# Group velocity definition

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1. Nov 13, 2015

### EmilyRuck

In the propagation of non-monochromatic waves, the group velocity is defined as

$v_g = \displaystyle \frac{d \omega}{d k}$

It seems here that $\omega$ is considered a function of $k$ and not viceversa.
But in the presence of a signal source, like an antenna in the case of electro-magnetic wave or a string in the case of sound waves, the actual independent quantity is $\omega$! And $k$ is a consequence of a vibration of angular frequency $\omega$ which propagates in a certain medium.
With $k = k(\omega)$, we would have

$\displaystyle \frac{dk(\omega)}{d \omega}$

$[\mathrm{seconds} / \mathrm{meters}]$ would be the measure units. The velocity could be simply taken as the reciprocal.

Why the dependent ($k = k(\omega)$) and the independent ($\omega$) variables has been exchanged in the above definition of $v_g$? It was just for the sake of measure-units?

If I wanted to find the derivative of - say - $y = \tan x$, it would be quite strange (if not incorrect) to take $dx / dy$: I would take $dy / dx$ (and then I would use its reciprocal, if I need it). Why here is not so strange instead?

2. Nov 13, 2015

### PWiz

No, it would not be strange at all. You can always differentiate an independent variable with respect to a dependent variable and still get completely valid results: $dy=sec^2 x dx$. Just rearrange the differentials in any way you like to get what you want. What can happen if you do this is is that you don't get the derivative in terms of the variable you differentiated with respect to, and this is not always convenient.

3. Nov 13, 2015

### PeroK

Mathematically, there is not so much difference as you might imagine. If you define $y = tan(x)$, then automatically you are defining $x = tan^-1(y)$ (on any interval where the original function is 1-1). Which one "depends" on the other is a matter of notation and perspective.

Normally you think of velocity as a function of time: $v = v_0 + at$, say. But, equally, this means time is a function of velocity: $t = \frac{v-v_0}{a}$ and hence $\frac{dt}{dv} = \frac{1}{a}$

4. Nov 18, 2015

### EmilyRuck

Thank you for both your answers! Now the definition of group velocity seems to be less weird.

5. Nov 19, 2015

### jasonRF

I realize this is a late reply, but this may help (or may not!). If you consider a wave packet with a spectrum of wave vectors $a(\mathbf{k})$, then it can be represented by (the real part of)
$$\psi(\mathbf{r},t) = \int d^3\mathbf{k} \, \, a(\mathbf{k}) e^{i(\omega(\mathbf{k}) t - \mathbf{k\cdot r})}$$
A wave packet will have a small range of wave vectors around $\mathbf{k}_0$ so that $a(\mathbf{k}) \approx 0$ except where $\mathbf{k}$ is near $\mathbf{k}_0$. In this case we can Taylor expand the dispersion relation, setting
$$\omega(\mathbf{k}) \approx \omega(\mathbf{k}_0) + \nabla_{\mathbf{k}}\omega(\mathbf{k}_0) \cdot(\mathbf{k-k_0})$$
where
$$\nabla_{\mathbf{k}} = \hat{\mathbf{x}}\frac{\partial}{\partial k_x} + \hat{\mathbf{y}}\frac{\partial}{\partial k_y} + \hat{\mathbf{z}}\frac{\partial}{\partial k_z}.$$

The wave packet can then be written,
$$\begin{eqnarray*} \psi(\mathbf{r},t) & = & e^{i(\omega(\mathbf{k}_0) t - \mathbf{k_0\cdot r})} \int d^3\mathbf{k} \, \, a(\mathbf{k}) e^{i(\nabla_{\mathbf{k}}\omega(\mathbf{k}_0) t - \mathbf{r}) \cdot (\mathbf{k-k_0})} \\ & = & A(\mathbf{r}-\nabla_{\mathbf{k}}\omega(\mathbf{k}_0) t )\, e^{i(\omega(\mathbf{k}_0) t - \mathbf{k_0\cdot r})} \end{eqnarray*}$$
where
$$\begin{eqnarray*} A(\mathbf{r}-\nabla_{\mathbf{k}}\omega(\mathbf{k}_0) t) & = & \int d^3\mathbf{k} \, \, a(\mathbf{k}) e^{i(\nabla_{\mathbf{k}}\omega(\mathbf{k}_0) t - \mathbf{r}) \cdot (\mathbf{k-k_0})} \end{eqnarray*}$$
is the envelope of the wave packet traveling at a velocity $\nabla_{\mathbf{k}}\omega(\mathbf{k}_0)$ which we identify as the group velocity.

