Electromagnetic field theory. Q and A

Pragmatist said:
For example if I add two waves each with total amplitudes of "1", then the result is not necessarily a single wave of amplitude "2", the actual result depends on the relative phases/frequencies/polarizations etc. In that sense it's more a "simple sum of complex parts", rather than a "simple sum", period, if you see what I mean.
Yes, I see what you mean. This what I was trying to get at with the "ripply ocean" vs. "one big wave" examples, but it bears stating explicitly.
 
Zombified said:
Yes, I see what you mean. This what I was trying to get at with the "ripply ocean" vs. "one big wave" examples, but it bears stating explicitly.

Thanks, I agree it's much clearer spelled out explicitly.

And while I'm at it, thinking about it, the issue of "interaction" between photons also could do with a little clarification. I'm thinking of a specific case where for example you have a plane polarized wave (which passes an aligned polarizer with 100% certainty) as the sum of two circularly polarized components in quadrature phase, each of which would only have a low statistical probability of passing the polarizer on their own. I think it's worth pointing out that although we can speak of a single composite entity (like the plane wave) as being an "interaction" between separate photons, it is still possible to decompose it back into into the original components. In that sense "interaction" is something of a convenience term, we can say that photons "interact" in the sense that a composite may have properties not explicitly "contained" in any of the individual components, but we can also say that they DON'T interact in the sense that we can (usually) recover those components unchanged from the composite. So, more accurately, they can "interact" in one sense but the components never lose their individual identities in the composition (in the more general low energy cases).
 
Pragmatist said:
I'm thinking of a specific case where for example you have a plane polarized wave (which passes an aligned polarizer with 100% certainty) as the sum of two circularly polarized components in quadrature phase, each of which would only have a low statistical probability of passing the polarizer on their own.
Yes, good clarification. This, actually, is more what I think of when I say superposition.

When I described interaction between photons above in terms of quantum mechanics, I'm talking about scattering photons off each other (the classical analogy would be bouncing billiard balls off each other). This is a completely seperate phenomenon from adding waves together, and is only significant at very high energies. Normally, and certainly for radio or visible photons, photons go right through each other.
 
Associating an H-field with photons seems a bit odd to me now. Introducing relativity to classical mechanics "gets rid of" the magnetic field. By that I mean the magnetic field is simply a result of the "force" travelling at c, whereas Maxwell's equations treated it as instantaneous in many respects.

With Maxwell's equations, a current in one line produces a magnetic field which in turn induces a current in a second line. However, this can also be modelled assuming electric fields travel at c. The electrons in the second line observe a doppler shift (due to the finite speed of propagation) in the electric field caused by the moving electrons in the first line. This doppler shift result in a net force in the second line causing those electrons to accelerate. In both cases we induce a current in the second line, whether we calculate a magnetic field or not.

In Maxwell's equations the magnetic field appears to be a manifestation of doppler shift or relativistic effects. When we speak of the H-field associated with a photon what exactly does it represent if we can speak of an H-field in this case at all.

Walt
 
In QED, photons don't really make E or H (or D or B) fields, they just make photon fields. E and H are classical approximations of the effects the photon field has.

Magnetic fields can just be considered the effect of a Lorentz transformation on electrical fields. This doesn't have anything to do with the speed of propagation of the field, though. If the current through a wire in constant, E and H aren't changing, so there's no change to propagate.

A current-carrying wire is electrically neutral, so the line charge density of positive and negative charge carriers is the same. Thus, there is no net electric field. However, they have different velocities - they are going in different directions. So if you transform into another frame of reference, so that the wire is Lorentz-contracted, the charge densities will not transform identically. Now, in this new frame of reference, the charge densities will be different, and so the wire will appear to have a net non-zero electric charge. Thus, there is an electric field there, and a charge at rest in that frame of reference will experience an electrostatic force.

Looking back at the original frame, you've got an electrically neutral wire, and a moving charge that is accelerating due to some force. That's the magnetic force on the moving charge. It turns out to match what you'd expect.
 
