28 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, as \hat{\mathbf{K}} \cdot \mathbf{E}_0 = 0. 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}.
Section 10.2.1 - Boundary Conditions at an Interface Between Two Materials
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}.
Now, consider a rectangular loop along the interface. If we then consider \int_S (\nabla \times \mathbf{E}) \cdot d\mathbf{S} = \int_{\partial S} \mathbf{E} \cdot d\mathbf{l} = -\frac{\partial \Phi_B}{\partial t}. If we let the width of the rectangle approach 0, then \int_{\partial S} \mathbf{E} \cdot d\mathbf{l} = \mathbf{E}_1 \cdot d\mathbf{l} + \mathbf{E}_2 \cdot d\mathbf{l} = 0, which implies that \mathbf{E}_1 \cdot d\mathbf{l} = \mathbf{E}_2 \cdot d\mathbf{l}, or in other words, \mathbf{E}_1^\parallel = \mathbf{E}_2^\parallel.
Applying the same logic to the other Maxwell equation, we see that \mathbf{H}_1^\parallel = \mathbf{H}_2^\parallel.
Section 10.2.2 - Normal Incidence
Consider a monochromatic plane wave that is incident normally from material 1 onto material 2. That is, the wave vector \mathbf{k} is normal to the interface. In this case, we will let the interface be at z=0 on the x-y plane, and \hat{\mathbf{k}} = \hat{\mathbf{z}}.
Here, we can write the incident wave as \mathbf{E}_i (z, t) = E_i \hat{\mathbf{x}} \cos{k_1 z - \omega t}. Then, Maxwell's equations give us \mathbf{H}_i(z, t) = Y_1 E_i \hat{\mathbf{y}} \cos(k_1 z - \omega 2). By symmetry, we expect the reflected wave to be in the -z direction, with some phase \phi_r such that the wave is a function of \cos(-k_1 z - \omega t + \phi_r). Additionally, we expect the transmitted wave to be of the form \cos(k_2 z - \omega t + \phi_t). This gives for material 1
$$\begin{align} \mathbf{E}_1(z, t) &= \mathbf{E}_i(z, t) + \mathbf{E}_r(z, t) &= E_i \hat{\mathbf{x}} \cos(k_1 z - \omega t) + \mathbf{E}_r \cos(k_1 z - \omega t - \phi_r) \ \mathbf{H}_1(z, t) &= \mathbf{H}_i(z, t) + \mathbf{H}_r(z, t) &= Y_1 E_i \hat{\mathbf{y}} \cos(k_1 z - \omega t) + \mathbf{H}_r \cos(k_1 z - \omega t - \phi_r) \ \end{align}$$
In material 2, we see that
$$\begin{align} \mathbf{E}_2(z, t) &= \mathbf{E}_t \cos(k_2 z - \omega t - \phi_t) \ \mathbf{H}_2(z, t) &= \mathbf{H}_t \cos(k_2 z - \omega t - \phi_t) \ \end{align}$$
Now, break \mathbf{E}_r into components such that \mathbf{E}_r = E_{rx} \hat{\mathbf{x}} + E_{ry} \hat{\mathbf{y}}. Then, we see that
$$\begin{align} \mathbf{E}r(z, t) &= (E{rx} \hat{\mathbf{x}} + E_{ry} \hat{\mathbf{y}}) \cos(k_1 z - \omega t - \phi_r) \ \mathbf{H}r(z, t) &= -Y \hat{\mathbf{z}} \times (E{rx} \hat{\mathbf{x}} + E_{ry} \hat{\mathbf{y}}) \cos(k_1 z - \omega t - \phi_r) \ \end{align}$$
Evaluation at z = 0, we see that
$$\begin{align} \mathbf{E}1 &= \hat{\mathbf{x}} E_i \cos(-\omega t) + (E{rx} \hat{\mathbf{x}} + E_{ry} \hat{\mathbf{y}}) \cos(-\omega t - \phi_r) \ \mathbf{H}1 &= \hat{\mathbf{y}} Y_1 E_i \cos(-\omega t) + Y_1(-E{rx} \hat{\mathbf{y}} + E_{ry} \hat{\mathbf{x}}) \cos(-\omega t - \phi_r) \ \end{align}$$
