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

I have the following decomposition for the electric component of light:

$$\renewcommand{\vec}[1]{\mathbf{#1}}\vec{E}(\vec r)=\frac1{4\pi^2} \iint_\Omega \vec A(k_x, k_y) \mathrm{e}^{i \vec{k} \cdot \vec{r}} \, \mathrm{d}k_x \mathrm{d}k_y.$$

Similarly, the magnetic field is: $$\renewcommand{\vec}[1]{\mathbf{#1}}\vec{H}(\vec r)=\frac1{4\pi^2} \iint_\Omega \frac{\vec k}{\omega \mu_0} \times \vec A(k_x, k_y) \mathrm{e}^{i \vec{k} \cdot \vec{r}} \, \mathrm{d}k_x \mathrm{d}k_y.$$

Fine, this is the setting. Now I wish to compute the average Poynting vector $\langle \vec S \rangle$:

$$\langle \vec S \rangle = \frac12 \operatorname{Re} [\vec E(\vec r) \times \vec H(\vec r)^*].$$ Is there a way to express $\langle \vec S \rangle$ in a nice form? Integrals of functions of $\vec A$ for example. I get horrendous expressions with convolutions that don't give me a nice compact formula.

I know that this might actually be more mathematics than physics, but it might be possible that the physical boundary conditions give a better solution.

share|cite|improve this question
I'll check Jackson's electrodynamics book and see if it says anything relevant, though if you are set on expressing $\langle S\rangle$ in terms of the momentum-space vector potential, it's possible you might not be able to do any better than a convolution. – David Z Mar 2 '12 at 6:34
@David: I want an expression that is the least computationally expensive. If that has a convolution, then so be it :-). – Jonas Teuwen Mar 2 '12 at 12:01
Convolution sucks, computationally; but there are always fourier transform tricks to save time. – Colin K Mar 17 '12 at 19:43

A way to simplify the expressions would be to calculate $\bf S$ in momenta space: $$ \newcommand{\dd}{\mathrm{d}} \newcommand{\bb}[1]{{\bf #1}} \bb{S}_p({\bf k}'') = \int e^{-i\bb{k}''\bb{r}}\dd \bb{r}\left(\dfrac{1}{2}\mathrm{Re}(\bb{E}(\bb{r})\times\bb{H}(\bb{r})^*)\right)\\ =\dfrac{1}{4}\int e^{-i\bb{k}''\bb{r}}\dd \bb{r}\left(\bb{E}(\bb{r})\times\bb{H}(\bb{r})^*+\bb{E}(\bb{r})^*\times\bb{H}(\bb{r})\right)\\ =\dfrac{1}{4}\dfrac{1}{16\pi^4 \omega \mu_0}\int \dd \bb{r} \int \dd \bb{k}\int \dd \bb{k}'e^{-i\bb{k}''\bb{r}}[\left((\bb{A}(\bb{k})e^{i\bb{k}\bb{r}})\times(\bb{k}'\times \bb{A}(\bb{k'})^*e^{-i\bb{k'}\bb{r}})+\\ +(\bb{A}(\bb{k})^*e^{-i\bb{k}\bb{r}})\times(\bb{k}'\times \bb{A}(\bb{k'})e^{i\bb{k'}\bb{r}})\right)] $$ Then we notice that the integration over $\bb{r}$ kills the exponents and provides $\delta$-function terms: $$ \bb{S}_p({\bf k}'')=\dfrac{1}{4}\dfrac{1}{4\pi^2 \omega \mu_0} \int \dd \bb{k}\int \dd \bb{k}'[\left(\bb{A}(\bb{k})\times(\bb{k}'\times \bb{A}(\bb{k'})^*)\delta(\bb{k}-\bb{k}'-\bb{k}'')+\\ +\bb{A}(\bb{k})^*\times(\bb{k}'\times \bb{A}(\bb{k'}))\delta(\bb{k}'-\bb{k}-\bb{k}'')\right)]\\ =\dfrac{1}{4}\dfrac{1}{4\pi^2 \omega \mu_0} \int \dd \bb{k}'\left[\bb{A}(\bb{k}'+\bb{k}'')\times(\bb{k}'\times \bb{A}(\bb{k'})^*)+\bb{A}(\bb{k}'-\bb{k}'')^*\times(\bb{k}'\times \bb{A}(\bb{k'}))\right] $$

You could potentially simplify this slightly further, were there any relations between $\bb{A}$ and $\bb{k}$. Going back to the coordinate expression for $\bb{S}$ will require another integration and will be no simpler than the original form you had given.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.