Surprises of the Faraday Cage

https://sinews.siam.org/DetailsPage/TabId/900/ArtMID/2243/ArticleID/757/Surprises-of-the-Faraday-Cage.aspx

by leephillips • 10 years ago

388 66 10 years ago

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I, perhaps naively, thought it was a well understood phenomenon. With random references in sci-fi, and a quick appearance in Enemy of The State (movie), I thought it had all been locked down.

It's somehow inspiring to learn how much is there for us to learn.

oneloop8• 10 years ago

What does "well understood" mean exactly? Ask the average inhabitant of Hacker News and he will tell you that "Faraday Cage the name of a phenomenon whereby a box of metal prevents electromagnetic fields inside", or something like that. But ask him how the strength of the electromagnetic field depends on the wire radius and mesh size, and he might not know the answer.

That said, the answer was known to Maxwell, as the author remarks, although not to the author himself. As is often the case, the problem is in the details. So in a sense, it is a well understood problem to people who know the details well enough.

I'm a physics graduate I did not know the answer, and I can assure you that the average physics graduate doesn't know the answer. In the year 2000 a graduate course in physics contains so much "advanced" physics that you end up learning a bit of a lot instead of a lot of a bit. My contemporaries and I know a lot of physics superficially, unfortunately. Time is limited, and in university you learn what you're fed.

But yes, I agree with the sentiment of your post, of course :-)

eternauta3k2• 10 years ago

A second-year Physics student can work out RF shielding using Kirchhoff's diffraction formula, or just looking at the wave equation in the k domain (spatial frequencies). Electrostatic shielding is the hard bit addressed in the article.

smallnamespace• 10 years ago

It's one thing to be able to know the classical equations of how something works (e.g. Maxwell's equations in this case, or Navier-Stokes is another good example), and quite another to understand the practical dynamics of the system as it evolves through time under boundary conditions that you care about (like the inside of a jet engine).

ChuckMcM2• 10 years ago

Never under estimate the power of asking questions. Even things that are "understood" can be, what I'll call, functionally misunderstood. Basically the understanding allows you to work around the question, but the understanding is wrong. It goes back forever, consider planetary motion, for one.

That said, I'm still trying to work out whether doing this analysis in Laplace space is sufficient.

CamperBob2• 10 years ago

I remember thinking I understood how a Van de Graaff generator worked. Simple enough, you'd think.

What I learned when I built one was that there is nothing simple or intuitive about electrostatics.

ajnin• 10 years ago

Pretty interesting read. I'd like to know the answer to the 3rd question at the end as well, microwave manufacturers must have known about this all along since the doors embed a chunky piece of metal with small holes and not a lightweight grid of thin wire as one might expect with the "traditional" understanding of the Fraday cage effect.

So did they figure out the theory independently ? Did they design the screens based on measurements ? Maybe there's a patent somewhere that may shed some light on this.

ianai• 10 years ago

A company making microwave ovens is not an academic. They could have had an intern figure it out. Without a reason to share such knowledge (aka financial incentive) a company will usually not.

jcoffland• 10 years ago

Engineers solve problems all the time either through experiment or applied science. Their goal is to make progress on the immediate project. Writing about it would be a sidetrack.

tfm• 10 years ago

Perhaps it was determined by trial and error! Or maybe the earlier oven iterations happened to get lucky. We don't know that embedding a flyscreen on the door of the food chamber was necessarily the look that people would have gone for!

Now of course there are statutory "radiation leakage" limits in most markets. One might easily imagine that engineers would take a few goes at implementing the Ezy-Look-Into oven sketched by the folks over at industrial design, measure the emissions levels with thinner wire screens, and shrug it back with several binders full of readings. When experiment and theory are in disagreement, the product manager is unlikely to fund too much research into picking apart Maxwell's Other Equations. Given the known configurations for "good enough" microwave shielding, presumably the design team gets to sign off on a suitable colour instead of insisting on mechanically-etched glass impregnated with nanowires.

colechristensen• 10 years ago

Radio propagation is quite a bit like magic, and if you're an engineer making a microwave, you probably really don't care that much about the theoretical underpinnings. The thought process goes something like this:

I have a loud radio source I want to keep contained in a box. I want people to be able to see into that box while it's on. I know that radio waves are blocked as long as the holes are smaller than some multiple or fraction of the wavelength of the radio source.

