Difference between revisions of "Construction of a Tabletop Michelson Interferometer"

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* [[Huygens Principle for a Planar Source|Determining the angle of the diffraction minimum for a circular aperture]]
 
* [[Huygens Principle for a Planar Source|Determining the angle of the diffraction minimum for a circular aperture]]
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== Determining Angle of First Diffraction Minimum ==
 
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We start off with Maxwell's Equation in the Lorentz gauge:
 
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:<math>\square^2A^\mu(\mathbf{r},t) = \square^2A^\mu (r)=\mu j^\mu (r)</math><br><br>
 
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where we use the metric signature (+,+,+,-) and<br><br>
 
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:<math>A^\mu = (\mathbf{A},\frac{\Phi} {c})</math>,
 
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:<math>\square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part^2}{\part t^2}</math>
 
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:<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math><br><br>
 
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Lorentz Gauge: <math>\part_\mu A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}-\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math><br><br>
 
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Introduce Green's function at<math> (\mathbf{r},t)=r \quad</math> from some impulse source at<math> r'=(\mathbf{r}',t') \quad</math><br><br>
 
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<math>\square^2_rG(r,r')=\delta^4(r-r')</math><br><br>
 
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Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r e^{-iq\cdot r} G(r,0)</math><br><br>
 
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Then <math> G(r,0)=\frac{1}{(2\pi)^2} \int d^4qe^{iq\cdot r} \tilde{G}(q)</math><br><br>
 
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Translational symmetry implies:<br><br>
 
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<math>G(r-r',0)=G(r,r') \quad </math><br><br>
 
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&there4;<math> G(r,r')=\frac{1}{(2\pi)^2}\int d^4q e^{iq\cdot (r-r')} \tilde{G} (q)</math><br> <math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4qe^{iq\cdot (r-r')}\tilde{G}(q)</math><br><br>
 
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<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4qe^{iq\cdot (r-r')}(-k^2+\frac{\omega^2}{c^2})</math>, where <math>q=(\mathbf{k},\frac{\omega}{c}) \quad</math><br><br>
 
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But, <math>\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q e^{iq\cdot (r-r')}</math><br><br>
 
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&there4;<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br><br>
 
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<math>G(r,r')=\frac{-1}{(2\pi)^4} \int d^4qe^{iq\cdot (r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br><br>
 
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Chose the "retarded" solution, such that the function is zero unless t>t'<br><br>
 
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<math>G(r,r')=\frac{1}{(2\pi)^4}\int d^3ke^{i\mathbf{k}\cdot (r-r')}\int d(\frac{\omega}{c}) \frac{e^{-i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta(t-t')</math><br><br>
 
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<math>=\frac{1}{(2\pi)^4}\int d^3ke^{i\mathbf{k}\cdot (r-r')}(2\pi i  \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math><br><br>
 
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<math>=\frac{-2\pi}{(2\pi)^4}\int_0 \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-1}^1 dze^{ik|\mathbf{r}-\mathbf{r'}|z}\Theta</math><br><br>
 
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<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{|\mathbf{r}-\mathbf{r'|}}\right)2\int_0 dk \sin(ck(t-t')) \sin(k|\mathbf{r}-\mathbf{r'}|)\Theta</math><br><br>
 
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<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta</math><br><br>
 
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But the term <math>\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math><br><br>
 
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&there4;<math>  G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}</math><br><br>
 
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Now to get the <math>G_1(r,r')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> r_3=z=0  \quad</math> we take the difference:<br><br>
 
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<math>G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'+2z'\hat{e_3}|}\right)</math><br><br>
 
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Now use Green's theorem:<br><br>
 
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Let  <math>F^\mu=A(r)\part_\mu G_1(r,r')-G_1(r,r')\part_\mu A(r)</math><br><br>
 
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<math>\int \part_\mu F_\mu d^4r= \int cdt \int d^3r[\part_\mu A \part^\mu G+A\part_\mu \part^\mu G_1-\part_\mu G \part^\mu A -G_1\part_\mu \part^\mu A]</math><br><br>
 
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But  <math>\part_\mu \part^\mu G_1(r,r')=\delta^4(r-r')</math><br><br>
 
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<math>\part_\mu \part^\mu A(r)= \mu j(r)</math>, let  <math>j(r)=0 \quad</math><br><br>
 
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<math>\int \part_\mu F_\mu d^4r=A(r')</math><br><br>
 
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Now invoke the divergence theorem on the half space <math>z>0 \quad</math>:<br><br>
 
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<math>A(r')=-\int d^2r\int cdt\left[A(r)\frac{\part}{\part z}G_1(r,r')-G_1(r,r')\frac{\part}{\part z}A(r)\right]</math>, where the last term is zero by the condition of<math>G_1(z=0,r')=0 \quad</math><br><br>
 
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<math>A(r')=-c\int dt\int d^2rA(r)\frac{\part}{\part z}G_1(r,r')</math><br><br>
 
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To do the t integral, I need to bring out the z derivative.  To do this, I first turn it into a z' derivative, using the relation: <br><br><br>
 
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<math>G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)</math>, where <math>\mathbf{r}''=\mathbf{r}'-2z'\hat{e_3}</math><br><br>
 
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<math>\frac{\part}{\part z}G_1(r,r')=\frac{-1}{4\pi}\left(\frac{\part}{\part z}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)\right)</math><br><br>
 
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&there4; <math>A(r')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2r\left(2\frac{A(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c})}{|\mathbf{r}-\mathbf{r}'|}\right)</math><br><br>
 
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At <math>z=0 \quad </math>, <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}</math><br><br>
 
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If<math>A(\mathbf{r},t) \quad</math> is independent of position, as in a plane wave propagating along the z axis, then:<br><br>
 
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<math>A(r')=\frac{-\part}{\part z'}\int_{z'}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\mathbf{0}},t'-\frac{z'}{c}\right)</math><br><br>
 
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This gives us uniform translation of waves at velocity c.  More generally: <br><br>
 
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<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{r}, t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\right)</math><br><br>
 
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<math>=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|c}\frac{-z'}{|\mathbf{r}-\mathbf{r}'|}\right)</math><br><br>
 
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<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\left(\frac{A\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^3}(-z')+\frac{1}{c}\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}(-z')\right)</math><br><br>
 
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In our case, we consider only those waves which drop off as <math>\frac{1}{r'} \quad</math>, so:<br><br>
 
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<math>A(r')=\frac{1}{2\pi}\int_{z=0} d^2r\left(\frac{1}{c}\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}(z')\right)</math><br><br>
 
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<math>A(r')=\frac{z'}{2\pi c}\int_{z=0} d^2r\left(\frac{\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|^2}\right)</math><br><br>
 
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In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>.  Also, <math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A}(\mathbf{r},0)e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math>.  So:<br><br>
 
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<math>A(r')=\frac{z'\dot{A_0}}{2\pi c}\int_{z=0} rdrd\phi \frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</math><br><br>
 
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== Special Case ==
 
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Picture an opaque screen with a circular aperture of radius a.<br><br>
 
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Let<math>\mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}</math><br><br>
 
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Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br><br>
 
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<math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br><br>
 
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<math>=r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin\theta'</math><br><br>
 
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<math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math>
 

Revision as of 23:43, 3 July 2009

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