# Why phase velocity and group velocity is not the same?

1. Nov 13, 2008

### KFC

I know the definition of group and phase velocity but why? Why they are different and what's the physics behind it to define such velocities?

2. Nov 13, 2008

### morrobay

3. Nov 14, 2008

### Andy Resnick

Loosely speaking, the group velocity represents how fast a wave packet propagates, while phase velocity refers to how fast a particular component of the packet propagates. The distinction becomes important in dispersive media (the refractive index varies with frequency) and for wave packets containing a large bandwidth.

4. Nov 14, 2008

### arildno

We may prove, for example for inviscid water waves, that the group velocity is the velocity by which energy is propagated throughout the medium (this does NOT contradict Andy's description!!).

An observable wave has, simplified, two main features:
A) a "wave-length" (say the typical distance between two peaks)
B) an amplitude.
Now, the phase velocity tells us the velocity by which the "peaks" move.
But, a peak without an amplitude is hardly observable!
That's where the group velocity can be said to "kick in":
Since amplitudes are directly related to the energy content, a previously invisible "peak" will rise once the energy to rise it comes there.

Thus, if the group velocity is higher than the phase velocity, as is the case with capillary waves, for example, then a typical image will be the coninual production of new waves IN FRONT of the previous wave front, whereas if the group velocity is lower than the phase velocity (as in typically gravity.driven water waves), then new waves will form in the wake of the wave train (as the energy has been "left behind" there)

5. Nov 15, 2008

### Phrak

Yes... But "what's the physics behind it..."

6. Nov 15, 2008

### Staff: Mentor

All waves that follow the differential wave equation and the principle of superposition can be formed into wave packets which can be described using the concepts of phase velocity and group velocity: water waves, sound waves, electromagnetic waves, the quantum-mechanical wave function, etc.

So "the physics behind it" is whatever physics makes some particular phenomenon behave as a wave.

7. Nov 15, 2008

### KFC

Thanks for the explanation. It's soft of clear now. Group velocity is easy to understand, but why we call the velocity of a specific component of wave 'phase' velocity? I mean why use 'phase'?

By the way, if the medium in which the wave is propagating is not dispersive, should the phase velocity equals to group velocity? Since wave velocity could be calculated with

$$v = \dfrac{\omega}{k}$$

If the wave only have two components so that the the group velocity be

$$v_g = \dfrac{\Delta\omega}{\Delta k}$$

If the medium is non-dispersion meidum so that $$\Delta\omega=0, \Delta k=0$$, how could we have $$v_p = v_g$$?

8. Nov 15, 2008

### Staff: Mentor

Because it's the ("horizontal") velocity at which a point on the wave with a specific (constant) phase travels. See the attached diagram.

Yes.

I would prefer to say that for a non-dispersive medium, $\Delta \omega$ is zero regardless of the value of $\Delta k$. If $\Delta k \ne 0$ this gives $v_g = \Delta \omega / \Delta k = 0$. If $\Delta k = 0$, then we have only one wave and it makes no sense to talk about "group velocity" as distinct from "phase velocity"!

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• ###### phase_vel.gif
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9. Nov 20, 2008

### Phrak

Thanks, jt.

I'm trying to reconcile particle mass, which is nearly synonymous with Vg < Vp, with the introduction of an internal degree of freedom. This internal degree of freedom obtains the same relationship, Vg < Vp where particles are classical.

Group velocity and wave velocity over the spectrum are not the same when w/k<>C, for some constant, C. Alternately we can have the condition where all waves have the property W/K=C (C=constant) over a spacetime manifold having a fifth cylindrical dimension, a la Kaluza-Klein. The relationship w/k<>C is reacquired, where w and k are wave and group velocities in the spacetime submanifold for any constant coordinate of the cylindrical dimension.

I'm curious as to where there may be an analogy within classical, non-relativistic physics, where a displacement degree of freedom results in the same relationships.

Last edited: Nov 20, 2008
10. Nov 20, 2008

### Staff: Mentor

Ugh. I can't believe I wrote this. :grumpy: If this were true, $\omega$ would have to be the same for all values of k. I'll start over.

In a non-dispersive medium, all waves travel at the same speed $v_p = \omega / k$, regardless of the value of k. That is, the dispersion relation is $\omega = v_p k$ where $v_p$ is a constant. Therefore the group velocity is $d \omega / dk = v_p$.

Not necessarily. In a non-dispersive medium, the two waves have $\omega_1 / k_1 = \omega_2 / k_2 = v_p$ (a constant), i.e. $\omega_1 = v_p k_1$ and $\omega_2 = v_p k_2$ for the same $v_p$. $\Delta \omega$ and $\Delta k$ can certainly be non-zero.

Last edited: Nov 20, 2008
11. Nov 21, 2008

### Phrak

The wave velocity of sound in a pipe made of material X can exceed the velocity of sound in material X.