I usually write about computer science, engineering, or music, but today I’m going to digress and talk about physics. In particular, I’ll explain how Maxwell’s equations predict the existence of electromagnetic waves.

I will assume you have a basic familiarity with vector calculus and partial differential equations, but no knowledge of electrodynamics.

There are two important vector fields you need to know about: the electric field \( \mathbf{E} \) and the magnetic field \( \mathbf{B} \). What is the significance of these two fields? Mathematically, they describe the forces that are exerted on charged particles due to the charges of other particles. This is called the Lorentz force, and is given by:

\[ \mathbf{F} = q \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right) \]

Here, \( \mathbf{v} \) is the velocity of the particle. Of course, this velocity depends on the choice of reference frame. Due to Faraday’s Law and Ampère’s Law (both described below), we’ll see that the choice of reference frame doesn’t actually matter (phew!). The factor \( q \) is the charge of the particle. For an electron, \( q \approx -1.602 \times 10^{-19} \) Coulombs. Notice that this value is negative, so electrons tend to accelerate in the opposite direction of the electric field.

Maxwell’s equations relate electric fields, magnetic fields, charges, and current. They come in various forms—here I’ll describe them in their microscopic, differential form. We’ll start with Gauss’s Law:

\[ \boldsymbol{\nabla} \cdot \mathbf{E} = \frac{ \rho }{ \epsilon_0 } \]

This states that for any point in space, the divergence of the electric field \( \mathbf{E} \) is proportional to the charge density \( \rho \) (the constant factor \( \epsilon_0 \) is called the permittivity of free space). More simply, Gauss’s Law tells us that charges create electric fields. There is also a similar law for the magnetic field \( \mathbf{B} \):

\[ \boldsymbol{\nabla} \cdot \mathbf{B} = 0 \]

In other words, magnetic “charges” do not exist; magnetic field lines are always loops. Speaking of magnetic fields, the next law states that, in addition to charges, a time-varying magnetic field can also affect the electric field:

\[ \boldsymbol{\nabla} \times \mathbf{E} = - \frac{ \partial \mathbf{B} }{ \partial t } \]

This is called Faraday’s Law. For example, if I pick an intertial reference frame \( X \) with a stationary electron and move a nearby magnet (which generates a magnetic field) through space, the magnetic field will be time-varying, so an electric field is induced due to Faraday’s Law. Let’s say this electric field imparts a Lorentz force \( f \) on the electron. From the reference frame \( Y \) of the magnet, the magnetic field appears stationary but the electron is moving and will feel the same Lorentz force \( f \) due to the second term in the Lorentz force. So the choice of reference frame does not matter.

The final equation is Ampère’s Law:

\[ \boldsymbol{\nabla} \times \mathbf{B} = \mu_0 \left( \mathbf{J} + \epsilon_0 \frac{ \partial \mathbf{E} }{ \partial t } \right) \]

The constant \( \mu_0 \) is called the permeability of free space. \( \mathbf{J} \) is the current density, which describes how current is locally flowing at a point in space. In English, Ampère’s Law states that a magnetic field is generated by electric current or by a time-varying electric field (the latter term is called “Maxwell’s correction”). Just as with Faraday’s Law, the derivative term will depend on the choice of reference frame, but in the end the Lorentz force always comes out to be the same.

Amazingly, from these simple laws, you can derive all of classical electrodynamics!

In a vacuum, there is no charge or current density. Then we can reduce Maxwell’s equations to:

\[ \boldsymbol{\nabla} \cdot \mathbf{E} = 0 \] \[ \boldsymbol{\nabla} \cdot \mathbf{B} = 0 \] \[ \boldsymbol{\nabla} \times \mathbf{E} = - \frac{ \partial \mathbf{B} }{ \partial t } \] \[ \boldsymbol{\nabla} \times \mathbf{B} = \mu_0 \epsilon_0 \frac{ \partial \mathbf{E} }{ \partial t } \]

Before we continue, we need the following vector calculus identity:

\[ \boldsymbol{\nabla} \times \left( \boldsymbol{\nabla} \times \mathbf{E} \right) = \boldsymbol{\nabla} \left( \boldsymbol{\nabla} \cdot \mathbf{E} \right) - \boldsymbol{\nabla}^2 \mathbf{E} \]

