How Electromagnetic Waves Pass Through Holes

[jonlg_uk]

Hi everyone I have a question that has been bothering me for a while now. I have basically built a Faraday cage using wire mesh. I know there is a relationship between the mesh hole size and the wave suppressing ability of the cage.

As I understand it an electromagnetic wave can be thought of as a series of peaks of energy. Each peak being separated from the next peak by a given wavelength...What has this got to do with the size of hole that a electromagnetic wave can travel through? Why would smaller waves, such as the 3G signal on my phone pass through the hole and larger waves not?

Can somebody explain it in layman's terms so I can form a graphical representation of what is actually happening?

I remember back in my university days, the professor was telling me how they managed to detect red LASER light when shining it through a hole of 1nm. He said it was the equivalent of squeezing an elephant through a keyhole. This got me thinking and has led my subsequent confusion.

I thank you in advance

Jon

 

[Blibbler]

Imagine if you will a fence running alongside a road.

The fence is made of metal strips a few centimetres wide, stuck vertically into the ground, with the strips a few centimetres apart.

Now imagine standing directly in front of the fence - it is half metal, half air. A vertically polarised EM wave could pass through the gaps; a horizontally polarised EM wave couldn't, in all probability, and waves of polarizations in between could pass with varying degrees (sorry) of success. Now get closer to the fence and look along its length to its farthest point. The closer you get to it the smaller the gaps in the fence will appear to be and, sure enough, the closer the fence will become to a mirror to any EM waves you shine on it.

So the answer to your question involves EM wave polarization, conductor to gap ratio and general geometry of the cage and antennas and the ratio of wavelength to geometry.

I hope this helps a bit.

 

[sophiecentaur]

@Blibber:
From your description of the fence (with vertical conductors) you appear to have got the polarisation thing the wrong way round. Vertically polarised waves will excite currents in the vertical conductors and these will re-radiate the wave back the way it came. This may be counter intuitive but the same thing applies to thin wire antennae - they behave as if they had a very large effective cross section to parallel polarised waves. Horizontally polarised waves, however, will not excite currents in the conductors (assuming their horizontal dimension is small compared with the wavelength) so a large proportion of the wave's energy will pass through.
The actual width of the slats, the angle of arrival and their thickness will have an effect on exactly how much HP power will pass through. Fine square mesh works better than slats, of course.

Unfortunately, when you require good screening, the small currents that will be induced around on the other side of the screen, through the holes, may radiate an inconveniently high level. It may seem small in percentage but the effect of interference on equipment tends to be more logarithmic than linear so you may still have signals leaking through at perhaps -60dB less than what you got outside and that can be an embarrassment. This is why good screened enclosures are so expensive. Doors, hatches and cable entry points need to be given special attention if you want really good screening.

 

[Blibbler]

@sophiecentaur

Many thanks for the correction - I'll look into it.

 

[Drakkith]

I think what the OP is asking is more related to the wavelength and not the polarization. Why does the wavelength matter when it comes to absorbing EM radiation or letting it pass?

 

[sophiecentaur]

He's discussing a grid and the spacing and size of the conductors, wrt wavelength, affects the way the currents will flow on the structure which will affect how much energy gets radiated on the other side.

 

[Drakkith]

He's discussing a grid and the spacing and size of the conductors, wrt wavelength, affects the way the currents will flow on the structure which will affect how much energy gets radiated on the other side.

Yes, but what does the wavelength have to do with it? Why does a shorter wavelength pass through holes smaller than a longer wavelength can?

 

[jonlg_uk]

@ Blibbler and sophiecentaur- Thanks the reply, although it was really interesting, it never quite answered my question.
As Drakkith said, I am specifically looking for the reason why shorter wavelengths may pass through the holes and longer wavelengths do not...

 

[jonlg_uk]

@ sophiecentaur

When I place my mobile phone inside my Faraday Cage, the signal is attenuated but it doesn't disappear completely. In cellular and most commercial applications signals are vertically polarized.

