In other words, for a given point in a transmitted wave, are the induced charge distributions on a receiving antenna the opposite of the antenna which created it?
We know that like charges repel and opposites attract. If some point of an EM wave was generated by a downward pointing e-field (vectors pointing from positive charges towards negative), then when that point reaches the receiving antenna, the charge carriers (electrons) would be forced to move in the direction opposite of the vector.
If some point of an EM wave was generated by a downward pointing e-field (vectors pointing from positive charges towards negative), then when that point reaches the receiving antenna, the charge carriers (electrons) would be forced to move in the direction opposite of the vector.
The premise is pretty simple: conductors interacting with external EM radiation (that is, radiation emitted from something else), re-radiate the energy (whatever hasn't been damped), but with the wave inverted.
conductors interacting with external EM radiation (that is, radiation emitted from something else), re-radiate the energy (whatever hasn't been damped), but with the wave inverted.
I assume re-radiation is a well-established explanation. We can further use this to explain the establishment of standing waves when EM radiation interacts with a conductive surface, with one of the nodes being at that surface.
I suspect this will sufficiently explain those behaviors, including the "three polarizer puzzle". All without having to resort to the insanity of QM probability distributions etc etc.
I assume re-radiation is a well-established explanation. We can further use this to explain the establishment of standing waves when EM radiation interacts with a conductive surface, with one of the nodes being at that surface.
re-radiation is not immediate process. It takes some time delay between EM wave consumption and radiation. This time delay exists between two events. And this time delay depends on matter type. This is why wave propagation in the matter is smaller than the speed of light in the vacuum. Each particle of matter store energy for some period of time and re-radiate it after some delay. It leads to lower wave propagation speed.
I suspect this will sufficiently explain those behaviors, including the "three polarizer puzzle". All without having to resort to the insanity of QM probability distributions etc etc.
I don't think that there is needs to involve QM, photons and other complicated stuff to explain "three polarizer" effect, in my opinion it can be done within classical (non quantum limit) theory with pure wave electrodynamics.
In my opinion the secret of "three polarizer" is that polarized film doesn't "filter photons", it consume and re-emits EM wave. And when you place 45 degree third polarized film between two other, it can consume half of EM wave passed through first polarized film and re-emit a new EM wave which will be 45 degree for the third polarized film. So, the third polarized film will pass a part of that EM wave. So, there is no quantum effect, this is a classic wave behavior.
We can't observe single radio frequency photons. Or even microwave photons. The reason is our antennas are utterly swamped beneath a sea of them generated within the antenna itself by Johnson noise, also known as blackbody radiation. We can only observe the superposition of enormous numbers of them, veritable tidal waves of them, large enough to overwealm this thermal noise. They exhibit wave behavior.
:wtf: Well, then, why has this "three polarizer effect" never, in the 134 years since Hertz, been seen by anyone working with radio or microwaves?
Might it be... :-// because polarizers simply don't "consume and re-emit a new EM wave"? It appears not!
Dirac first posed the "three polarizer paradox" of light in 1930. Here's a discussion https://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/ (https://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/) Light simply does not act like a classical wave.
Reflection, whether off a mirror or off the shorted end of a transmission line, occurs instantaneously. There is no time delay.
If there were any fixed time delay, it would result in a shift in phase.
If there were any random time delay, it would result in a loss of phase coherence.
No "absorption, delay, and re-emission" occurs.
Reflection and refraction are instantaneous and are predicted by classical wave equations.
Your illustration of standing wave is very nice.The premise is pretty simple: conductors interacting with external EM radiation (that is, radiation emitted from something else), re-radiate the energy (whatever hasn't been damped), but with the wave inverted.
Correct... but only if the conductor has the same voltage (electric potential) at all points along it. That can occur only if currents are flowing in the wire which create an EM field which exactly cancel the external EM field. This is how a mirror works. This is why a ground plane acts as a reflector, a mirror. This is also why an infinitely long wire (one far longer than the wavelength) does not act as a receiving antenna: it would reflect everything and receive nothing. It would create a standing wave, exactly as you illustrate. How does it do this?
A key fact you've failed to consider: current does not flow at infinite speed in a conductor, but only at the speed of light. This is true no matter how long the conductor is.
