Construction of a Tabletop Michelson Interferometer

Determining Angle for First Diffraction Minimum

We start off with Maxwell's Equation in the 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 \square^2A^\mu(\mathbf{x},t) = \square^2A^\mu (x)=(-\mu_1 j^\mu (x))}

Where: 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 = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part 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 (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'
$G(x,x')={\frac {1}{(2\pi )^{4}}}\int d^{3}ke^{-i\mathbf {k} (x-x')}\int d({\frac {\omega }{c}}){\frac {e^{i\omega (t-t')}}{({\frac {\omega }{c}}-k)({\frac {\omega }{c}}+k)}}\Theta$

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

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

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

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

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

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

Now to get the G₁(x,x') in the half-space with z>0 with the boundary condition G₁ at x₃=z=0 we take the difference:
$G_{1}(x,x')={\frac {-1}{4\pi }}\left({\frac {\delta (|\mathbf {x} -\mathbf {x} '|-c(t-t'))}{|\mathbf {x} -\mathbf {x} '|}}-{\frac {\delta (|\mathbf {x} -\mathbf {x} '-2z{\hat {e_{3}}}|-c(t-t'))}{|\mathbf {x} -\mathbf {x} '-2z{\hat {e_{3}}}|}}\right)$

Now use Green's theorem:
Let $\mathbf {F} =A(x)\mathbf {\nabla } G_{1}(x,x')-G_{1}(x,x')\mathbf {\nabla } A(x)$

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

But 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 \nabla^2G_1(x,x')=\delta^4(x-x')+\frac{1}{c^2}\frac{\part^2}{\part t^2}G_1(x,x')}

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

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 \int \nabla \cdot \mathbf{F} d^4x=A(x')+\frac{1}{c^2}\int d^4x\left[A\frac{\part^2}{\part t^2}G_1 - G_1\frac{\part^2}{\part t^2}A\right]}

The last term vanishes if G₁(x,x')and A(x) fall off sufficiently fast at t