Evanescent Waves in near field for aperture > lambda (diffraction)?

[Paul Colby]

No, I don't think #30 is the right track. One can assume a perfect conductor and an infinitely thin plane. Look at the tangent electric field. It's zero everywhere in the plane except inside the aperture. Inside the aperture it's not zero. One only needs to compute for this tangent field since the field everywhere else may be computed with a radiation integral involving a green function. One does this by treating the tangent electric field crossed with the plane normal as a magnetic current.

So, what one should be doing is computing the aperture field which means solving the boundary value problem. I really suggest you view this as adding together plane wave solutions such that the boundary conditions are met. Do you know what these conditions are?

 

[Paul Colby]

It occurs to me there may be a simple approach solution to the scattering of a plane wave from an aperture which would use what you've been posting on.

The first step is to start with a solvable problem, the infinite conducting plane reflecting a plane wave. The solution is $$ E(r)= E_i e^{-ik_i\cdot r} + E_r e^{-ik_r\cdot r},\text{ (1)} $$ where the constant polarization vector $E_r$ is chosen so that the total tangent field is zero on the plane $z=0$. Now, we drill a hole in our screen. Two things happen inside the hole. The first is there is now a tangent E-field. The second is there is no longer a surface current induced by the incident and reflected waves. Before we cut the hole the surface current from (1) is, $$J=\hat{z}\times\nabla\times E,\text{ (2)}$$The solution of the problem is obtained by adding a solution to the field equations which removes the surface current, $J(x,y)$, just within the hole. If one expands, $J(x,y)$, in terms of a Fourier series (like you've been doing) and adds in the z-dependence (as you've been doing) one gets just this field and the solution to the problem. Lot's of messy details left out here like sign choices for $k_z(k_x,k_y)$ but I'll leave that to you.

 

[PhDeezNutz]

It occurs to me there may be a simple approach solution to the scattering of a plane wave from an aperture which would use what you've been posting on.

The first step is to start with a solvable problem, the infinite conducting plane reflecting a plane wave. The solution is $$ E(r)= E_i e^{-ik_i\cdot r} + E_r e^{-ik_r\cdot r},\text{ (1)} $$ where the constant polarization vector $E_r$ is chosen so that the total tangent field is zero on the plane $z=0$. Now, we drill a hole in our screen. Two things happen inside the hole. The first is there is now a tangent E-field. The second is there is no longer a surface current induced by the incident and reflected waves. Before we cut the hole the surface current from (1) is, $$J=\hat{z}\times\nabla\times E,\text{ (2)}$$The solution of the problem is obtained by adding a solution to the field equations which removes the surface current, $J(x,y)$, just within the hole. If one expands, $J(x,y)$, in terms of a Fourier series (like you've been doing) and adds in the z-dependence (as you've been doing) one gets just this field and the solution to the problem. Lot's of messy details left out here like sign choices for $k_z(k_x,k_y)$ but I'll leave that to you.

That sound promising but difficult. But also enlightening and worthwhile. I will work on it for the remainder of the day and hopefully get something that’s respectable enough to work with.

I really can’t thank you enough for all your help.

 

[PhDeezNutz]

Again, the only reason the evanescent waves are required is to meet the boundary conditions on the plane. It's like fitting a square pulse. All wavelengths are needed to make this happen.

I think I understand this. The aperture function (or field in the aperture) is literally a square pulse (say for a single slit aperture) and therefore you need all frequencies to construct it.

$\hat{U} \left(k_x , k_y , z = 0 \right)= \iint A\left(x,y,0 \right) e^{-ik_x x} e^{- i k_y y} \, dx dy$

In the usual Kirchhoff Integral style we assume the field in the aperture is the same as the incident field (which is probably not true but it works).

So

$A\left(x,y,0\right)$ =

$0$ if $x \lt -a$

$A_0$ if $-a \leq x \leq a$

$0$ if $x \gt a$

​

If we take the Fourier transform of this aperture function (which is essentially a square wave) we get all frequencies. Is this what you meant?