If you accept this reality, then you can figure out how antennas work. Antennas work as antennas only if they are 1/4 wavelength long (or an odd multiple thereof, as described here https://en.wikipedia.org/wiki/Dipole_antenna (https://en.wikipedia.org/wiki/Dipole_antenna). This is what I attempted to suggest in the previous posting, but you didn't get: antennas resonate in-phase with the EM wave.)
And if you accept this reality, you can also come to understand why an infinitely long wire will not act as an antenna, but as a reflector. Each 1/4 wavelength along it, at each instant in time, the external EM field attempts to induce an opposite voltage, and an opposite current on it. The is exactly what you state in the quote above. But as time continues, these currents and resulting voltages will continue propagate down the wire in both directions, and at the speed of light, exactly cancelling one another at every point along the wire at every instant in time.
A quarter-wave antenna is different, only because no current can flow in it beyond it's ends. It is as simple as that!
As a consequence, a voltage develops across the antenna (and, 90 degrees out of phase with it) a current flows along it. The ends of an antenna are current nodes, and voltage anti-nodes. It is not out of phase with the external EM field, it is in phase with it.
Now, let's turn to linear polarizers. A simple quarter-wave dipole transmitting antenna emits a linearly polarized EM wave. A receiving antenna parallel to it receives it, and one perpendicular to it does not. Same for a second receiving antenna behind the first. No "three polarizer paradox" occurs. The receiving antennas do not alter the polarization of the transmitted EM wave.
A passive receiving antenna does also emit an EM wave, in phase with the one it receives. The two EM waves do interfere, constructively on-axis, and destructively off-axis, together creating a directional antenna. The elements are not acting as reflectors, but as in-phase re-radiators. But one may appear to act as a reflector if it is spaced 1/2 wavelength behind the transmitting antenna. (At a receiving antenna, they may appear out-of-phase depending only on the total difference in path length from the transmitting element to each passive element and then to the receiving antenna. That's why such an array is directional. But most directional antennas have multiple driven elements, with phase differences between them. Lots of examples here https://en.wikipedia.org/wiki/Category:Radio_frequency_antenna_types (https://en.wikipedia.org/wiki/Category:Radio_frequency_antenna_types))
The fact that all these work is proof your intuition "At first glance, it would seem to require that the charge distributions on the receiving antenna at a given point in time be the opposite of those which created that point at the original radiator." is incorrect. We observe exactly the opposite. (This fact is so fundamental it even has a name: Lorentz reciprocity https://en.wikipedia.org/wiki/Reciprocity_(electromagnetism) (https://en.wikipedia.org/wiki/Reciprocity_(electromagnetism)), reflecting its consequences.)
The "three polarizer paradox" does occur with light. It is one proof that light does not propagate purely as a wave. It is somehow localized... as if it were propagating as a particle, not as a wave. Sorry, but that cannot be rationalized by misunderstanding waves, nor how antennas work. nor by confusing antennas with their opposite: reflectors.
I have to apologize. I was so distracted into attempting to clarify misunderstandings on how antennas, reflectors etc work that I missed your underlying goal:My question actually stems from what I consider to be a completely unsatisfactory classical description of "simple" linear polarizing filters. ..this concept on the behavior of linear polarizing filters, I suspect this will sufficiently explain those behaviors, including the "three polarizer puzzle". All without having to resort to the insanity of QM probability distributions etc etc.
What's "unsatisfactory" about it, exactly?
msat, it appears you don't understand why the "three polarizer paradox" is a paradox: it is irreconcilable with how waves work. This paradox only appears when we start observing single photons. It doesn't appear when we're observing waves.
We can't observe single radio frequency photons. Or even microwave photons. The reason is our antennas are utterly swamped beneath a sea of them generated within the antenna itself by Johnson noise, also known as blackbody radiation. We can only observe the superposition of enormous numbers of them, veritable tidal waves of them, large enough to overwealm this thermal noise. They exhibit wave behavior.
That you'd propose to devise an experiment in which waves do not act like waves, well, reflects a fundamental misunderstanding of how waves work. I encourage you to continue to pursue understanding waves; it's very useful.