So what do I do? I think about what's easy and cheap to manufacture while being reliable. I try out a few things and measure the radio leakage. I pick the best solution out of the few I tried.

None of that really has anything to do with the subtleties of theory, the practice is you want something good at shielding that's good for the guys building it.

monochromatic• 10 years ago

That's probably a pretty good guess for what happened early on when designing microwave doors. And 99% of what's happened since is "the existing design is effective and cheap, so I'm not gonna try to fix what ain't broken."

DINKDINK• 10 years ago

Your post is a great description of the phenomenon that Tableb describes in his book Antifragility, as "Lecturing Birds how to fly" [1] Essentially the argument is: we learn by doing and formalize that knowledge in research. Birds don't need to be lectured on fluid mechanics(drag,lift,etc) to understand how to fly.

[1]http://www.yourparttimehrmanager.com/lecturing-birds-to-fly-...

mchouza• 10 years ago

Many theoretical and experimental studies were done. One example from the early 70s is http://ipnpr.jpl.nasa.gov/progress_report2/II/IIO.PDF

bjornsing3• 10 years ago

Wow! :) A quick read seems to indicate that the case where the "metal plate with holes in it" approaches a wire mesh (holes are large compared to spacing) is treated in "Reflectors for a Microwave Fabry-Perot Interferometer" published by W. Culshaw in 1959 [1].

It's still locked up behind a paywall though, some 57 years later! :( Anybody have access (or can afford the $13 / $33 to buy it)? I checked if I could access it through DeepDyve but, no.

(Perhaps this is the answer to Q1, that it really hasn't "remained unanalyzed for 180 years", but that our broken way of archiving scientific knowledge has hid the analysis?)

1. http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=112468...

joezydeco• 10 years ago

Or perhaps it was just easier and cheaper to fabricate a sheet-metal box with one side die-punched with a grid of holes.

I can totally see how making a cage of wires might use less material, but be magnitudes harder to fabricate correctly (electrically, mechanically, and aesthetically) and get right without any leaks.

mirimir2• 10 years ago

I also wonder about shielding in cellphones. I'm guessing that it's designed more-or-less empirically, based more on accepted practices than on theory.

Taniwha• 10 years ago

all of this shielding (including microwaves) is checked ... we all have to pass FCC emissions checks (very expensive) if we make consumer equipment, be it microwaves, cell phones or kid's toys

gtrubetskoy• 10 years ago

Another example in science that you'd think is well understood, but isn't is water. We still don't know why and how clouds form or why ice is slippery, we don't even know whether liquid water is strictly speaking a liquid and not a crystal some of the time.

mjevans• 10 years ago

My intuition on ice:

* Black ice is clear because it is ice with nearly ice water as lubricant (and black because heat absorption makes it more likely for dark ground to cause it first); I think everyone accepts this.

* Truly cold ice is more difficult. I suspect that it's easy for any piece that does stick to break off (under an uneven compression force), which provides a lot of small pieces that prefer to not stick to the ice it's self. Ice sticks to ice cube trays because it froze there and is thus force balanced to there. That's where I'd begin with theory and experiments if I wanted to take the time and effort to figure this out.

chadgeidel2• 10 years ago

We don't know why ice is slippery? I thought that was understood as hydroplaning (small amounts of ice melt and then you have a low-friction scenario where water separates the two surfaces). I've read that when ice is cold enough and the other surface (a rubber tire for example) is also below freezing point there is good traction.

Is there a description of this that a layman such as myself with a college degree in mathematics (heavy undergraduate physics) could understand?

nkurz• 10 years ago

It may be too basic for you, but I liked this 2005 overview from Physics Today: http://www.chemistry.northwestern.edu/documents/about/Rosenb...

The problem with the popularly held "pressure melts the ice" theory is that ice can still be very slippery even at temperature-pressure points that are not explained by that explanation.

It may be misleading to say that ice's slipperyness is "not understood", but it still is definitely an area of active research, with the understanding and theories evolving more rapidly than one might expect for something "obvious".

Wildgoose• 10 years ago

I distinctly remember a conversation with a friend when we were both working together around 1989 which was about the propagation of microwaves and the shielding in microwave ovens. He was an Electronics graduate from Durham University and he said that it wasn't properly understood.

Well, it looks like it is now!

