notation stuff

book/02_maxwell_eq.md

@@ -1,23 +1,43 @@
 # Maxwell's Equations
 
+## Definitions
+
+Variable | Name | Units
+---|---|---
+$\mathbf{E}$ | Electric field strength | $V/m$
+$\mathbf{D}$ | Electric flux density | $C/m^2$
+$\mathbf{H}$ | Magnetic field strength | $A/m$
+$\mathbf{B}$ | Magnetic flux density | $T$
+$\mathbf{J}$ | Current density | $A/m^2$
+$\rho$ | Charge density | $C/m^3$
+$Q$ | Charge | $C$
+$I$ | Current | $A$
+$V$ | Voltage | $V$
+
+## Maxwell's Equations
+
+$\mathbf{D}$ and $\mathbf{B}$ are auxilary fields which capture the impact of atomic dipole moments, magnetization, and more on a macroscopic scale.
+
 $$
 \begin{align}
-D &= \epsilon E \\
-B &= \mu H
+\mathbf{D} &= \epsilon \mathbf{E} \\
+\mathbf{B} &= \mu \mathbf{H}
 \end{align}
 $$
 
-Where $\epsilon$ is the electric permittivity and $\mu$ is the magnetic permeability. In prose I will sometimes refer to both $D$ and $E$ as the electic field and $B$ and $H$ as the magnetic field so don't get tripped up on that. Technically $D$ is the electric displacement field and $B$ is magnetic flux density.
+Some texts will keep everything in $\mathbf{E}$ and $\mathbf{H}$ microscopic fields by adding additional polarization ($\mathbf{P}$) and magnetization ($\mathbf{M}$) fields. There may be some benefits to such a representation when looking at semiconductor physics or material science but that is beyond my area of expertise and I will be using the macro fields.
+
+Note that in prose I will sometimes refer to both $\mathbf{D}$ and $\mathbf{E}$ as the electic field and $\mathbf{B}$ and $\mathbf{H}$ as the magnetic field so don't get tripped up on that.
 
 There are two ways to look at Maxwell's Equations: derivative form and integral form. Both can be useful however I think the derivative form tends to be a more concise way to look at things so we'll start with that.
 Don't worry about understanding all of these yet, we'll go through them one at a time.
 
 $$
 \begin{align}
-\nabla \cdot D &= \rho \\
-\nabla \cdot B &= 0 \\
-\nabla \times E &= - \frac{\delta B}{\delta t} \\
-\nabla \times H &= J + \frac{\delta D}{\delta t}
+\nabla \cdot \mathbf{D} &= \rho \\
+\nabla \cdot \mathbf{B} &= 0 \\
+\nabla \times \mathbf{E} &= - \frac{\delta \mathbf{B}}{\delta t} \\
+\nabla \times \mathbf{H} &= \mathbf{J} + \frac{\delta \mathbf{D}}{\delta t}
 \end{align}
 $$
 
@@ -25,14 +45,14 @@ While I like the derivative forms for conceptual understanding, often it makes m
 
 $$
 \begin{align}
-\oiint D \cdot dS &= \iiint \rho dV \\
-\oiint B \cdot dS &= 0 \\
-\oint E \cdot d\ell &= - \frac{d}{dt}\iint B \cdot dS \\
-\oint H \cdot d\ell &= \iint J \cdot dS + \frac{d}{dt} \iint D \cdot dS
+\oiint \mathbf{D} \cdot d\mathbf{s} &= \iiint \rho dv \\
+\oiint \mathbf{B} \cdot d\mathbf{s} &= 0 \\
+\oint \mathbf{E} \cdot d\mathbf{\ell} &= - \frac{d}{dt}\iint \mathbf{B} \cdot d\mathbf{s} \\
+\oint \mathbf{H} \cdot d\mathbf{\ell} &= \iint \mathbf{J} \cdot d\mathbf{s} + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s}
 \end{align}
 $$
 
-$dS$ is directed outward.
+$dS$ is directed outward for closed surfaces and uses the right hand rule when looped by a closed path.
 
 That's all there is to it. Isn't everything clear now?
 We'll break down each of these equations to understand their implications.
@@ -41,20 +61,20 @@ We'll break down each of these equations to understand their implications.
 
 $$
 \begin{align}
-\nabla \cdot D &= \rho \\
-\oiint D \cdot dS &= \iiint \rho dV \\
+\nabla \cdot \mathbf{D} &= \rho \\
+\oiint \mathbf{D} \cdot d\mathbf{s} &= \iiint \rho dv \\
 &= Q
 \end{align}
 $$
 
-In plain terms, this says that charge sources electric field.
+In plain terms, this says that charge sources the electric field.
 
