Changes

== Determining Angle for First Diffraction Minimum ==

== Determining Angle for First Diffraction Minimum ==

We start off with Maxwell's Equation in the Lorentz gauge:

We start off with Maxwell's Equation in the Lorentz gauge:

<math>\square^2A^\mu(\mathbf{x},t) = \square^2A^\mu (x)=(-\mu_1 j^\mu (x))</math>

<math>\square^2A^\mu(\mathbf{r},t) = \square^2A^\mu (r)=(-\mu_1 j^\mu (r))</math><br><br>

 

Where:<br><br>

Where:

<math>A^\mu = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part t^2}</math>

<math>A^\mu = (\mathbf{A},\frac{\Phi} {c}), \square^2=\part_\mu \part^\mu = \nabla^2 - \frac{1}{c^2} \frac{\part}{\part t^2}</math>

<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math><br><br>

<math>j^\mu = (\mathbf{j},c\rho), \part_\mu= (\mathbf{\nabla}, \frac{1}{c} \frac{\part}{\part t})</math><br><br>

Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math><br><br>

Lorentz Gauge: <math>A^\mu = 0 \rArr \mathbf{\nabla} \cdot \mathbf{A}+\frac{1}{c^2} \frac{\part\Phi}{\part t}=0</math><br><br>

Introduce Green's function at (x=t) from some impulse source at x'=(<b>x'</b>,t')<br><br>

Introduce Green's function at<math> (\mathbf{r},t)=r \quad</math> from some impulse source at<math> r'=(\mathbf{r}',t') \quad</math><br><br>

<math>\square^2_xG(x,x')=\delta^4(x-x')</math><br><br>

<math>\square^2_rG(r,r')=\delta^4(r-r')</math><br><br>

Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4x e^{iqx} G(x,0)</math><br><br>

Let <math> \tilde{G} (q) = \frac{1}{(2\pi)^2} \int d^4r e^{iqr} G(r,0)</math><br><br>

Then <math> G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqx} \tilde{G}(x,0)</math><br><br>

Then <math> G(q)=\frac{1}{(2\pi)^2} \int d^4qe^{iqr} \tilde{G}(r,0)</math><br><br>

In free space, translational symmetry implies:<br><br>

In free space, translational symmetry implies:<br><br>

<math>G(x-x',0)=G(x,x') \quad </math><br><br>

<math>G(r-r',0)=G(r,r') \quad </math><br><br>

&there4;<math> G(x,x')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(x-x')} \tilde{G} (q)</math><br> <math>\square^2_xG(x,x')=\frac{1}{(2\pi)^2}|int d^4qe^{-iq(x-x')}\tilde{G}(q)</math><br><br>

&there4;<math> G(r,r')=\frac{1}{(2\pi)^2}\int d^4q e^{-iq(r-r')} \tilde{G} (q)</math><br> <math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}|int d^4qe^{-iq(r-r')}\tilde{G}(q)</math><br><br>

<math>\square^2_xG(x,x')=\frac{1}{(2\pi)^2}\int d^4qe^{-iq(x-x')}(-k^2+\frac{\omega^2}{c^2})</math>, where <math>q=(\mathbf{k},\frac{\omega}{c}) \quad</math><br>

<math>\square^2_rG(r,r')=\frac{1}{(2\pi)^2}\int d^4qe^{-iq(r-r')}(-k^2+\frac{\omega^2}{c^2})</math>, where <math>q=(\mathbf{k},\frac{\omega}{c}) \quad</math><br><br>

But, <math>\square^2_xG(x,x')=\delta^4(x-x')=\frac{1}{(2\pi)^4}\int d^4q e^{-iq(x-x')}</math><br>

But, <math>\square^2_rG(r,r')=\delta^4(r-r')=\frac{1}{(2\pi)^4}\int d^4q e^{-iq(r-r')}</math><br><br>

&there4;<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br><br>

&there4;<math>\tilde{G}(q)=\frac{(2\pi)^2}{(2\pi)^4}\frac{1}{-q^2}= \frac{-1}{(2\pi)^2q^2}</math><br><br>

<math>G(x,x')=\frac{-1}{(2\pi)^4} \int d^4qe^{-iq(x-x')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br>

<math>G(r,r')=\frac{-1}{(2\pi)^4} \int d^4qe^{-iq(r-r')} \frac{1}{(k+\frac{\omega}{c})(k-\frac{\omega}{c})}</math><br><br>

Chose the "retarded" solution, such that the function is zero unless t>t'<br>

Chose the "retarded" solution, such that the function is zero unless t>t'<br><br>

