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I'm studying what happens to a wave packet if it propagates in a medium under linear optics conditions. In my equations I have the wave vector $k=\frac{\omega}{c}n(\omega)$ (often called $\beta$) which is developed in Taylor series as:

$\beta(\omega)\simeq\beta(\omega_0)+\left.\frac{d\beta}{d\omega}\right|_{\omega_0}(\omega-\omega_0)+\left.\frac{d^2\beta}{d\omega^2}\right|_{\omega_0}(\omega-\omega_0)^2=\beta(\omega_0)+\beta_1(\omega-\omega_0)+\beta_2(\omega-\omega_0)^2$

If we consider only $\beta_1$, we get:

$\beta_1=\left.\left[\frac{d}{d\omega}\frac{\omega}{c}n(\omega)\right]\right|_{\omega_0}=\frac{n(\omega_0)}{c}+\frac{\omega_0}{c}\left.\frac{dn}{d\omega}\right|_{\omega_0}$

where the first element is clearly the inverse of the phase velocity $v_p$ of a wave having frequency $\omega_0$, the second one is a "corrective" term and all together gives the inverse of the group velocity $\beta_1=\frac{1}{v_g}$. From RP Photonics "The first-order term contains the inverse group velocity (i.e., the group delay per unit length) and describes an overall time delay without an effect on the pulse shape". So, if I suppose $\beta_2=0$ I get a travelling pulse that won't be subjected to time broadening. It is also clear that, if $n$ does not depend on frequency (like in vacuum) we get $v_p=v_g$; vice versa, if $n=n(\omega) \Rightarrow v_g\neq v_p$, so the shape will move at a different velocity with respect its phase.

Now, my question is the following:

Suppose we have the second case, that is a Gaussian pulse whose envelope moves at a different group velocity with respect to the phase velocity as shown at this link: (https://en.wikipedia.org/wiki/File:Wave_packet_propagation_(phase_faster_than_group,_nondispersive).gif). For this to happen I have to suppose that $n$ depends on frequency so, from the previous equation, the two velocities are different. But if $n=n(\omega)$, this means that each wave composing the Gaussian packet will move at a different speed as shown in Figure 1 at this link (https://www.rp-photonics.com/group_velocity.html) where the phase fronts of different frequency components propagate with different velocities faster than group velocity and the shape of the pulse isn't broadening. From my book I read "The group velocity dispersion (GVD) is given by the dependence of $v_g$ from $\omega$: inside the envelope, different frequencies moves at different velocity and this is the cause of the time broadening of the pulse".

If velocities are different, why the pulse isn't broadening? And, if different velocities are possible without having a time broadening of the pulse, what is the physical cause of that broadening?

EDIT:

After a lot of research I have cleared up some of the doubts concerning the problem I mentioned above. I was helped a lot by this PDF (link: http://sharif.edu/~kavehvash/Group_Phase_Velocity.pdf) which explains what happens to the phase and group velocities according to the dispersion relation. In particular, two cases are distinguished:

  1. The dispersion relation is linear: this causes group velocity and phase velocity to be equal. This happens because in this condition it should be $n=n_0$ and from the formula of $\beta_1$ we get $v_g=v_p$.

  2. The dispersion relation is non-linear: in this case, the two velocities are always different from each other (unless you have special restrictive conditions). This is due to the fact that $n=n(\omega)$ and each component moves at a different speed. From the PDF: "A very important consequence of this is that our initial wave package broadens out with time because the partial waves forming it gradually move out of phase with each other" and this because, if $v_p=v_p(\omega)$, then $v_g=v_g(\omega)$. The only alternative for which $v_g$ does not depend on $\omega$ and does not broaden is that it is $0$, i.e. there is a standing wave.

If what I have written is correct I am left with the last doubt: is it possible to have $v_g\neq v_p$ without obtaining a time broadening as shown in the following two links?

Link 1: https://en.wikipedia.org/wiki/File:Wave_packet_propagation_(phase_faster_than_group,_nondispersive).gif

Link2 (Figure 1): https://www.rp-photonics.com/group_velocity.html

If yes, how?

