13
\$\begingroup\$

Clearly antennas are nothing but a device to radiate the electrical energy through electromagnetic waves.

Since visible light is also simply a certain range of frequencies, isn't it more easier to think of antennas as different shapes of "light" sources?

Like directional antenna is a hand held torchlight, high power means flood lights?

Why can't we simply state this in particle nature as it will be much simpler mathematically than the wave theory?

\$\endgroup\$
  • \$\begingroup\$ Antennas also have to receive E&M waves. \$\endgroup\$ – sstobbe Feb 5 at 18:59
  • 2
    \$\begingroup\$ Related on Physics SE: Can I use an antenna as a light source?. \$\endgroup\$ – The Photon Feb 6 at 0:08
  • \$\begingroup\$ @ThePhoton great coincidence. Simply shows it's a general question. But yeah, If only stackexchange provided a search option in the app😐 \$\endgroup\$ – user163416 Feb 6 at 1:31
22
\$\begingroup\$

For some cases, you can: If you have a large directional antenna, it might, from very far away, simply look like a beam-generating "flashlight" for radio waves. That breaks down very quickly if the wavelengths are not much, much smaller than all physical objects interacting with them.

We even use specific terms: If wavelengths are very small compared to all objects they meet and a few simple "macroscopic" formulas can describe their behaviour, we speak of optical (ray) propagation. When dealing with RF, we don't; RF doesn't behave like light, and thus, the usefulness of the analogy doesn't exist. So, no, we can't be "much simpler mathematically", because the easier model of what you know as light propagation simply doesn't work¹.

For most cases, you can't compare antennas to light sources.

First of all, the analogy with light sources doesn't work out fully: Your flashlight works with DC coming from a battery. Your waves coming out have frequencies beyond 10¹⁵ Hz. In an antenna, the method of generating the wave relies on the current going into the antenna already having the frequency to be emitted, and the antenna just acting as an impedance matching component between wave conductor and free space.

Then, the wave emitted from an antenna has some sort of wave front, which implies coherent phase! Your LED or light bulb doesn't have that, at all.

So, the light beam from a torch is simply physically very different from the beam from an antenna.


¹ Things are way more complicated for light than you think once you look very closely; a beam is not beam.

\$\endgroup\$
20
\$\begingroup\$

You are right, antennas and light sources are equivalent constructs. But the mathematics of light sources is not as simple as you seem to think.

The reason why most of the answers so far see them as different is just a matter of scale. While we would commonly call "RF" wavelengths of 1mm or above (300GHz) and "light" wavelengths of 1µm and below (300THz), with some concession for what lies in between (is it "low-infrared light" or "microwaves"?), the equations that govern their behavior are exactly the same: Maxwell's.

The problem is that such large difference of scales have consequences to how these interact with the world. While you can have discrete components interacting to generate a 1m RF signal, to generate a 100nm light signal you have to consider the interaction between electrons and their energy levels.

  • While a 10m tightly-focussed RF signal will propagate around a 1m metal disk with apparently no interaction, a narrow-focused 1µm light beam will be completely stopped in its tracks. While the first would be stopped by a mesh Faraday cage with 10cm openings, the second will pass unimpeded. Materials that are nearly completely transparent to one will completely stop the other and vice versa .

  • While you would need a rather massive antenna to focus a 10cm RF beam to achieve 90% power in a 1m spot at 1km, the equivalent lenses to do the same with 1µm light could fit in one hand.

  • While you can mostly ignore atmospheric effects (the interaction of the RF energy with air molecules) below 1GHz or so, atmospheric conditions will soon dominate above that and will become the main effect at light frequencies.

  • People that design optical lenses are well aware of the problems dealing with broadband signals (visible light occupies a whole octave from 380 to 740 nanometers, or 430–770 THz). Those are equivalent to the problems that broadband RF designers face, yet broadband RF seldom spans even 5% of the carrier frequency.

