Construction of a Tabletop Michelson Interferometer

Determining Angle for First Diffraction Minimum

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

Where: $A^{\mu }=(\mathbf {A} ,{\frac {\Phi }{c}}),\square ^{2}=\partial _{\mu }\partial ^{\mu }=\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial }{\partial t^{2}}}$ Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})}

Lorentz Gauge: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0}

Introduce Green's function at (x=t) from some impulse source at x'=(x',t')
$\square _{x}^{2}G(x,x')=\delta ^{4}(x-x')$

Let ${\tilde {G}}(q)={\frac {1}{(2\pi )^{2}}}\int d^{4}xe^{iqx}G(x,0)$

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

In free space, translational symmetry implies:
$G(x-x',0)=G(x,x')\quad$

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

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

$G(x,x')={\frac {-1}{(2\pi )^{4}}}\int d^{4}qe^{-iq(x-x')}{\frac {1}{(k+{\frac {\omega }{c}})(k-{\frac {\omega }{c}})}}$
Chose the "retarded" solution, such that the function is zero unless t>t'
$G(x,x')={\frac {1}{(2\pi )^{4}}}\int d^{3}ke^{-i\mathbf {k} (x-x')}\int d({\frac {\omega }{c}}){\frac {e^{i\omega (t-t')}}{({\frac {\omega }{c}}-k)({\frac {\omega }{c}}+k)}}\Theta$

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

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

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