frac{ \partial \Re\left[ \omega \right] }{ \partial k } \tag{1} $$

As the definitions above suggest, one can write the wave frequency and wavenumber in a form of continuity equation given by: $$ \frac{ \partial \mathbf{k} }{ \partial t } + \left( \mathbf{V}_{g} \cdot \nabla \right) \mathbf{k} = 0 \tag{2} $$

Another way of expressing the group velocity is that ...different k's propagate with velocity $\mathbf{V}_{g}$... [page 376 of Whitham, 1999] or $\mathbf{V}_{g}$ is ...the propagation velocity for k... [page 380 of Whitham, 1999]. So long as $\mathbf{V}_{g} \neq 0$, then one can show that $\lvert A \rvert^{2}$ propagates with velocity $\mathbf{V}_{g}$. Thus, in the absence of mass-transport and dissipation, the wave energy is carried at $\mathbf{V}_{g}$ [Whitham, 1999].


What are situations where the graphic shown can describe a physical thing? What kind of waves have this property and why is it useful?

An example is electromagnetic whistler waves in the solar wind. Their group speed can exceed their phase speed by up to a factor of two. That allows for the scenario where the phase speed is less than the solar wind speed but the group speed is larger. Thus, the wave can carry energy/momentum against the solar wind flow but the phase of the wave in an observation/stationary frame will be reversed (e.g., reversed polarization).

As for why it's useful, it's not really useful or not. It's a property of a phenomena. If the wave has a sufficiently large group speed, it can carry energy/momentum away from a source region even against the flow in which it may or may not be entrained.

How can I relate my intuition to the mathematics?

See my background descriptions above.


  1. Whitham, G. B. (1999), Linear and Nonlinear Waves, New York, NY: John Wiley & Sons, Inc.; ISBN:0-471-35942-4.
  • $\begingroup$ The math sounds very convincing, I like it. But how would you calculate $\mathbf k$? I suspect it would probably have to be an approximation but understanding how you would calculate that would really benefit my understanding of this. $\endgroup$ Commented Feb 22, 2021 at 21:19
  • $\begingroup$ @AccidentalTaylorExpansion - In data or in theory? In data, you measure a relevant quantity (e.g., magnetic field) at lots of spatial locations simultaneously and then do something like interferometry and other minimum variance techniques to get wave unit vector and group velocity. If you have good enough data, you can use something like cross-spectral analysis to get the phase, which you can use to get the wave vector with magnitude. $\endgroup$ Commented Feb 22, 2021 at 21:27
  • $\begingroup$ I meant more in theory. If you have some wave function $\psi(x,t)$ how would you extract $\mathbf k(x,t)$? I'm used to $\mathbf k$ being a parameter/variable just like $\mathbf x$. $\endgroup$ Commented Feb 22, 2021 at 21:44
  • $\begingroup$ @AccidentalTaylorExpansion - Ah okay, if the system is linear then it's just the Fourier transform and then apply the spatial nabla operator (i.e., gradient, divergence, curl, etc.). The reason this works is because, as you know, in the linear limit the exponent is just the wave phase so the spatial gradient only acts on the $\mathbf{k} \cdot \mathbf{x}$-term. $\endgroup$ Commented Feb 22, 2021 at 21:47

Let us consider for simplicity a wave packet that at $t=0$ has Gaussian shape: $$ f(x,0)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}=\int\frac{dk}{2\pi}e^{ikx}e^{-\frac{k^2\sigma^2}{2}}=\int\frac{dk}{2\pi}e^{ikx}f_k,\\ f_k=e^{-\frac{k^2\sigma^2}{2}}. $$ If the package propagates in a homogeneous media with dispersion relation $\omega(k)$, its time evolution can be written in s traightforward way: $$ f(x,t)=\int\frac{dk}{2\pi}e^{i(kx-\omega_kt)}f_k= \int\frac{dk}{2\pi}e^{i(kx-\omega_kt)}e^{-\frac{k^2\sigma^2}{2}} $$ The last integral cannot be evaluated for arbitrary $k$, but we can make approximation $$ \omega_k\approx \frac{d\omega_k}{dk}|_{k=0}\times k=v_g k,$$ in which case we readily obtain $$ f(x,t)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-v_g t)^2}{2\sigma^2}}=f(x-v_g t,0). $$ This is a wave propagating with speed $v_g$ without changing its shape. In dispersionless media, e.g., the one described by a wave equation $$ \frac{\partial f(x,t)}{\partial x^2}-\frac{1}{c^2}\frac{\partial f(x,t)}{\partial t^2}=0,$$ all solutions behave like that ($v_{ph}=v_g=c$), but in general this is not the case (e.g., if the equation contains additional derivatives in respect to position or time, such as, e.g., Korteveg - de Vries equation). Group velocity thus constitutes an attempt to expand our intuition from linear dispersionless media to dispersive (but stil linear) media.


  • I considered above the simples possible case, but generalizations can be done to non-Gaussian wave packets and wave-packets centered in k-space at points other than $k=0$. In this sense the correct definition of group velocity is $$ v_g(k_0)=\frac{\partial \omega_k}{\partial k}|_{k=k_0}, $$ i.e., specifying the point at which the derivative is centered (see, e.g., the Wikipedia derivation).
  • The group velocity is often defined in terms of partial (rather than full) derivatives or a a gradient, since we typically deal with media having more than one dimension: $$ \mathbf{v}_g(\mathbf{k}_0)=\nabla_\mathbf{k}\omega_\mathbf{k}|_{\mathbf{k}=\mathbf{k}_0}, $$

When we describe phonons we using the dispersion relation (angular velocity vs wavevector). And the slope of this graph can tell you how fast this phonons move (group velocity).


My preferred explanation is the so-called "stationarity of phase" argument. For a wave packet, each individual Fourier component with given $k$ has a different phase at a given position $x$: $\phi(k)=\omega(k) t - kx$. The group velocity tracks the position of the maximum of the wave packet as the function of time. The intuition is that the maximum of the packet occurs at a position where all the components are more or less in phase around the component $k_{0}$ of maximum amplitudes (the amplitudes then add up constructively), i.e. when $\phi(k) \approx cte$ around $k_{0}$ or $\frac{\partial \phi(k)}{\partial k}= 0$ . This yields the result $x=\frac{\partial \omega(k)}{\partial k}t$.


Think of $v_{group}$ as the average of the wave package. There is also phase velocity and you can see it on wiki search it moves differently by definition "The phase velocity of a wave is the rate at which the wave propagates in some medium."

But anyway, you can build your intuition based on maths. It's truly a beautiful subject. I hope you will be the next John Nash.


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