# The Wave Equation

I know that you all may be falling asleep by this point in the lessons, but WAKE UP IT’S TIME FOR THE BEAUTIFUL, FANTASTIC, ONE AND ONLY WAVE EQUATION! The wave equation is extremely important in a wide variety of contexts not limited to optics, such as in the classical wave on a string, or Schrodinger’s equation in quantum mechanics. In the case of these lessons, though, the wave equation has important implications in almost every aspect of calculation in optics, as it gives us the form of the waves that we are working with as the solution. Now that everyone understands its importance, let’s derive it!

DERIVATION

Considering first the greatly simplified case of a homogeneous, isotropic, nonconducting, source free, dielectric medium, Maxwell’s equations reduce to:

\begin{equation} \nabla \cdot E = 0 \end{equation}

\begin{equation} \nabla \times E = -\frac{\partial{B}}{\partial{t}} \label{maxwell}\end{equation}

\begin{equation} \nabla \cdot B = 0 \end{equation}

\begin{equation} \nabla \times B = \epsilon\mu\frac{\partial{E}}{\partial{t}} \end{equation}

Then, to motivate the next step, we will consider that we would like to be working with only the electric field to make our analyses much simpler. In order to do that, we would like to eliminate B in Equation \eqref{maxwell} by taking the curl of both sides. In this case, we would have:

\begin{align} \nabla \times (\nabla \times E) & = \nabla \times (-\frac{\partial{B}}{\partial{t}})\\ & = -\frac{\partial}{\partial{t}}(\nabla \times B) \\ & = -\epsilon\mu\frac{\partial^2E}{\partial{t}^2} \label{wave_eq}\end{align}

where Equation \eqref{wave_eq} is the most general form of the wave equation, and this is the form most commonly used when dealing with anisotropic media and near-field optics. However, we can use the vector identity for $\nabla \times \nabla$ in order to rewrite the wave equation in terms of scalar wave equations for each vector component of E. The wave equation is then given as:

\begin{equation} \nabla^2E = \mu\epsilon\frac{\partial^2E}{\partial{t}^2} \label{far_field_waveeq}\end{equation}

We can also just as easily find the wave equation for the magnetic field by eliminating “E” in the above equations. The magnetic field wave equation is given as:

\begin{equation} \nabla^2H = \mu\epsilon\frac{\partial^2H}{\partial{t}^2} \end{equation}

It it usually sufficient to start with the electric field representation of the wave equation; however some select circumstances may be more disposed to using the magnetic field representation.

It is important to keep in mind that the Laplacian operator acts on scalar fields, meaning that the wave equation is really three separate equations for x, y, z equations (or whatever your chosen coordinates are, although a change of coordinates would require a change to the wave equation itself).

Finally, a question you may have been asking yourself is if we can derive the speed of light, or is the speed of light just a measured quantity? It turns out we can actually derive the speed of light using our knowledge of the general form of the wave equation. The general form of the wave equation (applicable to any application) is given as:

\begin{equation} \nabla^2f(r,t) = \frac{1}{v^2}\frac{\partial^2}{\partial{t}^2}f(r,t) \end{equation}

where “$v$” is the phase velocity of the wave. Thus, in our present case, the speed of the electromagnetic wave (i.e. of light) would be given as:

\begin{align} v & = \frac{1}{\mu\epsilon} \\ & = \frac{1}{\mu_0\epsilon_0} \\ & = 2.9979 \times 10^8 m/s\end{align}

where we have made the approximation that $\mu_r$ in $\mu = \mu_r\mu_0$ is equal to 1 as well as $\epsilon_r$ in $\epsilon = \epsilon_r\epsilon_0$, and we have taken $\mu_0 = 1.256 \times 10^{-6} H/m$ and $\epsilon_0 = 8.854 \times 10^{-12} F/m$.

