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

@@ -5,34 +5,34 @@
 Consider an empty space. Then, it is evident that
 
 $$\begin{align}
-\div \vb{E} &= 0 \\
-\div \vb{H} &= 0
+    \nabla \cdot \mathbf{E} &= 0 \\
+    \nabla \cdot \mathbf{H} &= 0
 \end{align}$$
 
 $$\begin{align}
-\curl \vb{E} + \frac{\partial \vb{B}}{\partial t} &= 0 \\
-\curl \vb{H} - \frac{\partial \vb{D}}{\partial t} &= 0
+\nabla \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} &= 0 \\
+\nabla \times \mathbf{H} - \frac{\partial \mathbf{D}}{\partial t} &= 0
 \end{align}$$
 
-As $\vb{B} = \mu_0 \vb{H}$ and $\vb{D} = \varepsilon_0 \vb{E}$ in a vacuum, the third and fourth equations can be rewritten as
+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
 
 $$\begin{align}
-\curl \vb{E} + \mu_0 \frac{\partial \vb{H}}{\partial t} &= 0 \\
-\curl \vb{H} - \varepsilon_0 \frac{\partial \vb{E}}{\partial t} &= 0
+\nabla \times \mathbf{E} + \mu_0 \frac{\partial \mathbf{H}}{\partial t} &= 0 \\
+\nabla \times \mathbf{H} - \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} &= 0
 \end{align}$$
 
 We can take the curl of both equations and then substitute to see that
 
 $$\begin{align}
-\curl \curl \vb{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \vb{E}}{\partial t^2} &= 0 \\
-\curl \curl \vb{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \vb{H}}{\partial t^2} &= 0
+\nabla \times \nabla \times \mathbf{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0 \\
+\nabla \times \nabla \times \mathbf{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} &= 0
 \end{align}$$
 
 We can apply a vector identity to see
 
 $$\begin{align}
--\nabla^2 \vb{E} + \mu_0 \epsilon_0 \frac{\partial^2 \vb{E}}{\partial t^2} &= 0 \\
--\nabla^2 \vb{H} + \mu_0 \epsilon_0 \frac{\partial^2 \vb{H}}{\partial t^2} &= 0
+-\nabla^2 \mathbf{E} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} &= 0 \\
+-\nabla^2 \mathbf{H} + \mu_0 \varepsilon_0 \frac{\partial^2 \mathbf{H}}{\partial t^2} &= 0
 \end{align}$$
 
 ### Section 10.1.1 - The Wave Equation and Plane Waves
@@ -43,34 +43,34 @@ Consider some function $f(s)$. If $s = x - vt$ or $x + vt$, it is trivial to see
 
 **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, $v = c$.
+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 $\vb{E}_0$ and $\vb{H}_0$ to be in the plane normal to the direction of motion. That is, electomagnetic plane waves are transverse, not longitudinal.
+**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 $\vu{k}$, 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 $\vu{k}\vdot\vb{E}_0 = 0$.
+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 $\vb{H}(\vb{r}, t) = \sqrt{\frac{\epsilon_0}{\mu_0}} \vu{k} \times \vb{E}_0 f(\vu{k} \vdot \vb{r} - vt)$. Similarly, $\vu{k} \vdot \vb{H} = 0$.
+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 $\vb{S} = \vb{E} \times \vb{H} = c \epsilon_0 E_0^2 f^2(\vu{k} \vdot \vb{r} - vt) \vu{k}$. We can also see that $\epsilon_0 E^2 = \mu_0 H^2$ at any given time.
+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 $\vb{D} = \epsilon \vb{E}$ 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.
+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* $\vb{k}$ as $\vb{k} = k \vu{k}$, so that $\omega = kc$. Then,
+**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,
 
 $$\begin{align}
-\vb{E}(\vb{r}, t) &= \vb{E_0} \cos(\vb{k} \vdot \vb{r} - \omega t) \\
-\vb{H}(\vb{r}, t) &= \sqrt{\frac{\epsilon_0}{\mu_0}} \vu{k} \times \vb{E}_0 \cos(\vb{k} \vdot \vb{r} - \omega t)
+\mathbf{E}(\mathbf{r}, t) &= \mathbf{E_0} \cos(\mathbf{k} \cdot \mathbf{r} - \omega t) \\
+\mathbf{H}(\mathbf{r}, t) &= \sqrt{\frac{\varepsilon_0}{\mu_0}} \hat{\mathbf{k}} \times \mathbf{E}_0 \cos(\mathbf{k} \cdot \mathbf{r} - \omega t)
 \end{align}$$
 
 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 $\vb{S}$, momentum density $\vb{g}$, and momentum current density $-\overleftrightarrow{\vb{T}}$
+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
 
@@ -78,13 +78,13 @@ Monochromatic plane waves with frequency $\omega$ in a simple linear material ar
 
 ### 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.
+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:
 
 $$\begin{align}
-\vb{E}(\vb{r}, t) &= E_{x0} \vu{x} \cos(kz - \omega t) + E_{y0} \vu{y} \sin(kz - \omega t) \\
-\vb{H}(\vb{r}, t) &= \sqrt{\frac{\varepsilon}{\mu}} (E_{x0} \vu{y} \cos(kz - \omega t) - E_{y0} \vu{x} \sin(kz - \omega t))
+\mathbf{E}(\mathbf{r}, t) &= E_{x0} \hat{\mathbf{x}} \cos(kz - \omega t) + E_{y0} \hat{\mathbf{y}} \sin(kz - \omega t) \\
+\mathbf{H}(\mathbf{r}, t) &= \sqrt{\frac{\varepsilon}{\mu}} (E_{x0} \hat{\mathbf{y}} \cos(kz - \omega t) - E_{y0} \hat{\mathbf{x}} \sin(kz - \omega t))
 \end{align}$$
 
 If $E_{x0} = E_{y0}$, the wave is said to be circularly polarized.
@@ -93,8 +93,8 @@ If $E_{x0} = E_{y0}$, the wave is said to be circularly polarized.
 
