7.7 KiB
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
$$\begin{align} \nabla \cdot \mathbf{E} &= 0 \ \nabla \cdot \mathbf{H} &= 0 \end{align}$$
$$\begin{align} \nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} &= 0 \ \nabla \times \mathbf{H} - \frac{\partial \mathbf{D}}{\partial t} &= 0 \end{align}$$
As \mathbf{B} = \mu_0 \mathbf{H} and \mathbf{D} = \varepsilon_0 \mathbf{E} in a vacuum, the third and fourth equations can be rewritten as
$$\begin{align} \nabla \times \mathbf{E} + \mu_0 \frac{\partial \mathbf{H}}{\partial t} &= 0 \ \nabla \times \mathbf{H} - \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} &= 0 \end{align}$$
We can take the curl of both equations and then substitute to see that
$$\begin{align} \nabla \times \nabla \times \mathbf{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0 \ \nabla \times \nabla \times \mathbf{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} &= 0 \end{align}$$
We can apply a vector identity to see
$$\begin{align} -\nabla^2 \mathbf{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0 \ -\nabla^2 \mathbf{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} &= 0 \end{align}$$
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 \mathbf{E} = \mathbf{E}_0 f(s) and \mathbf{H} = \mathbf{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 \mathbf{E}_0 and \mathbf{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 \hat{\mathbf{k}}, we can write f(s) = f(\hat{\mathbf{k}} \cdot \mathbf{r} - vt), such that \mathbf{E}(\mathbf{r}, t) = \mathbf{E}_0 f(\hat{\mathbf{k}}\cdot\mathbf{r} - vt), where \hat{\mathbf{k}}\cdot\mathbf{E}_0 = 0.
From this, we can see that \mathbf{H}(\mathbf{r}, t) = \sqrt{\frac{\varepsilon_0}{\mu_0}} \hat{\mathbf{k}} \times \mathbf{E}_0 f(\hat{\mathbf{k}} \cdot \mathbf{r} - vt). Similarly, \hat{\mathbf{k}} \cdot \mathbf{H} = 0.
Additionally, we can compute \mathbf{S} = \mathbf{E} \times \mathbf{H} = c \varepsilon_0 E_0^2 f^2(\hat{\mathbf{k}} \cdot \mathbf{r} - vt) \hat{\mathbf{k}}. We can also see that \varepsilon_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 \mathbf{D} = \varepsilon \mathbf{E} and \mathbf{B} = \mu \mathbf{H}. However, this only holds true at a fixed frequency \omega. For multiple frequencies, we see that \mathbf{D}(\omega) = \varepsilon(\omega)\mathbf{E}(\omega) and \mathbf{B}(\omega) = \mu(\omega)\mathbf{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 \mathbf{k} as \mathbf{k} = k \hat{\mathbf{k}}, so that \omega = kc. Then,
$$\begin{align} \mathbf{E}(\mathbf{r}, t) &= \mathbf{E_0} \cos(\mathbf{k} \cdot \mathbf{r} - \omega t) \ \mathbf{H}(\mathbf{r}, t) &= \sqrt{\frac{\varepsilon_0}{\mu_0}} \hat{\mathbf{k}} \times \mathbf{E}_0 \cos(\mathbf{k} \cdot \mathbf{r} - \omega t) \end{align}$$
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 \mathbf{S}, momentum density \mathbf{g}, and momentum current density -\overleftrightarrow{\mathbf{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 \mathbf{E} = \mathbf{E}_0 f(\mathbf{k} \cdot \mathbf{r} - ct) is linearly polarized in the direction of \mathbf{E}_0. That is, the direction of polarization is the direction of \mathbf{E}, and if that direction is unchanging, the wave is linearly polarized.
Notably, an elliptically polarized wave can be described as follows:
$$\begin{align} \mathbf{E}(\mathbf{r}, t) &= E_{x0} \hat{\mathbf{x}} \cos(kz - \omega t) + E_{y0} \hat{\mathbf{y}} \sin(kz - \omega t) \ \mathbf{H}(\mathbf{r}, t) &= \sqrt{\frac{\varepsilon}{\mu}} (E_{x0} \hat{\mathbf{y}} \cos(kz - \omega t) - E_{y0} \hat{\mathbf{x}} \sin(kz - \omega t)) \end{align}$$
If E_{x0} = E_{y0}, the wave is said to be circularly polarized.
Section 10.2 - Reflection and Refraction of Plane Electromagnetic Waves at a a Planar Interface
This section will focus on plane monochromatic waves incident from material 1 onto material 2, where both materials are homogenous insulators and the surface between the two materials is smooth (on the scale of the wavelength).
In this case, we must re-consider Maxwell's equations. We know from previous sections that \nabla \cdot \mathbf{E} = \frac{\mathbf{\rho_e}}{\varepsilon_0} and \nabla \cdot \mathbf{H} = \frac{\mathbf{\rho_m}}{\mu_0}. We also know that \nabla \cdot \mathbf{D} = \rho_{ef} and \nabla \cdot \mathbf{B} = \rho_{mf}.
Consider the boundary between the two materials. If we consider \nabla \cdot \mathbf{D}, and take the integral over a Gaussian pillbox on the boundary, we can apply divergence theorem to see that \int_V \nabla \cdot \mathbf{D} dV = \int_{SofV} \mathbf{D} \cdot \hat{\mathbf{n}} dS = \rho_{efree}. If we assume the materials are insulating, we do not expect to find any electrical charge, so \rho_{efree} = 0. Thus, we can say that \int_{SofV} D \cdot \hat{\mathbf{n}} = 0, so \mathbf{D}_1 \cdot \hat{\mathbf{n}} + \mathbf{D}_2 \cdot \hat{\mathbf{n}} = \mathbf{D_1} \cdot \hat{\mathbf{z}} + \mathbf{D}_1 \cdot (-\hat{\mathbf{n}}) = 0. Then, we can say that \mathbf{D}_1 \cdot \hat{\mathbf{n}} = \mathbf{D}_2 \cdot \hat{\mathbf{n}}, or in simpler terms, \mathbf{D}_1^\perp = \mathbf{D}_2^\perp.
Applying the same logic to \mathbf{B}, we see that \mathbf{B}_1^\perp = \mathbf{B}_2^\perp. Note that due to the existence of polarization and magnetization, we cannot say the same regarding \mathbf{E} or \mathbf{H}.