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
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
We can take the curl of both equations and then substitute to see that
We can apply a vector identity to see
Section 10.1.1 - The Wave Equation and Plane Waves
Definition. The equation \([\frac{\partial^2}{\partial x^2} - \frac{1}{\nu^2} \frac{\partial^2}{\partial t^2}] f(x, t) = 0\) is well-known to mathematicians (see Differential Equations), and is known as the wave equation. In physics, the speed of the wave is \(\nu = c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}\).
Consider some function \(f(s)\). If \(s = x - \nu t\) or \(x + \nu t\), 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, \(\nu = c\).
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 \(f(s) = f(\hat{\mathbf{k}} \cdot \mathbf{r} - \nu t)\), such that \(\mathbf{E}(\mathbf{r}, t) = \mathbf{E}_0 f(\hat{\mathbf{k}}\cdot\mathbf{r} - \nu t)\), where \(\hat{\mathbf{k}}\cdot\mathbf{E}_0 = 0\).
From this, we can see that \(\mathbf{H}(\mathbf{r}, t) = \sqrt{\frac{\varepsilon_0}{\mu_0}} \hat{\mathbf{k}} \times \mathbf{E}_0 f(\hat{\mathbf{k}} \cdot \mathbf{r} - \nu t)\). Similarly, \(\hat{\mathbf{k}} \cdot \mathbf{H} = 0\).
Additionally, we can compute \(\mathbf{S} = \mathbf{E} \times \mathbf{H} = c \varepsilon_0 E_0^2 f^2(\hat{\mathbf{k}} \cdot \mathbf{r} - \nu t) \hat{\mathbf{k}}\). 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,
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 \(k = \frac{\omega}{\nu}\), and \(\nu = \frac{1}{\sqrt{\mu \varepsilon}}\).
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:
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
In material 2, we see that
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
Evaluation at \(z = 0\), we see that
In material 2, we see that
In this case, applying boundary conditions is as simple as matching components to obtain two (out of four) equations:
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
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
If we assume \(\mu_1 = \mu_2\), we can rewrite the equations in terms of wave numbers \(k_i\), where \(k_i = \frac{\omega}{\nu_i} = \omega \sqrt{\mu_i \varepsilon_i}\)
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
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, \(k_i = k_r = \frac{\omega}{\nu}\) as they describe propagation in the same media, so \(\theta_r = \theta_i\). As \(n \propto \frac{1}{\nu}\), 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:
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
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
Alongside \(E_i = Z_1 H_i\) and \(H_t = Z_2 H_t\), we can see that
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
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
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
Applying our identity for the curl of a curl and knowing the divergence of the electric and magnetic fields vanishes, we see that
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
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
This then lets us solve the first equation (again) for \(\kappa^2\), where we see that
Our final equations them become
Note that when \(\sigma \ll \varepsilon \omega\), the wave number collapses to \(k = \frac{\omega}{\nu}\), 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
We know that in material \(2\) (\(x > 0\)),
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 \(\omega t = 0\):
Now, substitute \(\omega t = \frac{\pi}{2}\):
We can then assume \(\mu_1 \approx \mu_2\). This modifies the second equations, leaving us with
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
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 \(I_t = \langle \mathbf{S}_t \cdot \hat{\mathbf{z}} \rangle = \frac{\sqrt{k_2^2 + \kappa^2}}{k_1} Y_1 E_t^2 \langle \cos(\omega t + \phi_t) \cos(\omega t + \phi_t + \phi) \rangle e^{-2\kappa z} = \frac{1}{2} Y_1 E_t^2 e^{-2\kappa z}\).
Section 10.3 - Electrodynamic Interactions between Waves and Mattter
Section 10.3.1 - Response Functions and Fourier Transforms
This section will cover the relations \(\mathbf{D} = \varepsilon \mathbf{E}\), \(\mathbf{B} = \mu \mathbf{H}\), and \(\mathbf{J} = \sigma \mathbf{E}\). However, as the logic is the same for each, I will focus only on the equation for the electric flux density.
We know that the permeability of a material is not a constant, but instead can depend on the electric field at all previous times. We then define a response function \(\varepsilon_R(t - t')\) such that
Definition. For this to make physical sense, we say that \(\varepsilon_R(\tau) = 0\) when \(\tau < 0\). That is, the electric flux density is only impacted by the past electric field in the material.
A common response function is \(\varepsilon_R(t - t') = \varepsilon \delta(t - t')\).
To make this worse, we know that the fields in a material are frequency-dependent. that is, \(\mathbf{D}(\mathbf{r}, t) = \varepsilon_0 \mathbf{E}(\mathbf{r}, t)\) becomes \(\mathbf{D}_\omega(\mathbf{r}, t) = \varepsilon(\omega) \mathbf{E}_\omega(\mathbf{r}, t)\).