Zombified said:
Magnetic fields can just be considered the effect of a Lorentz transformation on electrical fields. This doesn't have anything to do with the speed of propagation of the field, though. If the current through a wire in constant, E and H aren't changing, so there's no change to propagate.
Ah, like an idiot I did the math right, and got the name wrong. Makes sense, since no where in the math does frequency come into account. D'oh.
Originally posted by Pragmatist
Anyway, to put it simplistically, near the antenna, the (many) photons are tightly superposed and the resultant "fields" don't have to have exactly the same impedance as an individual photon. But the antenna is transmitting a spherical wavefront (ignoring complexities of polar response), and as the photon flux heads outwards the same number of photons emitted in a certain limited range of time are distributed over the surface of an ever expanding sphere, so the individual photons are further apart and so their superposition is less obvious. There comes a point where the individual photons are not tightly superposed and their individual impedance (that of free space) becomes more obvious. So the reason the E/H ratio or impedance appears to change as you head outward from the antenna is because the further out you go, the more individual "photon like" the whole thing appears, whereas at close range the composite "mass" of photons appears to behave more like a classical "field". [/B]
Pragmatist, if the antenna was actual transmitting a spherical front I don't think you get different near-field and far-field patterns. I disagree a with 'the reason the E/H ratio or impedance appears to change as you head outward from the antenna is because the further out you go, the more individual "photon like" the whole thing appears'. You want notice anything more individual photon like until power density is sufficient low, however where the far-field pattern approximations work is independant of the power from the antenna. As 69dodge pointed out, we don't need to look at individual photon behaviour to see. In fact we can see the non-spherical wave-front produced by an antenna by modelling it as a sum of infinitesimal point sources producing many spherical fronts.

Walt
 
Walter Wayne said:
Pragmatist, if the antenna was actual transmitting a spherical front I don't think you get different near-field and far-field patterns. I disagree a with 'the reason the E/H ratio or impedance appears to change as you head outward from the antenna is because the further out you go, the more individual "photon like" the whole thing appears'. You want notice anything more individual photon like until power density is sufficient low, however where the far-field pattern approximations work is independant of the power from the antenna. As 69dodge pointed out, we don't need to look at individual photon behaviour to see. In fact we can see the non-spherical wave-front produced by an antenna by modelling it as a sum of infinitesimal point sources producing many spherical fronts.

Walt

I said it was a simplistic explanation, probably OVER simplistic! :)

I was trying to avoid getting into things like virtual photons which I thought may confuse some readers, but let me restate it in a slightly more rigorous fashion.

The EM "field" in the vicinity of the antenna is considered in QED to be an effect of the interactions of virtual photon exchanges. That of course begs two questions, firstly, "what is a virtual photon?" and secondly "interactions between what?". The first is almost impossible to answer since in effect what is called a virtual photon is defined largely in terms of its effects and interactions in QED, there is no direct physical model I am aware of, although in a classical sense I guess it would be analogous to an EM wave with the E and H fields out of phase (real classical EM waves always have E and H in phase). The second is easier, because in effect a field is defined by measurement, the interaction is between the source (antenna) and the measuring device. So if we measure the "field" (meaning generically E/H or EM field) in the vicinity of an antenna we are actually (to a large extent) measuring the virtual photon interaction between the measuring device and the electrons in the antenna - subject to one caveat - that we must also consider interactions between the real measurement point and all other possible measurement points as well - typical QED style phenomenon, "follow all paths".

As I explained in an earlier post there is some confusion between the terms "interaction" and "superposition", when I mentioned superposition before I was thinking inclusively in terms of interactions.

Now when we talk about how fields change with distance from an antenna in practice we are talking about how our MEASUREMENTS of fields change with distance from an antenna (i.e. there is always an implied measuring device that interacts with the photons, both real and virtual).