<math>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</math><br><br>

<math>G(r,r')=\frac{1}{(2\pi)^4}\int d^3ke^{-i\mathbf{k}(r-r')}\int d(\frac{\omega}{c}) \frac{e^{i\omega(t-t')}}{(\frac{\omega}{c}-k)(\frac{\omega}{c}+k)}\Theta</math><br><br>

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

<math>=\frac{1}{(2\pi)^4}\int d^3ke^{-i\mathbf{k}(r-r')}(2\pi i  \frac{e^{ick(t-t')}-e^{-ick(t-t')}}{2k})\Theta</math><br><br>

<math>=\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</math><br><br>

<math>=\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{r}-\mathbf{r'}|z}\Theta</math><br><br>

<math>=\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</math><br><br>

<math>=\frac{-1}{(2\pi)^2}\left(\frac{1}{2i|\mathbf{r}-\mathbf{r'|}}\right)2\int dk sin(ck(t-t')) sin(k|\mathbf{r}-\mathbf{r'}|)\Theta</math><br><br>

<math>=\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</math><br><br>

<math>=\frac{1}{(2\pi)^2}\frac{2}{|\mathbf{r}-\mathbf{r}'|}\frac{2\pi}{4} \left[2\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))-2\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))\right]\Theta</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>

But the term <math>2\delta(|\mathbf{r}-\mathbf{r}'|+c(t-t'))\rightarrow 0 \quad\forall\quad t>t'</math><br><br>

&there4;<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>

&there4;<math>  G(r,r')=\frac{-1}{4\pi}\quad \frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}</math><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>

Now to get the <math>G_1(r,r')\quad </math> in the half-space with z>0 with the boundary condition <math>G_1\quad </math> at<math> r_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(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}'-2z\hat{e_3}|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'-2z\hat{e_3}|}\right)</math><br><br>

Now use Green's theorem:<br><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(r)\mathbf{\nabla}G_1(r,r')-G_1(r,r')\mathbf{\nabla}A(r)</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^4r= \int cdt \int d^3r[\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>

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

But  <math>\nabla^2G_1(r,r')=\delta^4(r-r')+\frac{1}{c^2}\frac{\part^2}{\part t^2}G_1(r,r')</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(r)=\mu j(r)+\frac{1}{c^2}\frac{\part^2}{\part t^2}A(r)</math>, let  <math>j(r)=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^4r=A(r')+\frac{1}{c^2}\int d^4r\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 <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>

The last term vanishes if <math>G_1(r,r')and A(r)\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^4r=A(r')</math><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>

<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(r')=-\int d^2r\int cdt\left[A(r)\frac{\part}{\part t}G_1(r,r')-G_1(r,r')\frac{\part}{\part z}A(r)\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>

<math>A(r')=-c\int dt\int d^2rA(r)\frac{\part}{\part z}G_1(r,r')</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(r,r')=\frac{-1}{4\pi}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)</math>, where <math>\mathbf{r}''=\mathbf{r}'-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(r,r')=\frac{1}{4\pi}\left(\frac{\part}{\part z}\left(\frac{\delta(|\mathbf{r}-\mathbf{r}'|-c(t-t'))}{|\mathbf{r}-\mathbf{r}'|}-\frac{\delta(|\mathbf{r}-\mathbf{r}''|-c(t-t'))}{|\mathbf{r}-\mathbf{r}''|}\right)\right)</math><br><br>

&there4; <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>

&there4; <math>A(r')=\frac{-1}{4\pi}\frac{\part}{\part z'}\int_{z=0} d^2r\left(2\frac{A(\mathbf{r},t'-\frac{\mathbf{r}-\mathbf{r}'}{c}}{\mathbf{r}-\mathbf{r}'}\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>

At <math>z=0 \quad </math>, <math>|\mathbf{r}-\mathbf{r}'|=\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>

If<math>A(\mathbf{r},t) \quad</math> is independent of <math>\mathbf{r} \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>

<math>A(r')=\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 translation of waves at velocity c.  More generally: <br><br>

This gives us uniform translation 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>

<math>A(r')=\frac{-1}{2\pi}\int_{z=0} d^2r\frac{\part}{\part z'}\left(\frac{A\left(\mathbf{r}, t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)}{|\mathbf{r}-\mathbf{r}'|}\right)</math><br><br>

<math>=\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)</math><br><br>

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

<math>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)</math><br><br>

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

In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>.  Also, <math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A_0}\frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'}</math>