Numerical Evaluation of the Kirchhoff Integral (Flux Pattern) (Units?)

[PhDeezNutz]

Homework Statement

First I'd like to take the (known) analytic approximation of Kirchhoff Integral and plot it in the far field. I'd imagine the scalar quantity $\psi$ is most closely related to the Electric Field Vector $\vec{E}$ so that is what I'll focus on for the second part.

Then I'd like to evaluate the Kirchhoff Integral in a way that is completely numerical.

I would then like to compare the two approaches graphically. As it is right now they are orders of magnitude off from each other.

The set up I'd like to entertain is a plane wave normally incident on a circular aperture in the $xy-plane$ and graph the flux through a $z = constant$ plane.

Relevant Equations

The Kirchhoff Integral is most generally formulated as

$\psi \left( \vec{r} \right) = \int\limits_{Aperture} \left(G \frac{\partial{\psi \left(\vec{r}'\right)}}{\partial{z'}} - \psi \left(\vec{r}'\right) \frac{\partial{G}}{\partial{z'}}\right) \, \rho' d\rho' d\phi'$

$\psi \left( \vec{r}' \right) = \left. \psi_0 e^{ikz'} \right|_{z'=0} = \psi_0 \Rightarrow \left. \frac{\partial{\psi \left(\vec{r}'\right)}}{\partial{z'}}\right|_{z'=0} = ik \psi_0$

Naturally there are vector equivalents of the Kirchhoff Integral. Taken from Jackson (10.113)

$\vec{E} \left( \vec{r} \right) = \frac{ie^{ikr}}{r} a^2 E_0 \cos \alpha \left( \vec{k} \times \vec{\epsilon}_2 \right) \frac{J_1 \left( \sin \theta \right)}{\sin \theta}$Where I just let $\alpha = 0$ for normally incident.

The result I get is (Thumbnail)

Now for the second approach and result

$\left| \vec{r} - \vec{r}' \right| = \sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2 - 2 z z' + z'^2}$

So The Green's Function for Free Space becomes

$G\left(\vec{r} , \vec{r}' \right) = \frac{e^{ik \sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2 - 2 z z' + z'^2}}}{\sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2 - 2 z z' + z'^2}}$

After analytically taking the derivatives the integral reduces to (broken up in 3 parts for simplicity) (and evaluated at $z' = 0$)

$\psi_1 = ik \psi_0 \int\limits_{Aperture} \frac{e^{ik \sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2}}}{\sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2}}\, \rho' d\rho' d\phi'$

$\psi_2 = ik \psi_0 \int\limits_{Aperture} \frac{ze^{ik \sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2}}}{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2}\, \rho' d\rho' d\phi'$

$\psi_3 = - \psi_0 \int\limits_{Aperture} \frac{z e^{ik \sqrt{\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2}}}{\left(\rho^2 + \rho'^2 - 2\rho \rho' \cos \left(\phi - \phi' \right) + z^2\right)^{\frac{3}{2}}}\, \rho' d\rho' d\phi'$

When I plot it against xy (like in the last picture I get) (Thumbnail)



I assume I didn't do anything stupid like make the wavelengths different, distances different, etc I think my I'm off by a constant of some sort but I don't know which one makes sense my units are right as far as i can tell.

In the first picture and second picture respectively we have a peak of

$2.8 \times 10^{-15}$

$5.2 \times 10^{-13}$Off by two orders of magnitude.

Can anyone point out where I messed up on my units If that is indeed the case.

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[PhDeezNutz]

@etotheipi I figured it out. I forgot some factors. I'll do a summary of my mistakes later.

Here are the results. Virtual perfect agreement to the point that if you overlayed the graphs you would only see one graph.

The left is the analytical approximation. The right is the result of brute numerical integration.