Phase velocity is defined as $v_p=\frac{\omega}{k}$ and is described in various textbooks as being the speed at which the phase of a wave propagates. If you have a wave train that is modulated by an envelope, then while the group velocity gives you the speed of the envelope the phase velocity gives you the speed of the wave within the envelope.

Do $v_p$ and $v_g$ have any meaning if we are only considering one sinusoidal wave and not a superposition of such waves? I have been told that the speed of a single sinusoidal wave is its phase velocity, and that phase velocity is what we're really referring to when we talk about a wave's speed.

However, how can phase velocity be greater than light if this is the case? The standard response seems to be that the group velocity is what determines the rate of information transfer, and it is this velocity that cannot exceed $c$. However, in special relativity a simple limit is set on the speed of any particle, and photons must travel only at speed $c$. If $v_p>c$ then does this not mean that the physical wave within the envelope is superluminal, and so the photons that comprise it are also superluminal?

  • $\begingroup$ I'm not going to vote to close yet, but the question In superluminal phase velocities, what is it that is traveling faster than light is highly related, if not a duplicate. The accepted answer to that question is an answer to this question. $\endgroup$ Commented May 17, 2019 at 21:23
  • $\begingroup$ The accepted answer basically says that the phase velocity represents nothing physical. I'm not sure I accept that, because for a single wave the phase velocity is the wave velocity, and this is clearly something physical because it is the velocity of the photons of light (assuming the wave is EM). $\endgroup$ Commented May 18, 2019 at 1:44

2 Answers 2


The short answer is: group velocity and phase velocity are just terms that help describe how frequency depends on wavelength in a material, and in specific instances can help give us information about how wave propagate in said material. However, at the end of the day, they're just mathematical quantities that aren't under any special obligation to have a neat physical interpretation.

Now, for the slightly longer answer. As you might already be aware, purely sinusoidal waves are in reality a poor way of modeling real signals, since they're infinite both in time and space. Luckily for us, we can express any real life signal that has some spatial confinement as an integral of sinusoidal functions, and these sinusoidal functions are in many ways easier to handle. The tool that lets us do this is the Fourier transform, which basically says that given an arbitrary wave $\alpha(x,t)$ that depends on position and time, we can rewrite it as

$$\alpha(x,t)=\int_{-\infty}^{\infty}A(k)e^{i(kx-\omega t)}dk$$

Where $k$ is the wavenumber (basically the reciprocal of wavelength), $A(k)$ is the Fourier transform of the waveform at $t=0$ (which basically tells us how much of each wavelength the initial signal packet contains), and $\omega=\omega(k)$ is some function of the wavenumber (notation here shamelessly stolen from the wikipedia page on group velocity). So far, this is pure math-- all we've done is write a function in a different way. Now, remembering that $e^{i\theta}=cos(\theta)+isin(\theta)$, you might realize that the integrand looks like an infinite sinusoidal wave traveling to the right at velocity $\omega / k$ for any given value of $k$ that we happen to be integrating over. This speed is the phase velocity $v_p$, and since $\omega$ is a function of $k$, $v_p$ is as well.

The important thing to note is that there isn't necessarily a clean physical interpretation of this quantity, since the thing we physically observe is the integral of the sinusoids, not any individual components of this integral. About all we can say in general about the phase velocity is that it tells us how fast the crest of an infinite sinusoid of definite frequency would travel in our medium. But infinite sinusoids don't really transfer information, given that they're already present everywhere, so the phase velocity doesn't tell us anything about the rate of information transfer in any generality. So, it's perfectly possible for $v_p$ to be greater than $c$ for some specific value of $k$ as long as $\omega (k)$ is a function such that no signal can propagate faster than $c$.

That being said, there are a few specific cases where phase velocity does have a physical interpretation. Namely, if $\omega/k$ is a constant, then waves will travel at the phase velocity undistorted so that the phase velocity is in fact the rate of information transfer. Aside from EM waves in a vacuum, this is rarely the case in physics-- $\omega$ is rarely proportional to $k$ and thus the phase velocity ceases to have a single value or simple physical meaning.

Finally, group velocity is defined as $\frac{\partial \omega}{\partial k}$ and so it doesn't really have much meaning for a single sinusoidal wave since derivatives depend on values around a point, not just at it. The group velocity is useful if our $\omega (k)$ is nearly linear, in which case $v_g$ gives the approximate rate of information transfer (this is exact if the dispersion is exactly linear, as with EM waves in a vacuum). Like before, this isn't true for all materials and almost every material will exhibit non-linear dispersion if pushed into an extreme enough regime. It can also be useful if the packet doesn't contain a large spread of frequencies or doesn't travel a long distance (basically, it's useful whenever we can readily approximate $\omega(k)$ as its first order Taylor expansion in the integral above).

TL;DR- In general, how a wave propagates through a medium is a very complex function that both depends on the medium and the shape of the wave. However, for some simple cases, the phase velocity and group velocity can point us in the right direction and save a lot of unnecessary work.

  • $\begingroup$ Thank you for the elaborate answer - it's helped clear up a lot of my confusion. There is one thing I'm not sure about though: what is information, and how does the rate of transfer of information relate to the speed of photons in the EM wave case? $\endgroup$ Commented May 18, 2019 at 13:40
  • $\begingroup$ @Pancake_Senpai The definition of information can be a little tricky to navigate and overall should probably be its own question. However, for the purposes of this problem I'd say that it's good enough to think of a signal carrying information as one where the receiver can use it to mark a certain instance in time as "special". So an infinite sinusoid (or any infinitely periodic signal) doesn't carry information because there is no way to distinguish one crest from any of the infinite number of ones that have gone by in the past. $\endgroup$ Commented May 18, 2019 at 15:25
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    $\begingroup$ @Pancake_Senpai As to the relationship with the speed of photons-- in actuality the wave packets describing individual photons are always traveling at $c$ because they're traveling through the vacuum between matter. When a material has some other dispersion relation, it's really a way of describing the average interactions that photons will have going through this matter as they are absorbed, re-emitted, and scattered. $\endgroup$ Commented May 18, 2019 at 15:34
  • $\begingroup$ Thank you for your answers. $\endgroup$ Commented May 19, 2019 at 13:24

Phase velocity and group velocity can exceed the speed of light in the case of anomalous dispersion.

Phase velocity is the velocity at which the phase of any one frequency component of the wave travels. You are asking what this physically means, and the answer is the crest of the wave, the phase velocity is at which the crest of the wave travels.

Anomalous dispersion is when the refraction index increases with increasing wavelength.

It does not mean that any information would travel faster then the speed of light.

What is very important that in such media, the speed of the wavefront is what really matters, and that can never exceed the speed of light in vacuum.

  • $\begingroup$ Don't you mean the speed of the wave packet is what matters? $\endgroup$
    – S. McGrew
    Commented May 18, 2019 at 4:30

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