In material 2, we see that
$$\begin{align} \mathbf{E}2(z, t) &= (E{tx} \hat{\mathbf{x}} + E_{ty} \hat{\mathbf{y}}) \cos(-\omega t - \phi_t) \ \mathbf{H}2(z, t) &= Y_2(E{tx} \hat{\mathbf{y}} - E_{ty} \hat{\mathbf{x}}) \cos(-\omega t - \phi_t) \ \end{align}$$
In this case, applying boundary conditions is as simple as matching components to obtain two (out of four) equations:
$$\begin{align} \mathbf{E}y &: E{ry} \cos(-\omega t - \phi_r) &= E_{ty} \cos(-\omega t - \phi_t) \ \mathbf{H}x &: Y_1 E{ry} \cos(-\omega t - \phi_r) &= -Y_2 E_{ty} \cos(-\omega t - \phi_t) \end{align}$$
However, this would imply that Y_1 = -Y_2, which is not possible for ordinary materials. As such, we set E_{ry} = 0, which then implies E_{ty} = 0 (or vice-versa). Then, with E_r = E_{rx} and E_t = E_{tx}, we see that the other two component-wise equations yield
$$\begin{align} E_i \cos(\omega t) + E_r \cos(\omega t - \phi_r) &= E_t \cos(\omega t - \phi_t) \ Y_1 E_i \cos(\omega t) - Y_1 E_r \cos(\omega t - \phi_r) &= Y_2 E_t \cos(\omega t - \phi_t) \ \end{align}$$
From this, we can apply the identity \cos(\omega t - \phi) = \cos(\omega t) \cos \phi + \sin(\omega t) \sin \phi to split each equation into two equations that must hold for any t. Then, comparing the E \sin \omega t and H \sin \omega t equations lead us to the conclusion that \sin \phi_r = \sin \phi_t = 0. Thus, we see that E_i + E_r = E_t and Y_1 (E_i - E_r) = Y_2 E_t. We can solve this system to see
$$\begin{align} E_t &= \frac{2Y_1}{Y_1 + Y_2} E_i \ E_r = \frac{Y_1 - Y_2}{Y_1 + Y_2} E_i \end{align}$$
If we assume \mu_1 = \mu_2, we can rewrite the equations in terms of wave numbers k_i, where k_i = \frac{\omega}{v_i} = \omega \sqrt{\mu_i \varepsilon_i}
$$\begin{align} E_t &= \frac{2k_1}{k_1 + k_2} E_i \ E_r = \frac{k_1 - k_2}{k_1 + k_2} E_i \end{align}$$
Definition. We can also define the index of refraction for a material n as n_i = \sqrt{\frac{\varepsilon_i \mu_i}{\varepsilon_0 \mu_0}}.
When \mu_1 = \mu_2 = \mu_0, we can write as n_i = \sqrt{\frac{\varepsilon_i}{\varepsilon_0}}. Then, we can write the reflected and transmitted amplitudes as
$$\begin{align} E_t &= \frac{2n_1}{n_1 + n_2} E_i \ E_r &= \frac{n_1 - n_2}{n_1 + n_2} E_i \end{align}$$
We can also define the average power incident on the interface as I_(in) = \langle \mathbf{S}_i \cdot \hat{\mathbf{n}} \rangle = \langle (\mathbf{E}_i \times \mathbf{H}_i) \cdot \hat{\mathbf{n}} \rangle. We know that \mathbf{E}_i \times \mathbf{H}_i is in the direction of \hat{\mathbf{n}}, so I_{in} = \langle (\mathbf{E}_i \times (Y_1 \hat{\mathbf{k}} \times \mathbf{E}_i) \cdot \hat{\mathbf{z}}) \rangle = \frac{1}{2} Y_1 E_i^2. We can further define I_r = -\frac{1}{2} Y_1 E_t^2 and I_t = \frac{1}{2}Y_2 E_t^2.
