# Maxwell Equations

## Differential Form

Maxwell's equations are,

$\ \nabla \cdot \vec{D} = \rho$

$\ \nabla \cdot \vec{B} = 0$

$\ \nabla \times \vec{E} = \frac{- \partial \vec{B}}{\partial t}$

$\ \nabla \times \vec{H} = \frac{\partial \vec{D}}{\partial t} + \vec{J}.$

Where the vectors are defined in the usual way. Also,

$\ \vec{D} = \epsilon_0 \vec{E} + \vec{P}$ and,

$\ \vec{B} = \mu_0 \vec{H} + \vec{M}.$

Taking a Fourier transform with respect to the time variable, we have,

$\ \nabla \cdot \vec{\hat{D}} = \rho$

$\ \nabla \cdot \vec{\hat{B}} = 0$

$\ \nabla \times \vec{\hat{E}} = -\dot{\imath} \omega \vec{\hat{B}}$

$\ \nabla \times \vec{\hat{H}} = \dot{\imath} \omega \vec{\hat{D}} + \vec{\hat{J}}.$

See Electromagnetics" for more.

## Integral Form

The integral form is also commonly given.

$\ \oint \nabla \cdot \vec{D} dV = \oint \vec{D} \cdot d\vec{S} = Q_{enclosed}$

$\ \oint \nabla \cdot \vec{B} dV = \oint \vec{B} \cdot d\vec{S} = 0$

$\ \int \nabla \times \vec{E} \cdot d\vec{S} = \oint \vec{E} \cdot d\vec{l} = \frac{- \partial }{\partial t} \int \vec{B}\cdot d\vec{S}$

$\int \nabla \times \vec{H} \cdot d\vec{S}= \oint \vec{H} \cdot d\vec{l} = \frac{\partial }{\partial t}\int \vec{D} \cdot \vec{S} + I$

## Power Flow into a Volume

The Poynting vector gives the direction of power flow. The Poynting vector is given by,

$\ \vec{\mathcal{S}} = \vec{E} \times \vec{H} .$

It has the units, watts per square meter.

The power flow through a differential area id given by $\ -\vec{\mathcal{S}}\cdot d\vec{S}$. The power flow through a volume is then given by summing up the power flow through the area enclosing that volume, thus:

$\ P = -\oint \vec{\mathcal{S}} \cdot d\vec{S} = -\int \nabla \cdot \vec{\mathcal{S}} dV .$

Expanding by Maxwell's equations,

$\ P = -\int \nabla \cdot \vec{E} \times \vec{H} dV$

$\ = -\int \vec{H} \cdot (\nabla \cdot \vec{E}) - \vec{E} \cdot (\nabla \cdot \vec{H}) dV$

$\ = -\int \vec{H} \cdot \left(-\frac{\partial}{\partial t} (\mu_0 \vec{H} + \vec{M})\right) - \vec{E} \cdot \left(\frac{\partial}{\partial t} (\epsilon_0 \vec{E} + \vec{P}) + \vec{J}\right) dV$

$\ = \int \vec{H} \cdot \left(\frac{\partial \mu_0 \vec{H}}{\partial t} + \frac{\partial \vec{M}}{\partial t} \right) + \vec{E} \cdot \left(\frac{\partial\epsilon_0 \vec{E}}{\partial t} + \frac{\partial \vec{P}}{\partial t}\right) + \vec{E} \cdot \vec{J} dV$

$\ = \int \frac{1}{2}\left(\mu_0 \frac{\partial |\vec{H}|^2}{\partial t} + \epsilon_0 \frac{\partial\ |\vec{E}|^2}{\partial t}\right) + \vec{H} \cdot \frac{\partial \vec{M}}{\partial t} + \vec{E} \cdot \frac{\partial \vec{P}}{\partial t} + \vec{E} \cdot \vec{J} dV .$

The first summand is the power density of the vacuum electromagnetic field; the second and third summands, when integrated, are the power dissipation by magnetic and electric dipoles; the fourth summand, when integrated is power dissipation by current.

Reference: ECE525 Lasers and Detectors. notes taken in class, 2010. Instructor: Joyce Poon.