For me, this way of representing a wave packet provides a reasonable motivation why representing the dispersion relation as $\omega = \omega(\mathbf{k})$ makes sense.

jason

Last edited: Nov 19, 2015
6. Nov 19, 2015

### jasonRF

While doing the dishes I realized I should have added a little more detail. In general we would have a spectrum $b(\mathbf{k},\omega)$ that must obey the dispersion relation of the medium. The wave packet would then be
$$\psi(\mathbf{r},t) = \int d\omega \, \int d^3 \mathbf{k} \, \, b(\mathbf{k},\omega) e^{i(\omega t - \mathbf{k\cdot r})} \delta(\omega - \omega(\mathbf{k}))$$
where th delta function enforces the dispersion relation. The integral over frequency is easy with the delta function. If we identify $b(\mathbf{k},\omega(\mathbf{k}))= a(\mathbf{k})$ then we have the case that I started my earlier post with.

$$\psi(\mathbf{r},t) = \int d\omega \, \int d^3 \mathbf{k} \, \, b(\mathbf{k},\omega) e^{i(\omega t - \mathbf{k\cdot r})} \delta(\mathbf{k}-\mathbf{k}(\omega))$$
Then we Taylor expand $\mathbf{k}(\omega) \approx \mathbf{k}(\omega_0) + \mathbf{r\cdot} \frac{\partial \mathbf{k}(\omega_0)}{\partial \omega} (\omega-\omega_0)$ and end up with a result that looks like ,
$$\begin{eqnarray*} \psi(\mathbf{r},t) & = & e^{i(\omega_0 t - \mathbf{k}(\omega_0)\mathbf{\cdot r})} \int d\omega \, \, b(\mathbf{k}(\omega),\omega) e^{i\left(t - \mathbf{r\cdot}\frac{\partial\mathbf{k}(\omega_0)}{\partial \omega}\right)\left(\omega-\omega_0\right)} \\ & = & e^{i(\omega_0 t - \mathbf{k}(\omega_0)\mathbf{\cdot r})} F\left(t - \mathbf{r\cdot}\frac{\partial\mathbf{k}(\omega_0)}{\partial \omega}\right). \end{eqnarray*}$$
This last representation has the wave packet envelope $F\left(t - \mathbf{r\cdot}\frac{\partial\mathbf{k}(\omega_0)}{\partial \omega}\right)$, where the derivative of the wave vector is a reciprocal group velocity of sorts. FOr me the other representation is nicer to think about, but either would do.

jason

7. Nov 24, 2015

### EmilyRuck

Your answers are always very useful and clear. This may be very detailed, but I need some time to understand it. I might ask you some clarifications in the next days! In the meanwhile, thank you :)

8. Nov 25, 2015

### PFfan01

In a recent study, this classical definition of group velocity is questioned for its violation of Fermat's principle in anisotropic media.
Can. J. Phys. 93: 1510–1522 (2015) dx.doi.org/10.1139/cjp-2015-0167

9. Nov 25, 2015

### jasonRF

I don't have access to that article, but based on the abstract it is about isotropic media moving at relativistic speeds. I guess the movement provides an anisotropy. My understanding of relativity is probably not up to the level of that paper anyway.

Yes, there are times when the approximations I made are not valid (eg the group velocity as calculated from the standard formula is larger than c) but then higher order derivatives will matter. The standard formula is certainly valid in a lot of anisotropic media, including magnetized plasmas.

jason

10. Nov 29, 2015

### vanhees71

I've the impresssion, this Wang likes to write confusing articles about well-understood classical physics, particularly electromagnetic waves. The article in question is also available via arXiv, but I haven't read it in detail anyway, because the abstract sounds as if it's a waste of time (also note that the talk about photons is at best misleading if not wrong).

Of course, as jasonRF's two very clear postings show, the description of a wave packet with help of the group velocity is an approximation which is valid for wave packets of very small spread in wave number, i.e., the $a(\vec{k})$ should be narrowly peaked. This is, however not a sufficient condition for this approximation to be valid. One can derive it more carefully by using the saddle-point approximation of the Fourier integral, and this reveals that it only makes sense as long as you are not too close to a resonance. This explains, why the group velocity of an electromagnetic wave can well be greater than the speed of light (in a region of socalled anomalous dispersion) without violating any relativistic "speed limit", because in such cases the group velocity looses its physical sense, because the approximation by the lowest-order saddle-point approximation breaks down in such cases, because the full Fourier integral reveals that the wave packet gets very much deformed with time and the description of its propagation as just a wave packet which only slowly varies the shape of the envelope with the group velocity as the propagation velocity of its peak becomes meaningless.

