Waves, Superposition, Dispersion

Let's see what happens when you add 2 waves together that have different wavelengths and different frequencies, and are moving in the same direction. Note: we are referring here to 2 waves each of which is a function of position, and is traveling in some direction so it is also a function of time. Such a wave is always represented as a $\sin$ and/or $\cos$ function (or equivalently, using exponential notation) and these waves exist for all values of $x$: they are non-localized. A wave that is localized is a more complicated thing that needs Fourier analysis to analyze, and we will save that for a later time.

So, we will need 2 wave functions: $$\begin{align} y_1 &= A\sin({k_1x-\omega_1t})\nonumber\\ y_2 &= A\sin({k_2x-\omega_2t})\nonumber\\ \end{align}\nonumber\\$$ To add these we first define $\Delta k \equiv k_2 - k_1$ and $\bar k \equiv \half(k_2+k_1)$ and solve for $k_1$ and $k_2$ to get $$\begin{align} k_1 &= \bar k - \half\Delta k\nonumber\\ k_2 &= \bar k + \half\Delta k\nonumber\\ \end{align}\nonumber\\$$ Similarly, we define $\Delta \omega \equiv \omega_2 - \omega_1$ and $\bar \omega \equiv \half(\omega_2+\omega_1)$ and solve for $\omega_1$ and $\omega_2$ to get $$\begin{align} \omega_1 &= \bar \omega - \half\Delta \omega\nonumber\\ \omega_2 &= \bar \omega + \half\Delta \omega\nonumber\\ \end{align}\nonumber\\$$

Then we can rewrite the 2 wave functions as $$\begin{align} y_1 &= A\sin({k_1x-\omega_1t})\nonumber\\ &= A\sin([\bar k-\half\Delta k]x-[\bar\omega-\half\Delta\omega]t)\nonumber\\ &= A\sin([\bar kx-\bar\omega t] - \half[\Delta k\cdot x-\Delta\omega\cdot t])\nonumber\\ \end{align}\nonumber\\$$ Similarly, we have $$\begin{align} y_2 &= A\sin({k_2x-\omega_2t})\nonumber\\ &= A\sin([\bar k+\half\Delta k]x-[\bar\omega+\half\Delta\omega]t)\nonumber\\ &= A\sin([\bar kx-\bar\omega t] + \half[\Delta k\cdot x-\Delta\omega\cdot t])\nonumber\\ \end{align}\nonumber\\$$ To make it easy to see how to add, let's define $\alpha \equiv \bar kx-\bar\omega t$ and $\beta \equiv \half(\Delta kx-\Delta\omega t)$ and write the 2 wave functions as $$\begin{align} y_1 &= A\sin(\alpha-\beta)\nonumber\\ y_2 &= A\sin(\alpha+\beta)\nonumber\\ \end{align}\nonumber\\$$ Adding them together gives $$\begin{align} y &= y_1+y_2=2A\cos(\beta)\sin(\alpha)\nonumber\\ &= 2A\cos(\half[\Delta k\cdot x-\Delta\omega\cdot t])\sin(\bar kx-\bar\omega t)\nonumber\\ &= 2A\cos(\half\Delta k[x-\frac{\Delta\omega}{\Delta k}t])\sin(\bar kx-\bar\omega )\label{beat2}\\ \end{align}\nonumber\\$$ This is exactly the result we got when we added 2 waves with 2 different wavelengths: the resulting wave goes like the average of the wavelengths, with an amplitude modulated at $\half$ the difference. But here, since these are 2 traveling waves going in the same direction, we have the resulting wave propagating with the average wavelength and frequency, with an amplitude that is also moving at the difference in wave number and angular frequency. This amplitude is usually called the "group", and it has a velocity (called the "group velocity") whcih can be seen clearly by rewriting equation $\ref{beat2}$ as $$y= 2A\cos(\half\Delta k[x-v_gt])\sin(\bar kx-\bar\omega )\nonumber$$ where we have replaced $$x-\frac{\Delta\omega}{\Delta k}t\nonumber$$ with $$v_g = \frac{\Delta\omega}{\Delta k}\label{vg}$$ So in summary, each wave has a phase velocity given by equation $\ref{vp}$, $v=\omega/k$, and we call this the "phase velocity" because it shows the velocity of each point on the wave, which is the wave phase. The sum of the 2 waves is called the group, with a group velocity given by equation $\ref{vg}$.

