nutshell/book/02_maxwell_eq.md

# 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.

$$
\begin{align}
\mathbf{D} &= \epsilon \mathbf{E} \\
\mathbf{B} &= \mu \mathbf{H}
\end{align}
$$

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.

$$
\begin{align}
\nabla \cdot \mathbf{D} &= \rho \\
\nabla \cdot \mathbf{B} &= 0 \\
\nabla \times \mathbf{E} &= - \frac{\delta \mathbf{B}}{\delta t} \\
\nabla \times \mathbf{H} &= \mathbf{J} + \frac{\delta \mathbf{D}}{\delta t}
\end{align}
$$

While I like the derivative forms for conceptual understanding, often it makes more sense when evaluating real world problems to look at volumes and surfaces rather than divergence and curl at any point. This is where the integral forms tend to be useful.

$$
\begin{align}
\oiint \mathbf{D} \cdot d\mathbf{s} &= \iiint \rho dv \\
\oiint \mathbf{B} \cdot d\mathbf{s} &= 0 \\
\oint \mathbf{E} \cdot d\mathbf{\ell} &= - \frac{d}{dt}\iint \mathbf{B} \cdot d\mathbf{s} \\
\oint \mathbf{H} \cdot d\mathbf{\ell} &= \iint \mathbf{J} \cdot d\mathbf{s} + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s}
\end{align}
$$

$dS$ is directed outward 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.

## Gauss's Law

$$
\begin{align}
\nabla \cdot \mathbf{D} &= \rho \\
\oiint \mathbf{D} \cdot d\mathbf{s} &= \iiint \rho dv \\
&= Q
\end{align}
$$

In plain terms, this says that charge sources the electric field.

## Gauss's Law for Magnetism

$$
\begin{align}
\nabla \cdot \mathbf{B} &= 0 \\
\oiint \mathbf{B} \cdot d\mathbf{s} &= 0
\end{align}
$$

The divergence of the magnetic field is zero. This means that magnetic monopoles cannot exist. In other words the magnetic field must always form closed loops.

## Ampere's Law

$$
\begin{align}
\nabla \times \mathbf{E} &= - \frac{\delta \mathbf{B}}{\delta t} \\
\oint \mathbf{E} \cdot d\mathbf{\ell} &= - \frac{d}{dt}\iint \mathbf{B} \cdot d\mathbf{s}
\end{align}
$$

This means that a changing magnetic field will cause circulation in the electric field.

## Faraday's Law

$$
\begin{align}
\nabla \times \mathbf{H} &= \mathbf{J} + \frac{\delta \mathbf{D}}{\delta t} \\
\oint \mathbf{H} \cdot d\mathbf{\ell} &= \iint \mathbf{J} \cdot d\mathbf{s} + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s} \\
&= I + \frac{d}{dt} \iint \mathbf{D} \cdot d\mathbf{s}
\end{align}
$$

There are two parts to this which are sometimes introduced individually since for slowly time varying fields $\delta \mathbf{D} / \delta t$ is negligible.

The first part means that magnetic field will always circulate electric current.

The second means that changes in the electric field will circulate the magnetic field.

## Static Simplifications
Near DC, a the $d/dt$ terms of Ampere's Law and Faraday's Law go to zero giving the following equations:

$$
\begin{align}
\nabla \times \mathbf{E} = \mathbf{0} \\
\nabla \times \mathbf{H} = \mathbf{J}
\end{align}
$$

The main takeaway from this is that with the electric field becomes conservative. This means that the path integral between two points only depends on the two points and is independent of the path taken. 

Note that voltage is defined as:

$$
\begin{align}
V = \int_p \mathbf{E} \cdot d\mathbf{\ell}
\end{align}
$$

Conservation of electric field has huge implications for real world testing. Any time you use a multimeter you're relying on this since the path the wires take is the path $p$ of integration and is largely uncontrolled.
At higher frequencies we have to worry about the path of integration. This is one reason why coaxial probes are used for oscilloscopes but not multimeters.

## 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