11. Dec 3, 2015

### PFfan01

Yes, this Wang likes to write papers challenging some well-established concepts; for example, I see a another recent article, "Electromagnetic power flow, Fermat’s principle, and special theory of relativity", http://dx.doi.org/10.1016/j.ijleo.2015.06.053 [Broken], where Poyting vector as EM power flow in an anisotropic medium is also questioned.

However I think the group-velocity definition for a light pulse is a little confusing, because the light pulse consists of many photons. Just like a group of bicycle riders, with each having different velocities, how can we define the velocity of the group of bicycle riders?

Last edited by a moderator: May 7, 2017
12. Dec 4, 2015

### PFfan01

However, Milonni indicates that “group velocity ceases to have physical significance in the case of anomalous dispersion, when it can exceed c” is a longstanding misconception, because the group velocity in such a case still retains “its meaning as the velocity of nearly undistorted pulse propagation, as experiments have shown” [1].

[1] P. W. Milonni, Fast Light, Slow Light, and Left-Handed Light, IOP Publishing, London, 2005; see Sec. 1.5 Group velocity and Sec. 2.9 Six velocities.

13. Jan 28, 2016

### EmilyRuck

They are both a useful way to represent this phenomenon.
As promised, I tried to follow your computations and I agree with them. Apart from the last theoretical discussions (that are anyway interesting), I would like to make a pair of questions about the wave packet itself.
- Very often the wave packet contains some information to be transmitted. Why is this information contained in the envelope $F$ of the signal? Maybe because that envelope is the actual wave shape which was chosen by the transmitter?
- If the envelope has a shape which is determined by the function $F$, how can it be distorted by dispersion? The form

$F \left( t - \mathbf{r} \cdot \displaystyle \frac{\partial \mathbf{k}(\omega_0)}{\partial \omega}z \right)$

seems not to show this, even with the strangest values for the derivative. Maybe the dispersion is only visibile when the Taylor expansion is no longer possible, and so the $F$ expression will be different?

14. Jan 28, 2016

### jasonRF

If there was no envelope then you have a sinusoidal signal that exists for all time. There isn't any information in that.

Intuitively, if the different frequencies of your wave packet have significantly different group velocity, then the envelope will quickly change as it propagates. My favorite example is the electromagnetic waves produced by a lightning strike. The essentially delta-function in time produces a broad spectrum of waves; the kHz range of which propagate along geomagnetic field lines through the magnetosphere and bounce between the north and south hemispheres. These waves propagate through a plasma and the different frequencies have quite different group velocity. Receivers on the ground can measure these elctromagnetic waves; since they are in the kHz range, we can push the electric signals received through a speaker and hear the 'whistle' that is caused by the dispersion. Google 'whistler' and you will find audio recordings of these.

The approximation I made essentially assumed that the second derivative was negligible (weak dispersion), and that the wave packet had a small bandwidth. Even with these assumptions, if the wave packet travels far enough through the media (ie long enough time) then the second derivative term in the Taylor series will be non-negligible. But for small enough time periods, dispersion is small and the approximations I made completely throw out that physics of the packet deformation.

How do you calculate more interesting cases? If the wave packet is wide band - that is, $a(\mathbf{k})$ is not narrowly peaked - then the approximations I made are not valid. In this case, if the quadratic term in the Taylor series is large (large second derivative and time) then you cannot neglect it. For some shapes of wave packets (eg Gaussian) you may be able to keep the quadratic term of the Taylor series and compute the integral to see how the packet deforms as it propagates. If you cannot do the integral, then the stationary phase approximation can often be a way to understand the distortion to first order. Steepest descents would provide more accurate results in most cases.

As explained by verhees71, near a resonance things get much more complicated. People like Brillioun wrote papers on this topic about 100 years ago.

This topic sounds easy at first but gets very complicated very quickly. Jackson's electrodynamics book has a reasonable intro, but if you google Brillioun precursor or Sommerfeld precursor you will find a lot of interesting material.

jason