# 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{aligned}\nabla \cdot E&=4\pi \rho \\\nabla \cdot B&=0\\\nabla \times E&=-{\frac {1}{c}}{\frac {\partial B}{\partial t}}\\\nabla \times B&={\frac {4\pi }{c}}J+{\frac {1}{c}}{\frac {\partial E}{\partial t}}\\\end{aligned}}\,\! 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:

1. Electrical charge creates a divergence of in the electric field (which gives rise to a radial potential).
2. Magnetic fields do not diverge. There are no magnetic “charges" (that we know of).
3. A magnetic field that changes with time creates a circular electric potential that drives charges around loops.
4. A moving electrical charge (a “current") or an electric field that changes with time creates a circular magnetic potential.

## 1 Maxwell’s Equations in Free Space

In free-space, where $\rho =0$ (no charge) and $J=0$ (no current), Maxwell’s equations say

{\begin{aligned}\nabla \cdot E&=0,\\\nabla \cdot B&=0,\\\nabla \times E&=-{\frac {1}{c}}{\frac {\partial B}{\partial t}},\\\nabla \times B&={\frac {1}{c}}{\frac {\partial E}{\partial t}}.\\\end{aligned}}\,\! Using the vector identity

$\nabla \times (\nabla \times F)=\nabla (\nabla \cdot F)-\nabla ^{2}F,\,\!$ 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{aligned}\nabla ^{2}E-{\frac {1}{c^{2}}}{\frac {\partial ^{2}E}{\partial t^{2}}}&=0,\\\nabla ^{2}B-{\frac {1}{c^{2}}}{\frac {\partial ^{2}B}{\partial t^{2}}}&=0.\\\end{aligned}}\,\! Why are these called wave equations? To make it easier to see why, let’s make them one dimensional.

### 1.1 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

${\frac {\partial ^{2}E}{\partial x^{2}}}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}E}{\partial t^{2}}}=0.\,\!$ 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

${\frac {\partial E(u)}{\partial x}}={\frac {\partial E}{\partial u}}{\frac {\partial u}{\partial x}}={\frac {\partial E}{\partial u}}.\,\!$ Similarly, for the time derivative, we have

${\frac {\partial E(u)}{\partial t}}={\frac {\partial E}{\partial u}}{\frac {\partial u}{\partial t}}=\pm {\frac {1}{c}}{\frac {\partial E}{\partial u}}.\,\!$ Based on these identities, we have

{\begin{aligned}{\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{aligned}}\,\! 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!

To see how this generalizes to plane waves, see Electromagnetic Plane Waves.

### 1.2 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 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 $x=0$ .

The curl of a sine wave (in 2 or more dimensions) is maximized around $x=0$ .

Since we know that

$\nabla \times E=-{\frac {1}{c}}{\frac {\partial B}{\partial t}},\,\!$ 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 $E_{0}=B_{0}$ .