According to your explanation the reason for this is because the wavelength between each successive energy peak is greater than the horizontal distance between the vertical conductors. As a result current is not exciting in the vertical conductors, and no wave is re-radiate back the way it came, due to the fact that their horizontal distance separating the vertical conductors is much shorter than the wavelength. As a result a portion of the waves energy will pass through the cage.

Is this right? If so, what effect are the horizontal conductors having on the vertically polarized mobile phone waveform signal? Surely they should be blocking it like it like a polarized filter does?

 

[sophiecentaur]

@ sophiecentaur

When I place my mobile phone inside my Faraday Cage, the signal is attenuated but it doesn't disappear completely. In cellular and most commercial applications signals are vertically polarized.

According to your explanation the reason for this is because the wavelength between each successive energy peak is greater than the horizontal distance between the vertical conductors. As a result current is not exciting in the vertical conductors, and no wave is re-radiate back the way it came, due to the fact that their horizontal distance separating the vertical conductors is much shorter than the wavelength. As a result a portion of the waves energy will pass through the cage.

Is this right? If so, what effect are the horizontal conductors having on the vertically polarized mobile phone waveform signal? Surely they should be blocking it like it like a polarized filter does?

That diagram is just plain wrong. EM waves aren't like waves on a wiggling piece of string so the 'picket fence' analogy doesn't apply. Treat robotroom with care if that's the sort of thing they are saying. In fact, a 'polaroid' filter consists of molecules that have been stretched inside a plastic matrix and these behave like tiny wire dipoles and absorb light with a polarisation parallel to these dipoles. Thus, they block light with polarisation which is not like that diagram would suggest, at all.

The reason for the reflection of VP waves (VP means that the Electric field is vertical, btw) by vertical strips of conductor is because currents are induced in the long lengths of wire and re-radiate energy (the current is mostly induced on the side of arrival of the wave, too). It behaves just the same as a solid sheet. This is because a single (infinitely) thin wire 'intercepts' much more than its cross section might suggest; it has an effective cross section many times the wire thickness. However, if there is a big space (wrt the wavelength) between the vertical conductors then this will leave 'gaps' between the regions of influence of the wires and the screening gets less.

If there are no horizontal wires connecting the vertical conductors, HP waves will not induce any current so they will not affect the waves. The small horizontal extent of the wires would not allow any significant horizontal currents to be induced.

A mesh is more complicated but consider a mesh with very small holes compared with the RF wavelength. This will behave more or less like a sheet of metal because currents can flow along it in virtually any direction - finding their way across the surface even when they flow diagonally.

If the wavelength is very very short and the wires of a large mesh are very fine then most of the energy will go straight through as it will not interact with the conductors to produce currents. You can see through a chicken wire fence, which tells you that em of visible wavelengths gets through the holes undisturbed. For cm microwaves and a mesh size of a few cm, each hole becomes like an individual antenna. Wire dipoles and slots in metal sheets are equivalent and both behave as radio antennae. This is not very well known but an example is the antenna in your mobile phone that will almost certainly be a slot type (they don't use sticky up whip antennae any more do they?). If you take the back off some phones, you can actually see a system of L shaped slots, which is the antenna; have a look. Anyway, there are currents flowing around the inside of all these slots and energy will also be radiated towards the back of the mesh too.
Of course that's not the end of the story. Low frequency waves have a habit of inducing currents around the outside of a cage and producing nodes and antinodes. This can mean that a lot of current can be present at some points on its surface. If there happens to be a high resistance gap (seam or door) at this point, a relatively high voltage can appear, which can produce an E field inside the cage.

 

[Gobil]

I think what the OP wants to know is the relationship between the LONGITUDINAL component of the EM wave, the WAVELENGTH, and the TRANSVERSE component of the EM wave, THE ELECTRIC FIELD.

In vacuum, light travels in the direction of propagation, at velocity c.
It has an electric (and magnetic) field point in the transverse directions.
The velocity these fields propagate in the transverse direction is...also c!
The electric field changes sign over half a wavelength, meaning the E field also propagates half a wavelength.

Over a full wavelength, the transverse E field distance traveled is ALSO one wavelength.

