# 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 ${\displaystyle \epsilon _{0}}$ (it’s incorporated into the definition of charge) and ${\displaystyle \mu _{0}}$ becomes ${\displaystyle 1/c^{2}}$. In these units, Maxwell’s equations are written:

{\displaystyle {\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 ${\displaystyle E}$ is the electric field, ${\displaystyle B}$ is the magnetic field, ${\displaystyle J}$ is the current (charge per time), and ${\displaystyle \rho }$ 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 ${\displaystyle \rho =0}$ (no charge) and ${\displaystyle J=0}$ (no current), Maxwell’s equations say

{\displaystyle {\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

${\displaystyle \nabla \times (\nabla \times F)=\nabla (\nabla \cdot F)-\nabla ^{2}F,\,\!}$

the curl (${\displaystyle \nabla \times }$) of the last 2 Maxwell’s equations reduces to the following wave equations (using that the divergence of each field is 0):

{\displaystyle {\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, ${\displaystyle {\hat {x}}}$), the wave equation for the electric field (derived above as a solution of Maxwell’s Equations in free space) is written

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

Let’s suppose that ${\displaystyle E(x)=E(x\pm ct)}$, and we define a new variable ${\displaystyle u\equiv x\pm ct}$. The Chain Rule tells us that a spatial derivative of ${\displaystyle E}$ gives us

${\displaystyle {\frac {\partial E(u)}{\partial x}}={\frac {\partial E}{\partial u}}{\frac {\partial u}{\partial x}}={\frac {\partial E}{\partial u}}.\,\!}$

Similarly, for the time derivative, we have

${\displaystyle {\frac {\partial E(u)}{\partial t}}={\frac {\partial E}{\partial u}}{\frac {\partial u}{\partial t}}=\pm {\frac {1}{c}}{\frac {\partial E}{\partial u}}.\,\!}$

Based on these identities, we have

{\displaystyle {\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 ${\displaystyle E}$ and ${\displaystyle B}$, this means that any electromagnetic waveform that translates at speed ${\displaystyle c}$ 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 ${\displaystyle E}$ and ${\displaystyle B}$

From the previous section, we see that in free space, the Maxwell equations give us 2 wave equations. One propagates the ${\displaystyle E}$-field at the speed of light, and the other propagates the ${\displaystyle B}$-field at the speed of light, but we haven’t related the ${\displaystyle E}$ and ${\displaystyle B}$ 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 ${\displaystyle E}$ and ${\displaystyle B}$. Let’s suppose that ${\displaystyle E}$ and ${\displaystyle B}$ are both described by sine waves of the same frequency, but with some arbitrary phase relative to one another. Let’s say that ${\displaystyle {\vec {E}}=E_{0}\sin({\vec {x}}+ct)}$, so that ${\displaystyle {\vec {E}}=0}$ at ${\displaystyle x=0,t=0}$. Since a sine wave is anti-symmetric around 0, one can draw a loop encircling ${\displaystyle x=0}$, and since ${\displaystyle {\vec {E}}}$ is positive on one side of that loop, and negative on the other side, ${\displaystyle \nabla \times E}$ is not zero, and in fact, is maximal at ${\displaystyle x=0}$.

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

Since we know that

${\displaystyle \nabla \times E=-{\frac {1}{c}}{\frac {\partial B}{\partial t}},\,\!}$

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