# Group velocity with gaussian pulse

A gaussian pulse wave propagating in a non-dispersive medium with velocity $v$ can be expressed as

$$h(z,t) = e^{-a(t - z / v)^2}$$

as suggested in this answer. In the case of a dispersive medium its expression will vary (as specified too in the linked answer). Here is an attempt to derive the new expression for a medium with a generic dispersion relation $k = k(\omega)$.

For $z = 0$: $$h(t) = e^{-at^2}, H(\omega) = \frac{1}{2} \sqrt{\frac{\pi}{a}} e^{-\frac{\omega^2}{4a}}$$

$h(z,t)$ may be obtained by the following reverse Fourier Transform, where the phase of each frequency component is shifted by $k(\omega)z$:

$$h(z,t) = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} H(\omega) e^{j \omega t} e^{-j k (\omega) z} d\omega$$

1) Is it correct?

With the assumption that the $H(\omega)$ is very narrow in $\omega$, $k$ may be expressed with its Taylor expansion

$$k \simeq k(\omega_0) + (\omega - \omega_0) \left. \frac{dk}{d\omega} \right|_{\omega = \omega_0} = k_0 + \omega k_0'$$

where $\omega_0 = 0$ and $(dk/d\omega)_{\omega = \omega_0} = k_0'$. So, the integral becomes

$$h(z,t) = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} H(\omega) e^{j \omega t} e^{-j k_0 z} e^{-j \omega k_0' z} d\omega$$

$$h(z,t) = \frac{1}{4 \sqrt{\pi a}} e^{-j k_0 z} \int_{-\infty}^{+\infty} e^{-\frac{\omega^2}{4a}} e^{j \omega t} e^{-j \omega k_0' z} d\omega$$

2) How to proceed, in order to explicitly get the velocity of propagation along $z$ of the gaussian pulse?

• The notation $j$ for $\sqrt{-1}$ is mostly common in engineering, by the way. In physics the focus is often on the spatial more than the temporal part, so you use $e^{ikz}$ and $e^{-i\omega t}$, but luckily most formulas can be easily translated using $i=-j$. – Emilio Pisanty Feb 20 '17 at 15:09

## 1 Answer

Your manipulations are essentially correct, but they could use some fine-tuning in terms of some bits of perspective. Generally, it is more useful to consider a gaussian pulse with a nonzero carrier frequency, of the form $$h(z,0)=Ae^{-z^2/2\sigma^2}\cos(k_0z) =\frac{\sigma A}{2\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\sigma^2(k-k_0)^2/2}e^{ik z} \mathrm dk +\text{c.c.}, \tag1$$ where $\text{c.c.}$ means the complex conjugate of the preceding term. Note, in particular, two things:

• I am starting with a position-space initial condition, as opposed to the boundary condition in your example. This is more in line with what you do in optics, though if you have radio waves a time-dependent source can be more relevant as a condition.

• I am using the wavevector $k$ as the basic building block, and not $\omega$. This is important already in 1D, because it enables me to reach both positive and negative wavevectors, and it becomes even more important in higher dimensionalities, since $\mathbf k$ will be a vector and I want access to the whole space to get the spatial shape of my initial conditions right.

Having done this, the decomposition $(1)$ gives me my gaussian pulse as a weighted superposition of plane waves $e^{ikz}$, and I know how those will evolve in time: as $$e^{i(kz-\omega(k)t)},$$ where $v_\varphi=\omega(k)/k$ is the phase velocity of the waves, which is fixed by the wave equation and in particular by the wavelength dependence of the refractive index.

In a sense, I'm done: the time evolution of the pulse will be given by $$h(z,t) =\frac{\sigma A}{2\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\sigma^2(k-k_0)^2/2}e^{i(kz-\omega(k)t)} \mathrm dk +\text{c.c.}, \tag2$$ and there's nothing more to say, particularly because this depends on $\omega(k)$ in a way that cannot be simplified further without more knowledge of that dispersion relation.

In the specific case you posit, you want to consider a pulse with a bandwidth $1/\sigma$ that is narrow enough that the dispersion only has enough space to show a linear variation: $$\omega(k)=\omega(k_0)+(k-k_0)\left.\frac{\mathrm d\omega}{\mathrm dk}\right|_{k_0} + O\left(\sigma^2(k-k_0)^2\right),$$ and we neglect the quadratic term. When we do this, we get the term $$v_g(k_0)=\left.\frac{\mathrm d\omega}{\mathrm dk}\right|_{k_0},$$ which is known as the group velocity, and it is this (and not the phase velocity) that will drive the velocity of the propagation of the pulse. If you put this dispersion relation into $(2)$, you get $$h(z,t) =\frac{\sigma A}{2\sqrt{2\pi}}e^{i(-\omega(k_0)+k_0v_g(k_0))t}\int_{-\infty}^\infty e^{-\sigma^2(k-k_0)^2/2}e^{ik(z-v_g(k_0)t)} \mathrm dk +\text{c.c.}, \tag3$$ and here you can see that we don't actually have any meaningful dispersion, since all we have done is take the Fourier transform in $(1)$ and change $z$ for $z-v_g(k_0)t$: this means that we have \begin{align} h(z,t) &=\frac{A}{2}e^{i(-\omega(k_0)+k_0v_g(k_0))t}e^{ik_0(z-v_g(k_0)t)}e^{-(z-v_g(k_0)t)^2/2\sigma^2}+\text{c.c.} \\& =\frac{A}{2}e^{-(z-v_g(k_0)t)^2/2\sigma^2}\cos(k_0(z-v_g(k_0)t)+(k_0v_g(k_0)-\omega(k_0))t), \tag4 \end{align} which essentially gives us the same spatial profile, possibly with some additional temporal oscillations.