 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 $\div \vb{E} = \frac{\vb{\rho_e}}{\varepsilon_0}$ and $\div \vb{H} = \frac{\vb{\rho_m}}{\mu_0}$. We also know that $\div \vb{D} = \rho_{ef}$ and $\div \vb{B} = \rho_{mf}$.
+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 $\div \vb{D}$, and take the integral over a Gaussian pillbox on the boundary, we can apply divergence theorem to see that $\int_V \div \vb{D} dV = \int_{SofV} D \vdot \vu{n} dS = \rho_{efree}$. 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 $\int_{SofV} D \vdot \vu{n} = 0$, so $\vb{D}_1 \vdot \vu{n} + \vb{D}_2 \vdot \vu{n} = \vb{D_1} \vdot \vu{z} + \vb{D}_1 \vdot (-\vu{n}) = 0$. Then, we can say that $\vb{D}_1 \vdot{z} = \vb{D}_2 \vdot{z}$, or in simpler terms, $\vdot{D}_1^\perp = \vdot{D}_2^\perp$.
+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 $\vb{B}$, we see that $\vdot{B}_1^\perp = \vdot{B}_2^\perp$. Note that due to the existence of polarization and magnetization, we cannot say the same regarding $\mathbf{E}$ or $\mathbf{H}$.
+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 $\mathbf{H}$.

docs/physics/electrodynamics/8-quasi-static-electrodynamics-ac-circuitts.md

@@ -44,11 +44,11 @@ $$\cos \phi = \frac{R}{\sqrt{R^2 + X_C^2}}$$
 
 ### Section 8.1.2 - Quasi-Static Error for a Parallel Plate Capacitor
 
-Consider a parallel-plate capacitor. We know that within the capacitor, the electric flux is $\vb{D}(t) = \varepsilon_0 \vb{E}(t)$. With the charge on a plate given by $Q(t)$, we can say that $\vb{D}(t) = \frac{Q(t)}{\pi R^2}$.
+Consider a parallel-plate capacitor. We know that within the capacitor, the electric flux is $\mathbf{D}(t) = \varepsilon_0 \mathbf{E}(t)$. With the charge on a plate given by $Q(t)$, we can say that $\mathbf{D}(t) = \frac{Q(t)}{\pi R^2}$.
 
-We also know by Ampere's law that $\curl \vb{H} = \frac{\partial}{\partial t} \vb{D}$. From this, given circular platFrom this, applying Stokes to Ampere's Law, we see that
+We also know by Ampere's law that $\nabla \times \mathbf{H} = \frac{\partial}{\partial t} \mathbf{D}$. From this, given circular platFrom this, applying Stokes to Ampere's Law, we see that
 
-$$\vb{H} = \frac{\partial Q}{\partial t} \frac{s}{2 \pi R^2} \vu{\varphi}$$
+$$\mathbf{H} = \frac{\partial Q}{\partial t} \frac{s}{2 \pi R^2} \hat{\mathbf{\varphi}}$$
 
 This continues on in this manner, however, I've opted to skip most of the math.
 
@@ -60,11 +60,11 @@ For an inductor, the voltage drop across an inductor is directly proportional to
 
 $$\Delta V_L = L \frac{d^2Q}{dt^2} = L\frac{dI}{dt}$$
 
-Consider a circular current loop with a voltage source in the $x-y$ plane. Then, applying Faraday's law, $\int_{circle} = (\curl \vb{E}) \cdot \vu{z} dS = -\frac{\partial}{\partial t} \int_{circle} \vb{B} \cdot \vu{z} dS$.
+Consider a circular current loop with a voltage source in the $x-y$ plane. Then, applying Faraday's law, $\int_{circle} = (\nabla \times \mathbf{E}) \cdot \hat{\mathbf{z}} dS = -\frac{\partial}{\partial t} \int_{circle} \mathbf{B} \cdot \hat{\mathbf{z}} dS$.
 
-Apply Stokes' law to the left hand side to see that $\int_{circumference} \vb{E} \cdot d\vb{l} = -\frac{\partial \Phi_B}{\partial t}$.
+Apply Stokes' law to the left hand side to see that $\int_{circumference} \mathbf{E} \cdot d\mathbf{l} = -\frac{\partial \Phi_B}{\partial t}$.
 
-With $\Phi_B = LI$, we see the "back EMF" opposing the increasing current will be $\vb{\mathcal{E}} = -L \frac{\partial I}{\partial t}$.
+With $\Phi_B = LI$, we see the "back EMF" opposing the increasing current will be $\mathbf{\mathcal{E}} = -L \frac{\partial I}{\partial t}$.
 
 Consider a circuit with a voltage source, a switch, an inductor, and a resistor in series. Then, by KVL, $V_{cell} = L \frac{dI}{dt} + IR$.
 
@@ -84,9 +84,9 @@ $$\sin \phi = -\frac{X_L}{Z}; \cos \phi = \frac{R}{Z}; I_0 = \frac{V_0}{Z}$$
 
 ### Section 8.1.4 - Calculation of Inductance
 
-Recall that the back EMF $\vb{\mathcal{E}} = -L \frac{\partial I}{\partial T}$. We can calculate the work done by this force as follows.
+Recall that the back EMF $\mathbf{\mathcal{E}} = -L \frac{\partial I}{\partial T}$. We can calculate the work done by this force as follows.
 