We can then compute R = \frac{|I_r|}{I_{in}} = \frac{E_r^2}{E_i^2} = (\frac{Y_1 - Y_2}{Y_1 + Y_2})^2 to see the fraction of power reflected and I_t = \frac{I_t}{I_{in}} = \frac{Y_2 E_t^2}{Y_1 E_i^2} = \frac{4 Y_1 Y_2}{(Y_1 + Y_2)^2}. Note that mathematically, R + T = 1. That is, all the power of the incident wave is either reflected or transmitted. Additionally, when \mu_1 \approx \mu_2, we can replace Y with n.
Section 10.2.3 - Oblique Incidence
Now, we assume that the wave is incident to the interface at some angle \theta_i.
Definition. The interfacial plane is the interface. The plane of incidence is the plane defined by the incident wave vector \mathbf{k}_i and the normal vector of the interfacial plane \mathbf{z}
Theorem. Snell's Law. In the plane if incidence, the continuity of the electromagnetic field implies that for all times on the z=0 plane, the argument of the \cos(\mathbf{k}_{i,r,t} \cdot \mathbf{r} - \omega t) must be equal. That is, in our example, for \mathbf{r} = x \hat{\mathbf{x}} + y \hat{\mathbf{y}}, we see that \mathbf{k}_i \cdot \mathbf{r} = \mathbf{k}_t \cdot \mathbf{r} + \phi_r = \mathbf{k}_t \cdot \mathbf{r} + \phi_t. Then, \phi_r and \phi_t must vanish, and the wave vectors must satisfy k_{ix}x + k_{iy}y = k_{rx}x + k_{ry}y = k_{tx}x + k_{ty}y. This implies that k_{ix} = k_{rx} = k_{tx} and k_{iy} = k_{ry} + k_{ty}.
More generally, the requirement of continuity on the interface implies that the three wave vectors are coplanar and that components parallel to the interface must be equal, leading to k_i \sin \theta_i = k_r \sin \theta_r = k_t \sin \theta_t. Note that in this case, k_i = k_r = \frac{\omega}{v} as they describe propagation in the same media, so \theta_r = \theta_i. As n \propto \frac{1}{v}, we see that n \propto k, so we can rewrite Snell's Law as n_i \sin \theta_i = n_t \sin \theta_t. This gives us the following wave vectors:
$$\begin{align} \mathbf{k}_i &= k_i(\sin \theta_i \hat{\mathbf{x}} + \cos \theta_i \hat{\mathbf{z}}) \ \mathbf{k}_r &= k_i(\sin \theta_i \hat{\mathbf{x}} - \cos \theta_i \hat{\mathbf{z}}) \ \mathbf{k}_t &=(k_t \sin \theta_t \hat{\mathbf{x}} + k_t \cos \theta_t \hat{\mathbf{z}}) = (k_i \sin \theta_i \hat{\mathbf{x}} + k_t \cos \theta_t \hat{\mathbf{z}}) \end{align}$$
Additionally, if we write Snell's Law as \sin \theta_t = \frac{n_i}{n_t} \sin\theta_i, we c an see that it predicts the angle of refraction. Note that if \frac{n_i}{n_t} \sin\theta_i \geq 1, we see total internal reflection and there will be no transmitted waves.
Reflected and transmitted Fields for Oblique Incidence
Any incident oblique wave can be written as a superposition of two waves, one with transverse magnetic and electric fields respectively. A transverse electric (TE) wave has its electric field perpendicular to the plane of incidence. Respectively, a transverse magnetic (TM) wave has its magnetic field perpendicular to the plane of incidence.
Thus, if \hat{\mathbf{k}}_i = \sin \theta_i \hat{\mathbf{x}} + \cos \theta_i \hat{\mathbf{z}}, we can say that an incident wave with arbitrary polarization can be written as \mathbf{E}(\mathbf{r}, t) = (E_{TE}(-\hat{\mathbf{y}}) + E_{TM}\hat{\mathbf{k}} \times (-\hat{\mathbf{y}})) \cos(\mathbf{k}_i \cdot \mathbf{r} - \omega t).