(By the way, QM is not insane, nor is it sane. It is counterintuitive, but it's simply the way mother nature works. It proves our intuition is flat wrong. Physicists were forced to this realization a century ago. We can learn from them. Attempting to cling to intuition in the face of proof it's wrong... well, that wouldn't be sane, but is perquisite for the Flat Earth Society, "free energy" suppression and countless more conspiracy theories. No one in the EEVblog community is going there. Not because of any need for conformity here. Simply because: it don't work.)
If they overlap the displacement current will be clearly opposed simply from capacitive effect, I don't see any reason why it would suddenly change at multiple wavelengths. So yeah 360*x+180 phase delay, but still just 360*x group delay of course.
I don't think that there is needs to involve QM, photons and other complicated stuff to explain "three polarizer" effect, in my opinion it can be done within classical (non quantum limit) theory with pure wave electrodynamics.
In my opinion the secret of "three polarizer" is that polarized film doesn't "filter photons", it consume and re-emits EM wave. And when you place 45 degree third polarized film between two other, it can consume half of EM wave passed through first polarized film and re-emit a new EM wave which will be 45 degree for the third polarized film. So, the third polarized film will pass a part of that EM wave. So, there is no quantum effect, this is a classic wave behavior.
And let's involve a little math. For example polarized film A consume light on the input and re-emits it with horizontal polarization. Now let's see what will happens with other polarized films:
1) When you place polarized films B with 90 degree to A, the output on B will be:
cos(90°) = 0
2) When you place polarized film B with 45 degree to A and polarized film C with 45 degree to B, the output of polarized film C will be:
cos(45°) * cos(45°) = 0.7071 * 0.7071 = 0.5
As you can see, it's pure waveform effect. And there is no needs to involve quantum effects to describe that. :)
Why no resistor in the middle of the receiver?
Shrug, you are not measuring what you asked about in the original post then. A receiver antenna is terminated, not a resonator.
I'm thinking of working in the ~6MHz ISM band due to only having a 25MHz scope. This also makes locating the antennas kind of easy at the expense of having somewhat unwieldy distances to work with.
but I think you have a misunderstanding of how polarizers work. So once again, I'll link a microwave polarizer demonstration: https://sciencedemonstrations.fas.harvard.edu/presentations/microwave-properties (https://sciencedemonstrations.fas.harvard.edu/presentations/microwave-properties). In the experiment, the wave is nulled if the transmitted wave is vertically polarized and the polarizer's wire element are vertically oriented. Therefore, the only way to make sense of this classically is if the "filter" inversely re-radiates the absorbed energy (effectively 180 degrees out of phase), which cancels out at least part of the original wave
one can show that the radiation transmitted by the filter is polarized in a new direction which depends on the orientation of the grid
What I've been trying to ask about is the charge distribution on a conductor (such as an antenna) when an e-field is induced on it by external EM radiation.
I did not mention "phase delay" - I talked about inverted re-radiation which, when considering a pure sine wave, would essentially appear as being 180 degrees out of phase, but not that it actually is!
Your experiment would not work due to reasons I already outlines.
Attached is a graphic I made to describe the effect I'm positing
your picture has mistake. Since both antennas placed at the same E field they both will have the same polarity on their terminals. But your picture shows inverted polarity. This is incorrect.
Inverted polarity will be if you move antenna into center of your image (where phase delay is 180 degree).
If you look closely, you will see that, as far as voltage measurements go, the polarities WILL be the same. And it's because of these polarities that cause the redistribution of charge carriers to try to cancel these fields. By what other mechanism would there be a current in a receiving antenna?!
I'm trying to design an experiment to test it. You say I'm wrong, which may very well be true, but you have not provided supporting evidence.
Just telling me something is some way does not constitute proof.
I argued AGAINST QM being the only viable explanation for the 3 polarizer paradox. So we can drop the QM thing too. My entire intent is to provide a CLASSICAL explanation for the paradox.
You say it already exists, but I have yet to see one that addresses experimental results. What I'm proposing I believe WOULD explain experimental results.
But that voltage will cause current to flow, right? Which way will that current flow?
[
You cannot argue against QM, because there is no paradox and no needs to involve QM to explain such behavior, because this is classic wave behavior.
First polarizer absorb wave and re-emit only vertical polarization. When vertically polarized wave falls on second 90 degree polarizer, it will not be re-emited and will be dissipated as heat, because cos(90°) = 0. When it falls on 45 degree polarizer, it will be absorbed and re-emited at 45 degree, and this is why it will be passed through third polarizer, because cos(45°) * cos(45°) = 0.5.