 

[Paul Colby]

Yes. I might add that for some range of aperture size, assuming the aperture field is just the incident field i expect would give a reasonable approximation for the radiation pattern. In fact this approximation is useful in designing waveguide antennas.

 

[PhDeezNutz]

Yes. I might add that for some range of aperture size, assuming the aperture field is just the incident field i expect would give a reasonable approximation for the radiation pattern. In fact this approximation is useful in designing waveguide antennas.

I think the Kirchhoff approach

1) Assuming fields in the aperture being the same as the incident field

2) Cranking this field through the greens function integral

breaks down when the aperture is small compared to the wavelength. I think you then have to refer to Hans Bethe’s approach outlined in his 1944 paper “Theory of Diffraction by Small Holes”. Somewhere in that paper he mentions a corrective factor for total radiation as compared to the Kirchhoff Integral for small apertures.

I will try really hard to post more detailed graphs today.

 

[PhDeezNutz]

Interesting I'm getting a power spike in the near field before getting an evanescent trail.

In the far field I do get some slight oscillations over a large range. But over 3 orders of magnitude (units of aperture radius)

smallest point being $\left(1.592 \times 10^(5) \right)$ and largest point being $\left( 16 \times 10^(7) \right)$ (again units of aperture radius) I get $\approx 2 \times 10^{-4}$ percent error. The graph is not as elegant as I would have liked but it does indicate that we can reasonably conclude that power is conserved in the far field.

On the other hand, I think as far as computing power through the aperture goes only the power in the far field is important for application. And power is reasonably conserved in the far field despite having an ugly graph.

I'm conflicted.

 

[Paul Colby]

Interesting I'm getting a power spike in the near field before getting an evanescent trail.

The power density drops as the distance to the aperture increases. The radiation is spreading as you go outward after all. I’m not certain how this is represented in your calculation.

 

[PhDeezNutz]

Before I post my new graphs I'd like to hopefully clear up some misconceptions I might have about Kirchhoff-Fraunhofer (far field) and Kirchhoff-Fresnel diffraction (near field).

My understanding is that in any sort of Kirchhoff approach first and foremost the dimensions of the aperture must be greater than the original wavelength. Let $a$ be the radius of the aperture

$a \gg \lambda \Rightarrow k_0 a \gg 1$

Furthermore the distance from the aperture to the field point (call it $r$) must be greater than both the radius of the aperture and the wavelength.

$r \gg a \gg \lambda$

The funny thing is my numerical $z-\text{flux}$ pattern has a strong agreement with known far field and near field solutions when I set $k_0 a = 15$ and a very poor agreement when $k_0 a = 1000$. If anything you would think the opposite to be true with the stated conditions above.

This is how I set my range of values. As we know a higher Fresnel Number $F_1 = \frac{a^2}{\lambda r_{min}}$ corresponds to the near field and a lower Fresnel Number $F_2 = \frac{a^2}{\lambda r_{max}}$ corresponds to the Far Field

First choice of $k_0 a = 15$

$k_0 a = 15$
$a = 10 \pi$
$k_0 = \frac{k_0 a}{a}$
$\lambda = \frac{2 \pi}{k_0}$
$F_1 = 1$
$F_2 = 0.05$
$r_{min} = \frac{a^2}{\lambda F_1} = 2.4a$
$r_{max} = \frac{a^2}{\lambda F_2} \approx 50a$

Second choice of $k_0 a = 1000$

$k_0 a = 1000$
$a = 10 \pi$
$k_0 = \frac{k_0 a}{a}$
$\lambda = \frac{2 \pi}{k_0}$
$F_1 = 1$
$F_2 = 0.05$
$r_{min} = \frac{a^2}{\lambda F_1} \approx 160a$
$r_{max} = \frac{a^2}{\lambda F_2} \approx 3182a$

When I compare against known solutions ($S_z$ flux) in the far field $r_{max}$ for $k_0 a = 15$ values I get the following percent error 3.5% error max

When I compare against known solutions ($S_z$ flux) in the far field $r_{max}$ for $k_0 a = 1000$ values I get the following percent error $2.5 \times 10^9$% error max (huge difference)

