Changes

In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>.  Without loss of generality, we consider a harmonic solution with a particular frequency &omega; = kc.

In cylindrical coordinates, <math>d^2r=rdrd\phi \quad</math>.  Without loss of generality, we consider a harmonic solution with a particular frequency &omega; = kc.

:<math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A}(\mathbf{r},0)e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math><br><br>

:<math>\dot{A}\left(\mathbf{r},t'-\frac{|\mathbf{r}-\mathbf{r}'|}{c}\right)=\dot{A}(\mathbf{r},0)e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}</math><br><br>

:<math>A(r')=\frac{z'\dot{A_0}}{2\pi c}\int_{z=0} rdrd\phi \frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</math>

:<math>A(r')=\frac{z'\dot{A_0}}{2\pi c}\int_{z=0} rdrd\phi\, \frac{e^{-ik(t'c-|\mathbf{r}-\mathbf{r}'|)}}{|\mathbf{r}-\mathbf{r}'|^2}</math>

== Special Case ==

== Special Case ==

Picture an opaque screen with a circular aperture of radius a.<br><br>  

Picture an opaque screen with a circular aperture of radius a.<br><br>  

Let<math>\mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}</math><br><br>

Let<math>\mathcal{J}(r')=\int_0^a rdr\int_0^{2\pi} d\phi\, \frac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|}</math><br><br>

Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br><br>

Then <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'}\mathcal{J}(r')</math><br>

<math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br><br>

But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>

<math>=r'-\frac{2r\rho'\cos\phi}{2r'}, \frac{\rho'}{r'}=\sin\theta'</math><br><br>

::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>

<math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math>

::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}\frac{\rho'}{r'}\sin\theta'</math><br><br>

so that <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math>