- #1
Wox
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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 textbooks.
Concider a monochromatic plane wave
[tex]\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}[/tex]
Then the scattered field of this wave from a crystal is given by (in the kinematic approximation)
[tex]\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}[/tex]
where [itex]\bar{\Delta x}_{j}[/itex] the equilibrium positions of the atoms, [itex]\bar{\delta x}_{j}(t)[/itex] their thermal displacement, [itex]r_{e}[/itex] the classical electron radius and [itex]f_{j}[/itex] atomic scattering factors (which are complex numbers).
The Poynting vector and the derived intensity (time-averaged norm of the Poynting vector) are defined as
[tex]\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}[/tex]
where [itex]T[/itex] 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 [itex]K=M_{K}e^{i\phi_{K}(t)}[/itex] then this is written as
[tex]
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
[/tex]
At least, that's how I would do it. But several textbooks write something else:
[tex]
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
[/tex]
which allows use to write the simple expression (time average of the [itex]cos^{2}(\ldots)[/itex] is [itex]1/2[/itex])
[tex]
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}
[/tex]
The [itex]\left<K\right>_{t}^{2}[/itex] 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 [itex]K[/itex] and the [itex]cos^{2}(\ldots)[/itex] separately. Can someone give me a hint or maybe a reference on this?
Concider a monochromatic plane wave
[tex]\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}[/tex]
Then the scattered field of this wave from a crystal is given by (in the kinematic approximation)
[tex]\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}[/tex]
where [itex]\bar{\Delta x}_{j}[/itex] the equilibrium positions of the atoms, [itex]\bar{\delta x}_{j}(t)[/itex] their thermal displacement, [itex]r_{e}[/itex] the classical electron radius and [itex]f_{j}[/itex] atomic scattering factors (which are complex numbers).
The Poynting vector and the derived intensity (time-averaged norm of the Poynting vector) are defined as
[tex]\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}[/tex]
where [itex]T[/itex] 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 [itex]K=M_{K}e^{i\phi_{K}(t)}[/itex] then this is written as
[tex]
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
[/tex]
At least, that's how I would do it. But several textbooks write something else:
[tex]
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
[/tex]
which allows use to write the simple expression (time average of the [itex]cos^{2}(\ldots)[/itex] is [itex]1/2[/itex])
[tex]
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}
[/tex]
The [itex]\left<K\right>_{t}^{2}[/itex] 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 [itex]K[/itex] and the [itex]cos^{2}(\ldots)[/itex] separately. Can someone give me a hint or maybe a reference on this?