# The Unapologetic Mathematician

## Plane Waves

We’ve derived a “wave equation” from Maxwell’s equations, but it’s not clear what it means, or even why this is called a wave equation. Let’s consider the abstracted form, which both electric and magnetic fields satisfy:

$\displaystyle\frac{\partial^2F}{\partial t^2}-c^2\nabla^2F=0$

where $\nabla^2$ is the “Laplacian” operator, defined on scalar functions by taking the gradient followed by the divergence, and extended linearly to vector fields. If we have a Cartesian coordinate system — and remember we’re working in good, old $\mathbb{R}^3$ so it’s possible to pick just such coordinates, albeit not canonically — we can write

$\displaystyle\frac{\partial^2F_x}{\partial t^2}-c^2\nabla^2F_x=0$

where $F_x$ is the $x$-component of $F$, and a similar equation holds for the $y$ and $z$ components as well. We can also write out the Laplacian in terms of coordinate derivatives:

$\displaystyle\frac{\partial^2f}{\partial t^2}-c^2\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}+\frac{\partial^2f}{\partial z^2}\right)=0$

Let’s simplify further to just consider functions that depend on $x$ and $t$, and which are constant in the $y$ and $z$ directions:

$\displaystyle\frac{\partial^2f}{\partial t^2}-c^2\frac{\partial^2f}{\partial x^2}=\left[\frac{\partial^2}{\partial t^2}-c^2\frac{\partial^2}{\partial x^2}\right]f=0$

We can take this big operator and “factor” it:

$\displaystyle\left[\left(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\right)\left(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\right)\right]f=0$

Any function which either “factor” sends to zero will be a solution of the whole equation. We find solutions like

\displaystyle\begin{aligned}\left[\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\right]A(x-ct)&=A'(x-ct)\frac{\partial(x-ct)}{\partial t}+cA'(x-ct)\frac{\partial(x-ct)}{\partial x}\\&=A'(x-ct)(-c+c)=0\\\left[\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\right]B(x+ct)&=B'(x+ct)\frac{\partial(x+ct)}{\partial t}-cB'(x+ct)\frac{\partial(x+ct)}{\partial x}\\&=B'(x+ct)(c-c)=0\end{aligned}

where $A$ and $B$ are pretty much any function that’s at least mildly well-behaved.

We call solutions of the first form “right-moving”, for if we view $t$ as time and watch as it increases, the “shape” of $A(x-ct)$ stays the same; it just moves in the increasing $x$ direction. That is, at time $t_0+\Delta t$ we see the same thing at $x$ that we saw at $x-c\Delta t$$c\Delta t$ units to the left — at time $t_0$. Similarly, we call solutions of the second form “left-moving”. In each family, solutions propagate at a rate of $c$, which was the constant from our original equation. Any solution of this simplified, one-dimensional wave equation will be the sum of a right-moving and a left-moving term.

More generally, for the three-dimensional version we have “plane-wave” solutions propagating in any given direction we want. We could do a big, messy calculation, but note that if $k$ is any unit vector, we can pick a Cartesian coordinate system where $k$ is the unit vector in the $x$ direction, in which case we’re back to the right-moving solutions from above. And of course there’s no reason we can’t let $A$ be a vector-valued function. Such a solution looks like

$\displaystyle A(r,t)=\hat{A}(k\cdot r-ct)$

The bigger $t$ is, the further in the $k$ direction the position vector $r$ must extend to compensate; the shape $\hat{A}$ stays the same, but moves in the direction of $k$ with a velocity of $c$.

It will be helpful to work out some of the basic derivatives of such solutions. Time is easy:

\displaystyle\begin{aligned}\frac{\partial}{\partial t}A(r,t)&=\frac{\partial}{\partial t}\hat{A}(k\cdot r-ct)\\&=\hat{A}'(k\cdot r-ct)\frac{\partial}{\partial t}(k\cdot r-ct)\\&=-c\hat{A}'(k\cdot r-ct)\end{aligned}

Spatial derivatives are a little trickier. We pick a Cartesian coordinate system to write:

\displaystyle\begin{aligned}\frac{\partial}{\partial x}A(r,t)&=\frac{\partial}{\partial x}\hat{A}(k\cdot r-ct)\\&=\hat{A}'(k\cdot r-ct)\frac{\partial}{\partial x}(k\cdot r-ct)\\&=k_x\hat{A}'(k\cdot r-ct)\\\frac{\partial}{\partial y}A(r,t)&=k_y\hat{A}'(k\cdot r-ct)\\\frac{\partial}{\partial z}A(r,t)&=k_z\hat{A}'(k\cdot r-ct)\end{aligned}

We don’t really want to depend on coordinates, so luckily it’s easy enough to figure out:

\displaystyle\begin{aligned}\nabla\cdot A(r,t)&=k\cdot\hat{A}'(k\cdot r-ct)\\\nabla\times A(r,t)&=k\times\hat{A}'(k\cdot r-ct)\end{aligned}

which will make our lives much easier to have worked out in advance.

February 8, 2012 - Posted by | Analysis, Differential Equations