Derivation of Group Velocity

Draft of derivation of group velocity versus phase velocity

A counter intuitive result is that sometimes waves can have a phase velocity faster than the speed of light. This can happen in wave-guides and plasmas for example. A natural question is how does the not violate one of the fundamental axioms of relativity ; that nothing can move faster than the speed of light. The canned answer is that relativity implies that information cannot move faster than the speed of light, and that in these situations information moves at the speed of its group velocity, something which is generally scrictly less than the speed of light. In this post I want to define group velocity, and show how you might derive that information, in this case a wave packet, moves at the group velocity.

Imagine you are trying to send a laser pulse down a fibre optic cable, your communication takes the form of turning the laser on and off. We will represent this by multiplying our traveling wave by a gaussian packet. We will treat travelling waves as complex exponentials of the form $f(x,t)=exp(i(kx-\omega t))$ . So our starting wave is then as follows.

$$ \psi(x)=\frac{1}{\sqrt{2\pi {\sigma}^2}}e^{-\frac{x^2}{2{\sigma}^2}}e^{ik_0x} $$

To model the time dependance of the wave we will model the dispersion of the wave by approximating the frequency as a function of wave vector, and then taylor expanding around the primary wavelength of the wave.

$$ \omega(k)=\omega_0+k'\frac{\partial \omega}{\partial k}+\mathcal{O}(k'^2) \\ k'=k-k_0 $$

The plan for the following is first we fourier transform the wave-packet. Then we introduce the time dependence of the wave-packet, which is simple to describe in fourier space, and then we will fourier transform back.

$$ \hat{\psi}(k)=\int_{-\infty}^\infty dx \frac{1}{\sqrt{2\pi {\sigma}^2}}e^{-\frac{x^2}{2{\sigma}^2}}e^{ik_0x}e^{-ikx} \\ \frac{-x^2}{2\sigma^2}-ik'x=-(\frac{x}{\sqrt{2\sigma^2}}+\frac{ik'\sigma}{\sqrt{2}})^2-\frac{ {k'}^2\sigma^2}{2} \\ \hat{\psi}(k)=\frac{\sigma}{\sqrt{2\pi}}e^{\frac{-\sigma^2}{2}(k-k_0)^2} $$

The time dependence in fourier space is accomplished by multiplying the function by $e^(-i\omega(k)t)$ . After that we will then fourier transform back.

$$ \begin{split} \psi(x,t) & =e^{-i\omega_0t}\frac{\sigma}{\sqrt{2\pi}}\int^\infty_{-\infty}dk e^{ikx} e^{-i\omega'(k-k_0)t} e^{-\frac{\sigma^2}{2}{k-k_0}^2} \\ & = e^{i(k_0x-w_0t)}\frac{\sigma}{\sqrt{2\pi}}\int^\infty_{-\infty}dk' e^{ik'(x-\omega't)} e^{-\frac{\sigma^2}{2}k'^2} \\ & = \frac{1}{\sqrt{2\pi\sigma^2}} e^{i(k_0x-\omega_0 t)} e^{-\frac{1}{2\sigma^2}(x-\omega't)^2} \end{split}$$

The gaussian integrals in fourier transforms were evaluated by completing the square. In the last fourier transform we used $-\frac{\sigma^2}{2}k'^2 + ik'(x-\omega't) = -(\frac{\sigma}{\sqrt{2}}+\frac{i}{\sigma\sqrt{2}}(x-\omega't))^2-\frac{1}{2\sigma^2}(x-\omega't)^2$ Additionally the gaussian integral was used. Examining the final functional form we can see that both the group velocity and the phase velocity appear; the wave runs through the wave packet at the phase velocity, however the packet itself moves with the group velocity.