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I Divergence and transversal extension integral definitions

  1. Apr 6, 2016 #1
    Hi. I am reading a paper about gaussian beams and the author says that gaussian beams have simultaneously minimal divergence and minimal transversal extension. In order to prove it, the author states that

    [itex]\mathrm{divergenece} \propto \int_{-\infty}^{+\infty} \frac{d\,k_{x}}{2\pi} \int_{-\infty}^{+\infty} \frac{d\,k_{y}}{2\pi} (k_{x}^{2}+k_{y}^{2})|A(k_{x},k_{y})|^2[/itex]

    and

    [itex]\mathrm{transversal \, extension} \propto \int_{-\infty}^{+\infty} d\,x \int_{-\infty}^{+\infty} d\,y (x^{2}+y^{2})|E|^2[/itex],

    where

    [itex]E = \int \frac{d\,k_{x}}{2\pi} \int \frac{d\,k_{y}}{2\pi} A(k_{x},k_{y}) \exp[i\,k_{x}\,x + i\,k_{y}\,y+i \, \sqrt{k^2-k_{x}^2-k_{y}^2} \, z] [/itex]

    is the electric field amplitude and [itex]A(k_{x},k_{y})[/itex] is the amplitude distribution.

    I look it up and I couldn't find these integral definitions of divergence and transversal extension. Does anybody have a source on these? Or could at least give me an idea about how they came to be defined this way?

    Thank you very much.
     
  2. jcsd
  3. Apr 6, 2016 #2

    blue_leaf77

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    The divergence of a beam of light is best characterized by the width of the spatial frequency spectrum ##|A(k_x,k_y)|^2##. The reason being the spatial frequency ##(k_x,k_y)## is proportional to the angle the corresponding plane wave component (a pair of spatial frequency coordinates ##(k_x,k_y)## describes a single plane wave component out of the plane wave bundle making up the beam) deviates from the beam's propagation direction. The wider ##|A(k_x,k_y)|^2## is, the more plane wave components propagating at larger angles are contained within that beam.

    The particular equation you have there is actually the formula for the width of the spectrum in terms of its variance in a radial direction in ##(k_x,k_y)## plane, i.e. ##k^2=k_{x}^{2}+k_{y}^{2}##. Remember that if you have a continuous probability distribution ##P(x)## of a continuous variable ##x##, you can find its variance from the mean square and the mean value of ##P(x)## where the mean square is given by
    $$
    \langle x^2 \rangle = \int_{-\infty}^\infty x^2 P(x) dx
    $$
    ,the mean is
    $$
    \langle x \rangle = \int_{-\infty}^\infty x P(x) dx.
    $$
    and the variance ##\sigma## is
    $$
    \sigma = \langle x^2 \rangle-\langle x \rangle^2
    $$
    In the equation of the Gaussian beam above, you take the analogy between ##P(x)## and ##|A(k_x,k_y)|^2## and use the above formula to calculate the mean square and the mean value of ##|A(k_x,k_y)|^2##. The mean value is zero because ##|A(k_x,k_y)|^2## is symmetric and centered about ##k_x=k_y=0## and hence the divergence, represented as the variance of ##|A(k_x,k_y)|^2## is equal to its mean square
    $$
    \langle k^2 \rangle = \int_{-\infty}^{+\infty} \frac{d\,k_{x}}{2\pi} \int_{-\infty}^{+\infty} \frac{d\,k_{y}}{2\pi} k^2|A(k_{x},k_{y})|^2
    $$
    Exactly the same principle about calculating the variance like the one above, but this time applied to the beam in the position plane.
     
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