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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 15:37, 2 July 2009
, 15:37, 2 July 2009→Determining Angle for First Diffraction Minimum
Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math>
Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math>
Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')
Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')<br>
<math>\square^2_xG(x,x')=\delta^4(x-x')</math>
<math>\square^2_xG(x,x')=\delta^4(x-x')</math>
In free space, translational symmetry implies:<br>
In free space, translational symmetry implies:<br>
<math>G(x-x',0)=G(x,x')</math>
<math>G(x-x',0)=G(x,x') \quad </math>
∴<math> G(x,x')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(x-x')} \tilde{G} (q)</math><br>
∴<math> G(x,x')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(x-x')} \tilde{G} (q)</math><br> <math>\square^2_xG(x,x')=\frac{1}{(2\pi)^2}|int d^4qe^{-iq(x-x')}\tilde{G}(q)</math><br>
<math>\square^2_xG(x,x')=\frac{1}{(2\pi)^2}\int d^4qe^{-iq(x-x')}(-k^2+\frac{\omega^2}{c^2})</math>, where <math>q=(\mathbf{k},\frac{\omega}{c}) \quad</math><br>
But, <math>\square^2_xG(x,x')=\delta^4(x-x')=\frac{1}{(2\pi)^4}\int d^4q e^{-iq(x-x')}</math><br>
&there4<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br>