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Time-average Poynting vector of crystal scattering

  1. Jul 5, 2012 #1

    Wox

    User Avatar

    To calculate the intensity of the scattered radiation from a crystal after irradiating with X-rays, one can add up all electromagnetic fields of the oscillating electrons (calculated using the Liénard–Wiechert potential). Taking the time-average of the norm of the Poynting vector of the scattered em-field, leads to an expression for the resulting scattered intensity. If the incident radiation is an electromagnetic plane wave with electric field
    [tex]
    \bar{E}(t,\bar{x})=\bar{E}_{0}e^{-i(\bar{k}\cdot\bar{x}-\omega t)}
    [/tex]
    then the term [itex]\cos^{2}(\bar{k}\cdot\bar{x}-\omega t+c^{te})[/itex] is the only time dependent term of the Poynting vector and its average is [itex]1/2[/itex]. However, the electrons are moving in the atom, the atoms undergo thermal motion and the [itex]\bar{E}_{0}[/itex] is also time dependent (e.g. unpolarized light). Therefore the appear more time-dependent terms in the Poynting vector. This is implicitly dealt with in many text books by considering the time average of each of these processes separately. The time average of the electron motion leads to the atomic scattering factor, the time average of the atomic thermal motion leads to the Debye Waller factor and the time average of [itex]\bar{E}_{0}[/itex] leads to the polarization factor. But why can these be averaged independently? Mathematically, it is like saying that "the integral of a product of functions" is "the product of the integrals of these functions". When is this a valid approximation?
     
  2. jcsd
  3. Jul 5, 2012 #2
    If you are looking at normal Bragg peaks, the only thing that counts is that thermally displaced atoms contribute less to the scattered intensity.

    If you are looking at other positions in reciprocal space you have to be more careful. And as you say, the usual approximations do not hold. The end result is what is called "thermal diffuse scattering" that can give information about phonons and other things.

    http://www.ideals.illinois.edu/handle/2142/18627

    Even at Bragg peaks you can get odd effects. These are fairly exotic, so you don't really have to worry about them.

    http://www.esrf.eu/UsersAndScience/Publications/Highlights/2003/HRRS/HRRS13/
     
  4. Jul 6, 2012 #3

    Wox

    User Avatar

    Thermal diffuse scattering and the lowering of the Bragg peak intensities due to thermal motion (Debye Waller factor) is all derived from stating that the scattered intensity (which is the time-averaged norm of the Poynting vector) given by
    [tex]
    I=c^{te}\left< K.K^{\ast}\right>_{t}
    [/tex]
    where
    [tex]
    K=\sum_{j}f_{j}e^{i\bar{Q}\cdot(\bar{x}_{j}+ \bar{\delta x}_{j}(t))}
    [/tex]
    with [itex]\bar{x}_{j}[/itex] the equilibrium position of the atom, [itex]\bar{\delta x}_{j}(t)[/itex] the thermal displacement and [itex]K^{\ast}[/itex] the complex conjugate. However, I don't see how one can arrive at this from the definition of the Poynting vectors and its intensity:
    [tex]
    I=\left< \|\bar{P}\|\right>_{t}=\frac{1}{\mu_{0}}\left<\| \mathcal{R}e(\bar{E})\times\mathcal{R}e(\bar{B})\|\right>_{t}
    [/tex]
    If we know that the instantanious scattered field is given by
    [tex]
    \bar{E}(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)}
    [/tex]
    [tex]
    \bar{B}(t,\bar{x})=\frac{1}{\omega} \bar{k}_{\text{scat}}\times \bar{E}(t,\bar{x})
    [/tex]
    [tex]
    \bar{k}_{\text{scat}}=\|\bar{k}\|\hat{x}
    [/tex]
    where [itex]\bar{k}[/itex] the wave vector of the incoming wave and [itex]\bar{k}_{\text{scat}}[/itex] the wave vector of the scattered wave, then the intensity is given by

    [tex]
    I=c\epsilon_{0}.\frac{r_{e}^{2}}{\bar{x}^{2}}.\bar{E}_{0}^{2} .(1-(\hat{x}\cdot\hat{E}_{0})^{2}). \left< S^{2}\right>_{t}
    [/tex]
    [tex]
    S=\mathcal{R}e(K)cos(\bar{k}_{\text{scat}}\cdot \bar{x}-\omega t)-\mathcal{I}m(K)sin(\bar{k}_{\text{scat}}\cdot\bar{x}-\omega t)
    [/tex]
    Therefore, one should be able to write
    [tex]
    \left< S^{2}\right>_{t}=c^{te}\left< K.K^{\ast}\right>_{t}
    [/tex]
    Since [itex]\left< S^{2}\right>_{t}[/itex] is written as
    [tex]
    \left< S^{2}\right>_{t}=\left< (\mathcal{R}e(K))^{2}\cos^{2}( \ldots)\right>_{t}+\left< (\mathcal{I}m(K))^{2}\sin^{2}(\ldots)\right>_{t}-2\left< \mathcal{R}e(K)\mathcal{I}m(K)\cos(\ldots)\sin( \ldots)\right>_{t}
    [/tex]
    this can only be possible when we write
    [tex]
    \left< (\mathcal{R}e(K))^{2}\cos^{2}( \ldots)\right>_{t}\approx\left< (\mathcal{R}e(K))^{2}\right>_{t}\left< \cos^{2}( \ldots)\right>_{t}=\frac{1}{2}\left< (\mathcal{R}e(K))^{2}\right>_{t}
    [/tex]
    [tex]
    \left< (\mathcal{I}m(K))^{2}\sin^{2}(\ldots)\right>_{t} \approx\left< (\mathcal{I}m(K))^{2}\right>_{t}\left<\sin^{2}( \ldots)\right>_{t}=\frac{1}{2}\left< (\mathcal{I}m(K))^{2}\right>_{t}
    [/tex]
    [tex]
    -2\left< \mathcal{R}e(K)\mathcal{I}m(K)\cos(\ldots)\sin( \ldots)\right>_{t} \approx -2\left< \mathcal{R}e(K)\mathcal{I}m(K)\right>_{t}\left< \cos(\ldots)\sin( \ldots)\right>_{t}=0
    [/tex]
    so that
    [tex]
    \left< S^{2}\right>_{t}= \frac{1}{2}\left< (\mathcal{R}e(K))^{2} + (\mathcal{I}m(K))^{2}\right>_{t}=\frac{1}{2}\left< K.K^{\ast} \right>_{t}
    [/tex]
    However, I'm not sure how to justify the approximations made (I have never seen them made explicitly in text books, but according to the reasoning from above, they are made implicitly). After all they state that the integral of a product is the product of integrals, as the time average is an integral over time.
     
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