This is why light didn't passed through 2 polarizers and passed through 3 polarizers.
You're just needs to understand how polarizer works. It rotate polarization. That's it.
Exactly the same thing happens with RF polarizer grid. When you place vertical grid, electrons can move only in vertical direction (along grid wires) and cannot move in horizontal direction. This is why re-emited wave will have only vertical polarization.
For both antennas current flow direction will be identical at one lambda distance. If it's not, you can be sure that the distance between antennas is not multiple of wavelength.
And this is known for at least 100 years.
The explanation given in that page (and ALL others that I've read) is that the energy is absorbed and converted to heat. If that's true, the rest of the experiment CANNOT be explained that way. What *could* explain it, however, is if the polarizer does indeed re-radiate the energy parallel with the filter wire elements but only if it is INVERTED (basically 180 degrees out of phase for a symmetrical wave).
In a transmitter, the amplifier forces the different charges in a dipole, and these charges are responsible for creating the electric field around the antenna and the magnetic field around the antenna due to the flow of charges.
A receiving antenna CANNOT be explained the same way! In a receiver, there is not an amplifier moving the charges around to each leg of a dipole.
Very much appreciate your diagrams, msat. "A picture is worth a thousand words" and communicate what you're thinking very clearly. :-+
Question: How fast can a voltage, or a current, or any electrical signal, travel down a wire: instantly, or only at the speed of light?
Your diagrams suggest you do agree that electric field waves travel through space at a finite speed, the speed of light, not instantly. But your diagrams also suggest you assume voltages travel down a wire (an antenna) instantly, at an infinite speed. Specifically, your diagram shows a receiving antenna having the same voltage along its entire length, and the voltage at the end of a receiving antenna wire arriving instantly a quarter-wavelength away at its terminal. That can only occur if the voltage at its end propagates instantly down the wire. Do you believe it does? Or if it does not, how would your diagram change?
Paradoxically, your diagram also shows a transmitting antenna emitting an electric field which looks (correctly) like a sine wave along its length, highest near the ends of the antenna and zero at its center terminals. This (correctly) implies it was created by an alternating voltage on the transmitting antenna wire which is greatest at its ends, and is near zero at its terminals. This can only occur if (and because) the voltage along the length of the transmitting antenna wire does not travel at infinite speed, but only at the speed of light. (If the voltage travelled instantly at infinite speed down the transmitting antenna wire, the entire wire would be at the same voltage at every instant in time, and would emit along its length a square wave, not a sine wave.)
Question: How fast can a voltage, or a current, or any electrical signal, travel down a wire: instantly, or only at the speed of light?
I think there's an error in your interpretation of what I'm positing, as it doesn't require any action that exceeds the speed of light to the best of my knowledge.
I think there's an error in your interpretation of what I'm positing, as it doesn't require any action that exceeds the speed of light to the best of my knowledge.
No. Your knowledge is wrong. According to your picture the voltage at the end of dipole is the same as voltage on antenna terminals. It means that wave propagation speed between antenna terminals and the end of wires is infinite and breaks speed of light limitation.
In reality the dipole doesn't have constant voltage across dipole length. There is always fixed time delay between you apply voltage to the terminals and when that voltage appears at the end of dipole wires. This phase delay will be added between electric field change and transmitter voltage change for transmit antenna. And the same this phase delay will be added between electric field change and voltage change for receiving antenna.
You didn't take this time delay into account. That's your mistake. You're needs to understand that charges cannot move across antenna length immediately it needs some time delay. This is very important to take this into account, otherwise your model for charge distribution will be completely incorrect.
the same for a transmit and receive antenna that are in phase. I'm merely positing that they are opposite - and I think there's pretty good evidence that it's true because it possibly explains other phenomenon.
your mistake is that you ignore wave propagation delay in the antenna wire. For half wavelength dipole it leads to a phase delay about 90 degree. If you sum phase delay for transmitter and receiver antenna phase delay will be about 90+90 = 180 degree. If you add 180 + 180, you will get 0 degree phase offset on receiving antenna. You can also add 360 degree phase delay for wave propagation in the space between antennas and you will get 0+360 = 0 degree phase offset.
