Electrodynamics Quizzes 7-8

docs/physics/electrodynamics/10-electromagnetic-waves.md

@@ -45,7 +45,7 @@ Consider some function $f(s)$. If $s = x - vt$ or $x + vt$, it is trivial to see
 
 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.
+**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 $H_0 = E_0 \sqrt{\frac{\varepsilon_0}{\mu_0}}$.
 
@@ -95,6 +95,163 @@ This section will focus on plane monochromatic waves incident from material 1 on
 
 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 $Z = \sqrt{\frac{\mu}{\epsilon}} = Y^{-1}$.
+
+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.*