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  • $\begingroup$ I saw this yesterday and though of trying to answer, but I think that the best is to see it visually, so I wanted to create some .gif with 2 and 3 frequencies for you to see the beating and how a constant $v_g$ causes slippage of the phase under the envelope, but if $v_g$ is linear (equivalent of constant GDD) then you have broadening. Try to picture this with just 2 frequencies first, for both $v_p=v_g$ (no slippage of phase) and $v_p\neq v_g$ (phase slips, crests move slower than phase). Then adding a third and getting the case where $v_g=const.$ and the case where not. $\endgroup$ Sep 16, 2021 at 16:17
  • $\begingroup$ a quick answer to "is it possible to have $v_g \neq v_p$?" Yes, take the example of a pulse propagating through a medium which has GVD=0 at that wavelength. In the case of fused silica (a common glass in optics) it is around $2\mu m$. (well, you still have higher order terms, but as higher order terms usually play a much smaller role in propagation, you would see that for example a 300 fs pulse, centered at $2\mu m$ passing through 5 mm of fused silica would suffer no broadening but a big phase shift). $\endgroup$ Sep 16, 2021 at 16:24
  • $\begingroup$ Thank you @JoséAndrade! I really appreciate your help. Fortunately (for me), I solved the problem today and I'll now answer here to my question. $\endgroup$
    – Tech
    Sep 16, 2021 at 22:02

2 Answers 2

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In order not to have time broadening and, at the same time, to have $v_g\neq v_p$ the frequencies that make up the pulse must be extremely close (for two frequencies: $\omega_1\simeq \omega_2$).

I solved the problem in the following way: in order to have $v_g\neq v_p$ we must have $v_p=v_p(\omega)$ and here my problem arose because this means that each frequency propagates at a different speed, so we would have temporal broadening of the pulse. However, if the frequencies are assumed to be close together, they will have almost equal phase velocity, so there is no broadening, but at the same time the dispersion relation can be non-linear so we can have $v_g\neq v_p$. Besides, the assumption that it is $\omega_1\simeq \omega_2$ is necessary because, if we assume that it is $\beta_2=0$, then we are leaving out the higher order terms in Taylor's expansion and this means that $\omega-\omega_0$ must be small. The easiest example is surely that of two neighbouring frequencies generating beat: if the dispersion relation is non-linear we will have $v_g\neq v_p$, but the frequencies will be close enough to have $\frac{\omega_1}{k_1}=\frac{\omega_1}{k_1}$ (same phase velocity, so no broadening). Certainly, in a real system, such frequencies can never be infinitely close, but in a real system this also depends on the duration of the pulse and the propagation length.

In summary, in a system that has a non-linear dispersion relation, it is possible to obtain a pulse whose phase velocity is different from its group velocity but, at the same time, in order for its envelope to remain the same during propagation, its component frequencies must be very close to each other.

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  • $\begingroup$ Well, Like I said, this is not entirely true. You can have a 10 nm width (a lot of frequencies) pulse propagate in a medium where $v_g \neq v_p$. Like I said in the comment. As long as you have no $\beta_n$ where $n>1$ then there is no broadening, only slippage of the phase under the envelope. Only higher order terms can broaden your pulse. As an example I gave, when propagating at the zero-dispersion-wavelength of a material ($\beta_2=0$) you do not have broadening, but $v_g \neq v_p$. (1/2) $\endgroup$ Sep 17, 2021 at 12:36
  • $\begingroup$ (2/2) A $\omega$ vs $\kappa$ graph with a straight line with non-zero slope means that for any pulse, the frequencies will all have different phase velocities, but group velocity is constant at value of slope. For the case where the slope is $c$, then phase and group velocities are the same. In all these cases there are no higher order terms, so no pulse broadening, only slippage. All you need to find is a material where at some bandwidth the slope will be approximately linear. $\endgroup$ Sep 17, 2021 at 12:47
  • $\begingroup$ Do my comments make sense? You can have a pulse which spans more than an octave (highest frequency 2x the value of the lowest) and if there are no $n=2$ and higher order terms, then you can easily have a pulse propagating at $v_g$ while the phase propagates at a completely different one (at $v_p$ of central frequency) with no broadening whatsoever. The phase relationship between all frequencies remains completely linear and their interference means a slippage of the phase but no broadening. $\endgroup$ Sep 17, 2021 at 17:51
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Sorry for my poor english. My native language is french.

It is certain that if you limit the spectral ranges of the wave packet, you will reduce the dispersion. But this is not very efficient!

If I understood your question correctly, the only strict solution to your problem would be to have a dispersion relation with a zero second derivative and therefore of the form $k=\frac{\omega-\omega_0}{v_g}$. In this case, we would have a constant group speed and no spreading of the wave packet.

I do not know of a system with a relation of this type but an approximation could be to place oneself on an inflection point of the relation of dispersion (minimum of the group velocity). In this case, we would have a relation of the previous form by neglecting term of the third order.

A classic example is that of gravity waves on the surface of water taking into account surface tension. There is effectively a minimum of the group speed for a wavelength of a few cm. In this case, a weak spreading of the wave packet can be expected.

Note that in general, the spreading of the wavelength leads to a separation of the various wavelengths: after a storm, the long wavelengths arrive on the shore before the small ones.

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