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}} 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')

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^2_xG(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 $G_{1}(x,x')\quad$ in the half-space with z>0 with the boundary condition $G_{1}\quad$ at$x_{3}=z=0\quad$ 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 $\nabla ^{2}G_{1}(x,x')=\delta ^{4}(x-x')+{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}G_{1}(x,x')$

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

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

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

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

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

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

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

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:


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

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

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

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

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

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

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

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

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

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