Since terminals on both antennas are in phase and frequency is the same, this is just impossible to get different charge distribution on both antennas.
We are discussing two half-wave dipole antennas, spaced N wavelengths apart, N = 0,1,2,3...
However, the voltage across the terminals of the transmitting and receiving antennas are anti-phase, 180 degrees out of phase, with one another.
3) if the terminals are tied to a transmitter, R<0, the voltage across the terminals is anti-phase with the current through the terminals, adds to the standing wave current through them, adding power to the antenna. It is a transmitting antenna.
R = (Zm / (2*pi)) * ((EulerGamma + log(2 * k * l) - Ci(2 * k * l) +
cos(2 * k * l) / 2 * (EulerGamma + log(k * l) + Ci(4 * k * l) - 2 * Ci(2 * k * l)) + sin(2 * k * l) / 2 * (Si(4 * k * l) -
2 * Si(2 * k * l))));
X = (Zm / (2 * pi)) * (Si(2 * k * l) + sin(2 * k * l) / 2 * (EulerGamma + log(k * l) + Ci(4 * k * l) -
2 * Ci(2 * k * l) - 2 * log(l / r)) + cos(2 * k * l) / 2 * (2 * Si(2 * k * l) - Si(4 * k * l)));
1) probe the voltage on the ends of the transmitting and receiving antennas (which we know will be in-phase), or
2) probe the voltage on their terminals (we know will be anti-phase, but he doesn't, so he will be misled).
Repeating, the current across the terminals of the transmitting and receiving antennas are in-phase. However, the transmitter (R<0) has an output voltage that increases that current, while the receiver input resistance (R>0) produces a voltage which decreases that current. The voltage across their terminals are 180 degrees out of phase.
Observe identical current is flowing the same direction in both the battery and the resistor. However, the voltages across the battery and the resistor are also identical but are in opposite directions.
If there is anything above you disagree with, please say so!
If you know second Kirchhoff's rule, you know that the sum of voltage drop will be zero if there is no EMF source in circuit.
The same you will be unable to detect if antenna transmitting or receiving when you measure voltage on antenna terminals. Exactly the same voltage can be for transmitting antenna and for receiving.
Does KVL only apply to circuits which do not contain an EMF source?
Does KVL not apply to a circuit loop containing a battery?
Please explain your understanding of KVL to me. (It appears to be different than that in any textbook?) Thank you!
Really?! That's quite remarkable.
But... you animation does show a voltage across the antenna terminals.
Assuming two identical antennas, with one transmitting [a sine wave] and one receiving, spaced apart by some whole number multiple of the wavelength, and both probed identically and fed to an oscilloscope, would the waves be in phase, or 180 degrees out (or something else?!)?
Really?! That's quite remarkable.
Yes, really. In order to detect energy flow direction through antenna terminals, you're needs to connect antenna through directional coupler. Such approach is used for SWR meters to measure direct and backward energy flow. Measuring voltage on antenna terminals is not enough to detect RF energy flow direction.
Just because you will be unable to distinguish voltage from forward wave and voltage from backward wave, they both will be summed on antenna terminals due to wave superposition principle. Even if there is just a single direction wave flow, you can see wave on terminals, but you will be unable to detect direction of that wave flow through terminals.
So, if you measure just voltage on antenna terminal, you will be unable to distinguish where is the source of that voltage. You will be unable to detect if wave flows from antenna to transceiver or from transceiver to antenna.
You can distinguish it if you know wave flow direction. But you will be unable to detect wave flow direction by measuring voltage on antenna terminals. Both wave flow directions will looks exactly the same for your voltmeter on antenna terminals.
I can say more. You can just disconnect antenna from antenna terminals and your voltmeter will show you the same wave on terminals, but that wave doesn't flow into antenna because it is completely disconnected. :)
Of course you can: the simplest way to measure current flowing in any wire is to insert a small current shunt resistor and measure the phase of the voltage across it
Any directional coupler relies on precisely the principle we are discussing: by KVL, the voltage and current change in sign (change in phase by 180 degrees) if the direction of the power flowing through it is reversed.
