Difference between revisions of "Construction of a Tabletop Michelson Interferometer"
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But the term <math>2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math><br><br> | But the term <math>2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math><br><br> | ||
∴<math> G(x,x')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'|}</math><br><br> | ∴<math> G(x,x')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{x}-\mathbf{x}'|-c(t-t'))}{|\mathbf{x}-\mathbf{x}'|}</math><br><br> | ||
| − | Now to get the | + | Now to get the <math>G_1(x,x')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> x_3=z=0 \quad</math> we take the difference:<br><br> |
<math>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)</math><br><br> | <math>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)</math><br><br> | ||
Now use Green's theorem:<br><br> | Now use Green's theorem:<br><br> | ||
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<math>\nabla^2A(x)=\mu j(x)+\frac{1}{c^2}\frac{\part^2}{\part t^2}A(x)</math>, let <math>j(x)=0 \quad</math><br><br> | <math>\nabla^2A(x)=\mu j(x)+\frac{1}{c^2}\frac{\part^2}{\part t^2}A(x)</math>, let <math>j(x)=0 \quad</math><br><br> | ||
<math>\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]</math><br><br> | <math>\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]</math><br><br> | ||
| − | The last term vanishes if | + | The last term vanishes if <math>G_1(x,x')and A(x)\quad </math> fall off sufficiently fast at <math>t\rightarrow\infin</math>. They do. So:<br><br> |
| − | + | <math>\int \nabla \cdot \mathbf{F} d^4x=A(x')</math><br><br> | |
| − | <math>\int \nabla \cdot \mathbf{F} d^4x=A(x')</math | + | Now invoke the divergence theorem on the half space <math>z>0 \quad</math>:<br><br> |
| − | Now invoke the divergence theorem on the half space <math>z>0 \quad</math>: | ||
<math>A(x')=-\int d^2x\int cdt\left[A(x)\frac{\part}{\part t}G_1(x,x')-G_1(x,x')\frac{\part}{\part z}A(x)\right]</math>, where the last term is zero by the constriction of<math>G_1(z=0) \quad</math><br><br> | <math>A(x')=-\int d^2x\int cdt\left[A(x)\frac{\part}{\part t}G_1(x,x')-G_1(x,x')\frac{\part}{\part z}A(x)\right]</math>, where the last term is zero by the constriction of<math>G_1(z=0) \quad</math><br><br> | ||
| − | <math>A(x')=-c\int dt\int d^2xA(x)\frac{\part}{\part z}G_1(x,x')</math | + | <math>A(x')=-c\int dt\int d^2xA(x)\frac{\part}{\part z}G_1(x,x')</math><br><br> |
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> | 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> | ||
<math>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)</math>, where <math>\mathbf{x}''=\mathbf{x}'-2z'\hat{e_3}</math><br><br> | <math>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)</math>, where <math>\mathbf{x}''=\mathbf{x}'-2z'\hat{e_3}</math><br><br> | ||
<math>\frac{\part}{\part z}G_1(x,x')=\frac{1}{4\pi}\left(\frac{\part}{\part 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)</math><br><br> | <math>\frac{\part}{\part z}G_1(x,x')=\frac{1}{4\pi}\left(\frac{\part}{\part 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)</math><br><br> | ||
| − | ∴ <math>A(x')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2x\left(2\frac{A(\mathbf{x},t'-\frac{\mathbf{x}-\mathbf{x}'}{c}}{\mathbf{x}-\mathbf{x}'}\right)</math> | + | ∴ <math>A(x')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2x\left(2\frac{A(\mathbf{x},t'-\frac{\mathbf{x}-\mathbf{x}'}{c}}{\mathbf{x}-\mathbf{x}'}\right)</math><br><br> |
| + | At <math>z=0 \quad </math>, <math>|\mathbf{x}-\mathbf{x}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}</math><br><br> | ||
| + | If<math>A(\mathbf{x},t) \quad</math> is independent of <math>\mathbf{x} \quad</math>, then:<br><br> | ||
| + | <math>A(x')=\frac{-\part}{\part z'}\int_{z'=0}^\infin dS A\left(\mathbf{0},t-\frac{S}{c}\right)=A\left(\mathbf{\emptyset},t'-\frac{z'}{c}\right)</math><br><br> | ||
| + | This gives us uniform translations of waves at velocity c. More generally: <br><br> | ||
| + | <math>A(x')=\frac{-1}{2\pi}\int_{z=0} d^2x\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{x}, t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|}\right)</math><br><br> | ||
Revision as of 18:46, 2 July 2009
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}}
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')