So in vacuum, the extent of the transverse E field in space, is equal to the wavelength of the light. That´s why, if you want to affect this electric field (Faraday cage), you have to have a cage with gaps the order of the wavelength. Red light gets through a 1 nm gap because it is not absorbed by the gap completely, the fields are affected on both sides, but the beams are diffracted, or scattered. not necessarily absorbed.

Hoped this helps.

 

[Born2bwire]

Building off of sophiecentaur's explanation, you can think of the reflection of the electromagnetic wave as being the result of the waves excited by the currents induced on the surface of an object by the incident wave. The secondary waves produced by these currents will cancel the incident wave on the transmission side and behave as a reflected wave on the incident side of the surface.

So if we have a solid infinite sheet of a perfect conductor and a vertically polarized plane wave strikes this sheet, then the induced currents on the surface of the sheet run in the vertical direction too. Each of these currents can be thought of as a source for the reflected wave. If we look at two adjacent current strips, we notice that the phase difference between the currents is going to be infinitesimal. The result being that the waves produced by these two strips are not going to strongly interfere with each other. However, if we take two current strips that are separated by half a wavelength, we see that the two strips are 180 degrees out of phase. The waves sourced by these two strips will strongly interfere with each other.

Now we replace the solid sheet with the vertically running wires that sophiecentaur has explained. Since the induced currents originally ran in the vertical direction, the use of a set of wires instead of a solid sheet of metal does not affect the physics of the problem. In addition, if we remove a small number of these wires in a regular fashion we also find that it does not strongly affect the behavior because the wires adjacent to the holes will act as sources that are very much like the removed wires. It is only when we remove enough wires that the adjacent wires differ by a significant amount of phase that the strength of the reflection becomes diminished.

The rule of thumb for the allowed gap is usually around a quarter of a wavelength.

So the reason is centered around the behavior of wave physics and how well the source currents induced on the wire grid can approximate the ideal current distribution needed to perfectly reflect the wave.

There are other reasons why radiation can be transmitted. The primary reason is that a metal is not a perfect conductor like we assumed above. All metals have a finite conductivity which will impede the metal's ability to generate the currents we need to perfectly reflect the wave. One consequence is that the wave is able to penetrate into the metal some distance (related to the wavelength of the wave and the conductivity of the metal). If the wire is very thin compared to the wavelength of the incident wave then some of the wave will be able to transmit through the wire directly. There are also the physical defects inherent in any Faraday cage. Even seemingly minor points like adding a hole in the cage for a feed line can create areas where induced currents can couple into the interior and radiate the fields inside. This is similar to wait sophiecentaur was describing in the end of his post.

 

[Drakkith]

In vacuum, light travels in the direction of propagation, at velocity c.
It has an electric (and magnetic) field point in the transverse directions.
The velocity these fields propagate in the transverse direction is...also c!
The electric field changes sign over half a wavelength, meaning the E field also propagates half a wavelength.

Over a full wavelength, the transverse E field distance traveled is ALSO one wavelength.

So does the electric and magnetic field just not travel any further somehow? After the photon travels one wavelength the fields don't continue to propagate out?

 

[Gobil]

this also puzzled me, and the answer I received was this:

the fields change direction over half a wavelength, so it can only travel a certain distance during the time it is going in that direction, which is one half wavelength. the next question I had was this; are they photons? they answer is, not really, they are VIRTUAL photons, which represent the forces between particles, they travel at c, and are used to describe forces between subatomic particles, they don´t have a frequency. If you think of two electrons close to each other, there is a repelling force... virtual photons are being exchanged. If you have an isolated electron, and another electron miraculously appear next to it, there will appear a force between them. the force cannot travel faster than c, and we call it virtual phtotons, so this is my interpretation. I am open to correction.

 

[Born2bwire]

this also puzzled me, and the answer I received was this:

the fields change direction over half a wavelength, so it can only travel a certain distance during the time it is going in that direction, which is one half wavelength. the next question I had was this; are they photons? they answer is, not really, they are VIRTUAL photons, which represent the forces between particles, they travel at c, and are used to describe forces between subatomic particles, they don´t have a frequency. If you think of two electrons close to each other, there is a repelling force... virtual photons are being exchanged. If you have an isolated electron, and another electron miraculously appear next to it, there will appear a force between them. the force cannot travel faster than c, and we call it virtual phtotons, so this is my interpretation. I am open to correction.

