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Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 18:46, 2 July 2009
, 18:46, 2 July 2009→Determining Angle for First Diffraction Minimum
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 G<sub>1</sub>(x,x') in the half-space with z>0 with the boundary condition G<sub>1</sub> at x<sub>3</sub>=z=0 we take the difference:<br><br>
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>
<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 G<sub>1</sub>(x,x')and A(x) fall off sufficiently fast at <math>t\rightarrow\infin</math>. They do. So:<br>
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><br><br><br>
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>:<br><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')=-\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><br><br><br>
<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>