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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 16:08, 2 July 2009
, 16:08, 2 July 2009→Determining Angle for First Diffraction Minimum
<math>G(x-x',0)=G(x,x') \quad </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>\square^2_xG(x,x')=\frac{1}{(2\pi)^2}|int d^4qe^{-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><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>
<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>
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>
∴<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br>
∴<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br><br>
<math>G(x,x')=\frac{-1}{(2\pi)^4} \int d^4qe^{-iq(x-x')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br>
<math>G(x,x')=\frac{-1}{(2\pi)^4} \int d^4qe^{-iq(x-x')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br>
Chose the "retarded" solution, such that the function is zero unless t>t'<br>
Chose the "retarded" solution, such that the function is zero unless t>t'<br>
<math>G(x,x')=\frac{1}{(2\pi)^4}\int d^3ke^{-i\mathbf{k}(x-x')}\int d(\frac{\omega}{c}) \frac{e^{i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta</math><br>
<math>G(x,x')=\frac{1}{(2\pi)^4}\int d^3ke^{-i\mathbf{k}(x-x')}\int d(\frac{\omega}{c}) \frac{e^{i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta</math><br><br>
<math>=\frac{1}{(2\pi)^4}\int d^3ke^{-i\mathbf{k}(x-x')}(2\pi i \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k)})\Theta
<math>=\frac{1}{(2\pi)^4}\int d^3ke^{-i\mathbf{k}(x-x')}(2\pi i \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math><br><br>
<math>=\frac{-2\pi}{(2\pi)^4}\int \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-i}^i dze^{-ik|\mathbf{x}-\mathbf{x'}|z}\Theta</math><br><br>
</math>
<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{2i|\mathbf{x}-\mathbf{x'|}}\right)2\int dk sin(ck(t-t')) sin(k|\mathbf{x}-\mathbf{x'})\Theta</math><br><br>