The observed electromagnetic wave is the result of the observation of a large set of photons. If I have a detector that is observing a continuous incident wave then that represents a continuous absorption of photons (the rate of the photons incident upon the detector being proportional to the magnitude of the electromagnetic fields). It is not very instructive to talk about the electromagnetic field of a single photon because the fields are the observable of the photon states and only when we have a large number of photons do we get a statistically consistent electromagnetic field. That isn't to say that we cannot have inconsistent fields in a sense. The granularity in the photons in a low power/high frequency wave gives rise to shot noise. So a well spatially localized electromagnetic pulse can be roughly correlated to a well spatially localized set of photon states.

Virtual photons do have a frequency, this is an important aspect in using virtual photons to explain the Casimir and van der Waals intermolecular forces. In terms of electromagnetic force, the virtual photons are only used as a mechanism in static fields. In the DC case, there are no real photons but one can use the model of virtual photons to explain the Coulombic force.

 

[jonlg_uk]

That diagram is just plain wrong. EM waves aren't like waves on a wiggling piece of string so the 'picket fence' analogy doesn't apply.

Thanks for your replies, this is what I am struggling with, what do electromagnetic wave look like when traveling from a vertical antenna?

An electromagnetic wave is a traverse wave. It's wavelength is the distance between two successive peaks separated in time. The peaks arrive one after each other. How I picture the wave doesn't tie into your description.
In order to understand your description I have to picture the faraday cage being perpendicular to the wavefront rather than head on, facing the antenna.

 

[sophiecentaur]

It doesn't "look like" anything. Any wiggly lines you may want to draw on a piece of paper are only 'graphs' to show the strength and direction of the electric and magnetic fields as they vary in space and time.
The wave that is launched from an electric dipole has its electric field parallel with the wires and, if you took an instant measurement of the fields at anyone time (A photograph of the graph) they would, as you say, vary from a maximum 'up', through a zero, down to a maximum 'down' and back through zero again as you move out from the dipole. The distance between successive positive maxima is called the wavelength. In time, the wave progresses outwards and if you measured the field at any point, as it varies in time, you would again get a variation up and down and back again. The rate at which this happens is called the frequency. If you multiply the wavelength by the frequency, that will tell you the speed at which the wave progresses. NO NET FORWARD MOVEMENT of fields or matter takes place.

btw, it is bad enough trying to discuss most aspects of electromagnetism, sensibly, using photons and it's even worse if you try to describe what goes on if you want to use photons and waves at the same time. You want waves, then talk in terms of waves. Sort that out first and then let QM come into the game - but only when you're quite sure you have classical wave ideas well sussed.

 

[Drakkith]

So would it be safe to say the EM radiation is a "disturbance" of an EM field that propagates outward but only transfers energy and interacts through specific quanta called photons? Or is that just way off?

 

[sophiecentaur]

I think that is absolutely right and, furthermore, there is very little else you can say, incontrovertibly, about photons. We'll be back on that other thread if we're not careful!

 

[Drakkith]

Interesting. That's a lot less confusing to me now. Not sure if that's a good thing or a bad thing!

 

[jonlg_uk]

I still can't see why the wavelength has anything to do with it graphically in my mind. I understand what you have say sophiecentaur, but still picture like this and can't understand why the wavelength has anything to do with the mesh dimensions...

P.S Sorry for being stupid and not understanding this.

 

[Gobil]

in my opinion there is a few things wrong with your graph.

1) it seems like you think the E field is getting bigger as it propagates, in a medium with a single refractive index, let's say vacuum, this is not the case. The E field remains the same as is propagates forward.

2) the second thing, and what I think is bothering you, is your wavelength. In vacuum it extends in the vertical direction-- one wavelength. so the transverse distance of the e-field is equal to one wavelength of the light.

so, this is why you need to have a cage, with the separation on the order of one wavelength or smaller, to affect the EM wave. simple as that.

