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Square root of volume in fourier expansion of the vector potential

  1. Oct 10, 2012 #1
    Hi. I just wondered why we use a [itex] 1/\sqrt{V}[/itex] in the fourier expansion of the vector potential. A regular 3 dimensional fourier expansion is just

    [tex] f(\vec r) = \sum_{\vec k} c_\vec{k} e^{i \vec k \cdot \vec r}[/tex]

    but as the solution to the equation

    [tex] (\frac{\partial ^2}{\partial t^2} - \nabla^2 ) \vec A(\vec r,t) = 0[/tex]

    one usually writes

    [tex] \vec A(\vec r,t) = \frac{1}{\sqrt{V}}\sum_{\vec k} \vec A_{0 \vec{k}}(t) e^{i \vec k \cdot \vec r}.[/tex]

    What is the reason for this? Doesn't this also screw up the dimension of the euation when [itex]\vec A_{0 \vec k}[/itex] already has the same dimension as the vector potential since the plane wave solutions are

    [tex] \vec A_{0 \vec k}(t) e^{i \vec k \cdot \vec r}?[/tex]
     
  2. jcsd
  3. Oct 11, 2012 #2

    Jano L.

    User Avatar
    Gold Member

    It is just a convention. Perhaps you came across it in a book on quantum theory of radiation, where it is sometimes used because it leads to expression of the Poynting energy of the field in the box V in which V does not appear.

    If we used the standard convention

    [tex]
    \vec A(\vec r,t) = \sum_{\vec k} \vec A_{0 \vec{k}}(t) e^{i \vec k \cdot \vec r},
    [/tex]

    the integration over volume would introduce the prefactor V in the total Poynting energy.

    The dimension of [itex]\vec A(\vec r,t)[/itex] stays the same as in the standard convention, but the dimension of [itex]\vec A_{0\vec k}[/itex] is [itex][ A(\vec r,t) ] \mathrm{m}^{3/2} [/itex].
     
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