15.9: Electromagnetic Waves

Maxwell predicted the existence of electromagnetic waves, and these were generated experimentally by Hertz shortly afterwards. In addition, the predicted speed of the waves was $3 \times 10^{8}\, \text{m s}^{-1}$, the same as the measured speed of light, showing that light is an electromagnetic wave.

In an isotropic, homogeneous, nonconducting, uncharged medium, where the permittivity and permeability are scalar quantities, Maxwell's equations can be written

$$\boldsymbol{\nabla} \cdot \textbf{E} = \rho. \tag{15.9.1} \label{15.9.1}$$

$$\boldsymbol{\nabla} \cdot \textbf{H} = 0. \tag{15.9.2} \label{15.9.2}$$

$$\boldsymbol{\nabla} \times \textbf{H} = \epsilon \dot {\textbf{E}}. \tag{15.9.3} \label{15.9.3}$$

$$\boldsymbol{\nabla} \times \textbf{E} = - \mu \dot{ \textbf{H}}. \tag{15.9.4} \label{15.9.4}$$

These equations involve $\textbf{E}$, $\textbf{H}$, and $t$. Let us see if we can eliminate $\textbf{E}$ and hence find an equation in just $\textbf{H}$ and $t$.

Take the $\textbf{curl}$ of equation $\ref{15.9.3}$, and make use of equation 15.6.4:

$$\textbf{grad}\, \text{div} \textbf{H} - \nabla^2\textbf{H} = \epsilon \dfrac{\partial}{\partial t} \textbf{curl}\, \textbf{E} \tag{15.9.5} \label{15.9.5}$$

Substitute for $\text{div} \ \textbf{H}$ and $\textbf{curl}\, \textbf{E}$ from equations $\ref{15.9.2}$ and $\ref{15.9.4}$ to obtain

$$\nabla^2 \textbf{H} = \epsilon \mu \ddot{\textbf{H}}\tag{15.9.6} \label{15.9.6}$$

This is the equation in terms of just $\textbf{H}$ and $t$ that we sought.

Comparison with equation 15.1.2 shows that this is a wave of speed $1/ \sqrt{\epsilon \mu}$ (Verify that this has the dimensions of speed.)

In a similar manner the reader should easily be able to eliminate $\textbf{B}$ to derive the equation

$$\nabla^2 \textbf{E} = \epsilon \mu \ddot{\textbf{E}} \tag{15.9.7} \label{15.9.7}$$

In a vacuum, the speed is $1/ \sqrt{\epsilon_o \mu_o}$. With $\mu_o = 4\pi \times 10^{-7}\, \text{H m}^{-1}$ and $\epsilon_o = 8.854 \times 10^{-12} \text{F m}^{-1}$, this comes to $2.998 \times 10^8 \, \text{m s}^{-1}$.

Can we eliminate $t$ from the equations, and hence obtain a relation between just $\textbf{E}$ and $\textbf{H}$? If you do, you will obtain

$$\dfrac{E}{H} = \sqrt{\dfrac{\mu}{\epsilon}}, \tag{15.9.8} \label{15.9.8}$$

which, in a vacuum, or free space, becomes

$$\dfrac{E}{H} = \sqrt{\dfrac{\mu_o}{\epsilon_o}} = 377\, \Omega, \tag{15.9.9} \label{15.9.9}$$

which is the impedance of a vacuum, or of free space. Since the SI units of $\textbf{E}$ and $\textbf{H}$ are, respectively V m^-1 and A m^-1, it is easy to verify that the units of impedance are V A^-1, or $\Omega$.