Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My professor worked out the Poynting Vector without using the Poynting Theorem-I can't remember how. I can't seem to figure out where the cross product would come in.

share|cite|improve this question
The Poynting vector is pretty much a definition. We can't guess how your professor might have done it, but perhaps if you include what you've tried here, we could give you some guidance? – David Z May 15 '13 at 2:13

You can derive the Poynting vector from the Lorentz force law, $\vec F = Q(\vec E + \vec v \times \vec B)$ like this:

(using the work-energy theorem)

$dW = dq(\vec E + \vec v \times \vec B)\cdot dl = dq(\vec E + \vec v \times \vec B)\cdot \vec v dt$


$\frac{dW}{dt} = \int_\tau(\vec E + \vec v \times \vec B)\cdot \vec v\rho d^3r = \int_\tau (\vec E \cdot \vec J)d^3r$.

Ampere's law says:

$\vec J = \frac{1}{\mu_0}(\vec \nabla \times \vec B) - \epsilon_0 \frac{\partial \vec E}{\partial t}$


$\vec E \cdot J = \vec E \cdot[\frac{1}{\mu_0}(\vec \nabla \times \vec B) -\epsilon_0\frac{\partial \vec E}{\partial t}] =$

(using BAC-CAB) identity)

$\frac{1}{\mu_0}[-\vec \nabla \cdot(\vec E \times \vec B) + \vec B \cdot(\vec \nabla \times \vec E)] - \vec E \cdot \epsilon_0\frac{\partial \vec E}{\partial t} =$

(using Faraday's law)

$\frac{1}{\mu_0}[-\vec \nabla \cdot(\vec E \times \vec B) - \vec B \cdot \frac{\partial \vec B}{\partial t}] - \vec E \cdot \epsilon_0\frac{\partial \vec E}{\partial t} =$

$-\frac{1}{2}\frac{\partial}{\partial t}(\epsilon_0 E^2 + \frac{B^2}{\mu_0}) - \nabla \cdot \frac{\vec E \times \vec B}{\mu_0}$.

Substituting this in the expression for $\frac{dW}{dt}$ and using the divergence theorem we get:

$\frac{dW}{dt} = -\frac{d}{dt}\int\frac{1}{2}(\epsilon_0 E^2 + \frac{B^2}{\mu_0})d^3r - \oint\frac{(\vec E \times \vec B)}{\mu_0}$.

But $\frac{\epsilon_0 E^2}{2}$ $\frac{B^2}{2\mu_0}$ are just the energies per unit volume of the electric and magnetic fields, respectively. Setting the surface integral on the right hand side to 0 for the time being, what the left hand side of the equation is saying is that the rate that the charges in the volume gain energy is the same as the rate that the electric and magnetic fields in the volume lose energy.

If the sum of the electric and magnetic fields is constant in time, however, then $-\frac{d}{dt}\int\frac{1}{2}(\epsilon_0 E^2 + \frac{B^2}{\mu_0})d^3r = 0$. If $\frac{dW}{dt}$ is not identically 0, however, this must mean that there is a flux of energy into or out of the volume to sustain the increase or decrease in energy. This flux is precisely the integrand in right hand side, or $\frac{(\vec E \times \vec B)}{\mu_0}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.