What I was trying to do was simplify a picture of this process by observing that the density of ALL photons decreases with the inverse square of distance. Therefore, near the source, the photon density is higher than further out. In addition to the actual PHOTON density, there is also an INTERACTION density as well, interaction density falls off as though the density were a function of points on the surface of an expanding sphere (to a crude approximation - that is NOT exact as far as I am aware, but it gives a generally understandable picture). And I am generalising by ignoring non spherical polar radiation patterns simply because it would unneccessarily complicate the picture - in other words my ideal antenna is a point source. The actual observed "field" at any point consists effectively of two components, a set of virtual photon interactions (some of them two way with the antenna) and a set of real photon interactions (from the antenna to the measuring device only). As you move outward from the source the density of both decreases so in effect there is less coupling between the measuring device and the source (lower virtual photon density) and there is also a lower real photon density as well. But because this is a QM phenomenon, in addition to the actual measuring point we can consider a series of other points on the surface of the same imaginary sphere as being "virtual measurement points". So there are also "sideways" (in a manner of speaking) virtual photon interactions between the real measurement point and the virtual ones. So the actual "field" is a highly entangled "mess" of virtual photon interactions between all kinds of points in space. Therefore the REAL photon density falls off simply as the square of distance, but the virtual photon density (which is proportional to the interaction density) is a complex power series in r. Obviously the contribution from higher order terms in r will fall off more rapidly with distance from the source - this is conceptually equivalent to saying that as the photon density decreases (both real and virtual), the coupling between photons decreases, the coupling between the real measurement point and the antenna decreases, and the coupling between all possible "virtual" measurement points decreases as well and therefore there are fewer interactions, thus less virtual photons.

That's the best way I can think of describing it! No doubt there are better explanations.

Anyway, at some point which we arbitrarily call the near field/far field "boundary" the density of interactions involving virtual photons (and therefore the generic "fields" which are the classical expression of those interactions) become effectively insignificant in RELATION to the density of real photons. This is why the power level doesn't matter, the issue is not a question of ABSOLUTE photon density but rather the RELATIVE proportion of virtual photon to real photon density. A virtual photon interaction is transient and virtual photons don't propagate, real photons do propagate. So analogously, the real photons are hidden in the mess of virtual photons near to the antenna, but further out, the interactions thin out, so does the virtual photon density and thus the real photons start to stand out as separate entities. When we measure a point in the far field, the coupling to the antenna is insignificant and so there are few or no virtual photon exchanges with the antenna, all that is left to measure is the flux of real photons which are not coupled to the antenna.

Zombified's kinetic gas analogy is a good one. In a dense gas there are many interactions between molecules of the gas and also between those molecules and an arbitrary point at the centre of the gas. As the pressure is reduced and the gas becomes more rarefied, there are statistically fewer interactions between gas molecules and the centre point (analogous to our antenna) and ALSO between individual molecules of gas. At some point the dynamic characteristics of a specific molecule in isolation start to become more significant than the dynamics of the interaction between that molecule and others. Hence the "true nature" of the dynamics of the isolated molecule starts to become more obvious. It's the same with our real photons.

I don't know if all that makes things clearer or confuses them further! :)

With respect to the issue of each point being a source of secondary wavefronts (Huygens' principle) that is another classical approximation. It is true in a classical sense that you could model a generic "field" as being composed of an infinite series of harmonic oscillators, therefore each point of the field is a "source" (oscillator). But we also know that this model doesn't work in practice because it is the same model that led to the ultraviolet catastrophe in statistical thermodynamics, which Plank resolved by quantizing the oscillators. Therefore an infinite field of continuous sources doesn't work in practice, it is only a classical first approximation, a quantized field of discrete sources would work however. And if you replace "discrete harmonic oscillator" with "virtual photon" you end up pretty much with the QED description.

Of course it does work as an approximation to the geometry of a real antenna. I'm not saying all antennas create spherical patterns, just that the idea of interaction density is easier to picture using a point source where complex geometries are not an issue.

The problem with all these explanations is that we have numerous different models for the same phenomena, classical, relativistic, and QM. There is much confusion because when we are familiar with all the models there is a tendency to keep bits from one model that seem "intuitively" correct and then misapply them to other models - rather like when we think of a photon as being comprised of classical E/H fields.
 