Quote from: msat on Yesterday at 07:47:48 pm (https://www.eevblog.com/forum/index.php?topic=289474.msg3688540#msg3688540)...we might (ahem) reflect (intentional pun, shamefully admitted) on the effect of the Earth ground plane on the numerical values of R21 and X21 you've calculated between two isolated half-wave dipoles. I think it reduces the numerical values of both R21 and X21 by a factor approaching 2, but has no effect on your conclusions as to their spacing.I'm thinking of working in the ~6MHz ISM band due to only having a 25MHz scope. This also makes locating the antennas kind of easy at the expense of having somewhat unwieldy distances to work with.
Realize the wavelength at 6 MHz is 50 meters. Your proposed antennas will be 25 meters long, and have to be suspended above the ground plane (the earth) to work. You propose to space them an integer number of wavelengths apart. This experiment will span a football field. The long coaxial cables from your receivers to the oscilloscope will be expensive...
cbutlera, thank you! :-+ Your contribution here is a gift, and is spot-on.
Of course you can: the simplest way to measure current flowing in any wire is to insert a small current shunt resistor and measure the phase of the voltage across it
No. If you insert shunt resistor, the standing wave mode will be changed and your measurement will be not relevant. Because when you add shunt, you will measure different circuit with different wave modes.
And you will be unable to use resistor shunt to measure amplitude of two waves moving in opposite direction in the same wire. Which is needed to distinguish forward power flow from a backward power flow.Any directional coupler relies on precisely the principle we are discussing: by KVL, the voltage and current change in sign (change in phase by 180 degrees) if the direction of the power flowing through it is reversed.
directional coupler is not intended to measure sign and phase. It measure amplitude of two waves which are flowing through cable simultaneously in two opposite directions - forward and backward wave. And these measurements are always positive, because amplitude cannot be negative.
There is no sense to measure phase between two waves in the cable because such phase difference is always changed in time from 0 to 360 degree in a loop, just because waves are moving towards each other.
You cannot simulate wave interference and other wave properties and wave effects of two waves with using DC circuit.
Transmitter impedance doesn't affect antenna impedance. Since energy loss in the transmitter and feeder line doesn't matter here, we can just assume that transmitter Zout = 0 + j0 Ω for simplicity...
If we use dipole length = 0.4775072*lambda and r=0.001, the dipole impedance will be Z = 63.665 + j0.000 Ω.
And since reactance on antenna terminals is X=0.000 Ω, the current and voltage on antenna terminals will be in phase. Isn't it? ;)
If you measure voltage on antenna terminals, you will be unable to detect energy flow direction and unable to detect current source - antenna or feeder...
Since transmitter working at resonant frequency of antenna, reactance on antenna terminals will be X = 0 Ω. It means that RF energy is not returned back to the transmitter and remains in antenna (antenna holds it in form of reactive field oscillations).
The same you will be unable to detect if antenna transmitting or receiving when you measure voltage on antenna terminals. Exactly the same voltage can be for transmitting antenna and for receiving.
One of my "conservative" engineer co-workers stated that he didn't "believe in quantum mechanics". I told him that he would have to give up using solid-state electronicsHe'll also have to give up believing in the stability of the atoms making up his body. According to classical physics, their electrons should radiate all their energy away and spiral into the nucleus in a very small fraction of a second.
This would seem to imply that if those filters in the link I just posted were superconductors, then they would no longer act as polarizers.In that case they would reflect all of the energy they didn't pass through.
- the small shunt resistor develops a small voltage across it which has zero degrees phase difference from the voltage on the line,
This is a marvel of internal inconsistencies!
First, what voltage? If X=0.000 Ω, then V = 0 and its phase is... simply undefined! Your point is meaningless.
Really! Well, I guess that senseless statement makes sense if you do attach the antenna to a X = 0 Ω transmitter or receiver, nothing will be transmitted or received, and 0 == 0. But that's not a very useful case, now is it?
So indeed your assumption "we can just assume that transmitter Zout = 0 + j0 Ω for simplicity..." is indeed simple: it means nothing happens at all! What can be simpler than that? :-// Umm, nothing!
I don't think most of the stuff that's been said here about transmitting and receiving antennas is all that relevant, because the wires in that setup are not acting as resonant dipoles. They're more like the "infinitely long wires" that were mentioned in another post. So figuring out the phase relationships in dipoles at resonance isn't going to help you much.
