# Time-average Poynting vector

1. Jul 4, 2012

### Wox

I'm having a problem in expressing the time-average norm of the Poynting vector of the scattered electromagnetic field from a crystal, as expressed in several text books.

Concider a monochromatic plane wave
$$\begin{split} \bar{E}_{\text{mono}}(t,\bar{x})&=\bar{E}_{0}e^{i(\bar{k}\cdot\bar{x}-\omega t})\\ \bar{B}_{\text{mono}}(t,\bar{x})&=\frac{1}{\omega}\bar{k}\times \bar{E}_{\text{mono}}(t,\bar{x}) \end{split}$$
Then the scattered field of this wave from a crystal is given by (in the kinematic approximation)
$$\begin{split} \bar{E}_{\text{scat}}(t, \bar{x})&=\frac{r_{e}}{\|\bar{x}\|}.\|\bar{E}_{0}\| .K.((\hat{x}\cdot\hat{E}_{0})\hat{x}-\hat{E}_{0}).e^{i(\bar{k}_{\text{scat}}\cdot \bar{x}-\omega t)}\\ \bar{B}_{\text{scat}}(t,\bar{x})&=\frac{1}{\omega}\bar{k}_{\text{scat}}\times \bar{E}_{\text{scat}}(t,\bar{x})\\ K&=\sum_{j=1}^{N}{f_{j}e^{-i\bar{Q}\cdot(\bar{\Delta x}_{j}+\bar{\delta x}_{j}(t))}}\\ \bar{Q}&=\bar{k}_{\text{scat}}-\bar{k}\\ \bar{k}_{\text{scat}}&=\|\bar{k}\|.\hat{x}(t) \end{split}$$
where $\bar{\Delta x}_{j}$ the equilibrium positions of the atoms, $\bar{\delta x}_{j}(t)$ their thermal displacement, $r_{e}$ the classical electron radius and $f_{j}$ atomic scattering factors (which are complex numbers).

The Poynting vector and the derived intensity (time-averaged norm of the Poynting vector) are defined as
$$\begin{split} \bar{P}(t,\bar{x})=&\frac{1}{\mu_{0}}\mathcal{R}e( \bar{E})\times \mathcal{R}e( \bar{B})\\ I(\bar{x})=&\left<\|P\|\right>_{t}=\frac{1}{T}\int_{0}^{T} \|P\|dt \end{split}$$
where $T$ a period of time which is much longer than the period of the atomic vibrations and the period of the monochromatic plane wave. If we write $K=M_{K}e^{i\phi_{K}(t)}$ then this is written as
$$I(\bar{x})=c\epsilon_{0}.\frac{r_{e}^{2}}{\bar{x}^{2}}.\bar{E}_{0}^{2} .M_{K}^{2}.(1-(\hat{x}\cdot\hat{E}_{0})^{2}).\frac{1}{T}\int_{0}^{T} cos^{2}(\bar{k}\cdot\bar{x}-\omega t+\phi_{K}(t))dt$$
At least, that's how I would do it. But several textbooks write something else:
$$I(\bar{x})= c\epsilon_{0}.\frac{r_{e}^{2}}{\bar{x}^{2}}.\bar{E}_{0}^{2} .\left<K\right>_{t}^{2}.(1-(\hat{x}\cdot\hat{E}_{0})^{2}).\frac{1}{T}\int_{0}^{T} cos^{2}(\bar{k}\cdot\bar{x}-\omega t)dt$$
which allows use to write the simple expression (time average of the $cos^{2}(\ldots)$ is $1/2$)
$$I(\bar{x})=c\epsilon_{0}.\frac{r_{e}^{2}}{\bar{x}^{2}}.\bar{E}_{0}^{2} .\left<K\right>_{t}^{2}.\frac{(1-(\hat{x}\cdot\hat{E}_{0})^{2})}{2}$$
The $\left<K\right>_{t}^{2}$ term leads to the structure factor, the Laue interference function and the Debye-Waller factor. But I have no idea how the time average (an integral) can be done for the factor $K$ and the $cos^{2}(\ldots)$ separately. Can someone give me a hint or maybe a reference on this?

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