 ## Gauss's Law for Magnetism
 
 $$
 \begin{align}
-\nabla \cdot B &= 0 \\
-\oiint B \cdot dS &= 0
+\nabla \cdot \mathbf{B} &= 0 \\
+\oiint \mathbf{B} \cdot d\mathbf{s} &= 0
 \end{align}
 $$
 
@@ -64,8 +84,8 @@ The divergence of the magnetic field is zero. This means that magnetic monopoles
 
 $$
 \begin{align}
-\nabla \times E &= - \frac{\delta B}{\delta t} \\
-\oint E \cdot d\ell &= - \frac{d}{dt}\iint B \cdot dS
+\nabla \times \mathbf{E} &= - \frac{\delta \mathbf{B}}{\delta t} \\
+\oint \mathbf{E} \cdot d\mathbf{\ell} &= - \frac{d}{dt}\iint \mathbf{B} \cdot d\mathbf{s}
 \end{align}
 $$
 
@@ -75,12 +95,13 @@ This means that a changing magnetic field will cause circulation in the electric
 
 $$
 \begin{align}
-\nabla \times H &= J + \frac{\delta D}{\delta t} \\
-\oint H \cdot d\ell &= \iint J \cdot dS + \frac{d}{dt} \iint D \cdot dS
+\nabla \times \mathbf{H} &= \mathbf{J} + \frac{\delta \mathbf{D}}{\delta t} \\
+\oint \mathbf{H} \cdot d\mathbf{\ell} &= \iint \mathbf{J} \cdot d\mathbf{s} + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s} \\
+&= I + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s}
 \end{align}
 $$
 
-There are two parts to this which are sometimes introduced individually since for slowly time varying fields $\delta D / \delta t$ is negligible.
+There are two parts to this which are sometimes introduced individually since for slowly time varying fields $\delta \mathbf{D} / \delta t$ is negligible.
 
 The first part means that magnetic field will always circulate electric current.
 
@@ -91,8 +112,8 @@ Near DC, a the $d/dt$ terms of Ampere's Law and Faraday's Law go to zero giving
 
 $$
 \begin{align}
-\nabla \times E = 0 \\
-\nabla \times H = J
+\nabla \times \mathbf{E} = \mathbf{0} \\
+\nabla \times \mathbf{H} = \mathbf{J}
 \end{align}
 $$
 
@@ -102,7 +123,7 @@ Note that voltage is defined as:
 
 $$
 \begin{align}
-V = \int_p E \cdot d\ell
+V = \int_p \mathbf{E} \cdot d\mathbf{\ell}
 \end{align}
 $$
 
@@ -111,4 +132,48 @@ At higher frequencies we have to worry about the path of integration. This is on
 
 ## Magnetic Hysteresis
 
-## 
+## Free Space Propagation
+
+In free space we can assume $J = 0$ since there are no charge carriers.
+Similarly we can assume $\rho = 0$. While this is true in vacuum, it may not be valid in air, however superposition lets us ignore any constant fields when looking at propagation.
+
+This leaves us with a simplified set of equations
+
+$$
+\begin{align}
+\nabla \cdot \mathbf{E} &= 0 \\
+\nabla \cdot \mathbf{H} &= 0 \\
+\nabla \times \mathbf{E} &= - \mu_0 \frac{\delta \mathbf{H}}{\delta t} \\
+\nabla \times \mathbf{H} &= \epsilon \frac{\delta \mathbf{E}}{\delta t}
+\end{align}
+$$
+
+Note that $\epsilon$ has not been simplified to $\epsilon_0$ so this is valid in a dielectric as well as in true free space.
+
+We will need a fun identity to make this work
+
+$$
+\begin{align}
+\nabla \cdot \left( A \times B \right) &= B \cdot \left( \nabla \times A \right) - A \cdot \left( \nabla \times B \right)
+\end{align}
+$$
+
+Additionally, the following identity will come in handy
+$$
+\begin{align}
+\frac{d}{dt}(H \cdot H) &= \frac{d}{dt}(H^2) \\
+&= 2 H \frac{dH}{dt} \\
+\frac{d}{dt}(E \cdot E) &= 2 E \frac{dE}{dt}
+\end{align}
+$$
+
+$$
+\begin{align}
+\nabla \times E &= - \mu_0 \frac{\delta H}{\delta t} \\
+H \left( \nabla \times E \right) &= - \mu_0 H \cdot \frac{\delta H}{\delta t} \\
+\nabla \cdot (E \times H) + E \cdot (\nabla \times H) &= -\mu_0 H \cdot \frac{dH}{dt} \\
+\nabla \cdot (E \times H) + E \cdot \left(\epsilon \frac{\delta E}{\delta t}\right) &= -\mu_0 H \cdot \frac{dH}{dt}
+\end{align}
+$$
+
+TODO: finish this part