I do find it interesting that my friend knew this, but many senior scientists, even at Oxford University, were not aware of it.

A good illustration of the split between theoretical science and its practical application!

MichailP• 10 years ago

Analysis of such problems is usually done using numerical electromagnetic tools, based on finite element method (FEM), finite difference time domain method (FDTD), method of moments (MOM) or hybrid methods. Use of such tools is not a straightforward task, and it takes a lot of effort to create accurate models. Numerical methods are used because apart from very few problems almost all problems of practical interest (such as microwave ovens mentioned in the article) can only be solved numerically.

madengr2• 10 years ago

The Faraday cage shields perfectly against electrostatic fields, not electromagnetic fields. That's where the skin depth is important.

oneloop8• 10 years ago

Very not static

https://upload.wikimedia.org/wikipedia/commons/c/c0/Cage_de_...

madengr2• 10 years ago

I suppose we have a different notion of shielding. If you look at this page, you'll see the shielding effectiveness dropping with frequency, hence the skin depth.

http://www.ets-lindgren.com/iSeries-71

Panoramix2• 10 years ago

So, intuitively speaking, why does the field decreases with increasing wire radius?

I also don't understand the factor epsilon*log(r). Doesn't that contradict the above statement? (smaller radius should lead to a larger field, not smaller)

castratikron• 10 years ago

I've always thought of it like this (could be wrong):

An electric field moves charge. Charge has mass, so it takes work to change it's momentum. When the electric field wave enters the metal, the charge is moved around by the field. So a lot of the energy in the wave is turned into motion. If the wires were thicker, there would be more charge and mass to move around, so more of the energy of the wave is lost moving the charge around by the time it exits the metal.

oneloop8• 10 years ago

epsilon log(r) is the "shielding". I suspect that the field inside will be attenuated by a factor of 1/(epsilon log(r)).

Panoramix2• 10 years ago

epsilon is the gap, so either way it should be on opposite side of the log(r). Smaller gap, stronger shielding but smaller radius weaker shielding. Am I going nuts?

ChuckMcM2• 10 years ago

I'm just saying " Intuitively, sinusoidal oscillation in one direction corresponds to exponential decay in the direction at right angles in the complex plane. A contour integral estimate of Fourier coefficients exploits this decay to prove exponential accuracy."

Isn't particularly intuitive to me :-). But the answer to the three questions is that engineers are not generally mathematicians, and once something meets the requirements they move on to the next problem. Shielding with wire mesh can be tested with a field strength meter and no math, so if the cage isn't shielding enough you adjust it until it does, and then move on.

danielmorozoff• 10 years ago

This is a fantastic paper. Truly shows a deeper intuition can be gained if a rigorous mathematical description is made. Upon reading it, it seems so obvious how treating wires as point charges would give flaws, and the surprise that the radius is more important than the interwire distance, after seeing the results make sense. Remarkable work.

plesner• 10 years ago

L1. There are gaps out there.

I had a similar experience. At one point I decided that I wanted to understand, step by step, how Ada Lovelace's Bernoulli program works[1]. There was a few steps that didn't seem right to me but I was sure if there were any bugs in there they would be well-known and mentioned somewhere I could find. This being one of the most iconic and historically significant programs ever written. Indeed I found more than one person claiming that the program was known to be correct.

It took a long time before I could believe that it wasn't just me not getting it, there were indeed a few things out of place and apparently there was just no mention of it anywhere on the web or in any of the research into the analytical engine I could get my hands on.

[1] http://h14s.p5r.org/2012/12/bernoulli.html

e12e2• 10 years ago

Reminds me of Alan Kay's frequently mentioned nugget about Kepler, that he waited for years to try and fit orbits to ellipsis, because, being so close to circles and ovals, surely all the people doing this stuff before had tried ellipsis, and found them wanting? (No, they hadn't).

Gibbon1• 10 years ago

I've been lead to understand that some issues may have been before Tycho Brahe no one had made accurate measurements, especially of the entire tracks across the sky, not just 'interesting' places.

reeboo• 10 years ago

Intuitively, sinusoidal oscillation in one direction corresponds to exponential decay in the direction at right angles in the complex plane. A contour integral estimate of Fourier coefficients exploits this decay to prove exponential accuracy.

