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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.

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  • $\begingroup$ 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. $\endgroup$
    – David Z
    Mar 2, 2012 at 6:34
  • $\begingroup$ @David: I want an expression that is the least computationally expensive. If that has a convolution, then so be it :-). $\endgroup$
    – JT_NL
    Mar 2, 2012 at 12:01
  • $\begingroup$ Convolution sucks, computationally; but there are always fourier transform tricks to save time. $\endgroup$
    – Colin K
    Mar 17, 2012 at 19:43

1 Answer 1

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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.

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