In the case of a transverse electric field, \mathbf{E} is perpendicular to the plane of incidence. We know that \mathbf{E}_r \perp \mathbf{k}_r and \mathbf{E}_t \perp \mathbf{k}_r. Knowing our definition of \mathbf{r}, we can say that E_{rz} = \tan \theta_i E_{rx} and E_{tz} = -\tan \theta_t E_{tx}. Continuity of \mathbf{D}^\perp requires that \varepsilon_1 E_{rz} = \varepsilon_2 E_{tz}, and continuity of \mathbf{E}^\parallel requires that E_{rx} = E_{tx}. We can then see that \varepsilon_1 \tan \theta_i = -\varepsilon_2 \tan \theta_t E_{tx}, which we can substitute E_{rx} = E_{tx} to see that \varepsilon_1 \tan \theta_i = \varepsilon_2 \tan \theta_t. However, this gives us an equation for \tan \theta_t which contradicts with an equation that is not derived in the textbook. This implies that the x and z components of the electric field vanish.
All that is left is the $y$-direction, for which continuity requires that E_i + E_r = E_t. The continuity of the $x$-component of the magnetic fields thus requires that [Y_1 (\hat{\mathbf{k}}_i \times E_i \hat{\mathbf{y}} + \hat{\mathbf{k}}_r \times E_r \hat{\mathbf{y}}) - Y_2 (\hat{\mathbf{k}}_t \times E_t \hat{\mathbf{y}})] \cdot \hat{\mathbf{x}} = 0. We can use the scalar triple product identity to simplify this to [Y_1 (\hat{\mathbf{k}}_i E_i + \hat{\mathbf{k}}_r E_r) - Y_2(\hat{\mathbf{k}}_t E_t)] \cdot \hat{\mathbf{z}} = 0. This can be simplified to tell us that Y_i (E_i - E_r) \cos(\theta_i) = Y_2 E_t \cos \theta_t. These can be solved to find that E_t = \frac{2Y_1 \cos \theta_i}{Y_1 \cos \theta_i + Y_2 \cos \theta_t} E_i and E_r = \frac{Y_1 \cos \theta_i - Y_2 \cos \theta_t}{Y_1 \cos \theta_i + Y_2 \cos \theta_t}E_i.
We can again calculate R = \frac{|I_r|}{I_{in}} = \frac{E_r^2}{E_i^2} = (\frac{Y_1 \cos \theta_i - Y_2 \cos \theta_t}{Y_1 \cos \theta_i + Y_2 \cos \theta_t})^2. Similarly, we see that T = \frac{4Y_1 \cos(\theta_i) \cdot Y_2 \cos(\theta_t)}{(Y_1 \cos \theta_i + Y_2 \cos \theta_t)^2}. Once again, I_r + I_t = I_{in} and R + T = 1.
Now, consider the case of a transverse magnetic field. In this case, we know that \mathbf{H_{i, r, t}} = H_{i, r, t}\hat{\mathbf{y}}, and thus \mathbf{E} = Z \mathbf{H} \times \hat{\mathbf{k}}, where Z = \sqrt{\frac{\mu}{\varepsilon}} = Y^{-1}.
Thus, we can say that
$$\begin{align} \mathbf{E}_i &= Z_1 H_i \hat{\mathbf{y}} \times \hat{\mathbf{k}}_i &= Z_1 H_i (\cos \theta_i \hat{\mathbf{x}} - \sin \theta_i \hat{\mathbf{z}}) \ \mathbf{E}_r &= Z_1 H_r \hat{\mathbf{y}} \times \hat{\mathbf{k}}_r &= -Z_1 H_r (\cos \theta_i \hat{\mathbf{x}} + \sin \theta_i \hat{\mathbf{z}}) \ \mathbf{E}_t &= Z_2 H_t \hat{\mathbf{y}} \times \hat{\mathbf{k}}_t &= Z_2 H_t (\cos \theta_t \hat{\mathbf{x}} - \sin \theta_t \hat{\mathbf{z}}) \ \end{align}$$
Then, continuity of \mathbf{H}^\parallel implies that H_i + H_r = H_t, and continuity of the electric fields implies that Z_1 (H_i - H_r) \cos \theta_i = Z_2 H_t \cos \theta_t.
