# Maxwell's equations

(for more exhaustive treatment of Maxwell's equations, please refer to wikipedia:Maxwell's Equations

## Microscopic Maxwell's equations

This version of Maxwell's equations is called 'microscopic', as it is in differential form and they are satisfied at any point as long as we know the local charge density $\rho$ and current density $\mathbf{J}$.

$\nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}$
$\nabla \cdot \mathbf{B} = 0$
$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$
$\nabla \times \mathbf{B} = \mu_0\left(\mathbf{J} + \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} \right)$

## Macroscopic Maxwell's equations

This version is called 'macroscopic', because we use $\textbf{D}$ and $\textbf{H}$ which are defined as a combination of electric (magnetic) field and polarization (magnetization) vector, respectively. Polarization (magnetization) is a macroscopic quantity because it is given as electric (magnetic) dipole moment per unit volume at a given point.

By 'macroscopic', we mean that the quantity only makes sense as an average over a small volume surrounding the specific position. Density is a good example of macroscopic variable, and we would be in trouble if we interpret density as a microscopic variable. For example, microscopic density of salt will fluctuate between zero (vacancy in solid) and almost infinity (nucleus of either Na+ or Cl-), depending on the microscopic position.

$\nabla \cdot \mathbf{D} = \rho_\text{f}$
$\nabla \cdot \mathbf{B} = 0$
$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$
$\nabla \times \mathbf{H} = \mathbf{J}_\text{f} + \frac{\partial \mathbf{D}} {\partial t}$

where $\rho_\text{f}$ and $\mathbf{J}_\text{f}$ refers to free charge density and free current density, respectively.

In terms of only E and B,

$\nabla \cdot \mathbf{E} = \frac1{\epsilon_0} (\rho_\text{f}-\nabla\cdot\mathbf{P})$
$\nabla \cdot \mathbf{B} = 0$
$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$
$\nabla \times \mathbf{B} = \mu_0\left(\mathbf{J}_\text{f}+\nabla\times\mathbf{M} + \epsilon_0\frac{\partial \mathbf{E}}{\partial t}+\frac{\partial \mathbf{P}}{\partial t}\right)$

### Maxwell's eq in l.i.h. medium

Since

$\mathbf{D}=\epsilon\mathbf{E}$
$\mathbf{B}=\mu\mathbf{H}$
$\mathbf{J}=\sigma\mathbf{E}$

we have

$\nabla \cdot \mathbf{E} = \frac{\rho_\text{f}}{\epsilon}$
$\nabla \cdot \mathbf{B} = 0$
$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$
$\nabla \times \mathbf{B} = \mu\sigma\mathbf{E} + \mu\epsilon\frac{\partial \mathbf{E}}{\partial t}$

## Maxwell's equations for isotropic nonconducting media

In this case, the four equations become very symmetric in terms of E and H field:

$\nabla \cdot \mathbf{E} = 0$
$\nabla \cdot \mathbf{H} = 0$
$\nabla \times \mathbf{E} = -\mu\frac{\partial \mathbf{H}} {\partial t}$
$\nabla \times \mathbf{H} = \epsilon\frac{\partial \mathbf{E}} {\partial t}$

Also, when we assume the fields in the form of plane harmonic waves (replace ${\partial/\partial t}$ with $-i\omega$, and $\nabla$ with $i\mathbf{k}$):

$\mathbf{k} \cdot \mathbf{E} = 0$
$\mathbf{k} \cdot \mathbf{H} = 0$
$\mathbf{k} \times \mathbf{E} = \mu\omega\mathbf{H}$
$\mathbf{k} \times \mathbf{H} = -\epsilon\omega\mathbf{E}$

Above equations lead to the relationship between magnitudes of the fields:

$H={\epsilon\omega\over k}E = \epsilon v E = \sqrt{\epsilon\over\mu}E \equiv {E \over Z}$