The Non-Radiating Antenna

https://physicsworld.com/a/introducing-the-non-radiating-antenna/

by ithkuil • 5 years ago

75 22 5 years ago

jbay8082• 5 years ago

By the principle of reciprocity, such an antenna also shouldn't couple at all to the far field of a distant radiating source... Right?

So this should work well for a near-field coupling device, but not as some kind of distant silent receiver.

The "macroscopic atom" analogy makes me wonder, because atoms can absorb traveling photons just fine. But then again, only as well as they also radiate.

petschge• 5 years ago

If it only receives the near field, can I used that to subtract the near field from a close-by regular antenna? That should make it more immune to local influences and improve the SNR of a far away signal?

jpmattia4• 5 years ago

Unless I'm missing something, I don't think reciprocity helps you much here. There is a contained radiation field when the antenna is excited; there are no fields far away. Now when you start with no fields far away, it doesn't tell you much about the excitation near the antenna.

You can envision some radiation pattern from far away antennas which have overlap with the near-field modes. As a consequence, some amount of energy will be received by this antenna. Naturally it depends on how well the near-field mode is excited as to how much coupling occurs.

jbay8082• 5 years ago

What I mean by reciprocity is, if distant far-field antenna B can transfer energy to our contained-field antenna A, then shouldn't our contained-field antenna A also be able to transfer energy to the far-field antenna B? Wouldn't that be inconsistent with A not having a far-field emission?

If A doesn't produce any field at B, then it doesn't seem possible that the field B produces at A could be absorbed by A. Thermodynamics should guarantee this, otherwise you could use it (in a roundabout way) to violate the second law.

MrYellowP• 5 years ago

Serious:

They've discovered a new form of metaball?

How do I calculate this?

denton-scratch• 5 years ago

I misread that, and was put in mind of "spaghetti with metaballs".

alanbernstein• 5 years ago

"Metaball" to me refers to the organic-looking, morphing, implicit surface used in computer graphics. What's the connection to this?

IshKebab• 5 years ago

So that was a really bad explanation but it sounds like it makes evanescent waves? Otherwise I don't see how it could work. Also I don't really understand how you can make a symmetric point source sort of thing that makes evanescent waves.

Very unclear article. A single animation would have helped massively.

madengr2• 5 years ago

RF & antenna engineer here. I’d like to know the input impedance at resonance. If you take two closely spaced, parallel dipoles and drive them out of phase, you get the same effect of nulling the far field. The issue is the impedance drops substantially.

Now take one dipole and insert a 1/2 wavelength delay (an additional resonator), and you end up with the same. So what they have done is synthesize the second dipole with the disk, so two dipoles along a common axis rather than parallel.

Dipole near fields drop off at 1/r^3, so yes, this is a method of near field coupling for power transfer or localized beating.

jpmattia4• 5 years ago

I’d like to know the input impedance at resonance.

The lack of radiated energy means the real part of the impedance must be zero (or close to it?) There's some energy being stored, but no guesses as to sign of the imaginary part.

beervirus• 5 years ago

Right, this looks a lot like a quadrupole.

Animats• 5 years ago

Where does the energy go? Can you use this as a dummy load? What gets hot then?

madengr2• 5 years ago

There will be losses in the dipole conductor and resonator dielectric, so you have a finite input impedance, but that’s where the energy should go; into heat. You could use it as a dummy load over very narrow bandwidth.

jpmattia4• 5 years ago

I'm spitballing, but there is a small amount of energy stored in the field around the structure. Since there's no dissipation, it would not make for a decent dummy load.

exporectomy• 5 years ago

Can someone explain if the introduction about quantum states is in any way the same effect or just an analogy? Could it be that discrete energy levels in atoms exist because those are the non-radiating states and they're non-radiating for a somehow-similar reason?

jpmattia4• 5 years ago

an someone explain if the introduction about quantum states is in any way the same effect or just an analogy?

Not the same effect, and not an analogy; it's more about the history of non-radiating accelerating charges. When it was first understood that hydrogen was composed of a positive nucleus and an orbiting electron, there was quite a bit of head scratching as to how the electron orbit was stable and didn't just radiate its orbital energy away.

Bohr's model was somewhat ad hoc, but it said that the only allowed orbits would fit integral numbers of the electron's wavelength and would not radiate its energy away. The model got the right numbers for the orbital energies so it explained the spectral lines of hydrogen, though details of the hydrogenic orbitals would have to wait until Schrodinger's equation.

Gravityloss• 5 years ago

A picture would be nice.

pontifier• 5 years ago

My understanding is that coaxial cable, and twisted pair cables exist because they also don't radiate significant energy to the far field.

What new understanding should I be getting from this new design?

eitland2• 5 years ago

Now you got me thinking:

That maybe opened a new perspective to me, thinking of the a coaxial as an antenna shielded by the coat, instead of just a different way to have two conductors in one cable. I don't know if this difference means anything since the signals that travel though it are typically waves around 0.

This maybe interesting to someone: While I worked in the field with a radio technician I'm fairly certain I heard that the (or maybe just one?) way they make radio work in tunnels is by stretching a coaxial signal cable with small slits through it. That kind of fits very well with how I understood it above.

(While I have studied basic electronic engineering I'm not a radio technician myself so if this doesn't sound right, please correct or fill in. Also, according to the technician I worked with antennas are really interesting, at one point he pointed out that he had laughed at what later became his favourite antenna because his first thought was that it "obviously" could not work.)

R0b0t1• 5 years ago

It's not an antenna since it doesn't release the signal, it's a waveguide. Instead of vaccuum propogating EM fields it's metal.

rapjr9• 5 years ago

I wonder if this has wireless security applications along the lines of this paper:

https://dl.acm.org/doi/pdf/10.1145/3300061.3300120

Possibly the reduced far field radiation could enhance the security, kind of a "cone of silence" for radio, making long distance attacks even less likely.

The Non-Radiating Antenna

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