Why does my program work better for $k_0 a = 15$? You would think it would work better for $k_0a = 1000$Another concern I have is that depending on how many grid points I use to integrate and generate the points on the graph I can get very different numbers. When using a $10 \times 10 \times 10$ grid I get up to $3 \times$ more power than when i use a $20 \times 20 \times 20$ grid. Conventional wisdom says more points makes for a more accurate integral but how do I know when I'm getting close? BTW I used trapezoidal integration.
Here is the near field calculated power for $k_0 a = 15$ for $20 \times 20 \times 20$ 3D integration grid

Here is the near field calculated power for $k_0 a = 15$ for $10 \times 10 \times 10$ 3D integration grid

The former has 4 times as many points (surface integral $n \times n$ as opposed to $n \times n \times n$) and roughly $\frac{1}{3}$'rd power.

 

[Paul Colby]

I’m not 100% certain I’m following what you’re doing. First off where are the 3D integrations coming from? What shape are these grids? Etc.

Now in general trapezoid integration is a good choice. However I suspect as $ka$ gets bigger the integrand becomes more oscillatory. Your grids may be under sampled in this limit.

 

[PhDeezNutz]

I’m not 100% certain I’m following what you’re doing. First off where are the 3D integrations coming from? What shape are these grids? Etc.

Now in general trapezoid integration is a good choice. However I suspect as $ka$ gets bigger the integrand becomes more oscillatory. Your grids may be under sampled in this limit.

You are absolutely right. As we both know Fraunhofer diffraction is characterized by a Bessel function of the first kind. At the very least we want the grid to be fine enough to capture the first zero (most of the radiation is captured in the main lobe).

I checked my grid for $k_0 a = 1000$ and it's not fine enough to capture the first zero of the Bessel Function (first kind) leading to an under sampling and over estimation in that limit...exactly as you said.

 

[PhDeezNutz]

Smaller $k_0 a$ doesn't seem to lead to this problem. Luckily small $k_0 a$ is exactly what I want to test against Bethe Diffraction (small hole long wavelength) (with the corrective factor offered by Bethe from the Kirchhoff Integral) so I'm good from that perspective. That said if I really want my Kirchhoff Integral program to be robust I should probably add unit tests to ensure the grid is fine enough to capture the first zero of the Bessel function...which doesn't seem trivial at first glance but I'll save that for later.

 

[Paul Colby]

It should be a simple matter to compute the first Bessel zero before constructing your grid. Why not base the grid size and spacing from this?

 

[PhDeezNutz]

It should be a simple matter to compute the first Bessel zero before constructing your grid. Why not base the grid size and spacing from this?

Assuming in the far field

$\left|\vec{E}\right| \sim \frac{J_1 \left(ka \sin \theta\right)}{ka \sin \theta}$

and we wanted to devise a linear 3D grid

x = linspace(-rmax,rmax,n)
y = linspace(-rmax,rmax,n)
z = linspace(-rmax,rmax,n)

[X,Y,Z] = meshgrid(x,y,z)

that captures the first zero in the $z_{chosen} = rmax$ plane. How would we do this? To find the zeros we would first have to create a grid and hopefully that grid would be fine enough to contain that first zero.

I was only able to confirm that my earlier grid was not fine enough by looking up the first zero of $J_1 \left( ka x\right)$ by looking it up on wolfram alpha. I figured if $x$ was fine enough to capture the first zero then so would $\sin \theta$ (I don't know how true that is).

 

[PhDeezNutz]

Also another question if you have the time.

According to Hans Bethe http://www.physics.miami.edu/~curtright/Diffraction/Bethe1944.pdf small holes (compared to wavelength) diffract substantially less power than large holes (compared to wavelength). I will refer to the small hole scenario as the Bethe limit.