-$$\frac{dW}{dt} = I(\vb{\mathcal{E}}) = IL \frac{dI}{dt} = \frac{1}{2} L \frac{d}{dt}(I^2)$$
+$$\frac{dW}{dt} = I(\mathbf{\mathcal{E}}) = IL \frac{dI}{dt} = \frac{1}{2} L \frac{d}{dt}(I^2)$$
 
 Then, integrating both sides, we see that
 
@@ -96,9 +96,9 @@ We also know from Section 3.2 that the work needed to create a magnetic field is
 
 $$W = \frac{1}{2} \mu_0 \int_V H^2 dV$$
 
-Now, consider a long air-filled solenoid with $n$ turns per unit length and cross-sectional area $A$. Then, we know the flux through a cross-section of the solenoid will be $\Phi = BA = \mu_0 n I A$. Then, the back-EMF for one loop of the solenoid can be given by $\vb{\mathcal{E}}_{1 loop} = - \frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt}$. Then, the total induced EMF will be $\vb{\mathcal{E}} = nl\vb{\mathcal{E}}_{1 loop}$, where $l$ is the length of the solenoid. Thus, by the definition of back-EMF,
+Now, consider a long air-filled solenoid with $n$ turns per unit length and cross-sectional area $A$. Then, we know the flux through a cross-section of the solenoid will be $\Phi = BA = \mu_0 n I A$. Then, the back-EMF for one loop of the solenoid can be given by $\mathbf{\mathcal{E}}_{1 loop} = - \frac{d\Phi}{dt} = -\mu_0 n A \frac{dI}{dt}$. Then, the total induced EMF will be $\mathbf{\mathcal{E}} = nl\mathbf{\mathcal{E}}_{1 loop}$, where $l$ is the length of the solenoid. Thus, by the definition of back-EMF,
 
-$$\vb{\mathcal{E}} = -nl \mu_0 nA \frac{dI}{dt} = -L\frac{dI}{dt}$$
+$$\mathbf{\mathcal{E}} = -nl \mu_0 nA \frac{dI}{dt} = -L\frac{dI}{dt}$$
 
 So, $L = \mu_0 n^2 Al$
 
@@ -106,15 +106,15 @@ We can also compute thsi by energy. We know that $W = \frac{1}{2} \mu_0 \int_V H
 
 Now, consider a coaxial cable. That is, consider a solid cylinder of radius $a$ that conducts current in the $+z$ direction. The circuit is completed by a thin cylindical shell outside of the conductor yet still with radius $a$. We also assume that current density is uniform within the cylindrical conductor.
 
-Recall that from Ampere's Law, for $s \in (0, a)$, we have $\vb{H}(s) = \frac{I_enc}{2\pi s}\vu{\varphi}$, with $I_enc = \frac{I\pi s^2}{\pi a^2}$. Thus, $\vb{H} = \frac{Is}{2\pi a^2}\vu{\varphi}$. Then,
+Recall that from Ampere's Law, for $s \in (0, a)$, we have $\mathbf{H}(s) = \frac{I_enc}{2\pi s}\hat{\mathbf{\varphi}}$, with $I_enc = \frac{I\pi s^2}{\pi a^2}$. Thus, $\mathbf{H} = \frac{Is}{2\pi a^2}\hat{\mathbf{\varphi}}$. Then,
 
 $$W = \frac{1}{2} \mu_0 \int_V H^2 dV = \frac{1}{2} \mu_0 l \int_0^a 2\pi s ds (\frac{Is}{2\pi a^2})^2 = \frac{I^2}{2} \frac{\mu_0 l}{2\pi} \frac{1}{4}$$
 
 This implies that $L = \frac{\mu_0 l}{8\pi}$.
 
-We can also solve this via flux. We know that $\vb{B}(s) = \frac{u_0 Is}{2\pi a^2}\vu{\varphi}$. Then,
+We can also solve this via flux. We know that $\mathbf{B}(s) = \frac{u_0 Is}{2\pi a^2}\hat{\mathbf{\varphi}}$. Then,
 
-$$\Phi = \int_0^a \vb{B} \vdot \vu{n} dS = \int_0^a \frac{\mu_0 Is}{2\pi a^2} l ds = \frac{\mu_0 I l}{2\pi} \frac{1}{2}$$
+$$\Phi = \int_0^a \mathbf{B} \cdot \hat{\mathbf{n}} dS = \int_0^a \frac{\mu_0 Is}{2\pi a^2} l ds = \frac{\mu_0 I l}{2\pi} \frac{1}{2}$$
 
 This is off by a factor of $2$. Instead, multiply by a fator of $f(s) = \frac{s^2}{a^2}$ to see
 
@@ -126,9 +126,9 @@ This becomes complicated, and is thus omitted.
 
 ### Section 8.1.5 - Quasi-static Error for a Solenoidal Inductor
 
-In section 5.2, we learned that in a long solenoid, $\vb{H} = nI \vb{z}$, where $I$ is the current and $n$ the number of turns per uniut length. This, however, was dependent of the current being constant. Now, let current be represented as $I(t) = I_0 \cos (\omega t + \phi)$. Now, $\vb{H}(t) = nI(t) \vu{z}$.
+In section 5.2, we learned that in a long solenoid, $\mathbf{H} = nI \mathbf{z}$, where $I$ is the current and $n$ the number of turns per uniut length. This, however, was dependent of the current being constant. Now, let current be represented as $I(t) = I_0 \cos (\omega t + \phi)$. Now, $\mathbf{H}(t) = nI(t) \hat{\mathbf{z}}$.
 
-This implies magnetic flux with a density of $\vb{B} = \mu_0 n I(t) \vu{z}$. Given the inductor with radius $s$, this will cause changing flux $\Phi_B(t) = \vb{B} * A = \mu_0 n I(t) \pi s^2$.
+This implies magnetic flux with a density of $\mathbf{B} = \mu_0 n I(t) \hat{\mathbf{z}}$. Given the inductor with radius $s$, this will cause changing flux $\Phi_B(t) = \mathbf{B} * A = \mu_0 n I(t) \pi s^2$.
 