Definition. Given some function \(f(t)\), we can say that the Fourier transform of \(f(t)\), represented by \(\tilde{f}(\omega)\), is defined as
We then can define the inverse Fourier transform such that
We also know that \(\delta(x - x') = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i(x-x')t}dt\). Then, we can write
We know that \(\delta(\omega - \omega') = \frac{1}{2\pi} \int_{-\infty}^{\infty}e^{i(\omega - \omega') t'}dt'\), which lets us rewrite \(\mathbf{J}\) as
Theorem. Convolution Theorem. By taking the Fourier transform of both sides, we see that \(\tilde{\mathbf{D}} = \tilde{\varepsilon}(\omega) \tilde{\mathbf{E}}(\omega)\).
Section 10.3.2 - Classical Models for Permittivity and Conductivity
Assuming we are in a material, we know that \(\varepsilon = \varepsilon_0 (1 + \chi_e)\). If we move to a response function, we see that
Definition. The \(\epsilon_0 \chi_{eR}(t-t')\) function is the susceptibility response function.
As we are restricted to linear relations to the electric field, we can use the constructive equation for \(\mathbf{D}\) to see that
This specifically excludes materials in which the flux density depends on higher powers of the electric field. In the limit where the field is time-independent, this reduces to the original constructive equations. This implies that in the static (time-independent) limit,
Consider an electron bound to its nucleus by a spring-like restoring force. We see that \(F_{spring} = -Kx\) for some \(K\) (uppercase, in contrast to mechanics), and damped by some force \(F_{damping} = -m \gamma \dot{x}\), and driven by some force \(F_{field} = q_e E_0 e^{-i \omega t}\). Then, we see that
We assume a solution of the form \(x(t) = x_0(\omega) \exp(-i \omega t)\), and see that
Then, we can find that there exists a dipole moment \(\tilde{p}(t) = qx(t) = \frac{\frac{q^2}{m}E_0}{\omega_0^2 - \omega^2 -i\omega\gamma}\), where the real part of the expression is the physical dipole moment.
If instead of being subject to an oscillatory force from a changing electric field, assume we have a delta function that imparts an impulse at \(t=0\). then, we set \(x(0) = 0\), and \(\dot{x}(0) = \frac{qE\Delta t}{m}\). Our differential equation then becomes \(\ddot{x} + \gamma \dot{x} + \omega_0^2 x=0\).
We can assume a solution of the form \(x(t) = A e^{-\beta t} \cos(\omega t - \phi)\). Applying initial conditions, we see that \(\phi = \frac{\pi}{2}\) and \(A = \frac{qE\delta t}{\omega m}\). Substituting into the differential equation, we see that \(\beta = \frac{\gamma}{2}\) and \(\omega = \omega_r = \sqrt{\omega_0^2 - (\frac{\gamma}{2})^2}\).
Thus, a harmonic oscillator with resonance frequency \(\omega_0\), damping coefficient \(\gamma\), and mass \(m\), given initial momentum \(q E \delta t\), we will see a damped oscillation \(x(t) = \frac{qE\Delta t}{m \omega_r} \exp(-\frac{\gamma}{2}t) \sin(\omega_r t)\), where \(\omega_r\) is defined as above.
Now, adjust the initial momentum for the problem to be applied at \(t = t_i\) to see that
We can then apply a summation for an arbitrary time-dependent electric field.
We can then tweak this to find the Green function, which is the solution to
The Green function then becomes
Then, for an arbitrary electric field \(E_x(t)\), the solution to \((\omega_0^2 + \gamma \frac{d}{dt} + \frac{d^2}{dt^2})x(t) = E_x(t)\) can be writteen as
We know that \(p(t) = qx(t)\) represents the dipole moment. We can then see that
We can substitute the inverse Fourier transform for the electric field to see that
Applying the relation \(\int_{-\infty}^{\infty} G(t-t')e^{-i\omega t'} dt' = -e^{i \omega t} \tilde{G}(\omega)\), we see that
This then implies that \(\tilde{\mathbf{p}}(\omega) = -\frac{q^2}{m}\tilde{G}(\omega)\mathbf{\tilde{E}}(\omega)\).
We can calculate that \(\tilde{G}(\omega) = \frac{1}{\omega_0^2 - \omega^2 - i \gamma \omega}\). Then, if we have \(N\) molecules per unit length, each with \(f_i\) electrons with resonant frequency \(\omega_i\) and damping parameter \(\gamma_i\), we see that
We can use this to then calculate the Fourier transform of the electric flux density, which tells us that
I've skipped the effects on optics, as well as the Drude response function, for brevity.