Below you can see 2 waves drawn in blue with (slightly) different angular frequency and wavelengths, so each wave will have a different phase velocity given by equation $\ref{vp}$. The yellow dot on each wave represents some constant point in order to make it easy to see the different phase velocities.

Below the 2 blue waves is the sum of both waves, in red, which shows the "beat" pattern (the group), with a wavelength given by the average of the 2 blue waves, and the amplitude has the beat modulation.

Hit the "Start" button to start the simulation. You will see the 2 blue waves start traveling at constant phase velocities, and you will see the amplitude modulation (the group) also traveling as per equation $\ref{beat2}$. This traveling group with a group velocity given by equation $\ref{vg}$. The up and down arrow buttons allow you to change the relative wavelength and period of the waves to see how that effects the group. For instance, if you make $\Delta\lambda$ small or $\Delta\T$ large, then you can get an arbitrarily large group velocity $v_g$ and this will be clear when you run the simulation.

Non-localized waves

Note that so far here we have been considering waves that are described by a wave function $y(t)$ such as $$y(t) = A\sin(kx-\omega t)\nonumber$$ where $v_p=\omega/k$.

This wave function is defined for all space, and has an infinite extent (all $x$). This is called a "de-localized" wave.

Can the phase velocity $v_p=\omega/k$ exceed that for light? Special relativity requires that nothing can exceed the speed of light, $c$, but put another way, it says that no signals can travel faster than $c$. If we define the signal velocity $v_s$, then special relativity says $v_s\lt c$. So the question we are really asking here is whether the phase velocity can also be the signal velocity, and we are asking it in the context of a non-localized wave.

So how would you send a signal (aka information) in a traveling wave moving with phase velocity $v_p$? Because if you can send information in a traveling wave at the velocity $v_p$, then $v_p=v_s\le c$ by special relativity.

To address this, imagine you were looking at this wave coming in, using some kind of detector. What you would be measuring could be the amplitude $A$, the wavelength $\lambda$, the frequency $f$ (or period $T$), or the phase contant. And what you would see is that all of these values are constant. And that means that no information is being sent other than the first moment when someone turned it on and you measured the first values. After that, there's no change in the wave, so no information, consistent with the rules for information theory: a constant value has no information, because you can predict the next value, and a random input would have maximum information because you could not predict the next value from looking at all the previous values. So as a rule, an incoming de-localized wave with some phase velocity $v_p$ carries no information, $v_p\ne v_s$. Since there's no constraint on $v_p$ due to special relativity, it could in fact be greater than the speed of light, $c$.

However, for non-localized light waves in a vacuum, $v_p=\omega/k=c$. If we add 2 light waves with different frequencies together, then using equation $\ref{vg}$ we can write $$\Delta\omega = \omega_2-\omega_1 = v_pk_2-v_pk_1 = v_p\Delta k\nonumber$$ so the group velocity will be $$v_g = \frac{\Delta\omega}{\Delta k}=\frac{v_p\Delta k}{\Delta k}=v_p\nonumber$$ and so the phase and group velocity are equal for non-localized waves at the same frequency in a vacuum. However, in dispersive media, where the index of fraction (and hence the velocity) are functions of the incoming frequency, then we will have a different phase, group, and even signal velocity.

Localized waves

This is an entirely different ball of wax. For localized waves, we need to use the mathematics of Fourier analysis

As you will notice from the above, it's possible to make the group velocity $v_g$ greater than the phase velocity $v_p$ of either wave, and additionally, as $\Delta k\to 0$, we can make $v_g$ arbitrarily larger. Even larger than the speed of light.