From this, we can solve for H_r and H_t to see that
$$\begin{align} H_r &= \frac{Z_1 \cos \theta_i - Z_2 \cos \theta_t}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_i \ H_t &= \frac{2Z_1 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_t \end{align}$$
Alongside E_i = Z_1 H_i and H_t = Z_2 H_t, we can see that
$$\begin{align} E_r &= \frac{Z_2 \cos \theta_t - Z_1 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_i \ E_t &= \frac{2Z_2 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_t \end{align}$$
The relative signs of E_r and H_r was chosen to agree with their relationship for normal incidence.
Again, we can define R and T as
$$\begin{align} R &= (\frac{Z_2 \cos \theta_t - Z_1 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t})^2 \ T &= \frac{4Z_1 Z_2 \cos \theta_i \cos \theta_t}{(Z_1 \cos \theta_i + Z_2 \cos \theta_t)^2} \end{align}$$
Interestingly, if Z_2 \cos \theta_t = Z_1 \cos \theta_i for a wave with any polarization, the reflection's transverse magnetic component vanishes. If \mu_1 \approx \mu_2, we see that this condition becomes n_1 \cos \theta_t = n_2 \cos \theta_i, which using Snell's law, becomes
\sqrt{1-(\frac{n_1}{n_2})^2 \sin^2 \theta_i} = \frac{n_2}{n_1} \cos \theta_i
Definition. This is satisfied when \frac{n_2}{n_1} = \tan \theta_i. This angle, for which there is no reflected TM wave, is called Brewster's Angle.
Section 10.2.4 - Reflection from Conductors
We know that in a conductor, with current density \sigma, we can state that \mathbf{J}_e = \sigma \mathbf{E}. This then reintroduces the current term in Maxwell's equation, so that \nabla \times \mathbf{H} = \sigma \mathbf{E} + \varepsilon \frac{\partial \mathbf{E}}{\partial t}.
Then, if we curl the curls, we see that
$$\begin{align} \nabla \times \nabla \times \mathbf{E} &= -\mu \frac{\partial}{\partial t}(\nabla \times \mathbf{H}) &= -\mu \frac{\partial}{\partial t}(\sigma \mathbf{E} + \varepsilon \frac{\partial \mathbf{E}}{\partial t}) \ \nabla \times \nabla \times \mathbf{H} &= \sigma \nabla \times \mathbf{E} + \varepsilon \frac{\partial}{\partial t}(\nabla \times \mathbf{E}) &= -\mu\sigma \frac{\partial}{\partial t}\mathbf{H} + \varepsilon \frac{\partial}{\partial t}(-\mu \frac{\partial}{\partial t}\mathbf{H}) \ \end{align}$$
Applying our identity for the curl of a curl and knowing the divergence of the electric and magnetic fields vanishes, we see that
$$\begin{align} \nabla^2 \mathbf{E} &= -\mu \sigma \frac{\partial \mathbf{E}}{\partial t} - \mu \varepsilon \frac{\partial^2 \mathbf{E}}{\partial t^2} \ \nabla^2 \mathbf{H} &= -\mu \sigma \frac{\partial \mathbf{H}}{\partial t} - \mu \varepsilon \frac{\partial^2 \mathbf{H}}{\partial t^2} \ \end{align}$$
We see that if \sigma = 0, we recover the equations for propagation in a linear material.
We can solve these equations manually, but that would make us sad. Instead, we assume solutions of the form \mathbf{E} = E_0 \hat{\mathbf{x}} \cos(kz - \omega t) e^{-\kappa z}.