 By Faraday's Law,
 
@@ -136,13 +136,13 @@ $$E_{induced} = -N \frac{\partial \Phi_B}{\partial{t}}$$
 
 This can be rearranged to see
 
-$$2\pi s \vb{E}_{induced}(t) = -\frac{\partial \Phi_b(t)}{\partial t} = -\mu_0 n \pi s^2 \frac{\partial}{\partial t} I(t)$$
+$$2\pi s \mathbf{E}_{induced}(t) = -\frac{\partial \Phi_b(t)}{\partial t} = -\mu_0 n \pi s^2 \frac{\partial}{\partial t} I(t)$$
 
-This can be used to find $\vb{E} = -\mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \vu{\varphi}$, equivalent to a flux density $\vb{D}(t) = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \vu{\varphi}$. We can then apply Ampere's law to see that
+This can be used to find $\mathbf{E} = -\mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \hat{\mathbf{\varphi}}$, equivalent to a flux density $\mathbf{D}(t) = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial}{\partial t}I(t) \hat{\mathbf{\varphi}}$. We can then apply Ampere's law to see that
 
-$$\curl \vb{H} = \frac{\partial \vb{D}(t)}{\partial t} = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial^2}{\partial t^2}I(t) \vu{\varphi}$$
+$$\nabla \times \mathbf{H} = \frac{\partial \mathbf{D}(t)}{\partial t} = -\varepsilon_0 \mu_0 n \frac{s}{2} \frac{\partial^2}{\partial t^2}I(t) \hat{\mathbf{\varphi}}$$
 
-We can work backwards to find that $\vb{H}(s, t) = (1 - \frac{\mu_0 \varepsilon_0 s^2 \omega^2}{4}) H_{z0}(t)\vu{z}$. Then, the same conditions should apply as in 8.1.2. That is, $\omega$ should be low enough or the d evice small enough that light can easily propagate across the device during one period of oscillation.
+We can work backwards to find that $\mathbf{H}(s, t) = (1 - \frac{\mu_0 \varepsilon_0 s^2 \omega^2}{4}) H_{z0}(t)\hat{\mathbf{z}}$. Then, the same conditions should apply as in 8.1.2. That is, $\omega$ should be low enough or the d evice small enough that light can easily propagate across the device during one period of oscillation.
 
 ## Section 8.2 - Circuits with Resistance, Capacitance and Inductance and a Sinusoidal EMF
 
@@ -213,24 +213,24 @@ $$u = \frac{1}{2}(\varepsilon_0 e^2 + \mu_u H^2)$$
 
 In a capacitor, we cannot directly state the energy. However, we can state its rate of change:
 
-$$\frac{du_E}{dt} = \vb{E} \vdot \frac{d\vb{D}}{dt}$$
+$$\frac{du_E}{dt} = \mathbf{E} \cdot \frac{d\mathbf{D}}{dt}$$
 
-In the case where $\vb{D} = \varepsilon \vb{E}$, that is, in the presence of a simple dielectric,
+In the case where $\mathbf{D} = \varepsilon \mathbf{E}$, that is, in the presence of a simple dielectric,
 
-$$u_E = \frac{1}{2} \varepsilon E^2 = \frac{1}{2}\vb{E} \vdot \vb{D}$$
+$$u_E = \frac{1}{2} \varepsilon E^2 = \frac{1}{2}\mathbf{E} \cdot \mathbf{D}$$
 
 This is mirrored in solenoids / inductors, where
 
-$$\frac{du_M}{dt} = \vb{H} \vdot \frac{d\vb{B}}{dt}$$
+$$\frac{du_M}{dt} = \mathbf{H} \cdot \frac{d\mathbf{B}}{dt}$$
 
-In the case where $\vb{B} = \mu \vb{H}$, that is, in the presence of a simple magnetic material,
+In the case where $\mathbf{B} = \mu \mathbf{H}$, that is, in the presence of a simple magnetic material,
 
-$$u_M = \frac{1}{2}\vb{H} \vdot \vb{B}$$
+$$u_M = \frac{1}{2}\mathbf{H} \cdot \mathbf{B}$$
 
 Lastly, this continues for electromagnetic fields.
 
-$$\frac{du_{EM}}{dt} = \vb{E} \vdot \frac{d\vb{D}}{dt} + \vb{H} \vdot \frac{d\vb{B}}{dt}$$
+$$\frac{du_{EM}}{dt} = \mathbf{E} \cdot \frac{d\mathbf{D}}{dt} + \mathbf{H} \cdot \frac{d\mathbf{B}}{dt}$$
 
 In simple electromagnetic materials, we see that
 
-$$u_{EM} = \frac{1}{2}(\vb{E} \vdot \vb{D} + \vb{H} \vdot{B}) = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$$
+$$u_{EM} = \frac{1}{2}(\mathbf{E} \cdot \mathbf{D} + \mathbf{H} \cdot \mathbf{B}) = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$$

docs/physics/electrodynamics/9-conservation-laws.md

@@ -6,70 +6,70 @@ The laws of physics exhibit temporal, spatial, and angular symmetry. That is, th
 
 The conservation of chargee is another symmetry-based conservation law, derived from "gauge invariance". In this section, this is conservation law is assumed to be valid.
 