He argues

$H_{Kirchhoff} \sim ka^2 H_0$

$H_{Bethe} \sim k^2 a^3 H_0$

Assuming the same goes for $E$ we have

$E_{Kirchhoff} \sim ka^2 E_0$

$E_{Bethe} \sim k^2 a^3 E_0$

making

$S_{Kirchhoff} \sim k^2 a^4$

$S_{Bethe} \sim k^4 a^6$
I can readily accept $S_{Bethe} \sim k^4 a^6$ because after all this is the scattering cross section of a small object. What I am unable to conclude is $S_{Kirchhoff} \sim k^2 a^4$ from the far field formulas (The first can be found in Zangwill and the others can be inferred).

$\vec{E} \left(r, \theta, \phi \right) = -\frac{i}{2} E_0 \left(ka \right)^2 \left( \frac{e^{ikr}}{kr}\right) \left[ \frac{2 J_1 \left(k a \sin \theta \right)}{ka \sin \theta}\right]\left(\sin \phi \hat{\theta} + \cos \phi \cos \theta \hat{\phi} \right)$

$\vec{H} \left(r, \theta, \phi \right) = -\frac{i}{2} H_0 \left(ka \right)^2 \left( \frac{e^{ikr}}{kr}\right) \left[ \frac{2 J_1 \left(k a \sin \theta \right)}{ka \sin \theta}\right]\left(\cos \phi \cos \theta \hat{\theta} + \sin \phi \hat{\phi} \right)$

$vec{S} \left(r , \theta, \phi \right) = \frac{1}{2} \left( \frac{E_0}{4 Z_0}\right) \left(k^4 a^4 \right)\left( \frac{1}{k^2 r^2}\right) \left[ \frac{2 J_1 \left(k a \sin \theta \right)}{ka \sin \theta}\right]^2 \left( \sin^2 \phi + \cos^2 \phi \cos^2 \theta \right) \hat{r} = \frac{1}{2 Z_0} \left| \vec{E} \right|^2 \hat{r}$

I am able to conclude

$S_{Kirchhoff} \sim k^2 a^4$

If I assume $\frac{J_1 \left( k a \sin \theta\right)}{ka \sin \theta}$ is unitless but that does not make sense to me.

But when I multiply my Kirchhoff Power Solution by the corrective factor $k^2 a^2$ I get very good agreement between the corrected Kirchhoff Solution (adjustment for small apertures) and the Bethe's solution;

They are on the same order of magnitude for varying $ka$ up to around $0.7$ (because we're dealing with small holes). They vary by 20% or so.

So the corrective factor $k^2a^2$ is corroborated numerically, I just don't know how to make sense of it mathematically.

 

[PhDeezNutz]

Actually I think I get it. Assuming $ka$ is small (As is the case with the Bethe limit) using the Bessel Function Multiplication Theorem (wikipedia)

$\frac{1}{ka} \frac{J_1 \left( ka \sin \theta\right)}{ka \sin \theta} = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\frac{\left(1 -k^2a^2 \right) \sin \theta}{2}\right)^n J_{1+n} \left( \sin \theta \right)$

$\Rightarrow$

$J_1 \left(ka \sin \theta \right) \approx ka J_1 \left(\sin \theta\right)$ to first order

Meaning
$\frac{J_1 \left( k a \sin \theta\right)}{ka \sin \theta} \approx \frac{ka J_1 \left(\sin \theta\right)}{ka \sin \theta} = \frac{J_1 \left(\sin \theta \right) }{\sin \theta }$ which I guess is constant in $ka$ ( better terminology than unitless).

 

[Paul Colby]

I was only able to confirm that my earlier grid was not fine enough by looking up the first zero of J1(kax) by looking it up on wolfram alpha. I figured if x was fine enough to capture the first zero then so would sin⁡θ (I don't know how true that is).

The zeros of the Bessel are fixed, so the first zero is, $$J_1(3.8137..)=0.$$
From this we get, $$kax\sin\theta \approx 3.8137$$.
$ka$ and $\sin\theta$, are known. So, how many $x$'s does one need?

I only just recently circumvented the pay wall for the paper you referenced in #1. Looks like you're generalizing this to 3D.