In other words, the curl of the curl is equal to the gradient of the divergence minus the vector Laplacian (the proof follows directly from the definition of the vector Laplacian). For this identity, I chose the letter \( \mathbf{E} \) because we will be taking the curl of the curl of the electric field:

\[ \boldsymbol{\nabla} \times \left( \boldsymbol{\nabla} \times \mathbf{E} \right) \]

But first, let’s take the curl of Faraday’s Law to get this in terms of the magnetic field:

\[ \boldsymbol{\nabla} \times \left( \boldsymbol{\nabla} \times \mathbf{E} \right) = - \frac{ \partial \left( \boldsymbol{\nabla} \times \mathbf{B} \right) }{ \partial t } \]

Now we apply our vector identity to the left-hand side:

\[ \boldsymbol{\nabla} \left( \boldsymbol{\nabla} \cdot \mathbf{E} \right) - \boldsymbol{\nabla}^2 \mathbf{E} = - \frac{ \partial \left( \boldsymbol{\nabla} \times \mathbf{B} \right) }{ \partial t } \]

By Gauss’s Law in a vacuum, we can eliminate the leftmost term:

\[ \boldsymbol{\nabla}^2 \mathbf{E} = \frac{ \partial \left( \boldsymbol{\nabla} \times \mathbf{B} \right) }{ \partial t } \]

Now we can use Ampère’s Law in a vacuum on the right-hand side:

\[ \boldsymbol{\nabla}^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{ \partial^2 \mathbf{E} }{ \partial t^2 } \]

This is called the Helmholtz wave equation. It states that the second spatial derivative (the vector Laplacian) of the electric field is proportional to the second time derivative, where the proportionality constant is \( \mu_0 \epsilon_0 \).

In a classical setting (no relativistic or quantum effects), Maxwell’s equations always hold—and we’ve just shown how they imply the Helmholtz wave equation in a vacuum. Let’s solve the wave equation. Once we have a solution, we’ll check that it agrees with Maxwell’s equations (not all do).

One solution is \( \mathbf{E} = \mathbf{0} \) (which implies \( \mathbf{B} = \mathbf{0} \)), but it’s a bit prosaic to call that a wave. Still, a quick look at Maxwell’s equations reveals that this solution does not violate the laws of classical electrodynamics, so we can have \( \mathbf{E} = \mathbf{B} = \mathbf{0} \) in a vacuum. In other words, darkness is physically possible, which is great because otherwise I would have difficulty sleeping at night.

The nice thing about Maxwell’s equations is that they are linear in \( \mathbf{E} \) and \( \mathbf{B} \). In other words, if we find multiple solutions to these equations, their superposition is also a solution. This fact is very useful in harmonic analysis. In particular, we can represent almost all functions as the superposition of basic sinusoidal waves with the Fourier series. Conveniently, the form of the Helmholtz wave equation suggests a sinusoid might be a solution. If it works and agrees with Maxwell’s equations, then the superposition principle allows us to predict the existence of all kinds of electromagnetic waves.

With that in mind, let’s try a solution of the form \( \mathbf{E} = \hat{x} E_0 \cos \left( \omega t - k z - \phi \right) \), for some unknown \( E_0 \) (amplitude), \( \omega \) (frequency), \( k \) (wavenumber), and \( \phi \) (phase shift). We say that this wave is \( \hat{x} \)-polarized and propagates in the \( \hat{z} \) direction.

First, we’ll try it with the wave equation:

\[ \boldsymbol{\nabla}^2 \mathbf{E} = \mu_0 \epsilon_0 \frac{ \partial^2 \mathbf{E} }{ \partial t^2 } \] \[ \boldsymbol{\nabla}^2 \left( \hat{x} E_0 \cos \left( \omega t - k z - \phi \right) \right) = \mu_0 \epsilon_0 \frac{ \partial^2 }{ \partial t^2 } \left( \hat{x} E_0 \cos \left( \omega t - k z - \phi \right) \right) \] \[ - \hat{x} E_0 k^2 \cos \left( \omega t - k z - \phi \right) = - \hat{x} \mu_0 \epsilon_0 E_0 \omega^2 \cos \left( \omega t - k z - \phi \right) \] \[ k^2 = \mu_0 \epsilon_0 \omega^2 \] \[ \frac{k}{\omega} = \sqrt{\mu_0 \epsilon_0} \]

So \( \mathbf{E} = \hat{x} E_0 \cos \left( \omega t - k z - \phi \right) \) is a solution to the wave equation as long as \( \frac{k}{\omega} = \sqrt{\mu_0 \epsilon_0} \). The quotient of the frequency \( \omega \) and the wavenumber \( k \) is the speed \( c \) of the wave, so we can see that in a vacuum \(c = \frac{\omega}{k} = \frac{1}{\sqrt{ \mu_0 \epsilon_0 }} \). In SI units, that gives 299,792,458 m/s, which seems just right.