I couldn't get past this part. What is the author saying here?

oneloop8• 10 years ago

Some aspects of electromagnetism are better discussed in terms of complex analysis. I believe the first sentence is referring to this: https://en.wikipedia.org/wiki/Euler%27s_formula The function e^(i x) is a (complex) sinusoidal oscillation when x is a real number, but is a (complex) exponential funciton when x is an imaginary number. The phrasing "in the direction at right angles in the complex plane" refers to the fact that the real and imaginary axis are perpendicular to each other.

I'm not sure about the second phrase. A contour integral is an integral over a closed path on the complex plane, and there's a theorem that says that if the function and the path have certain properties, the result of this path integral is just some coefficients (called residues). But I'm not sure how that's connected to the rest of the conversation. https://en.wikipedia.org/wiki/Residue_theorem

evanpw• 10 years ago

The residue theorem implies that you can compute some integrals on the real line by closing up the contour in the complex plane and accounting for any poles that you've enclosed. The trick is to choose your contour so that the contribution to the integral of the new piece goes to zero as the contour gets larger and larger (think of a real piece which goes from -R to R, and a semicircle in the upper half-plane connecting those two points; as R->infty, the real part of the integral goes from -infty to infty). One reason the new piece may go to zero is that oscillating functions on the real line turn in to exponential decay as you go up or down along the imaginary axis (as you point out), so as the new piece of the contour moves up or down, its contribution to the integral gets exponentially smaller.

(This is hard to explain without pictures and formulas, but you can find some examples here: http://web.williams.edu/Mathematics/sjmiller/public_html/302...).

adrianratnapala4• 10 years ago

The idea is that for each Fourier component of our source, we want to evaluate the following integral:

A = \int[+-inf] dx (\sin x) / \sqrt(x^2 + y^2) = Im \int[+-inf] dx (\exp ix) / \sqrt{(x+iy)(x-iy)}

So this is an integral over a particular contour (the real axis) in the complex plane, with poles at +-iy. We can play the usual contour games and say it equals a different contour integral

  A =  \int(something far away that vanishes) 
     + \int(once around one of the poles)

The second integral is an exponential decay because you get two factors of $i$.

For discrete point-lattices you have an periodic array of delta functions rather than a single sinusoid. Summing the Fourier components thus gives a sum of exponentials, each one dopping off faster than the last. So I guess you get an overall function like 1 /(1 - e^x).

adrianratnapala4• 10 years ago

I'm still reading the article, so perhaps I am missing something. But it seems to me that Feynman was perfectly entitled to think in terms of point (really line) charges.

By Guass' law, an isolated cyclinder with constant voltage will look to the outside world exactly like a line of constant charge-density. One cylinder among many will be slighly different, because the corresponding line charge will have external voltages superposed. But as the radius of the cylinder approaches zero, those will vanish in proportion to the 1/r voltage from the central charge.

Now the author might have some other way of getting to the same result. But that doesn't mean Feynman's argument was wrong -- it was just different.

oneloop8• 10 years ago

Let me tell you how I understand this.

I think the most important part of the question is whether the field inside decreases exponentially or linearly with the distance between wires. To the extent that Feynman didn't incorrectly answer with "exponentially", he wasn't wrong.

However, whether the wires have constant charge or constant voltage (across the cross section) is not just a matter of the argument being "different". As the author explains, if you take the wires to have be point-like (in cross section) then Feynman is right in taking the charge to be constant. However, if you want to discuss the scenario where the wires are not point-like, then you have to pick: do you impose that your wires have constant charge or constant voltage? You take ideal conductors to have constant voltage across, and the charge distribution is whatever comes from solving the relevant equations.

But I'm open to be shown to be mistaken though :-)

adrianratnapala4• 10 years ago

It's absolutely true that fat wires will behave at least a little differently from the point-like wires.

But if you are looking for a physical intuition behind the general mathematical form, then the thin-wire limit where you start. Big-wire deviations are an advanced topic, fit for engineers.

N.B. there's a difference between "thin wire" and "point like". I am saying that real wires, with constant-voltage surfaces will _behave_ like point-like charges as they get smaller.

Animats• 10 years ago

Feynman wrote "The method we have just developed can be used to explain why electrostatic shielding by means of a screen is often just as good as with a solid metal sheet. Except within a distance from the screen a few times the spacing of the screen wires, the fields inside a closed screen are zero."