 

[sophiecentaur]

2) the second thing, and what I think is bothering you, is your wavelength. In vacuum it extends in the vertical direction-- one wavelength. so the transverse distance of the e-field is equal to one wavelength of the light.

The wave covers a whole plane, normal to the direction of propagation once you are at a great distance from the source. This doesn't seem to be what your statement 2 says.

When the wave hits a vertical wire, the effect of that wire is limited to around a quarter wave from the wire (the effect gradually drops off as you get further away). Like I said before, even an infinitely thin wire has an 'effective cross section' that is much wider than the wire itself. So a grid of wires, spaced by a quarter wave, will start to be a reasonable screen. The screening gets better as the wires a closer together, of course but you can see how the screening is affected by separation vs wavelength now (?).

 

[jonlg_uk]

The wave covers a whole planeThe screening gets better as the wires a closer together, of course but you can see how the screening is affected by separation vs wavelength now (?).

No I need a picture of how it works. The wavelengths as you say are the distance between successive maxima peaks separated by time...if that is still the case why should the separation distance between the vertical wires have anything to do with it?

This is what I understand so far...

When electromagnetic waves are polarized vertically they have a component of their electric fields aligned parallel to the vertical wires that induces the movement of electrons along the length of the wires. Since the electrons are free to move in this direction, the vertical wires behaves in a similar manner to the surface of a metal when reflecting light; and the wave is reflected backwards along the incident beam.

For waves with electric fields perpendicular to the wires, the electrons cannot move very far across the width of each wire; therefore, little energy is reflected, and the incident wave is able to pass through the mesh.

What I don't understand is why the separation distance between the vertical wires must be less than the wavelength of the wave. I can't picture why this is and its keeping me awake

 

[sophiecentaur]

You want a physical picture of waves produced by a dipole. The Web is full of them - just go hunting. You'll find one that suits you, I'm sure.

If you really want to introduce electron movement into this explanation then beware. Electrons drift (on average) through metals at speeds of less than 1mm per second. When AC is being considered - particularly RF frequencies then just how far do you think the electrons go before turning around and going the other way? In 1 second, with a relatively low frequency of 10MHz and ignoring the fact that the motion would be sinusoidal, the distance traveled each cycle would be 10^-7m. For the thinnest wire you could imagine, this would allow them to move across the wire just as easily as up and down it. All the basics of electromagnetism was sorted out long before Chadwick and his chums discovered the electron but everyone gets taught Electricity in terms of these things zapping along wires but it just ain't like that. Just stick to CURRENT. If you have a problem relating current with this tiny movement of electrons then you only have to realize that there are Ten to the power of 'lots and lots' of them available to move just a little distance to give you 1A of current. In a short length (i.e. across the wire), the induced current will be much less than in a long wire - for electromagnetic reasons and not because of the room available for electrons to move about.

Back to EM waves. It's easier to describe how a dipole works as a transmitter first. As the volts at the mid point alternate, a wave flows along the wire and bounces back from the ends (like a wave travels through some coax or along a twisted or parallel pair of wires). The current through the wire and the volts on it are almost completely out of phase by 90degrees so there are 'strong' E and H fields around the wire that do not propagate outwards at all - they just 'occupy' the space around the wire as a standing wave, dropping off in amplitude to almost nothing by one wavelength away. However, the phase between V and I is not exactly 90degrees. There is an in-phase component that can propagate and that is what carries the energy away as a progressive EM wave (the one we're talking about). A way of looking at what is going on is to think of the standing wave pattern as causing the traveling wave to radiate away. The 'region of influence' of the wire is far beyond the dimensions of the metal. (I have made this comment more than once before already; have you read it?)
The way a wire works as a receiving antenna is similar. A passing wave excited some currents in the ire and these currents produce a standing wave around it (as before) which causes the currents to build up to some maximum and start to affect the passing EM wave over a large area. (It takes a number of cycles for this to occur in the same way as it takes a while for resonance to build up with sounds). So a wire has an Effective Cross Section of about a quarter of a wavelength. It follows that wires spaced by about a quarter wavelength will behave very similar to a metal plate.

