Construction of a Tabletop Michelson Interferometer

Determining Angle for First Diffraction Minimum

We start off with Maxwell's Equation in the Lorentz gauge: $\square ^{2}A^{\mu }(\mathbf {r} ,t)=\square ^{2}A^{\mu }(r)=(-\mu _{1}j^{\mu }(r))$

Where:

$A^{\mu }=(\mathbf {A} ,{\frac {\Phi }{c}}),\square ^{2}=\partial _{\mu }\partial ^{\mu }=\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial }{\partial t^{2}}}$ $j^{\mu }=(\mathbf {j} ,c\rho ),\partial _{\mu }=(\mathbf {\nabla } ,{\frac {1}{c}}{\frac {\partial }{\partial t}})$

Lorentz Gauge: $A^{\mu }=0\Rightarrow \mathbf {\nabla } \cdot \mathbf {A} +{\frac {1}{c^{2}}}{\frac {\partial \Phi }{\partial t}}=0$

Introduce Green's function at$(\mathbf {r} ,t)=r\quad$ from some impulse source at$r'=(\mathbf {r} ',t')\quad$

$\square _{r}^{2}G(r,r')=\delta ^{4}(r-r')$

Let ${\tilde {G}}(q)={\frac {1}{(2\pi )^{2}}}\int d^{4}re^{iqr}G(r,0)$

Then $G(q)={\frac {1}{(2\pi )^{2}}}\int d^{4}qe^{iqr}{\tilde {G}}(r,0)$

In free space, translational symmetry implies:

$G(r-r',0)=G(r,r')\quad$

∴$G(r,r')={\frac {1}{(2\pi )^{2}}}\int d^{4}qe^{-iq(r-r')}{\tilde {G}}(q)$
$\square _{r}^{2}G(r,r')={\frac {1}{(2\pi )^{2}}}|intd^{4}qe^{-iq(r-r')}{\tilde {G}}(q)$

$\square _{r}^{2}G(r,r')={\frac {1}{(2\pi )^{2}}}\int d^{4}qe^{-iq(r-r')}(-k^{2}+{\frac {\omega ^{2}}{c^{2}}})$, where $q=(\mathbf {k} ,{\frac {\omega }{c}})\quad$

But, $\square _{r}^{2}G(r,r')=\delta ^{4}(r-r')={\frac {1}{(2\pi )^{4}}}\int d^{4}qe^{-iq(r-r')}$

∴${\tilde {G}}(q)={\frac {(2\pi )^{2}}{(2\pi )^{4}}}{\frac {1}{-q^{2}}}={\frac {-1}{(2\pi )^{2}q^{2}}}$

$G(r,r')={\frac {-1}{(2\pi )^{4}}}\int d^{4}qe^{-iq(r-r')}{\frac {1}{(k+{\frac {\omega }{c}})(k-{\frac {\omega }{c}})}}$

Chose the "retarded" solution, such that the function is zero unless t>t'

$G(r,r')={\frac {1}{(2\pi )^{4}}}\int d^{3}ke^{-i\mathbf {k} (r-r')}\int d({\frac {\omega }{c}}){\frac {e^{i\omega (t-t')}}{({\frac {\omega }{c}}-k)({\frac {\omega }{c}}+k)}}\Theta$

$={\frac {1}{(2\pi )^{4}}}\int d^{3}ke^{-i\mathbf {k} (r-r')}(2\pi i{\frac {e^{ick(t-t')}-e^{-ick(t-t')}}{2k}})\Theta$

$={\frac {-2\pi }{(2\pi )^{4}}}\int {\frac {k^{2}dk}{k}}\sin \left({ck(t-t')}\right)2\pi \int _{-i}^{i}dze^{-ik|\mathbf {r} -\mathbf {r'} |z}\Theta$

$={\frac {-1}{(2\pi )^{2}}}\left({\frac {1}{2i|\mathbf {r} -\mathbf {r'|} }}\right)2\int dksin(ck(t-t'))sin(k|\mathbf {r} -\mathbf {r'} |)\Theta$

$={\frac {1}{(2\pi )^{2}}}{\frac {2}{|\mathbf {r} -\mathbf {r} '|}}{\frac {2\pi }{4}}\left[2\delta (|\mathbf {r} -\mathbf {r} '|+c(t-t'))-2\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))\right]\Theta$

But the term $2\delta (|\mathbf {r} -\mathbf {r} '|+c(t-t'))\rightarrow 0\quad \forall \quad t>t'$

∴$G(r,r')={\frac {-1}{4\pi }}\quad {\frac {\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))}{|\mathbf {r} -\mathbf {r} '|}}$

Now to get the $G_{1}(r,r')\quad$ in the half-space with z>0 with the boundary condition $G_{1}\quad$ at$r_{3}=z=0\quad$ we take the difference:

$G_{1}(r,r')={\frac {-1}{4\pi }}\left({\frac {\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))}{|\mathbf {r} -\mathbf {r} '|}}-{\frac {\delta (|\mathbf {r} -\mathbf {r} '-2z{\hat {e_{3}}}|-c(t-t'))}{|\mathbf {r} -\mathbf {r} '-2z{\hat {e_{3}}}|}}\right)$

Now use Green's theorem:

Let $\mathbf {F} =A(r)\mathbf {\nabla } G_{1}(r,r')-G_{1}(r,r')\mathbf {\nabla } A(r)$