-Recall the *continuity equation*, that is, $\frac{\partial \rho(\vb{r})}{\partial t} = - \div \vb{J}(\vb{r})$. That is, the charge density at any point in space is equal to to the divergence of the current. From this, we can integrate to find $Q(t) = \int_V \rho(\vb{r}, t) dV$, and $\frac{dQ}{dt} = -\int_{\partial V} \vb{J}(\vb{r}, t) \vdot \vu{n} dS$. This is a *local conservation law*, because it does not address situations in which charge decreases in one region and increases in another without the flow of current.
+Recall the *continuity equation*, that is, $\frac{\partial \rho(\mathbf{r})}{\partial t} = - \nabla \cdot \mathbf{J}(\mathbf{r})$. That is, the charge density at any point in space is equal to to the divergence of the current. From this, we can integrate to find $Q(t) = \int_V \rho(\mathbf{r}, t) dV$, and $\frac{dQ}{dt} = -\int_{\partial V} \mathbf{J}(\mathbf{r}, t) \cdot \hat{\mathbf{n}} dS$. This is a *local conservation law*, because it does not address situations in which charge decreases in one region and increases in another without the flow of current.
 
 ## Section 9.2 - Conservation of Electomagnetic Energy
 
 Consider a volume $V$ with surface $\partial V$, that encloses some magnetic and electric point charges. Then, for any electric charge $q_{ei}$ or magnetic charge $q_{mj}$, the force on each charge is
 
 $$\begin{align}
-    \vb{F}_i &= q_{ei} (\vb{E}_i + \vb{v}_i \times \vb{B}_i) \\
-    \vb{F}_j &= q_{mi} (\vb{H}_j - \vb{v}_j \times \vb{D}_j)
+    \mathbf{F}_i &= q_{ei} (\mathbf{E}_i + \mathbf{v}_i \times \mathbf{B}_i) \\
+    \mathbf{F}_j &= q_{mi} (\mathbf{H}_j - \mathbf{v}_j \times \mathbf{D}_j)
 \end{align}$$
 
-We also know that the rate at which energy changes due to changing fields is $\frac{dw_i}{dt} = \vb{F}_i \vdot \vb{v}_i$ and $\frac{dw_m}{dt} = \vb{F}_j \vdot \vb{v}_j$. This allows us to conclude that at any point $\vb{r}$ inside the volume, the rate at which the mechanical energy density changes is
+We also know that the rate at which energy changes due to changing fields is $\frac{dw_i}{dt} = \mathbf{F}_i \cdot \mathbf{v}_i$ and $\frac{dw_m}{dt} = \mathbf{F}_j \cdot \mathbf{v}_j$. This allows us to conclude that at any point $\mathbf{r}$ inside the volume, the rate at which the mechanical energy density changes is
 
-$$\frac{du_{mech}}{dt} = \sum_i \delta(\vb{r} - \vb{r}_i) \vb{F}_i \vdot \vb{v}_i + \delta(\vb{r} - \vb{r}_j) \vb{F}_j \vdot \vb{v}_j$$
+$$\frac{du_{mech}}{dt} = \sum_i \delta(\mathbf{r} - \mathbf{r}_i) \mathbf{F}_i \cdot \mathbf{v}_i + \delta(\mathbf{r} - \mathbf{r}_j) \mathbf{F}_j \cdot \mathbf{v}_j$$
 
-Since $(\vb{v}_i \times \vb{B}_i) \vdot \vb{v}_i = 0$ and $(\vb{v}_j \times \vb{D}_j) \vdot \vb{v}_j = 0$, and the current densities are given as $\vb{J}_e(\vb{r}) = \sum_i \vb{v}_i q_{ei}\delta(\vb{r} - \vb{r}_i)$ and $\vb{J}_m(\vb{r}) = \sum_j \vb{v}_j q_{mj}\delta(\vb{r} - \vb{r}_j)$, we can rewrite this as
+Since $(\mathbf{v}_i \times \mathbf{B}_i) \cdot \mathbf{v}_i = 0$ and $(\mathbf{v}_j \times \mathbf{D}_j) \cdot \mathbf{v}_j = 0$, and the current densities are given as $\mathbf{J}_e(\mathbf{r}) = \sum_i \mathbf{v}_i q_{ei}\delta(\mathbf{r} - \mathbf{r}_i)$ and $\mathbf{J}_m(\mathbf{r}) = \sum_j \mathbf{v}_j q_{mj}\delta(\mathbf{r} - \mathbf{r}_j)$, we can rewrite this as
 
-$$\frac{du_{mech}}{dt} = \vb{J}_e(\vb{r}) \vdot \vb{E}(\vb{r}) + \vb{J}_m(\vb{r}) \vdot \vb{H}(\vb{r})$$
+$$\frac{du_{mech}}{dt} = \mathbf{J}_e(\mathbf{r}) \cdot \mathbf{E}(\mathbf{r}) + \mathbf{J}_m(\mathbf{r}) \cdot \mathbf{H}(\mathbf{r})$$
 
 Combine this with the Maxwell equations to remove current densities, we see that
 
-$$\frac{du_{mech}}{dt} = (\curl \vb{H} - \frac{\partial \vb{H}}{\partial t}) \vdot \vb{E} + (-\curl \vb{E} - \frac{\partial \vb{B}}{\partial t}) \vdot \vb{H}$$
+$$\frac{du_{mech}}{dt} = (\nabla \times \mathbf{H} - \frac{\partial \mathbf{H}}{\partial t}) \cdot \mathbf{E} + (-\nabla \times \mathbf{E} - \frac{\partial \mathbf{B}}{\partial t}) \cdot \mathbf{H}$$
 
 With a vector identity, this simplifies to
 
-$$\frac{du_{mech}}{dt} = -\curl (\vb{E} \times \vb{H})$$
+$$\frac{du_{mech}}{dt} = -\nabla \times (\mathbf{E} \times \mathbf{H})$$
 
-**Definition**. We call $\vb{S} = \vb{E} \times \vb{H}$ the *Poynting vector*.
+**Definition**. We call $\mathbf{S} = \mathbf{E} \times \mathbf{H}$ the *Poynting vector*.
 