The wave re-emitted by filter is 180 phase degree with original incident wave, so it cancel original wave due to wave superposition principle.
This would seem to imply that if those filters in the link I just posted were superconductors, then they would no longer act as polarizers.
In that case they would reflect all of the energy they didn't pass through.
Usually the intuitive, non-math explanation of receiving antennas is avoided in introductory texts. Instead, they steer around the problem by only explaining transmitters and the emission of EM waves. Yet the absorption of EM waves by receiving antennas is quite fascinating. It's also a minor 'hole in physics' which needs some serious filling. (Phd engineering students take note!)
In order to absorb EM waves, antennas must always emit EM waves at the same time. (To receive, we must transmit.) This process involves interference patterns and wave-cancellation.
Whenever some EM plane-waves strike a dipole antenna, they induce a voltage across the two antenna elements, and they produce a current along the antenna. Yet an antenna with a current must be producing a magnetic field. And an antenna with a voltage across it must produce an e-field. These induced fields won't just sit still: our receiving antenna starts transmitting! The dipole antenna emits an EM sphere-wave which spreads in all directions. But this pattern of outgoing waves is superposed onto the incoming plane-waves, producing an interference pattern. In addition, the phase of the emitted wave is opposite that of the incoming plane-waves.
Far downstream from the receiving antenna, the plane-wave is being cancelled by the antenna's sphere-wave. Our receiving antenna casts a shadow, it "punches a hole" in the pattern of incoming plane-waves. If we could see this shadow, it would resemble a bulls-eye pattern: a dark central disk surrounded by circles of nodes and antinodes. It's an interference pattern, but it's different from typical patterns: some energy has gone missing. It does have the expected dark nodes, and high antinodes, but the average total energy is less than the energy in the two original waves. That's how EM absorption works. The receiving antenna has launched a wave which "sucks in" energy from the incoming wave. The reception of EM waves is a wave-cancellation process.
This wave-cancellation explains how light-absorption works; how atoms absorb photons, how objects cast shadows, and how antennas receive radio wattage. But it has odd features. Normally if we add two EM wave-patterns, the resulting intereference pattern has the same energy as the two waves being combined. (The extra energy found in the "peaks" has just been moved out of the "troughs," so total energy is always conserved.) Yet this conservation rule is broken when the sphere-waves emitted by a receiver are added to the plane-waves passing by. Instead, some net energy has gone missing. This energy ends up inside the receiving antenna. It's the net absorbed energy, the "received energy," and in the simplest case would heat up the load-resistor attached to the antenna wires. (Or, if no load resistor was present, it would heat up the antenna itself.)
This entire wave-absorption process has consequences. For example, the receiving antenna has thrown off waves in all direction. It absorbed some energy, but it also reflected energy backwards, and to the sides. These "scattered waves" are required by the receiving process, and are always present in both RF physics and in optics. At best, an antenna can only receive 50% of the waves with which it interacts. It must scatter another 50% away. (Notice that when an atom absorbs photons, it always must emit other photons as part of the process. This is absolutely required by the physics, but rarely mentioned in QM courses.)
Have you ever sat down and tried to grasp the operation of Yagi/Uda antennas? Or the log-periodic, or the Rhombic with it's weird little load-resistor? Or, how about the "Effective Aperture" concept, where receiving antennas can behave much larger than their physical size? Nearfield NSOM microscopes and evanescent waves for wireless power? Or, examine the tiny ferrite coil inside old-school AM radios, how can such a small device absorb significant energy, and why does it always have a variable capacitor attached? Look at crystal radios: they always stop working when the tuning circuit is opened (because the LC resonator was never just a filter that removes unwanted stations!) Terk and Select-a-Tenna sell weird products which violate basic physics?!
All these apparently-odd situations require that we understand the wave-cancellation process described above. They only seem obscure and mysterious. It's all because we usually haven't been taught the basics we need to dissect them. The majority of introductory textbooks do us a great disservice: they insist that we only learn about wave-emission. Then we're supposed to automatically understand everything about wave-absorption ...because reciprocity?
Uh, no.
Uhhhh.... Isn't this EXACTLY what I've been saying throughout this entire thread which you've been disagreeing with me about?!?!