With more than 2 waves, things get more complicated, and so we can make use of Fourier analysis. So imagine that we have a bunch of waves, with wave functions $\psi_i$ where $i$ is the index that goes from $1$ to the number of waves, $N$. Then when you add up all of the waves, you get a function $f(x,t)$ given by $$f(x,t) = \sum_{i=1}^N \psi_i\nonumber$$ As $N$ gets large, we can replace the sum by an integral, and write the function $f(x,t)$ evaluated at $f(x,0)$ (t=0) as a Fourier integral $$f(x,0) = \int_{-\infty}^{\infty} A(k)e^{ikx}dk\nonumber$$ By the principle of superposition, the wave function $f(x,t)$ will be given by $$f(x,t) = \int_{-\infty}^{\infty} A(k)e^{i(kx-\omega t)}dk\label{fourier1}$$ where $\omega=\omega(k)$ is some function of the wave number. For light in a vacuum, $c=\omega/k$ so $\omega(k)=ck$.

Equation $\ref{fourier1}$ describes a "wave packet", $f(x,t)$, that is moving to the right (along increasing $x$). We don't know what $\omega(k)$ is, but if the wave packet is "peaked" around some central value $k_0$, then we can define $\omega_0=\omega(k_0)$ and expand $\omega(k)$ using a Taylor expansion: $$\begin{align} \omega(k) &\to \omega(k_0) + (k-k_0)\frac{\partial\omega}{\partial k}\rvert_{k_0}\nonumber \\ &= \omega_0 + (k-k_0)\omega'(k_0)\label{taylor1}\end{align}$$ where $\omega'(k_0)= \frac{\partial \omega}{\partial k}\rvert_{k_0}$

Substituting this into equation $\ref{fourier1}$ gives $$\begin{align} f(x,t) &= \int_{-\infty}^{\infty} A(k)e^{i(kx-[\omega_0+(k-k_0)\omega') t)}dk\nonumber\\ &=\int_{-\infty}^{\infty} A(k)e^{ikx}e^{-i\omega_0t}e^{-ik\omega't}e^{ik_0\omega't}dk\nonumber\\ &=\int_{-\infty}^{\infty} A(k)e^{ikx}e^{ik_0x}e^{-ik_0x}e^{-i\omega_0t}e^{-ik\omega't}e^{ik_0\omega't}dk\nonumber\\ &=e^{i(k_0x-\omega_0t)}\int_{-\infty}^{\infty} A(k)e^{i(k-k_0)x}e^{-i(k-k_0)\omega't}dk\nonumber\\ &=e^{i(k_0x-\omega_0t)}\int_{-\infty}^{\infty} A(k)e^{i(k-k_0)(x-\omega't)}dk\label{fourier2}\\ \end{align}$$ The first part of $f(x,t)$ describes a wave propagating with phase velocity given by $v_p=\omega_0/k_0$, and the 2nd part describes a wave propagating along the $x$ direction with a group velocity given by $$v_g = \frac{\partial x}{\partial t} =\omega'=\frac{\partial\omega}{\partial k}\rvert_{k_0}\label{groupexp}$$ Note: this works if we have $N$ waves that are grouped around a central wavelength $k_0$, and if the function $\omega(k)$ is mostly linear (call this the "Linear" region) so that we can ignore the 2nd derivative $\partial^2\omega/\partial k^2$.

In the next simulation we will add more than 2 waves. Hit the "START" button to start the simulation. The yellow dot shows the motion of each wave. You can click on either "Linear" (the default), or "NonLinear" to change $\omega(k)$. In "Linear" mode, the red wave shows a clear grouping, with each group moving along at the group velocity given by equation $\ref{groupexp}$. Here the group velocity moves faster than the phase velocity of the individual waves, but that's only a function of the slope of $\omega(k)$, seen in the chart below.

If you click on "NonLinear", it changes $\omega(k)$ to be nonlinear, so that the next term in the expansion (equation $\ref{taylor1}$) is non-zero, and in this case large. You can see the effect of the grouping in the red wave: the groups are changing as the wave propagates, and there's no real group velocity because there's no real group. This means that in order to get a consistent and continuous group propagation, so that you can send information, you need the angular frequency $\omega$ to be linear in the wave number. This is something that is guaranteed when you form a wave packet from a Fourier sum.

In the plot below you can see $\omega$ vs $k$. The group velocity is defined as the slope in that plane, given by $\delta\omega/\delta k = \partial\omega/\partial k$ in the Linear region.