Poynting's theorem

From Physics wiki

Jump to: navigation, search

In differential form, Poynting's theorem relates the energy flux of the electromagnetic field to the rate of work done on the charges:

\frac{\partial u}{\partial t} + \boldsymbol{\nabla}\cdot \mathbf{S} = -\mathbf{J}\cdot\mathbf{E}\,,

where

u = \frac{1}{2\mu_0} B^2 + \frac{\epsilon_0}{2} E^2\,

is the energy density of the fields, and

\mathbf{S} = \frac{1}{\mu_0} \mathbf{E}\times\mathbf{B}\,

is the energy flux, or Poynting vector, of the fields. In it's integral form

\frac{\partial}{\partial t} \int_\mathcal{V}\!dV\, u + \int_{\mathcal{S}=\partial\mathcal{V}}\!dS\,\, \mathbf{S} \cdot \hat\mathbf{n} = -\int_{\mathcal{V}}\!dV\, \mathbf{J}\cdot\mathbf{E}\,.

Derivation

The rate of work done on a single charge by the electromagnetic field is

P = \mathbf{v}\cdot \mathbf{f} = q \mathbf{v}\cdot \left( \mathbf{E} + \mathbf{v}\times\mathbf{B}\right) = q \mathbf{v} \cdot \mathbf{E}\,,

so that for a charge distribution it is

P = \int_\mathcal{V}\!dV\, \mathbf{J}\cdot \mathbf{E}\,.

We may write this in terms of the fields using Maxwell's equations and using \boldsymbol{\nabla} \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\boldsymbol{\nabla} \times \mathbf{A}) - \mathbf{A} \cdot (\boldsymbol{\nabla} \times \mathbf{B}) ,

\int_\mathcal{V}\!dV\, \mathbf{J}\cdot \mathbf{E}\, =\int_\mathcal{V}\!dV\, \left(\frac{1}{\mu_0} \boldsymbol{\nabla}\times\mathbf{B} - \epsilon_0 \frac{\partial\mathbf{E}}{\partial t} \right)\cdot\mathbf{E}\,,
=\int_\mathcal{V}\!dV\, \left(-\frac{1}{\mu_0} \boldsymbol{\nabla} \cdot (\mathbf{E}\times\mathbf{B}) + \frac{1}{\mu_0} \mathbf{B} \cdot (\boldsymbol{\nabla}\times\mathbf{E}) - \epsilon_0 \frac{\partial\mathbf{E}}{\partial t} \cdot\mathbf{E} \right)\,,
=\int_\mathcal{V}\!dV\, \left(-\frac{1}{\mu_0} \boldsymbol{\nabla} \cdot (\mathbf{E}\times\mathbf{B}) - \frac{1}{\mu_0} \mathbf{B} \cdot \frac{\partial \mathbf{B}}{\partial t} - \epsilon_0 \frac{\partial\mathbf{E}}{\partial t} \cdot\mathbf{E} \right)\,,
=-\frac{\partial}{\partial t} \int_\mathcal{V}\!dV\, \left( \frac{1}{2\mu_0} B^2 + \frac{\epsilon_0}{2} E^2 \right) - \int_{\mathcal{S}=\partial\mathcal{V}}\!dS\, \frac{1}{\mu_0} (\mathbf{E}\times\mathbf{B})\cdot \hat\mathbf{n}\,.

As a differential statement,

\frac{\partial u}{\partial t} + \boldsymbol{\nabla}\cdot \mathbf{S} = -\mathbf{J}\cdot\mathbf{E}\,.
on to Maxwell stress tensor
Retrieved from "http://www.physics.thetangentbundle.net/wiki/Classical_electrodynamics/Poynting%27s_theorem"
Personal tools