Through calculation, we see that
$$\begin{align} \nabla^2 \mathbf{E} &= -k^2 E_0 \hat{\mathbf{x}} \cos(kz - \omega t) e^{-\kappa z} + 2 k \kappa E_0 \hat{\mathbf{x}} \sin(kz - \omega t) e^{-\kappa z} + \kappa^2 E_0 \hat{\mathbf{x}} \cos(kz - \omega t) e^{-\kappa z} \ \mu \sigma \frac{\partial \mathbf{E}}{\partial t} &= \mu \sigma \omega E_0 \hat{\mathbf{x}} \sin(kz - \omega t)^{-\kappa z} \ \mu \sigma \frac{\partial^2 \mathbf{E}}{\partial^2 t} &= -\mu \sigma \omega^2 E_0 \hat{\mathbf{x}} \cos(kz - \omega t)^{-\kappa z} \end{align}$$
We can substitute this into the wave equation. For the wave equation to hold true at all times, by matching terms of \sin and \cos we see that -k^2 + \kappa^2 + \mu \varepsilon \omega^2 = 0 and \mu \sigma \varepsilon - 2k \kappa = 0. We can solve the second for \kappa to see that \kappa = \frac{\mu \sigma \omega}{2k}. This then allows us to solve the first equation for k^2, where we see
k^2 = \frac{\mu \varepsilon \omega^2}{2}(1+\sqrt{1+(\frac{\sigma}{\varepsilon \omega})^2})
This then lets us solve the first equation (again) for \kappa^2, where we see that
\kappa^2 = \frac{\mu \varepsilon \omega^2}{2}(\sqrt{1+(\frac{\sigma}{\varepsilon \omega})^2} - 1)
Our final equations them become
$$\begin{align} \kappa = \omega \sqrt{\mu\varepsilon} \sqrt{\frac{\sqrt{1 + (\frac{\sigma}{\varepsilon \mu})^2} + 1}{2}} \ \kappa = \omega \sqrt{\mu\varepsilon} \sqrt{\frac{\sqrt{1 + (\frac{\sigma}{\varepsilon \mu})^2} - 1}{2}} \ \end{align}$$
Note that when \sigma \ll \varepsilon \omega, the wave number collapses to k = \frac{\omega}{v}, and we recover propagation in a vacuum. However, when \sigma \gg \varepsilon \omega, we see that k = \kappa = \sqrt{\frac{\mu \omega \sigma}{2}}. Here, the distance the wave propagates before decreasing by a factor of \frac{1}{e} is known as the skin depth d, where d = \frac{1}{\kappa} = \sqrt{\frac{2}{\mu \omega \sigma}}. This is significantly less than the wavelength \lambda = \frac{2\pi}{k}. Note that in this limit, the electromagnetic wave is heavily damped.
The magnetic field is simpler to solve. We know that \frac{\partial\mathbf{H}}{\partial t} = -\frac{1}{\mu}\nabla \times \mathbf{E} = \hat{\mathbf{y}} \frac{E_0 e^{-\kappa z}}{\mu}[k \sin(kz - \omega t) + \kappa \cos(kz - \omega t)]
Integrating, we see that \mathbf{H} = \hat{\mathbf{y}} = \frac{E_0 e^{-\kappa z}}{\mu \omega}[k \cos(kz - \omega t) - \kappa \sin(kz - \omega t)].
We can combine this to see \mathbf{H} = \hat{\mathbf{y}} e^{-\kappa z} \frac{\sqrt{k^2 + \kappa^2}}{\mu \omega} \cos(kz - \omega t + \phi) where \cos \phi = \frac{k}{\sqrt{k^2 + \kappa ^2}} and \sim \phi = \frac{\kappa}{\sqrt{k^2 + \kappa^2}}.
Reflection of EM waves Incident from an Insulator onto a Conductor
In material 1 (z < 0), we know that
$$\begin{align} \mathbf{E}_1 &= \hat{\mathbf{x}} E_i \cos (k_1 z - \omega t) + \hat{\mathbf{x}} E_r \cos (-k_1 z - \omega t + \phi_r) \ \mathbf{H}_1 &= Y_1\hat{\mathbf{y}} E_i \cos (k_1 z - \omega t) - Y_1 \hat{\mathbf{y}} E_r \cos (-k_1 z - \omega t + \phi_r) \end{align}$$
We know that in material 2 (x > 0),
$$\begin{align} \mathbf{E}_1 &= \hat{\mathbf{x}} E_t \cos (k_2 z - \omega t + \phi_t) e^{-\kappa z} \ \mathbf{H}_1 &= \hat{\mathbf{y}} frac{\sqrt{k_2^2 + \kappa^2}}{\mu_2 \omega} \cos (k_2 z - \omega t + \phi + \phi_t) e^{-\kappa z} \end{align}$$
Here, \phi = \tan^{-1} (\frac{\kappa}{k_2}) os a phase shift intrinsic to the conductor.