Notice that when an atom absorbs photons, it always must emit other photons as part of the process. This is absolutely required by the physics, but rarely mentioned in QM courses.
Uhhhh.... Isn't this EXACTLY what I've been saying throughout this entire thread which you've been disagreeing with me about?!?!
What I disagree is that polarizer filter cancels wave with specific phase offset. The polarizer doesn't deal with phase offset. It deals with polarization of wave.
The wave re-emitted by filter is 180 phase degree with original incident wave, so it cancel original wave due to wave superposition principle.
Notice that when an atom absorbs photons, it always must emit other photons as part of the process. This is absolutely required by the physics, but rarely mentioned in QM courses.
No. When electron absorbs a photon, it doesn't emit a new photon. On the contrary, original photon is collapsed and it's energy is transfered into movement of electron. Otherwise it will break conservation of energy principle.
I suggest you to forgot about photons and quantum physics and don't use it all, because it confuses you, there is no needs and no sense to involve such things as photons and quantum effects to explain these things. Quantum effects do not have a noticeable effect on the discussed things.
People knew how antenna and polarizer work many years before concepts such as photon and quantum physics were introduced, and these knowledge still works, QM doesn't cancel it. So there is no needs to involve photons to describe EM field behavior at non quanum limit :)
So which one is it?
And for some reason, in that entire quote, what you extracted from it is the bit about QM. And even still, the author wasn't wrong.
As best we know, any absorbed "photon" will eventually be re-emitted.
I somewhat unintentionally came across a stackexchange post a few weeks back (https://electronics.stackexchange.com/questions/132325/how-does-a-receiving-antenna-generate-current-voltage/146766 (https://electronics.stackexchange.com/questions/132325/how-does-a-receiving-antenna-generate-current-voltage/146766)), which I wanted to repost here sooner but decided not to. However, I do think there's a relevant response by user wbeaty which is worthwhile to quote hereQuoteUsually the intuitive, non-math explanation of receiving antennas is avoided in introductory texts. Instead, they steer around the problem by only explaining transmitters and the emission of EM waves. Yet the absorption of EM waves by receiving antennas is quite fascinating. It's also a minor 'hole in physics' which needs some serious filling. (Phd engineering students take note!)
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That's how EM absorption works. The receiving antenna has launched a wave which "sucks in" energy from the incoming wave. The reception of EM waves is a wave-cancellation process.
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The majority of introductory textbooks do us a great disservice: they insist that we only learn about wave-emission. Then we're supposed to automatically understand everything about wave-absorption ...because reciprocity?
Uh, no.
Pondering the experiment proposed by the original poster...Quote from: msat on Yesterday at 07:47:48 pm (https://www.eevblog.com/forum/index.php?topic=289474.msg3688540#msg3688540)...we might (ahem) reflect (intentional pun, shamefully admitted) on the effect of the Earth ground plane on the numerical values of R21 and X21 you've calculated between two isolated half-wave dipoles. I think it reduces the numerical values of both R21 and X21 by a factor approaching 2, but has no effect on your conclusions as to their spacing.I'm thinking of working in the ~6MHz ISM band due to only having a 25MHz scope. This also makes locating the antennas kind of easy at the expense of having somewhat unwieldy distances to work with.
Realize the wavelength at 6 MHz is 50 meters. Your proposed antennas will be 25 meters long, and have to be suspended above the ground plane (the earth) to work. You propose to space them an integer number of wavelengths apart. This experiment will span a football field. The long coaxial cables from your receivers to the oscilloscope will be expensive...
This reduction in both R21 and X21 arises from two causes:
1) elevation of these horizontal dipoles above the ground plane. This is likely to be very low. This is included in equation 11-7-11 on Kraus page 468, and approaches 2 as the elevation approaches zero if the ground is perfectly conductive or lossless.
2) at 7 MHz, the ground is likely to be low-loss dielectric, its relative permittivity given by equation 16-4-4 on Kraus page 718 and illustrated in Figure 16-6, so will do little to reduce this factor from 2. But even if it is not (in some regions of the desert southwest US, ground is dry to depths >50 m), it still has negligible effect on your conclusions.
I was idly pondering this, conclude it has no effect on your conclusions as to antenna spacing, but would appreciate your comments either confirming or correcting my thinking.