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Chapter 10 - Electromagnetic Waves
Section 10.1 - Time-Dependent Electromagnetic Fields in a Vacuum Satisfy the Wave Equation
Consider an empty space. Then, it is evident that
$$\begin{align} \div \vb{E} &= 0 \ \div \vb{H} &= 0 \end{align}$$
$$\begin{align} \curl \vb{E} + \frac{\partial \vb{B}}{\partial t} &= 0 \ \curl \vb{H} - \frac{\partial \vb{D}}{\partial t} &= 0 \end{align}$$
As \vb{B} = \mu_0 \vb{H} and \vb{D} = \varepsilon_0 \vb{E} in a vacuum, the third and fourth equations can be rewritten as
$$\begin{align} \curl \vb{E} + \mu_0 \frac{\partial \vb{H}}{\partial t} &= 0 \ \curl \vb{H} - \varepsilon_0 \frac{\partial \vb{E}}{\partial t} &= 0 \end{align}$$
We can take the curl of both equations and then substitute to see that
$$\begin{align} \curl \curl \vb{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \vb{E}}{\partial t^2} &= 0 \ \curl \curl \vb{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \vb{H}}{\partial t^2} &= 0 \end{align}$$
We can apply a vector identity to see
$$\begin{align} -\nabla^2 \vb{E} + \mu_0 \epsilon_0 \frac{\partial^2 \vb{E}}{\partial t^2} &= 0 \ -\nabla^2 \vb{H} + \mu_0 \epsilon_0 \frac{\partial^2 \vb{H}}{\partial t^2} &= 0 \end{align}$$
Section 10.1.1 - The Wave Equation and Plane Waves
Definition. The equation [\frac{\partial^2}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2}{\partial t^2}] f(x, t) = 0 is well-known to mathematicians (see Differential Equations), and is known as the wave equation. In physics, the speed of the wave is v = c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}.
Consider some function f(s). If s = x - vt or x + vt, it is trivial to see that f(x) satisfies the wave equation.
Definition. A plane wave is a solution to the Laplacian form of the last two Maxwell equations for empty space that also satisfy the one-dimensional wave equation. However, these solutions may not be valid electromagnetic waves as they are not guaranteed to satisfy the first two Maxwell equations.
Notably, the functions for \vb{E} = \vb{E}_0 f(s) and \vb{H} = \vb{H}_0 g(s) do not have to be equal. However, v = c.
Definition. A plane electromagnetic wave is a plane wave which satisfies the first two Maxwell equations. The divergence equations restrict \vb{E}_0 and \vb{H}_0 to be in the plane normal to the direction of motion. That is, electomagnetic plane waves are transverse, not longitudinal.
Additionally, the curl equations force f(s) = g(s), such that H_0 = E_0 \sqrt{\frac{\varepsilon_0}{\mu_0}}.
Definition. The quantity Y_0 = \sqrt{\frac{\varepsilon_0}{\mu_0}} is the vacuum admittance and its inverse, Z_0 = \sqrt{\frac{\mu_0}{\varepsilon_0}} is the vacuum impedance.
If we assume the direction of propagation can be written as \vu{k}, we can write f(s) = f(\vu{k} \vdot \vb{r} - vt), such that \vb{E}(\vb{r}, t) = \vb{E}_0 f(\vu{k}\vdot\vb{r} - vt), where \vu{k}\vdot\vb{E}_0 = 0.
From this, we can see that \vb{H}(\vb{r}, t) = \sqrt{\frac{\epsilon_0}{\mu_0}} \vu{k} \times \vb{E}_0 f(\vu{k} \vdot \vb{r} - vt). Similarly, \vu{k} \vdot \vb{H} = 0.
Additionally, we can compute \vb{S} = \vb{E} \times \vb{H} = c \epsilon_0 E_0^2 f^2(\vu{k} \vdot \vb{r} - vt) \vu{k}. We can also see that \epsilon_0 E^2 = \mu_0 H^2 at any given time.
Section 10.1.2 - Monochromatic Plane Waves
In any simple material, we like to say that \vb{D} = \epsilon \vb{E} and \vb{B} = \mu \vb{H}. However, this only holds true at a fixed frequency \omega. For multiple frequencies, we see that \vb{D}(\omega) = \epsilon{\omega}\vb{E}(\omega) and \vb{B}(\omega) = \mu(\omega)\vb{H}(\omega). This causes problems. As such, we will want to consider waves that are only composed of one frequency under Fourier decomposition.
Definition. A monochromatic plane wave is a plane wave in which the full Fourier series of f(x) has only one term. That is, f(x) is \sin(x) or \cos(x). We furthermore define a wave vector \vb{k} as \vb{k} = k \vu{k}, so that \omega = kc. Then,
$$\begin{align} \vb{E}(\vb{r}, t) &= \vb{E_0} \cos(\vb{k} \vdot \vb{r} - \omega t) \ \vb{H}(\vb{r}, t) &= \sqrt{\frac{\epsilon_0}{\mu_0}} \vu{k} \times \vb{E}_0 \cos(\vb{k} \vdot \vb{r} - \omega t) \end{align}$$
Notably, the frequency, or number of cycles per second, is f = \frac{\omega}{2\pi}, and wavelength \lambda = \frac{2\pi}{k}.
We can calculate the energy density u, energy current density \vb{S}, momentum density \vb{g}, and momentum current density -\overleftrightarrow{\vb{T}}
Section 10.1.3 - Monochromatic Plane Waves in a Linear Model
Monochromatic plane waves with frequency \omega in a simple linear material are similar to monochromatic plane waves in a vacuum, except when in a material, we know that the magnitude of the wave vector k = \frac{\omega}{v}, and v = \frac{1}{\sqrt{\mu \varepsilon}}.
Section 10.1.4 - Polarization of Monochromatic Plane Waves
Any plane wave described in such a way that \vb{E} = \vb{E}_0 f(\vb{k} \vdot \vb{r} - ct) is linearly polarized in the direction of \vb{E}_0. That is, the direction of polarization is the direction of \vb{E}, and if that direction is unchanging, the wave is linearly polarized.
Notably, an elliptically polarized wave can be described as follows:
$$\begin{align} \vb{E}(\vb{r}, t) &= E_{x0} \vu{x} \cos(kz - \omega t) + E_{y0} \vu{y} \sin(kz - \omega t) \ \vb{H}(\vb{r}, t) &= \sqrt{\frac{\varepsilon}{\mu}} (E_{x0} \vu{x} \cos(kz - \omega t) + E_{y0} \vu{y} \sin(kz - \omega t)) \end{align}$$