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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 17:48, 2 July 2009
, 17:48, 2 July 2009→Determining Angle for First Diffraction Minimum
<math>G_1(x,x')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'|}-\frac{\delta(|\mathbf{x}-\mathbf{x}'-2z\hat{e_3}|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'-2z\hat{e_3}|}\right)</math><br><br>
<math>G_1(x,x')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'|}-\frac{\delta(|\mathbf{x}-\mathbf{x}'-2z\hat{e_3}|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'-2z\hat{e_3}|}\right)</math><br><br>
Now use Green's theorem:<br>
Now use Green's theorem:<br>
Let<math>\mathbf{F}=A(x)\mathbf{\nabla}G_1(x,x')-G_1(x,x')\mathbf{\nabla}A(x)
Let <math>\mathbf{F}=A(x)\mathbf{\nabla}G_1(x,x')-G_1(x,x')\mathbf{\nabla}A(x)</math><br><br>
<math>\int \mathbf{\nabla} \cdot \mathbf{F}d^4x= \int cdt \int d^3x[\mathbf{\nabla}A \cdot \mathbf{\nabla}G+A\nabla^2G_1-\mathbf{\nabla}G \cdot \mathbf{\nabla}A -G_1\nabla^2A]</math><br><br>
</math>
But <math>\nabla^2G_1(x,x')=\delta^4(x-x')+\frac{1}{c^2}\frac{\part^2}{\part t^2}
G_1(x,x')</math><br><br>