This means, then, that if you want to include some actual dispersive physics, you need to go one step further, and you need to take the dispersion relation up to and including the second order, which now reads \begin{align} \omega(k) & =\omega(k_0)+(k-k_0)\left.\frac{\mathrm d\omega}{\mathrm dk}\right|_{k_0}+\frac12(k-k_0)^2\left.\frac{\mathrm d^2\omega}{\mathrm dk^2}\right|_{k_0} + O\left(\sigma^3(k-k_0)^3\right) \\& \approx\omega(k_0)+(k-k_0)v_g(k_0)+\frac12(k-k_0)^2\alpha_2(k_0). \end{align} (Here it appears I'm running against convention, since both Wikipedia and RP Photonics have this Group Velocity Dispersion as a derivative against $\omega$, but I'm not sure how they manage with a vector $\mathbf k$ in e.g. a non-isotropic medium. For more details see your local optics textbook; what follows is mostly for flavour.)

If you put this into the integral in $(2)$, it's still a perfectly doable gaussian integral, but of course now we have time in the quadratic term as well $$h(z,t) =\frac{\sigma A}{2\sqrt{2\pi}}e^{i(k_0v_g(k_0)-\omega(k_0))t}\int_{-\infty}^\infty e^{-\frac12\left(\sigma^2-i\alpha_2(k_0)t\right)(k-k_0)^2}e^{ik(z-v_g(k_0)t)} \mathrm dk +\text{c.c.}, \tag5$$ and this changes what the Fourier transform looks like: \begin{align} h(z,t) =\frac{ A}{2}\sqrt{\frac{\sigma^2}{\sigma^2+i\alpha_2(k_0)t}}e^{i(k_0v_g(k_0)-\omega(k_0))t} e^{ik_0(z-v_g(k_0)t)} \exp\left(-\frac{z^2/2}{\sigma^2+i\alpha_2(k_0)t}\right) +\text{c.c.} \tag6 \end{align} Here, as you can see, there isn't an immediate factorization into a cosine and a gaussian, but you can get one with some minor manipulations, essentially by multiplying and dividing by the conjugate of the gaussian's denominator: you can set $$\exp\left(-\frac{z^2/2}{\sigma^2+i\alpha_2(k_0)t}\right) = \exp\left(-\frac12\frac{\sigma^2-i\alpha_2(k_0)t}{\sigma^4+\alpha_2(k_0)^2t^2}z^2\right),$$ and assign the oscillatory part, as $e^{+i(\cdot)z^2}$, into the cosine, which gives you \begin{align} h(z,t) & =\frac{A/2}{\sqrt{1+i\alpha_2(k_0)t/\sigma^2}} \exp\left(-\frac{z^2/2}{\sigma^2+\alpha_2(k_0)^2t^2/\sigma^2}\right) \\& \quad \times \exp\left(+i\left( k_0(z-v_g(k_0)t) +\frac12\frac{\alpha_2(k_0)t}{\sigma^4+\alpha_2(k_0)^2t^2} z^2 +(k_0v_g(k_0)-\omega(k_0))t \right)\right) +\text{c.c.} \tag6 \end{align} Unfortunately, this is probably about as far as analytics can get you, because of the imaginary part in the square root, though if you're willing to do some ugly algebra there's probably some more that you can do.

Physics-wise, though, you're essentially done, and the important parts are already baked into the quadratic term inside the oscillatory exponential. To see its effect, the simplest thing to do is to graph this and let it loose: Mathematica code through Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/B1AA9.png"]

This is really the core of what (linear) dispersion will do to a pulse, and it comes in two important aspects:

• Most crucially, the pulse gets chirped: that is, the higher-frequency components get pushed to the front of the pulse, and the local wavelength in the back of the pulse gets longer. This is a natural consequence of dispersion, which at its essence simply says 'plane waves of different wavelenths travel at different speeds'; here you can see those plane waves getting out of sync with each other as they go.

• As a corollary of that, the pulse gets stretched temporally, because its components at different wavelengths no longer coincide well. This is another natural consequence of dispersion, and it can be a bummer if what you're after is a really short pulse. Fortunately, though, it can easily be undone, by simply passing it through a similar optical element with the opposite group delay dispersion, as is done to great effect in chirped pulse amplification.

That's probably plenty for now, so I'll leave it at that ;-).

• You gave a huge answer, thank you. Despite the delay, I tried to read it carefully. A first question about your text: in the beginning, why are you considering a $\mathrm{c.c.}$ term to be summed to the integral? Instead, there should be another integral with $e^{- \frac{\sigma^2 (k + k_0)^2}{2}}$. – BowPark Mar 9 '17 at 16:45
• @BowPark They're (most likely) one and the same, related through the change of variable $k\mapsto -k$. There is a chance I messed up the algebra, though - check and let me know. – Emilio Pisanty Mar 9 '17 at 19:01