Originally posted by Pragmatist
This is why the power level doesn't matter, the issue is not a question of ABSOLUTE photon density but rather the RELATIVE proportion of virtual photon to real photon density.
Ok, that makes sense.

And with the rough correspondence<blockquote>static field = virtual photons
radiation = real photons</blockquote>the classical analogue is: the near field is where the Coulomb field dominates; the far field is where the radiation dominates. (The Coulomb field drops off as 1/r<sup>2</sup>; the radiation only as 1/r. So the near field is near, and the far field is far.)

Googling for "near far field" turned up a nice page, which talks also about the case where the near field is mainly magnetic, rather than electric.
 
I see this goes nicely without me :).

I wonder where Coghill is ;).

Hans
 
I know this one isn't strictly field theory, but I guess one of you lot will know an answer or three anyway;

The old conundrum of is light a particle or a waveform (as it can be experimentally shown to be either) ... is that just light that is being referred to or the whole EM spectrum?
 
Benguin said:
I know this one isn't strictly field theory, but I guess one of you lot will know an answer or three anyway;

The old conundrum of is light a particle or a waveform (as it can be experimentally shown to be either) ... is that just light that is being referred to or the whole EM spectrum?
That's the entire EM spectrum. When you talk strictly about photons, there's no qualitative difference between radio, microwaves, IR, visible, UV, X-rays, or gamma rays. The differences appear when light interacts with matter, because different materials (or constituents thereof) interact with particular frequency bands.

Actually in quantum mechanics, all particles also have a wave-like nature, including those you normally think of as "solid" like electrons or atomic nuclei. Measuring diffraction (a wave phenomenon) for electrons using a specially made CRT tube is not an uncommon exercise for a physics student.
 
MRC_Hans said:
Q:Are antennas designed so their impedance matches the air impedance?

A: Not necessarily. Impedance matching is a question of efficiency of energy transfer. It is preferable if the antenna is matched to the air (~300 ohms), but it is essential that it is matched to the transmitter. Thus various practical considerations lead to the use of antennas with other impedances. Also, for long wavelengths, it may not be practical to build a half-wave antenna, so lower fractions are used. Likewise, directional antenna designs often give other impedances.
Sorry I'm sort of rereading backwards here.

I thought if you didn't match the impedance of the antenna to the impedance of the medium its in the antenna would effectively reflect some of the power you put into it back into your gear, which wastes power? Isn't the idea to match the transmitter to the antenna rather than the other way around?

I'm not a radio expert by any means, so I could be wrong...
 
Zombified said:
Sorry I'm sort of rereading backwards here.

I thought if you didn't match the impedance of the antenna to the impedance of the medium its in the antenna would effectively reflect some of the power you put into it back into your gear, which wastes power? Isn't the idea to match the transmitter to the antenna rather than the other way around?

I'm not a radio expert by any means, so I could be wrong...

It's one of those theory/practice dichotomies. Yes, any impedance mismatch will cause reflection of power, and yes, ideally, IN THEORY, the antenna should be matched to free space.

However, in practice, you have to remember that real antennas are not beautifully isolated mathematical constructs. For example a real antenna will have objects in its near field that will couple to it, so in practice no matter how well calculated the antenna is, the real effective impedance that is presented to the transmitter will be modified by near field coupling to things that have nothing to do with the antenna itself (like the ground, trees, buildings etc). Therefore in practice one would try to design an antenna optimally by taking major factors into account (i.e. the ground), but its real impedance once it is sited will almost certainly be different from what you expected. So in practice you would tend to tune up the feeder circuit (i.e. the link from the transmitter) to reduce VSWR locally (between transmitter and antenna) - practically that usually involves some form of impedance transformer like an antenna tuner.

Ultimately the method used by most RF engineers is to simply play with antenna parameters and siting until the received signal at some far field point is maximized and at each adjustment you would balance the transmitter to the antenna via the tuner which is inline between the transmitter and antenna.

