196
edits
Changes
Jump to navigation
Jump to search
Line 32:
Line 32: − + − + Line 40:
Line 40: − + + + + + + + +
Construction of a Tabletop Michelson Interferometer (view source)
Revision as of 18:13, 2 July 2009
, 18:13, 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>
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>
<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>
Now use Green's theorem:<br><br>
Let <math>\mathbf{F}=A(x)\mathbf{\nabla}G_1(x,x')-G_1(x,x')\mathbf{\nabla}A(x)</math><br><br>
Let <math>\mathbf{F}=A(x)\mathbf{\nabla}G_1(x,x')-G_1(x,x')\mathbf{\nabla}A(x)</math><br><br>
<math>\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]</math><br><br>
<math>\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]</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>\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 t
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>
<math>\int \nabla \cdot \mathbf{F} d^4x=A(x')</math><br>
Now invoke the divergence theorem on the half space <math>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>
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>
<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>