GENERAL TIME DOMAIN SOLUTION

Given the above examples for other places where the wave equation crops up, you may have already guess the solution to the wave equation. However, the first time I saw the wave equation I certainly did not know the form of the solution completely, so I will write out the solution explicitly here. The electric field that would solve Equation \eqref{far_field_waveeq} is the simple harmonic plane wave, given as:

\begin{equation} E(r,t) = E_0cos(\omega{t}-k\cdot{r}) \end{equation}

where $k$ is the wave vector and is commonly given as either $|k| = \frac{2\pi}{\lambda}$ or $|k| = \frac{n\omega}{c}$. Now, it is useful to pause here and note that plane waves are unphysical solutions, because as written they can propagate to infinity unaffected. In order to make them physical, some accounting for amplitude or phase change/damping must be included in the equation, which we will get to in short order. In the meantime, the reason you will find plane waves listed and used everywhere in order to represent the propagating EM-wave is because they are still very useful! Here’s are the reasons why you should trust me and still use them:

1. When used to describe physical situations such as the center of a well-collimated laser beam, the plane wave approximation is actually quite accurate. This is because of their enhanced coherence such that they can propagate for many Rayleigh ranges (we will explain what this is in a few sections) with negligible distortion such that they essentially go to infinity.
2. The plane wave expansion is a common technique to solve Maxwell’s Equations, and we will be doing this ourselves within this section (look at the section on the Helmholtz equation!). In particular, they are useful in situations with periodic media, such as photonic crystals.

Now, to make our lives easier, we will use the exponential form of the plane wave for the rest of this section, and really the rest of our dealings in optics. The exponential form of the plane wave is given as:

\begin{equation} E(r,t) = E_0e^{i(\omega{t}-k\cdot{r})} \end{equation}

where, since the electric field is a real quantity, we then just take the real part of this field.

Now, we can prove that the exponential form is a viable solution to the wave equation (Equation \eqref{far_field_waveeq}) by plugging in the solution as follows:

\begin{align} \nabla^2{E} & = E_0\nabla^2e^{i(\omega{t} – k\cdot{r})} \\ & = -k^2E_0e^{i(\omega{t}-k\cdot{r})} \\ & = -k^2E \end{align}

and

\begin{equation} \mu\epsilon\frac{\partial^2E}{\partial{t^2}} = -\omega^2\mu\epsilon{E} \end{equation}

In order to make these two sides equal, then, we have that

\begin{align} k^2 & = \omega^2\mu\epsilon \\ & = \frac{\omega^2n^2}{c^2}\end{align}

where we have used the fact that $\mu\epsilon = \frac{n^2}{c^2}$ (found further up on this page). Thus, the exponential solution does solve the wave equation!

Another important result that we can obtain from our definition of the electric field comes when we further analyze it using Maxwell’s equations. According to our original assumption of a source-free medium, the divergence of the electric field should be zero. As such, we have:

\begin{align} \nabla\cdot{E} & = 0 \\ & = \nabla\cdot[E_0e^{i(\omega{t}-k\cdot{r})}] \\ & = -ik \cdot E_0e^{i(\omega{t}-k\cdot{r})} \end{align}

which means that $k \cdot E_0 = 0$. Knowing that the dot product tells “how much of” two vectors overlap, since this dot product is zero, then this means that “$k$” and “$E$” do not overlap at all; in fact they are transverse to one another! The same result would occur for the magnetic field. Now the question is, how do the electric and magnetic field relate with respect to the $k$-vector, the direction of propagation? Can we find a right-handed orthogonal set? The answer is, as you might anticipate, oh you betcha! But let’s prove it like the good scientists we are using Maxwell’s Equations first:

\begin{equation} \nabla \times E = -\frac{\partial{B}}{\partial{t}} \end{equation}

\begin{equation} \nabla \times E_0e^{i(\omega{t}-k\cdot{r})} = -\frac{B_0e^(i(\omega{t}-k\cdot{r})}{\partial{t}} \end{equation}

\begin{equation} -ik\times E = -i\omega{B} \label{keb_orth}\end{equation}

Thus, we have the important result that $k\times{E} = \omega{B}$, which means that $B$ is indeed orthogonal to $k$ and $E$, giving us the proof for our incredible intuition that $k$, $E$, and $B$ form a right-handed orthogonal set. A few final considerations, note that $E$ and $B$ are in phase, since there are no multiplied exponentials that would make them out of phase. Also note that the magnitudes of the field are related by a simple constant value, namely $n/c$ (after a few simple steps of algebra with Equation \eqref{keb_orth}). Thus, the choice of $E$ or $B$ for representation of the propagating light wave truly is arbitrary, and you can easily convert from one to another using this relationship!