 With this vector, we can define
 
-$$\frac{du_{mech}}{dt} = -\curl \vb{S}$$
+$$\frac{du_{mech}}{dt} = -\nabla \times \mathbf{S}$$
 
 ### Section 9.2.2 - Energy Density for Linear Materials
 
-For a simple material, that is, one in which $\vb{D} = \epsilon \vb{E}$ and $\vb{B} = \mu \vb{H}$, we can express the rate of change of electomagnetic energy $\frac{\partial u_{em}}{\partial t}$ as $\frac{\partial u_{em}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} (\vb{E} \vdot \vb{D} + \vb{H} \vdot \vb{B})$. After integration, we can see that $u_{em} = \frac{1}{2}(\epsilon E^2 + \mu H^2)$.
+For a simple material, that is, one in which $\mathbf{D} = \varepsilon \mathbf{E}$ and $\mathbf{B} = \mu \mathbf{H}$, we can express the rate of change of electomagnetic energy $\frac{\partial u_{em}}{\partial t}$ as $\frac{\partial u_{em}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t} (\mathbf{E} \cdot \mathbf{D} + \mathbf{H} \cdot \mathbf{B})$. After integration, we can see that $u_{em} = \frac{1}{2}(\varepsilon E^2 + \mu H^2)$.
 
-**Definition**. A material is said to be an *anisotropic linear material* if $\vb{D} = \overleftrightarrow{\vb{\epsilon}} \vdot \vb{E}$ and $\vb{B} = \overleftrightarrow{\vb{\mu}} \vdot \vb{H}$. Then, the internal electromagnetic energy becones
+**Definition**. A material is said to be an *anisotropic linear material* if $\mathbf{D} = \overleftrightarrow{\mathbf{\varepsilon}} \cdot \mathbf{E}$ and $\mathbf{B} = \overleftrightarrow{\mathbf{\mu}} \cdot \mathbf{H}$. Then, the internal electromagnetic energy becomes
 
-$$u_{em} = \frac{1}{2}(\vb{E} \vdot \overleftrightarrow{\vb{\epsilon}} \vdot \vb{E}$ and $\vb{B} + \vb{H} + \overleftrightarrow{\vb{\mu}} \vdot \vb{H}) = \frac{1}{2}\sum_{i,j}(\epsilon_{ij}E_iE_j + \mu_{ij}H_iH_j)$$
+$$u_{em} = \frac{1}{2}(\mathbf{E} \cdot \overleftrightarrow{\mathbf{\varepsilon}} \cdot \mathbf{E}+ \mathbf{H} + \overleftrightarrow{\mathbf{\mu}} \cdot \mathbf{H}) = \frac{1}{2}\sum_{i,j}(\varepsilon_{ij}E_iE_j + \mu_{ij}H_iH_j)$$
 
 If the dyadics are symmetric, the energy functions uniquely specify the energy in terms of fields. In general, the polarization or magnetization of a material may depend on its past history and on time, and as such, the energy density for such materials cannot be expressed entirely in terms of the fields.
 
 ### Section 9.2.3 - Poynting Vector Examples
 
-Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\vb{H} = \frac{R}{2\pi s R}\vb{\varphi}$. Additionally, we know that
+Consider a long coaxial cable, bridged by a constant voltage $V$ on one side and a resistor $R$ on the other. Then, we know that between the conductors, $\mathbf{H} = \frac{R}{2\pi s R}\mathbf{\varphi}$. Additionally, we know that
 
-$$V=\int_a^b \vb{E} \vdot d\vb{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$
+$$V=\int_a^b \mathbf{E} \cdot d\mathbf{l} = \frac{Q}{2\pi \ell \varepsilon_0} \ln(\frac{b}{a}) \Rightarrow \frac{Q}{\ell} = \frac{2\pi\varepsilon_0}{\ln(\frac{b}{a})} V$$
 
-This then implies that $\vb{E} = \frac{V}{\ln(\frac{b}{a})s} \vu(s)$. We can then solve for both the energy density and Poynting vector, as well as $\vb{v} = \vb{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that
+This then implies that $\mathbf{E} = \frac{V}{\ln(\frac{b}{a})s} \hat{\mathbf{s}}$. We can then solve for both the energy density and Poynting vector, as well as $\mathbf{v} = \mathbf{S} / u$, the speed at which energy moves through the cable. With the impedance for a coaxial cable $Z_{C0} = \sqrt{\frac{\mu_0}{\varepsilon_0}} \frac{\ln(\frac{b}{a})}{2\pi}$, we see that
 
-$$\vb{v} = \frac{2c \vu{z}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$
+$$\mathbf{v} = \frac{2c \hat{\mathbf{z}}}{\frac{R}{Z_{C0}} + \frac{Z_{C0}}{R}}$$
 
 ---
 
 Consider a long cylindrical ohmic wire of radius $a$, length $L$, and resistivity $\rho$ along the $z$-axis. If this wire is carrying a constant current $I$, we know that inside the wire, $E_z = \rho J = \rho \frac{I}{\pi a^2}$. From Ampere's Law, $H_\phi = \frac{s}{2\pi a^2} I$. Then, inside the wire,
 
-$$\vb{S} = -s \frac{\rho I^2}{2 \pi^2 a^4} \vu{s}$$
+$$\mathbf{S} = -s \frac{\rho I^2}{2 \pi^2 a^4} \hat{\mathbf{s}}$$
 
