# Maxwell Equations

### Short Topical Videos

### Reference Material

- Feynman Lectures on Physics, Volume II, Ch. 18 Maxwell's Equations (Feynman, Caltech)
- Maxwell's Equations (Wikipedia)
- Wave equation & electromagnetic radiation (Scott Hughes, MIT)

### Related Topics

- Electromagnetic Plane Waves
- Lorentz transformations for the Lorentz invariant forms
- Plasma Frequency

<latex> \documentclass[]{article} \usepackage[top=1in,bottom=1in,left=1in,right=1in]{geometry} \usepackage{amsmath} \usepackage{graphicx} \usepackage{natbib}

\begin{document} \title{Maxwell's Equations for Electromagnetic Waves}

First, a word of caution. We're going to work in CGS units, and some of the reference material above use SI units. In CGS units, there is no $\epsilon_0$ (it's incorporated into the definition of charge) and $\mu_0$ becomes $1/c^2$. In these units, Maxwell's equations are written: \begin{eqnarray} \nabla\cdot E &= 4\pi\rho\\ \nabla\cdot B &= 0\\ \nabla\times E &= -\frac1c\frac{\partial B}{\partial t}\\ \nabla\times B &= \frac{4\pi}{c}J + \frac1c\frac{\partial E}{\partial t}\\ \end{eqnarray} where $E$ is the electric field, $B$ is the magnetic field, $J$ is the current (charge per time), and $\rho$ is the charge density. In words, these equations state: \begin{enumerate} \item Electrical charge creates a divergence of in the electric field (which gives rise to a radial potential). \item Magnetic fields do not diverge. There are no magnetic ``charges" (that we know of). \item A magnetic field that changes with time creates a circular electric potential that drives charges around loops. \item A moving electrical charge (a ``current") or an electric field that changes with time creates a circular magnetic potential. \end{enumerate}

\section{Maxwell's Equations in Free Space}

In free-space, where $\rho=0$ (no charge) and $J=0$ (no current), Maxwell's equations say \begin{eqnarray} \nabla\cdot E &= 0,\\ \nabla\cdot B &= 0,\\ \nabla\times E &= -\frac1c\frac{\partial B}{\partial t},\\ \nabla\times B &= \frac1c\frac{\partial E}{\partial t}.\\ \end{eqnarray} Using the vector identity \begin{equation} \nabla\times(\nabla\times F) = \nabla(\nabla\cdot F)-\nabla^2F, \end{equation} the curl ($\nabla\times$) of the last 2 Maxwell's equations reduces to the following wave equations (using that the divergence of each field is 0): \begin{eqnarray} \nabla^2E -\frac1{c^2}\frac{\partial^2 E}{\partial t^2} &= 0,\\ \nabla^2B - \frac1{c^2}\frac{\partial^2 B}{\partial t^2} &=0.\\ \end{eqnarray} Why are these called wave equations? To make it easier to see why, let's make them one dimensional.

\subsection{A One-Dimensional Wave Equation}

Along one dimension (let's say, $\hat x$), the wave equation for the electric field (derived above as a solution of Maxwell's Equations in free space) is written \begin{equation} \frac{\partial^2 E}{\partial x^2} - \frac1{c^2}\frac{\partial^2 E}{\partial t^2} = 0. \end{equation} Let's suppose that $E(x)=E(x\pm ct)$, and we define a new variable $u\equiv x\pm ct$. The Chain Rule tells us that a spatial derivative of $E$ gives us \begin{equation} \frac{\partial E(u)}{\partial x}=\frac{\partial E}{\partial u}\frac{\partial u}{\partial x}=\frac{\partial E}{\partial u}. \end{equation} Similarly, for the time derivative, we have \begin{equation} \frac{\partial E(u)}{\partial t}=\frac{\partial E}{\partial u}\frac{\partial u}{\partial t}=\pm\frac1c\frac{\partial E}{\partial u}. \end{equation} Based on these identities, we have \begin{eqnarray} \frac{\partial^2 E}{\partial x^2}&=\frac{\partial^2 E}{\partial u^2},\\ \frac{\partial^2 E}{\partial t^2}&=c^2\frac{\partial^2 E}{\partial u^2}.\\ \end{eqnarray} If you plug these into the wave equation at the top of this section, you'll see that it solves the equation. In the general 3D case for $E$ and $B$, this means that any electromagnetic waveform that translates at speed $c$ solves these equations. And I probably needn't point out that an electromagnetic waveform traveling at the speed of light is, well... light!

\emph{To see how this generalizes to plane waves, see Electromagnetic Plane Waves.}

\subsection{The Relative Orientation of $E$ and $B$}

From the previous section, we see that in free space, the Maxwell equations give us 2 wave equations. One propagates the $E$-field at the speed of light, and the other propagates the $B$-field at the speed of light, but we haven't related the $E$ and $B$ fields to one another. To do this, we need to re-examine the Maxwell equations for free space.

We'll defer a rigorous derivation of this until we've examined {\it plane waves} as a solution to these wave equations. In the meantime, here's a sketch of the arguments for the relative phases of $E$ and $B$. Let's suppose that $E$ and $B$ are both described by sine waves of the same frequency, but with some arbitrary phase relative to one another. Let's say that $\vec E=E_0\sin(\vec x+ct)$, so that $\vec E=0$ at $x=0,t=0$. Since a sine wave is anti-symmetric around 0, one can draw a loop encircling $x=0$, and since $\vec E$ is positive on one side of that loop, and negative on the other side, $\nabla\times E$ is not zero, and in fact, is maximal at $x=0$.

\begin{figure} \includegraphics[width=2in]{curlE.png} \caption{The curl of a sine wave (in 2 or more dimensions) is maximized around $x=0$.} \end{figure}

Since we know that \begin{equation} \nabla\times E = -\frac1c\frac{\partial B}{\partial t}, \end{equation} It must be true that $\frac{\partial B}{\partial t}$ is maximized at $x=0,t=0$. For $\sin(x+ct)$, this happens when $x=0,t=0$. So by this hand-wavy argument (which we can can make more rigorous later), $E$ and $B$ must be in-phase with one another (i.e. they hit 0 at the same time). Moreover, because $\nabla\times \vec E$ is perpendicular to $\vec E$ and perpendicular to the spatial gradient of $\vec E$ (which we've set to be in the $\hat x$ direction, the direction the wave moves with time), $\vec B$ must be perpendicular to $\vec E$ and perpendicular to the direction the wave propagates. Finally, if you work out the spatial and temporal derivatives at $x=0,t=0$, you'll see that $E_0=B_0$.