 

[PhDeezNutz]

I think we're all good as far as establishing evanescent behavior in the near field. I just used that paper to assure readers that we were supposed to get evanescent behavior. But the way you explained it is perfect (Square waves matching the boundary conditions).

The papers I'm reading now are

Hans Bethe's Seminal 1944 Paper

http://www.physics.miami.edu/~curtright/Diffraction/Bethe1944.pdf (see specifically the section "Comparison with the Kirchhoff Integral")

and a more accessible explanation

https://www.tandfonline.com/doi/abs/10.1080/02726343.2011.590960

I'm using the 2nd paper to calculate power through a small circular aperture and then compare it to what Bethe said in his paper. Namely that

$(Bethe) = (k^2 a^2) (Kirchhoff)$ in regards to power.
The 3D generalization of the Kirchhoff Integral is covered in Chapter 21.8 of Zangwill's Text and Chapter 10.7 of Jackson.

 

[PhDeezNutz]

Alright here are the final results of comparing Kirchhoff vs. Bethe in the small hole limit. I could probably stand to make the tick mark labels bigger for easier readability (I didn't know it was that hard to read until after I saved it as a picture).

1) Comparing power through the aperture for varying small $ka$

2) semi-log plot of power through the aperture Bethe vs. Kirchhoff

3) Percent Error Between the results, less than 20% across various values of small $ka$

The first point seems like an outlier in an otherwise upward percent error trend. We expect a stronger agreement with smaller $ka$ and that seems to hold except for the beginning outlier. I'm going to have another look at my code and see if I can fix it somehow.

@Paul Colby would it be alright If I PM'ed you my 50 page writeup of my efforts so far for a critique? Even if you could look at it for a mere 5-10 minutes I would appreciate it. I've been struggling with the Bethe Approach for about 10 months and I finally think I know enough to summarize my efforts. Hopefully it's effective in communicating the Bethe Approach (Chapter 3).

Paper 1 in #48 is where Bethe justifies his approach but Paper 2 gives a better idea of how to use it. Paper 2 deals with sources on both sides of the aperture, I then use those results to argue for sources on one side. I construct an argument (perfect reflection) for why the "short-circuited fields" are the same as the incident fields albeit with a different direction.

In a little bit I will try to use your approach in #47 to come up with a criteria for creating a fine enough grid to capture the first zero of $J_1 \left( 1000 \sin \theta\right)$ in the far field.

 

[Paul Colby]

would it be alright If I PM'ed you my 50 page writeup of my efforts so far for a critique?

Totally fine I'll enjoy reading it. Bethe's paper is of course a master piece. I have a similar paper by H. Levine and J. Schwinger that uses the variational approach. It appears in multiple publications. My copy is in "Theory of Electromagnetic Waves" a symposium circa 1951.

 

[PhDeezNutz]

Totally fine I'll enjoy reading it. Bethe's paper is of course a master piece. I have a similar paper by H. Levine and J. Schwinger that uses the variational approach. It appears in multiple publications. My copy is in "Theory of Electromagnetic Waves" a symposium circa 1951.

It truly is a master piece. I can't believe he devised a strategy almost 80 years ago to deal with diffraction by small holes. On one hand it can be shown the scattering cross section for large disks is $\sim k^2 a^4$ and the scattering cross section for small disks is $\sim k^4 a ^6$ so naturally we would expect small holes to diffract less power than large holes...that part is almost obvious. The thing that amazes me is that he knew how to approach the problem in a way that would reconcile this and because the paper is some 40 pages long, justifying the approach is not trivial. I used his approach and got pretty close to corroborating his claim that

$(Bethe) = (k^2 a^2) (Kirchhoff)$

How he knew his results would turn out so well 80 years ago is nothing short of amazing.

Also I just sent the write-up. I'm not going to lie, rather than using Bethe's paper directly I used this paper to guide my understanding of Bethe

https://www.tandfonline.com/doi/full/10.1080/02726343.2011.590960?scroll=top&needAccess=true&

Pretty clever way of deriving Hans Bethe's results from purely energy considerations instead of going into all sorts of math (that is beyond me at this point).