-Outside the wire, we know that $\vb{E} = \frac{\rho I}{\pi a^2} \vu{z}$ and $\vb{H} = \frac{I}{2\pi s}\vu{\varphi}$, so
+Outside the wire, we know that $\mathbf{E} = \frac{\rho I}{\pi a^2} \hat{\mathbf{z}}$ and $\mathbf{H} = \frac{I}{2\pi s}\hat{\mathbf{\varphi}}$, so
 
-$$\vb{S} = -\frac{\rho I}{\pi a^2} \frac{I}{2\pi s} \vu{s}$$
+$$\mathbf{S} = -\frac{\rho I}{\pi a^2} \frac{I}{2\pi s} \hat{\mathbf{s}}$$
 
-Notably, inside the wire, $\div \vb{S} = \frac{\rho I^2}{\pi^2 a^4}$, but is equal to $0$ outside of the wire.
+Notably, inside the wire, $\nabla \cdot \mathbf{S} = \frac{\rho I^2}{\pi^2 a^4}$, but is equal to $0$ outside of the wire.
 
 ---
 
@@ -79,50 +79,50 @@ Further examples are present but omitted.
 
 We know that the rate of change of momentum for any given particle is simply the force acting on it. To calculate this, recall the force density:
 
-$$\vb{f} = \sum_i \vb{F}_i\delta(\vb{r}-\vb{r}_i) + \sum_j \vb{F}_j\delta(\vb{r}-\vb{r}_j) = \sum_i q_{ei}\delta(\vb{r}-\vb{r}_i)(\vb{E} + \vb{v}_i \cross \vb{B}) + \sum_j q_{ej} \delta(\vb{r}-\vb{r}_i) (\vb{H} - \vb{v}_j \times \vb{D})$$
+$$\mathbf{f} = \sum_i \mathbf{F}_i\delta(\mathbf{r}-\mathbf{r}_i) + \sum_j \mathbf{F}_j\delta(\mathbf{r}-\mathbf{r}_j) = \sum_i q_{ei}\delta(\mathbf{r}-\mathbf{r}_i)(\mathbf{E} + \mathbf{v}_i \times \mathbf{B}) + \sum_j q_{ej} \delta(\mathbf{r}-\mathbf{r}_i) (\mathbf{H} - \mathbf{v}_j \times \mathbf{D})$$
 
 Converting to currents, we see that
 
-$$\vb{f}(\vb{r}) = \rho_e(\vb{r})\vb{E}(\vb{r}) + \vb{J}_e(\vb{r}) \times \vb{B}(\vb{r}) + \rho_m(\vb{r}) + \vb{H}(\vb{r}) - \vb{J}_m(\vb{r}) \times \vb{D}(\vb{r})$$
+$$\mathbf{f}(\mathbf{r}) = \rho_e(\mathbf{r})\mathbf{E}(\mathbf{r}) + \mathbf{J}_e(\mathbf{r}) \times \mathbf{B}(\mathbf{r}) + \rho_m(\mathbf{r}) + \mathbf{H}(\mathbf{r}) - \mathbf{J}_m(\mathbf{r}) \times \mathbf{D}(\mathbf{r})$$
 
 Substituting in Maxwell's Equations, we see that
 
-$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \varepsilon_0 (\div \vb{E})\vb{E} + (\curl \vb{E})\times\vb{D} + \mu_0(\div \vb{H})\vb{H} + (\curl \vb{H})\times\vb{B}$$
+$$\mathbf{f}(\mathbf{r}) + \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = \varepsilon_0 (\nabla \cdot \mathbf{E})\mathbf{E} + (\nabla \times \mathbf{E})\times\mathbf{D} + \mu_0(\nabla \cdot \mathbf{H})\mathbf{H} + (\nabla \times \mathbf{H})\times\mathbf{B}$$
 
-Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\vb{T}}$, so that
+Now, we claim that the right-hand side is the divergence of some tensor $\overleftrightarrow{\mathbf{T}}$, so that
 
-$$\vb{f}(\vb{r}) + \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = \div \overleftrightarrow{\vb{T}}$$
+$$\mathbf{f}(\mathbf{r}) + \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = \nabla \cdot \overleftrightarrow{\mathbf{T}}$$
 
-This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\div \overleftrightarrow{\vb{T}} = \div \overleftrightarrow{\vb{T}}_e + \div \overleftrightarrow{\vb{T}}_m$. Then, we  can state
+This tensor is the Maxwell Stress Tensor. We claim that the divergence of this tensor is composed of both an electric and magnetic part, so that $\nabla \cdot \overleftrightarrow{\mathbf{T}} = \nabla \cdot \overleftrightarrow{\mathbf{T}}_e + \nabla \cdot \overleftrightarrow{\mathbf{T}}_m$. Then, we  can state
 
 $$\begin{align}
-\div \overleftrightarrow{\vb{T}}_e &= \varepsilon_0 [(\div \vb{E})\vb{E} + (\curl \vb{E})\times \vb{E}] \\
-\div \overleftrightarrow{\vb{T}}_m &= \varepsilon_0 [(\div \vb{H})\vb{H} + (\curl \vb{H})\times \vb{H}]
+\nabla \cdot \overleftrightarrow{\mathbf{T}}_e &= \varepsilon_0 [(\nabla \cdot \mathbf{E})\mathbf{E} + (\nabla \times \mathbf{E})\times \mathbf{E}] \\
+\nabla \cdot \overleftrightarrow{\mathbf{T}}_m &= \varepsilon_0 [(\nabla \cdot \mathbf{H})\mathbf{H} + (\nabla \times \mathbf{H})\times \mathbf{H}]
 \end{align}$$
 