The reflected power from impedance mismatch between the antenna and free space tends to end up in the near field rather than reflecting back to the transmitter. Of course this can potentially cause problems which might need further sub optimal adjustments. For example, you could get arcing across a dipole or actual RF heating of nearby objects. But overall it tends to be something of a "fiddle with it until it works" situation.

The REAL secret of RF engineering is to make lots of impressive looking calculations, plot Smith Charts and generally LOOK as though you are doing something highly technical, complicated and magical (for the edification and benefit of lay onlookers) whilst simply randomly twiddling knobs until the darn thing actually works....! :D
 
Oh, I forgot to mention above that Hans is absolutely correct, that it is essential to match the antenna to the transmitter. The antenna tuner, which is usually the only adjustable part of a practical setup, is typically considered to be part of the antenna rather than the transmitter, although in practice it doesn't matter which way round you name it (i.e. matching antenna to transmitter is the same as matching transmitter to antenna). If you don't match the feed between transmitter and antenna, then power will be reflected back into the transmitter and will burn out the output stage drivers of the transmitter, or it will cause the feeder cables to heat up and even burn. At high powers you could even melt the metal connectors. Transmitter output stages are usually very sensitive and can easily be overloaded, typically a reverse voltage transient can easily exceed the breakdown voltage of the transmitter output transistors.
 
MRC_Hans said:
I see this goes nicely without me :).

I wonder where Coghill is ;).

Hans

Based on past experience, I would imagine that Coghill himself doesn't know where he is....! :D

Come on, Rog! You were complaining before that Hans hadn't kept his promise to start this thread - well he has. And since this was principally for your benefit you could at least extend the courtesy of contributing to it.
 
Pragmatist said:
Based on past experience, I would imagine that Coghill himself doesn't know where he is....! :D

Come on, Rog! You were complaining before that Hans hadn't kept his promise to start this thread - well he has. And since this was principally for your benefit you could at least extend the courtesy of contributing to it.

Now, now, after having seen the huge E/M gaffes rogbot has made you expect him to relish further exposure? Especially with his apparent beliefs about E/M exposure? ;)
 
Pragmatist said:
The REAL secret of RF engineering is to make lots of impressive looking calculations, plot Smith Charts and generally LOOK as though you are doing something highly technical, complicated and magical (for the edification and benefit of lay onlookers) whilst simply randomly twiddling knobs until the darn thing actually works....! :D
Well, yeah, that part I knew. :)
 
Pragmatist said:
*snip*
The REAL secret of RF engineering is to make lots of impressive looking calculations, plot Smith Charts and generally LOOK as though you are doing something highly technical, complicated and magical (for the edification and benefit of lay onlookers) whilst simply randomly twiddling knobs until the darn thing actually works....! :D
Yeah, we used to say that receiver thechology was 90% science and 10% magic, transmitter technology was 75% science and 25% magic, and antenna technology was 50% science and 60% magic (yeah, 50/60 ;)).

However, I think the mobile phone industry has moved those figures in favor of science, lately, otherwise I simply don't think they could get the power density and efficiency they have these days.

Hans
 
* BUMP *

It is a pity that Roger Coghill never showed up here. He is actually very silent for the time being. Maybe he is too busy pedling his magic water or the magnetic therapies at some important conference?
 
Do wireless phone accessories that claim to shield the head from RF radiation work?

Since there are no known risks from exposure to RF emissions from wireless phones, there is no reason to believe that accessories that claim to shield the head from those emissions reduce risks. Some products that claim to shield the user from RF absorption use special phone cases, while others involve nothing more than a metallic accessory attached to the phone. Studies have shown that these products generally do not work as advertised. Unlike "hand-free" kits, these so-called "shields" may interfere with proper operation of the phone. The phone may be forced to boost its power to compensate, leading to an increase in RF absorption. In February 2002, the Federal trade Commission (FTC) charged two companies that sold devices that claimed to protect wireless phone users from radiation with making false and unsubstantiated claims. According to FTC, these defendants lacked a reasonable basis to substantiate their claim.

From this site:
http://www.fda.gov/cellphones/qa.html#6a

Maybe Coghill would be interested in this.
 

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