Section 10.4 - Guided Waves and Transmission Lines
Section 10.4.1 - Coaxial Cables
We know that for a coaxial cable with radii \(R_1\) and \(R_2\), given that the distance between wavelengths \(R_2 - R_1\) is significantly less than some arbitrary length \(\Delta z\), which is less than some length scale over which the electric field changes \(\lambda\), the electric field is defined as \(\mathbf{E}(s) = \frac{Q'}{2\pi s \varepsilon} \hat{\mathbf{s}}\). We can then see the voltage between the two insulators is simply
Then, we know that capacitance is \(C = Q / \Delta V\), so capacitance per unit length becomes
Similarly, we know that the magnetic field is given by Ampere's law as \(\mathbf{H} = \frac{I}{2 \pi s}\), as we place the Ampere's loop has radius \(s\) around the inner conductor. Then, we see the flux is
This tells us that inductance per unit length, \(L' = \Phi / I \Delta Z\) can be given by
Now, for some length \(\Delta z\), we know that \(\Delta Q = Q' \Delta z\), and then by definition,
That is, the rate at which charge in a cylinder changes is equal to the current entering minus the current leaving. This then implies that
Since we know that \(Q' = C' V\) (as \(C' = Q' / V\)), we can take the time derivative to see that
Now, consider a cylindrical loop (a rectangle revolved around the center, excluding the conductor). Faraday's law tells us that \(\mathbf{E} \cdot d\mathbf{l} = -\frac{d\Phi}{dt}\), where \(\Phi = L' \Delta Z I\) is the flux through the loop. As the electric field is radial and conservative, we know that
We can take the limit to see that
This gives is the two telegrapher's equations:
We can then take the derivative of the first with respect to \(t\) and the derivative of the second with respect to \(z\), we see that
This tells us that
We can instead take the derivative of the first equation with respect to \(z\) and the derivative of the second with respect to \(t\) to see that
These are wave equations with velocity \(\nu = \frac{1}{\sqrt{L'C'}}\). In the case of light, \(\nu = \frac{1}{\sqrt{\mu \epsilon}}\).
We presume solutions of the form \(V(z, t) = V_0 \cos(kz - \omega t + \phi_V)\) and \(I(z, t) = I_0 \cos(kz - \omega t + \phi_I)\), with a shared velocity. Applying the telegrapher's equations, we see that \(\phi_I = \phi_V\) and \(\frac{V_0}{I_0} = -\sqrt{\frac{L'}{C'}} = -Z\).
We have a formula for \(C'\) and \(L'\), so we can write \(Z\) as
Often, modern coaxial cables have \(\ln \frac{R_2}{R_1} \approx 1\), and the permittivity of free space is \(\sqrt{\frac{\mu_0}{\epsilon_0}} = 377 \Omega\). Then, the impedance of a coaxial cable is often of order \(Z \approx \frac{377 \Omega}{2\pi \sqrt{\varepsilon_r}} \approx 60 \Omega\).
Definition. The inverse of impedance is called the admittance.
Section 10.4.2 - Parallel Conductor Transmission Lines
Consider a two-wire transmission line, with distance \(d\) between the lines. Then, the electric field between the lines becomes
Then, integrating from \(x = a\) to \(x = d - a\) (where \(a\) is the radius of each conductor), we see that
Note that the integrals diverge if \(a = 0\). We can then see that
We can also see that the magnetic field is given by
This then lets us calculate flux as \(\Phi_B = \mu_0 \Delta z \frac{I}{\pi} \ln{(\frac{d}{a} - 1)}\), which means unit inductance becomes
Again, the continuity equation tells us that \(\frac{\partial Q'}{\partial t} = -\frac{\partial I}{\partial z}\), and the capacitance relation lets us derive that
Then, we can use the back EMF to see that
This lets us once again calculate wave equations for \(V\) and \(I\), with \(\nu = c\)
Section 10.4.3 - Transmission Lines with Dissipation
Especially at high frequencies, we see that the dielectric functions become complex leading to a leakage current between insulators. We thus modify our telegrapher's equations to see that
Here, the \(R'\) is the resistance per unit length.
The wave equation for voltage then becomes
We now assume a solution of the form \(V(z, t) = V_0 \cos (kz - \omega t) e^{-\kappa z}\).
Applying the wave equation, we see that for this solution to work, \(\kappa^2 - k^2 = -\omega^2 L' C' + R' G'\) and \(2 k \kappa = (L' G' + R' C') \omega\).
From this, we can solve for \(k\) and \(\kappa\) by setting \(b = \omega^2 L' C' - R' G'\) and \(c = (L' G' + R' C') \omega\) so that \(k = \frac{1}{\sqrt{2}} \sqrt{b + \sqrt{b^2 + c^2}}\) and \(\kappa = \frac{c}{2k}\).
We can then solve for \(\omega = \omega(k)\), which ends up being a very sad equation.
The Heaviside condition is the condition in which we change the parameters so that the parameters \(\frac{G'}{C'} = \omega_C\) and \(\frac{R'}{L'} = \omega_L\) are equal. Then, we see \(\omega^2 = k^2 v_0^2\).