Feynman is writing about electrostatic shielding, not RF shielding.[1] Electrostatic shielding involves a static electric field, as from a Van de Graff generator or other high voltage DC source. RF shielding is quite different. For one thing, it's wavelength-dependent. This is well known; you can block RF with a mesh only where the holes are much smaller than the wavelength. This is in the ARRL Handbook. If this were not the case, wire screens would block light.

JPL's paper mentioned by someone else [2] has a simple formula for a plate with a uniform pattern of holes, derived from waveguide theory. Inputs are spacing between holes, size of hole, and wavelength. They tested this against a range of perforated metal samples. There's a note that if the hole size approaches the spacing, corrections need to be applied. But they didn't, apparently, have to do that for the samples they tested.

The misconception may have arisen because Feynman was writing in the 1960s, before RF in the gigahertz range was a common thing. Other than radar, everything used long wavelengths by modern standards. Treating RF as an electrostatic problem works if the wavelength is much greater than the hole size, because the electric field will be nearly uniform over the area of interest.

(I'm not an RF guy, but I used to work in an R&D facility that did a lot of RF work and had various Faraday cages, some solid metal, some mesh. A big practical problem is leakage at slots - covers, doors, etc. - that don't have RF-tight gaskets or where the gasket has been damaged. Slots can be longer than a wavelength, and thus no longer block RF. Microwave ovens, if they leak RF, usually do so at the door seals, not the window.)

[1] http://www.feynmanlectures.caltech.edu/II_07.html (Section 7.5) [1] http://ipnpr.jpl.nasa.gov/progress_report2/II/IIO.PDF

NamTaf• 10 years ago

Nothing much to add but I just wanted to say thank you for this post. It's concise, calls on obvious counter-examples (light not being blocked), conveys the points clearly and establishes sound conclusions. For lack of a better term, it was enjoyable to read.

bjornsing3• 10 years ago

But the OP is writing about electrostatic shielding too... What makes you think the below section from the OP does not apply to electrostatic shielding?

"The error is that Feynman’s wires have constant charge, not constant voltage. It’s the wrong boundary condition! I think that Feynman, like me and most others beginning to think about this problem, must have assumed that the wires may be taken to have zero radius. The trouble is, a point charge makes sense, but a point voltage does not. (Dirichlet boundary conditions for the Laplace equation can only be applied on sets of positive capacity.) Since the correct boundary condition cannot be applied at points, I’m guessing Feynman reached for one that could, intuiting that it would still catch the essence of the matter. This is a plausible intuition, but it’s wrong."

I'm not an expert in electrical field theory, but a constant voltage boundary condition seems very reasonable to me (and a constant charge one seems less reasonable), also in the electrostatic case.

andyidsinga• 10 years ago

next time someone suggests a faraday cage for rf work, I think I'll suggest a maxwell array instead :)

lifeisstillgood• 10 years ago

So, is there a good foundational course on RF? It is more and more important but I have almost no intuition (nor it seems the maths) to approach it beyond "ohh look, three bars of strength, I can rely on IP now"

aidenn0• 10 years ago

RF is a really big field and it depends on whether you want to approach it from a theoretical or practical direction.

I have a physics background, and all I can say about EM from that point of view is that learning EM without basic vector calculus is a bit like learning mechanics without basic integral calculus: it's more complicated than it would be with the more advanced math background.

One simple example is just maxwell's equations; compare the two forms in this table from wikipedia[1]. With about one semester extra of college math you can use the form on the right rather than the left.

1: https://en.wikipedia.org/wiki/Maxwell%27s_equations#Formulat...

IshKebab• 10 years ago

I tried to build a faraday cage out of a big tube of aluminium (~10mm thick) with similar caps. Machined so very thin gaps, and I even taped them up with copper tape. The weird thing is it doesn't work. I put a BLE transmitter in there and it reduced the signal strength by maybe 40 dB, but I was expecting perfect containment.

Can anyone explain that?

KarthikaCohen• 10 years ago

There will be a live chat with the author of the article, Nick Trefethen, on Thursday, August 18 from 12 noon - 1:00 p.m. ET through the comments page on the SIAM News site: https://sinews.siam.org/DetailsPage/TabId/900/ArtMID/2243/Ar... Be sure to join us and ask any questions of the author regarding this topic!

Surprises of the Faraday Cage

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