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 \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4x e^{iqx} G(x,0)}
Then 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 G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqx} \tilde{G}(x,0)}
In free space, translational symmetry implies:
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 G(x-x',0)=G(x,x') \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 G(x,x')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(x-x')} \tilde{G} (q)}
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')=\frac{1}{(2\pi)^2}|int d^4qe^{-iq(x-x')}\tilde{G}(q)}
, 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 q=(\mathbf{k},\frac{\omega}{c}) \quad}
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 \square^2_xG(x,x')=\delta^4(x-x')=\frac{1}{(2\pi)^4}\int d^4q e^{-iq(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 \tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^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 G(x,x')=\frac{-1}{(2\pi)^4} \int d^4qe^{-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'
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 G(x,x')=\frac{1}{(2\pi)^4}\int d^3ke^{-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}
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{-2\pi}{(2\pi)^4}\int \frac{k^2dk}{k} \sin\left({ck(t-t')}\right) 2\pi\int_{-i}^i dze^{-ik|\mathbf{x}-\mathbf{x'}|z}\Theta}
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)^2}\left(\frac{1}{2i|\mathbf{x}-\mathbf{x'|}}\right)2\int dk sin(ck(t-t')) sin(k|\mathbf{x}-\mathbf{x'}|)\Theta}
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)^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 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 2\delta(|\mathbf{x}-\mathbf{x}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'}
∴
Now to get the 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 G_1(x,x')\quad }
in the half-space with z>0 with the boundary condition 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 G_1\quad }
atFailed 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 x_3=z=0 \quad}
we take the difference:
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 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
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 \mathbf{\nabla} \cdot \mathbf{F}d^4x= \int cdt \int d^3x[\mathbf{\nabla}A \cdot \mathbf{\nabla}G+A\nabla^2G_1-\mathbf{\nabla}G \cdot \mathbf{\nabla}A -G_1\nabla^2A]}
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}
The last term vanishes if 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 G_1(x,x')and A(x)\quad }
fall off sufficiently fast at 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 t\rightarrow\infin}
. They do. So:
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')}
Now invoke the divergence theorem on the half space 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 z>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 A(x')=-\int d^2x\int cdt\left[A(x)\frac{\part}{\part t}G_1(x,x')-G_1(x,x')\frac{\part}{\part z}A(x)\right]}
, where the last term is zero by the constriction of
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')=-c\int dt\int d^2xA(x)\frac{\part}{\part 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:
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 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 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 \mathbf{x}''=\mathbf{x}'-2z'\hat{e_3}}
∴ 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}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2x\left(2\frac{A(\mathbf{x},t'-\frac{\mathbf{x}-\mathbf{x}'}{c}}{\mathbf{x}-\mathbf{x}'}\right)}
At 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 z=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 |\mathbf{x}-\mathbf{x}'|=\sqrt{r^2+z'^2}=S, dS=\frac{rdr}{\sqrt{r^2+z'^2}}}
If is independent of 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 \mathbf{x} \quad}
, then:
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{-\part}{\part z'}\int_{z'=0}^\infin dS A\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:
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\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{x}, t'-\frac{|\mathbf{x}-\mathbf{x}'|}{c}\right)}{|\mathbf{x}-\mathbf{x}'|}\right)}
![{\displaystyle \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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fcea9b36150eabba8a50f39f374da0533b34e68)