Let’s test our solution against Maxwell’s equations in a vacuum. Taking the divergence of \( \mathbf{E} \) yields \( 0 \) because the wave is \( \hat{x} \)-polarized but has no dependence on \( \hat{x} \). Therefore Gauss’s Law is satisfied. The rest of Maxwell’s equations involve \( \mathbf{B} \), so we must find a \( \mathbf{B} \) that works with our trial solution \( \mathbf{E} \).

Let’s start with Faraday’s law:

\[ \boldsymbol{\nabla} \times \mathbf{E} = - \frac{ \partial \mathbf{B} }{ \partial t } \] \[ \boldsymbol{\nabla} \times \left( \hat{x} E_0 \cos \left( \omega t - k z - \phi \right) \right) = - \frac{ \partial \mathbf{B} }{ \partial t } \] \[ \hat{y} E_0 k \sin \left( \omega t - k z - \phi \right) = - \frac{ \partial \mathbf{B} }{ \partial t } \]

This tells us that the time derivative of \( \mathbf{B} \) is a sinusoid polarized in the \( \hat{y} \) direction. Therefore:

\[ \mathbf{B} = \hat{y} E_0 \frac{k}{\omega} \cos \left( \omega t - k z - \phi \right) + \mathbf{C} \]

The vector \( \mathbf{C} \) is the constant of integration. From the Helmholtz equation we saw that \( \frac{k}{\omega} = \sqrt{\mu_0 \epsilon_0} \), so we can rewrite \( \mathbf{B} \) as:

\[ \mathbf{B} = \hat{y} E_0 \sqrt{\mu_0 \epsilon_0} \cos \left( \omega t - k z - \phi \right) + \mathbf{C} \]

Now let’s test \( \mathbf{E} \) and \( \mathbf{B} \) against Ampère’s Law:

\[ \boldsymbol{\nabla} \times \mathbf{B} = \mu_0 \epsilon_0 \frac{ \partial \mathbf{E} }{ \partial t } \] \[ \boldsymbol{\nabla} \times \left( \hat{y} E_0 \sqrt{\mu_0 \epsilon_0} \cos \left( \omega t - k z - \phi \right) + \mathbf{C} \right) = \mu_0 \epsilon_0 \frac{ \partial }{ \partial t } \left( \hat{x} E_0 \cos \left( \omega t - k z - \phi \right) \right) \] \[ - \hat{x} E_0 k \sqrt{\mu_0 \epsilon_0} \sin \left( \omega t - k z - \phi \right) = - \hat{x} \mu_0 \epsilon_0 \omega E_0 \sin \left( \omega t - k z - \phi \right) \] \[ \frac{k}{\omega} = \sqrt{ \mu_0 \epsilon_0 } \]

This is the same constraint that we got from the Helmholtz equation (so it works). Finally, we check Gauss’s Law for Magnetism:

\[ \boldsymbol{\nabla} \cdot \mathbf{B} = 0 \] \[ \boldsymbol{\nabla} \cdot \left( \hat{y} E_0 \frac{k}{\omega} \cos \left( \omega t - k z - \phi \right) + \mathbf{C} \right) = 0 \] \[ 0 = 0 \]

So our sinusoidal \( \mathbf{E} \) and \( \mathbf{B} \) fields agree with all of Maxwell’s equations. Since \( E_0 \), \( \omega \), and \( \phi \) were arbitrary (and we saw that \( k \) depends on \( \omega \)), any amplitude, frequency, and phase shift is valid (but the wave always has constant velocity \(c = \frac{1}{\sqrt{ \mu_0 \epsilon_0 }} \) in a vacuum). By the superposition principle, we can take any linear combination of waves of this form without violating Maxwell’s equations. This means, at the very least, any wave which can be decomposed into basic sinusoids (almost every wave) is valid. Furthermore, the polarization of the wave \( \hat{x} \) and the direction of propagation \( \hat{z} \) were also chosen arbitrarily—Maxwell’s equations do not have a preference, as long as they are perpendicular.

So Maxwell’s equations predict that a lot of electromagnetic waves are possible. Let’s experimentally verify our findings—oh, you’ve done that by reading this article!