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Huygens Principle for a Planar Source (view source)
Revision as of 15:25, 8 July 2009
, 15:25, 8 July 2009→Special Case
But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>
But <math>|\mathbf{r}-\mathbf{r}'|=\sqrt{(x-x')^2+(y-y')^2+z'^2}</math>
::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>
::<math>=\sqrt{r^2+r'^2+2r\rho^2\cos\phi}</math><br>
::<math>=r'-\frac{2r\rho'\cos\phi}{2r'}\frac{\rho'}{r'}\sin\theta'</math><br><br>
::<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><br><br>
so that <math>|\mathbf{r}-\mathbf{r}'|=r'-r\cos{\phi}\sin{\theta}'</math> and <math> \frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}\left(1+\frac{2r\sin\theta'\cos\phi}{r'}\right)</math><br><br>
In this particular case, we are dealing with far-field effects only, so <math>\frac{2r\sin\theta'\cos\phi}{r'}\rightarrow 0 </math> and <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx 1</math><br><br>
In this particular case, we are dealing with far-field effects only, so <math>\frac{2r\sin\theta'\cos\phi}{r'}\rightarrow 0 </math> and <math>\frac{1}{|\mathbf{r}-\mathbf{r}'|^2} \approx \frac{1}{r'^2}</math><br><br>
So, <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}'|}=\frac{e^{ikr'}}{r'^2}\int_0^a rdr\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math><br><br>
So, <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}'|}=\frac{e^{ikr'}}{r'^2}\int_0^a rdr\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math><br><br>
The integral <math>\int_0^{2\pi} d\phi\, e^{-ikr\sin{\theta}'\cos{\phi}}</math> is the integral representation of the zero order Bessel function of the first kind with <math> kr\sin{\theta}' \quad</math> as the argument. This gives us the equation:<br><br>
<math>\mathcal{J}(r')=\frac{e^{ikr'}}{r'^2}\int_0^a rdr 2\pi J_0(kr \sin{\theta}') </math><br><br>
To simplify the math, we make use of the fact that we can represent this Bessel functions as the derivative of a Bessel function of a different order. In general, the formula to compute this derivative is <br><br> <math>z^{v-k}J_{v-k}(z)=\left(\frac{1}{z}\frac{\part}{\part z}\right)^kz^vJ_v(z)</math><br><br>
In this case, we take <math>v=k=1 \quad</math> and <math>z=kr\sin{\theta}' \quad</math>. So<br><br>
<math>J_0(kr\sin{\theta}')=\left(\frac{1}{kr\sin{\theta}'}\frac{\part}{\part (kr\sin{\theta}')}\right)(kr\sin{\theta}')J_1(kr\sin{\theta}')=\frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><br><br>
This gives us the equation<math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{r'^2}\int_0^a rdr \frac{d}{k\sin{\theta}'dr} J_1(kr\sin{\theta}')</math><br><br>
We invoke the principle that the integral of a derivative is the function evaluated at the end points to give us the equation<br><br><math>\mathcal{J}(r')=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}\left[aJ_1(ka\sin{\theta}')-0J_1(0k\sin{\theta}')\right]=2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')</math><br><br> and <math>A(r')=\frac{z'\dot{A_0}}{2\pi c}e^{-ikct'} 2\pi\frac{e^{ikr'}}{k\sin{\theta}'r'^2}aJ_1(ka\sin{\theta}')=\frac{z'\dot{A_0}a}{c}\frac{e^{ikr'-ikct'}}{k\sin{\theta}'r'^2}J_1(ka\sin{\theta}'</math>