Electromagnetic waves

We will consider electric and magnetic fields in a vacuum, that is, where the charge density $\rho=0$ . From Maxwell’s equations we have

\begin{align*} \nabla \cdot \mathbf E &= 0 & \nabla \cdot \mathbf B &= 0 \\ \nabla \times \mathbf E &= -\frac{\partial \mathbf B}{\partial t} &\quad \nabla \times \mathbf B &= \mu_0 \epsilon_0 \frac{\partial \mathbf E}{\partial t}. \end{align*}

We now use a vector identity $\nabla \times (\nabla \times \mathbf A) = \nabla (\nabla \cdot \mathbf A) - (\nabla \cdot \nabla) \mathbf A$ . Plugging in $\mathbf E$ , we get

\begin{align*} \nabla (\nabla \times \mathbf E) &= \nabla(\nabla \cdot \mathbf E) - \nabla^2 \mathbf E \\ \nabla \times \left(-\frac{\partial \mathbf B}{\partial t} \right) &= -\nabla^2 \mathbf E. \end{align*}

We can pull the time derivative out of the left side of the equation since the curl is only spatial to get

\begin{align*} -\frac{\partial}{\partial t} (\nabla \times \mathbf B) &= -\nabla^2 \mathbf E \\ -\frac{\partial}{\partial t} \left(\mu_0 \epsilon_0 \frac{\partial \mathbf E}{\partial t} \right) &= -\nabla^2 \mathbf E \\ \mu_0 \epsilon_0 \frac{\partial^2 \mathbf E}{\partial t^2} &= \nabla^2 \mathbf E. \end{align*}

This is the wave equation in 3D!

Let’s do the same thing for $\mathbf B$ .

\begin{align*} \nabla (\nabla \times \mathbf B) &= \nabla(\nabla \cdot \mathbf B) - \nabla^2 \mathbf B \\ \nabla \times \left(\mu_0\epsilon_0\frac{\partial \mathbf E}{\partial t} \right) &= -\nabla^2 \mathbf B \\ \mu_0\epsilon_0 (\nabla \times \mathbf E) &= \nabla^2 \mathbf B \\ \mu_0\epsilon_0 \frac{\partial^2 \mathbf B}{\partial t^2} &= \nabla^2 \mathbf B. \end{align*}

So magnetic fields are also waves.

Maxwell’s equations tell us that time varying electric fields create magnetic fields, and vice versa. Electromagnetic waves involve electric and magnetic fields changing in time. As we will see, electric and magnetic waves create each other and can propagate forever in a vacuum. Generally, the magnitude of the corresponding fields is related by $B = \frac{E}{c}$ .

Traveling electromagnetic waves

Let’s write a traveling wave solution $\mathbf E(\mathbf r, t) = \mathrm{Re}(\mathbf E_0 e^{i(\mathbf k \cdot \mathbf r - \omega t + \phi)})$ where $\mathbf E$ is the direction of the electric field, $\mathbf k$ is the direction of wave propagation, $\mathbf r$ is where in space we are evaluating the function, and $\omega = c |\mathbf k|$ .

From Maxwell’s equations we say

TODO: This looks wrong. Go through the derivation properly.

\begin{align*} \nabla \times \mathbf E &= -\frac{\partial B}{\partial t} \\ \mathbf B &= \nabla \times \int \mathbf E\, dt \\ \mathbf B &= \nabla \times \frac{\mathbf E}{c}. \end{align*}

Then since the wave is propagating in the $\mathrm{dir}(\mathbf k) = \mathrm{dir}(\mathbf E \times \mathbf B)$ direction, we know $\mathrm{dir}(\mathbf B) = \mathrm{dir}(\mathbf k \times \mathbf E)$ , we see $\mathbf B = \frac 1c \hat{\mathbf k} \times \mathbf E$ .

Importantly, this relation only holds when $\mathbf E$ is a traveling wave of the form showed above. It is not the case generally.

Standing electromagnetic waves

We now consider the behavior of an electromagnetic wave near a conductor. Assume there is a perfect conductor in the $x,y$ plane. We know that the electric field within a conductor is zero $\mathbf E(z=0)=0$ . The single traveling wave solution we have considered cannot represent this.

Let’s assume that when the wave hits the conductor, it reflects back. The electric fields at the conductor must sum to zero to satisfy our boundary condition.

Assume our incident electric and magnetic waves are

\begin{align*} \mathbf E_i &= \frac{E_0}{2} \cos(\mathcal Kz - \omega t) \hat{\mathbf x} \\ \mathbf B_i &= \frac{E_0}{2c} \cos(\mathcal Kz - \omega t) \hat{\mathbf y}. \end{align*}

Then to satisfy $\mathbf E(z=0)=0$ , our reflected waves must be

\begin{align*} \mathbf E_r &= -\frac{E_0}{2} \cos(-\mathcal Kz - \omega t) \hat{\mathbf x} \\ \mathbf B_r &= \frac{E_0}{2c} \cos(-\mathcal Kz - \omega t) \hat{\mathbf y}. \end{align*}

The reflected $\mathbf E_r$ fields amplitude direction is opposite the incident $\mathbf E_i$ so that it cancels at $z=0$ , and its propagation direction is $-\hat{\mathbf z}$ , opposite the incident wave. The $\mathbf B_r$ field has the same amplitude direction as $\mathbf B_i$ but opposite propagation direction. This is necessary so that the waves satisfy Maxwell’s equations, i.e., so that $\mathrm{dir}(\mathbf E_r \times \mathbf B_r) = -\hat{\mathbf z}$ is the correct propagation direction.

The superposition of the incident and reflected waves gives us the total waves

\begin{align*} \mathbf E = \mathbf E_i + \mathbf E_r &= \frac{E_0}{2} (\cos(\mathcal Kz -\omega t) + \cos(-\mathcal K z-\omega t)) \hat{\mathbf x} \\ &= E_0 \sin(\omega t)\sin(\mathcal K z),\\ \mathbf B = \mathbf B_i + \mathbf B_r &= \frac{E_0}{2c} (\cos(\mathcal Kz -\omega t) + \cos(-\mathcal K z-\omega t)) \hat{\mathbf y} \\ &= \frac{E_0}{c} \cos(\omega t)\cos(\mathcal K z). \end{align*}

The superposition gives us two standing waves. Indeed, we could have started with the general standing wave solution and plugged in our boundary conditions to reach the same result.

Note that the $\mathbf E$ and $\mathbf B$ fields are $90\degree$ out of phase, so the relation we used earlier when considering the traveling waves does not apply.

Energy carried by an electromagnetic wave

Electric and magnetic fields have energy, but it is often more useful to think about the energy flowing in a wave. The Poynting vector describes the direction and rate of energy transfered by a wave in units of power per area, $\mathrm W \cdot \mathrm m^{-2}$ .