 

[jonlg_uk]

The way a wire works as a receiving antenna is similar. A passing wave excited some currents in the ire and these currents produce a standing wave around it (as before) which causes the currents to build up to some maximum and start to affect the passing EM wave over a large area. (It takes a number of cycles for this to occur in the same way as it takes a while for resonance to build up with sounds). So a wire has an Effective Cross Section of about a quarter of a wavelength. It follows that wires spaced by about a quarter wavelength will behave very similar to a metal plate.

My confusion is boiling down to the definition/illustration of the wave length in a vertically polarized wave...

 

[jonlg_uk]

This cross sectional image explains how a vertical antenna emits a wave.

http://www.measurement-testing.com/Electromagnetic-wave-transmission.jpg

Is it correct to say that if I was to place my faraday cage in the middle of that waveform, the only side of the cage that would allow wave to pass through are the sides that are perpendicular to the electric field? Assuming adequate wire separation distance

 

[sophiecentaur]

Actually, I was looking on the web and there are fewer diagrams of the fields around a dipole than I thought but -
this link or http://www.microwaves101.com/encyclopedia/absorbingradar1.cfm and even http://juluribk.com/2010/01/14/radiation-from-dipole/ show how Mr Hertz, himself, visualised what happens.

Someone beat me to it!

 

[sophiecentaur]

This cross sectional image explains how a vertical antenna emits a wave.

Is it correct to say that if I was to place my faraday cage in the middle of that waveform, the only side of the cage that would allow wave to pass through are the sides that are perpendicular to the electric field?


Assuming adequate wire separation distance

That diagram shows the Electric fields. They are parallel with the dipole (in the equatorial region). The 'holes' are where the field changes direction and there is a half wavelength between successive holes. Look at the rather jerky animation on one of my links, above. I have a feeling that you are confusing the Fields with the Waves. That diagram of yours explains it perfectly well.

The side of the cage that will be affected most will be the side which is parallel to the field that 'would have been there' and on the side facing the source. Of course, once the cage is there, there will be currents in it that will distort the field.
Think about it - you don't point an antenna at the transmitter. You put it parallel with the transmitting dipoles and at right angles to the direction of arrival. So the wires of the cage that count will be parallel to the incident E ('transverse') field.

Once the cage is in place, some currents will appear even around the sides, which is why just standing behind a flat screen of finite size will not give total screening. Currents find their way round the edges and radiate behind the screen.

 

[jonlg_uk]

...and on the side facing the source

Thanks sophiecentaur, you have really helped clear up a lot of confusion for me, the only thing I don't understand now is this ^^

If the side is facing the source head on then the electric field will not be inducing current in it?

OR do you mean the vertical edge(s) of the side facing the source?...now that I can understand...

 

[sophiecentaur]

Every face of the cage can be expected to have currents induced on it, actually, whatever the aspect, particularly when you have an open structure (imperfect).

To avoid confusion, I was discussing the face that is normal to the direction of the wave progress. This corresponds to the simplest treatment of the screen. Of course, the currents will be maximum in the plane of polarisation of the wave and currents will flow all over the face. Depending on the dimensions (in wavelengths) there will be an uneven distribution of currents with nodes and antinodes. Did I ever suggest that the currents would only be on a "vertical edge"?

 

[Born2bwire]

It should also be pointed out that a wavelength is both a temporal and spatial property. If we have a uniform plane wave, then we would see the phase of the fields vary in space over the distance of a wavelength in addition to the variation of the phase at a single point over the time of a single cycle. So the wavelength is not the distance between peaks in time (that would be the period), it is the spatial distance between peaks at a given point in time (assuming a monochromatic uniform wave).

 

[sophiecentaur]

I think you are taking a step too far with that one. WaveLENGTH is clearly a distance and Period is a time. They are different variables- hence the representation of a progressive wave as:

E = E₀Cos(ωt - kx)

And the well known relationship for wave speed:
c = λf

You can have an oscillation with no wave motion at all

E = E₀Cos(ωt)

Or a stationary 'picture' of a sinusoid:

E = E₀Cos(kx)