-We know that $\div(\vb{EE}) = (\div \vb{E})\vb{E} + (\vb{E} \vdot \nabla)\vb{E}$ and $\div(\overleftrightarrow{\vb{I}}f) = \grad f$. If we let $f = \frac{1}{2}\vb{E} \vdot \vb{E}$, we see that $\grad(\frac{1}{2}\vb{E} \vdot \vb{E}) = (\vb{E} \vdot \nabla)\vb{E} + (\curl \vb{E})\vb{E}$. Then, we see that
+We know that $\nabla \cdot(\mathbf{EE}) = (\nabla \cdot \mathbf{E})\mathbf{E} + (\mathbf{E} \cdot \nabla)\mathbf{E}$ and $\nabla \cdot(\overleftrightarrow{\mathbf{I}}f) = \nabla f$. If we let $f = \frac{1}{2}\mathbf{E} \cdot \mathbf{E}$, we see that $\nabla(\frac{1}{2}\mathbf{E} \cdot \mathbf{E}) = (\mathbf{E} \cdot \nabla)\mathbf{E} + (\nabla \times \mathbf{E})\mathbf{E}$. Then, we see that
 
 $$\begin{align}
-\overleftrightarrow{\vb{T}}_e &= \varepsilon_0 \vb{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\vb{I}}(\vb{E} \vdot \vb{E}) \\
-\overleftrightarrow{\vb{T}}_m &= \mu_0 \vb{HH} - \frac{\mu_0}{2} \overleftrightarrow{\vb{I}}(\vb{H} \vdot \vb{H})
+\overleftrightarrow{\mathbf{T}}_e &= \varepsilon_0 \mathbf{EE} - \frac{\varepsilon_0}{2} \overleftrightarrow{\mathbf{I}}(\mathbf{E} \cdot \mathbf{E}) \\
+\overleftrightarrow{\mathbf{T}}_m &= \mu_0 \mathbf{HH} - \frac{\mu_0}{2} \overleftrightarrow{\mathbf{I}}(\mathbf{H} \cdot \mathbf{H})
 \end{align}$$
 
-Knowing that $\overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}}_e + \overleftrightarrow{\vb{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum,
+Knowing that $\overleftrightarrow{\mathbf{T}} = \overleftrightarrow{\mathbf{T}}_e + \overleftrightarrow{\mathbf{T}}_m$, and that $u = \frac{1}{2}(\varepsilon_0 E^2 + \mu_0 H^2)$ is the energy density of the electromagnetic fields in a vacuum,
 
-$$\overleftrightarrow{\vb{T}} = \varepsilon_0 \vb{EE} + \mu_0 \vb{HH} - \overleftrightarrow{\vb{I}}u$$
+$$\overleftrightarrow{\mathbf{T}} = \varepsilon_0 \mathbf{EE} + \mu_0 \mathbf{HH} - \overleftrightarrow{\mathbf{I}}u$$
 
-Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\vb{g}(\vb{r}) = \vb{D}(\vb{r}) \times \vb{B}(\vb{r})$. Thus,
+Additionally, we denote the time rate of change of the momentum density of the electromagnetic fields as $\mathbf{g}(\mathbf{r}) = \mathbf{D}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})$. Thus,
 
-$$\vb{f}(\vb{r}) = \frac{\partial}{\partial t}\vb{g} = \div \overleftrightarrow{\vb{T}}$$
+$$\mathbf{f}(\mathbf{r}) = \frac{\partial}{\partial t}\mathbf{g} = \nabla \cdot \overleftrightarrow{\mathbf{T}}$$
 
 ---
 
-The Divergence Theorem states that $\int_V(\div \overleftrightarrow{\vb{T}}) dV = \int_{SofV} dS \vu{n} \vdot \overleftrightarrow{\vb{T}}$. We can prove this by expanding the left-hand side over a cube. Note that as $\overleftrightarrow{\vb{T}}$ is symmetric, $\vu{n} \vdot \overleftrightarrow{\vb{T}} = \overleftrightarrow{\vb{T}} \vdot \vu{n}$.
+The Divergence Theorem states that $\int_V(\nabla \cdot \overleftrightarrow{\mathbf{T}}) dV = \int_{SofV} dS \hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}}$. We can prove this by expanding the left-hand side over a cube. Note that as $\overleftrightarrow{\mathbf{T}}$ is symmetric, $\hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}} = \overleftrightarrow{\mathbf{T}} \cdot \hat{\mathbf{n}}$.
 
 Given a static field, the momentum does not change in time. That is,
 
-$$\frac{\partial}{\partial t} \vb{g} = \frac{\partial}{\partial t}(\vb{D} \times \vb{B}) = 0$$
+$$\frac{\partial}{\partial t} \mathbf{g} = \frac{\partial}{\partial t}(\mathbf{D} \times \mathbf{B}) = 0$$
 
-Then, we can see that $\vb{f}(\vb{r}) = \div \overleftrightarrow{\vb{T}}$. We can thus integrate over the volume to find force on an object.
+Then, we can see that $\mathbf{f}(\mathbf{r}) = \nabla \cdot \overleftrightarrow{\mathbf{T}}$. We can thus integrate over the volume to find force on an object.
 
-$$\vb{F} = \int_V \vb{f}(\vb{r}) dV = \int_V \div \overleftrightarrow{\vb{T}} dV = \int_{\partial V} dS \vu{n} \vdot \overleftrightarrow{\vb{T}} = \int_{\partial V} \overleftrightarrow{\vb{T}} \vdot \vu{n} dS$$
+$$\mathbf{F} = \int_V \mathbf{f}(\mathbf{r}) dV = \int_V \nabla \cdot \overleftrightarrow{\mathbf{T}} dV = \int_{\partial V} dS \hat{\mathbf{n}} \cdot \overleftrightarrow{\mathbf{T}} = \int_{\partial V} \overleftrightarrow{\mathbf{T}} \cdot \hat{\mathbf{n}} dS$$