The continuity of \mathbf{E}^\parallel and \mathbf{H}^\parallel gives us two equations. We can then substitute in \omega t = 0:
$$\begin{align} E_i + E_r \cos \phi_r &= E_t \cos \phi_t \ E_i - E_r \cos \phi_r &= \frac{\mu_1 v_1 \sqrt{k_2^2 + \kappa^2}}{\mu_2 \omega} E_t \cos(\phi_t + \phi) \ \end{align}$$
Now, substitute \omega t = \frac{\pi}{2}:
$$\begin{align} E_r \sin \phi_r &= E_t \sin \phi_t \ -E_r \sin \phi_r &= \frac{\mu_1 v_1 \sqrt{k_2^2 + \kappa^2}}{\mu_2 \omega} E_t \sin(\phi_t + \phi) \ \end{align}$$
We can then assume \mu_1 \approx \mu_2. This modifies the second equations, leaving us with
$$\begin{align} E_i - E_r \cos \phi_r &= \frac{\sqrt{k_2^2 + \kappa^2}}{k_1} E_t \cos(\phi_t + \phi) \ -E_r \sin \phi_r &= \frac{\sqrt{k_2^2 + \kappa^2}}{k_1} E_t \sin(\phi_t + \phi) \ \end{align}$$
Now, apply the sine and cosine additive identities and the definitions for \sin \phi and \cos \phi, alongside the definitions of \sin \phi_t and \cos \phi_t.
$$\begin{align} E_i - E_r \cos \phi_r &= E_t(\cos(\phi_t \frac{k_2}{k_1}) - \sin(\phi_t \frac{\kappa}{k_1})) \ -E_r &= E_t (\sin(\phi_t \frac{k_2}{k_1}) + \cos(\phi_t \frac{\kappa}{k_2})) \end{align}
Adding these equations lets us see that \tan(\phi_t) = -\frac{\kappa}{k_1 + k_2}. Applying the \sigma \rightarrow 0, we see \phi_t \rightarrow 0. Applying \sigma \gg \varepsilon \omega, we see that \phi_t \rightarrow -\frac{\pi}{4}.
Adding the first equations, we also wee that
E_t = \frac{2k_1 E_i}{\sqrt{(k_1 + k_2)^2 + \kappa^2}}
This also has the correct limits.
Furthermore, we can also obtain formulae for \tan \phi_r = \frac{-2k_1 \kappa}{k_1^2 - k_2^2 - \kappa^2}, and E_r = \frac{\sqrt(k_1^2 - k_2^2 - \kappa^2)^2 + (2 k_1 \kappa)^2}{(k_1 + k_2)^2 + \kappa^2} E_i. These collapse to the reflection at normal incidence values when \sigma \rightarrow 0. If \sigma \gg \varepsilon \mu, we see that \kappa \approx k_2 \gg k_1, as well as \tan \phi_r \rightarrow 0 and \frac{E_r}{E_i} \rightarrow 1 (complete reflection).
We can also compute energy currents. That is, I_i = \frac{1}{2} Y_1 E_i^2 = \frac{1}{2} v_1 \varepsilon_1 E_i^2, and I_r = \frac{1}{2}v_1 \varepsilon_1 E_r^2.
The transmitted wave is a bit more complicated. We see I_t = \langle \mathbf{S}_t \cdot \hat{\mathbf{z}} \rangle = \frac{\sqrt{k_2^2 + \kappa^2}}{k_1} Y_1 E_t^2 \langle \cos(\omega t + \phi_t) \cos(\omega t + \phi_t + \phi) \rangle e^{-2\kappa z} = \frac{1}{2} Y_1 E_t^2 e^{-2\kappa z}.