$\int \mathbf {\nabla } \cdot \mathbf {F} d^{4}r=\int cdt\int d^{3}r[\mathbf {\nabla } A\cdot \mathbf {\nabla } G+A\nabla ^{2}G_{1}-\mathbf {\nabla } G\cdot \mathbf {\nabla } A-G_{1}\nabla ^{2}A]$

But $\nabla ^{2}G_{1}(r,r')=\delta ^{4}(r-r')+{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}G_{1}(r,r')$

$\nabla ^{2}A(r)=\mu j(r)+{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}A(r)$, let $j(r)=0\quad$

$\int \nabla \cdot \mathbf {F} d^{4}r=A(r')+{\frac {1}{c^{2}}}\int d^{4}r\left[A{\frac {\partial ^{2}}{\partial t^{2}}}G_{1}-G_{1}{\frac {\partial ^{2}}{\partial t^{2}}}A\right]$

The last term vanishes if $G_{1}(r,r')andA(r)\quad$ fall off sufficiently fast at $t\rightarrow \infty$. They do. So:

$\int \nabla \cdot \mathbf {F} d^{4}r=A(r')$

Now invoke the divergence theorem on the half space $z>0\quad$:

$A(r')=-\int d^{2}r\int cdt\left[A(r){\frac {\partial }{\partial t}}G_{1}(r,r')-G_{1}(r,r'){\frac {\partial }{\partial z}}A(r)\right]$, where the last term is zero by the constriction of$G_{1}(z=0)\quad$

$A(r')=-c\int dt\int d^{2}rA(r){\frac {\partial }{\partial z}}G_{1}(r,r')$

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:


$G_{1}(r,r')={\frac {-1}{4\pi }}\left({\frac {\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))}{|\mathbf {r} -\mathbf {r} '|}}-{\frac {\delta (|\mathbf {r} -\mathbf {r} ''|-c(t-t'))}{|\mathbf {r} -\mathbf {r} ''|}}\right)$, where $\mathbf {r} ''=\mathbf {r} '-2z'{\hat {e_{3}}}$

${\frac {\partial }{\partial z}}G_{1}(r,r')={\frac {1}{4\pi }}\left({\frac {\partial }{\partial z}}\left({\frac {\delta (|\mathbf {r} -\mathbf {r} '|-c(t-t'))}{|\mathbf {r} -\mathbf {r} '|}}-{\frac {\delta (|\mathbf {r} -\mathbf {r} ''|-c(t-t'))}{|\mathbf {r} -\mathbf {r} ''|}}\right)\right)$

∴ $A(r')={\frac {-1}{4\pi }}{\frac {\partial }{\partial z'}}\int _{z=0}d^{2}r\left(2{\frac {A(\mathbf {r} ,t'-{\frac {\mathbf {r} -\mathbf {r} '}{c}}}{\mathbf {r} -\mathbf {r} '}}\right)$

At $z=0\quad$, $|\mathbf {r} -\mathbf {r} '|={\sqrt {r^{2}+z'^{2}}}=S,dS={\frac {rdr}{\sqrt {r^{2}+z'^{2}}}}$

If$A(\mathbf {r} ,t)\quad$ is independent of $\mathbf {r} \quad$, then:

$A(r')={\frac {-\partial }{\partial z'}}\int _{z'=0}^{\infty }dSA\left(\mathbf {0} ,t-{\frac {S}{c}}\right)=A\left(\mathbf {\emptyset } ,t'-{\frac {z'}{c}}\right)$

This gives us uniform translation of waves at velocity c. More generally:

$A(r')={\frac {-1}{2\pi }}\int _{z=0}d^{2}r{\frac {\partial }{\partial z'}}\left({\frac {A\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|}}\right)$

$={\frac {-1}{2\pi }}\int _{z=0}d^{2}r\left({\frac {A\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{3}}}(-z')+{\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|c}}{\frac {-z'}{|\mathbf {r} -\mathbf {r} '|}}\right)$

$A(r')={\frac {-1}{2\pi }}\int _{z=0}d^{2}r\left({\frac {A\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{3}}}(-z')+{\frac {1}{c}}{\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{2}}}(z')\right)$

In our case, we consider only those waves which degrade as ${\frac {1}{r}}\quad$, so:

$A(r')={\frac {-1}{2\pi }}\int _{z=0}d^{2}r\left({\frac {1}{c}}{\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{2}}}(z')\right)$

$A(r')={\frac {-z'}{2\pi c}}\int _{z=0}d^{2}r\left({\frac {{\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)}{|\mathbf {r} -\mathbf {r} '|^{2}}}\right)$

In cylindrical coordinates, $d^{2}r=rdrd\phi \quad$. Also, ${\dot {A}}\left(\mathbf {r} ,t'-{\frac {|\mathbf {r} -\mathbf {r} '|}{c}}\right)={\dot {A_{0}}}e^{-ik(t'c-|\mathbf {r} -\mathbf {r} '|)}$. So:

$A(r')={\frac {-z'{\dot {A_{0}}}}{2\pi c}}\int _{z=0}rdrd\phi {\frac {e^{-ik(t'c-|\mathbf {r} -\mathbf {r} '|)}}{|\mathbf {r} -\mathbf {r} '|^{2}}}$