Most of engineering is dealing with models, models that considerably simplify the problem at hand and have a range of validity (all models are wrong, some models are useful). That is why in the lower ranges of RF we deal with KCL, KVL, and Ohm's laws in our circuits instead of trying to solve them by direct application of Maxwell's equations. But go higher in frequency and now you have to switch to s-parameters and transmission lines as wires stop behaving as mere wires. Go higher still, into the "light" domain, and now using photons and electron energy transition levels becomes advisable.

But all of those models are just simplifications of Maxwell's equations with their narrow domain of applicability. But knowing this and where the models fail, can help prime our design intuition.

\$\endgroup\$
  • \$\begingroup\$ Let me be the first one to upvote this excellent answer and point out that it gives my very superficial answer are solid model complement. Thanks! \$\endgroup\$ – Marcus Müller Feb 5 at 19:29
  • 1
    \$\begingroup\$ @MarcusMüller Thanks!! Your answer is what prompted this one, there were some subtle aspects of it that I felt needed some elaboration. \$\endgroup\$ – Edgar Brown Feb 5 at 19:57
  • 1
    \$\begingroup\$ "...you can mostly ignore atmospheric effects... below 1GHz or so..." yikes! Tell that to radio astronomers and ham operators (water, ionosphere, respectively). ;-) Overall very nice answer! \$\endgroup\$ – uhoh Feb 6 at 13:35
8
\$\begingroup\$

Firstly, "light" on its own usually means "visible light". Antennas do not emit visible light.

We can more correctly say that light is EM radiation and antennas emit EM radiation.

Why can't we simply state this in particle nature as it will be much simpler mathematically

Is it? You've not cited any of the maths in your post. And for most purposes the wave pattern is what we want; it tells us where the radio waves can be recieved most strongly. For most communication frequencies radio waves aren't a light-like "beam", they diffract a lot.

\$\endgroup\$
  • \$\begingroup\$ Well, for starters, taking light as a particle alone would mean perfect radial radiation(in case of dipole), and frequency shift, polarization, refraction, reflection etc..can all become a lot simpler. Like polarization is throwing the ball with different spins. Reflection is like bouncing off.But I suppose, because we simply don't measure it...It behaves as a wave and unlike visible light...Doesn't fall into particle states? \$\endgroup\$ – user163416 Feb 5 at 16:31
  • \$\begingroup\$ "would mean perfect radial radiation(in case of dipole)" - I don't follow; the radiation pattern is measurable, changing the model doesn't change what it actually is? See Edgar's example for how important diffraction is. \$\endgroup\$ – pjc50 Feb 5 at 20:47
2
\$\begingroup\$

In some cases, one can. And surely in our world of meters light can be very reliably approximated as a ray. But so can an EM wave in the scales of 1000000000, with objects that are only in many thousands of kilometers.

But, life only looks simple for optics in our world. When we have to deal with light propagating through micrometer sized structures, arrays or conductors, the ray approximation is of no use. (Google plasmonics, photonics or photonic crystals etc. They use modes, resonances, more Maxwellian equations.) Just like it lacks the power to explain RF phenomena accurately in our world.

\$\endgroup\$
2
\$\begingroup\$

Why can't we simply state this in particle nature as it will be much simpler mathematically than the wave theory?

When we say a photon is a "particle" of light energy, we mean that only discrete amounts of energy can be absorbed from or emitted into the electromagnetic field.

But these particles don't move according to the rules of ballistics that apply to bullets or billiard balls. They move according to a wave equation that is essentially the same as the wave equation that describes classic electromagnetic propagation.

So there's no free lunch here. Electromagnetic "particles" are just as mathematically complex as the waves they replace.

\$\endgroup\$
1
\$\begingroup\$

Antennas can be treated as a light source, but it emits in a different way. If you are considering normal RF antenna, then these do not radiate visible light that carries information because the light has a much higher frequency than the antenna resonating frequency. A typical RF antenna (3 KHz and 300 GHz) is simply too large to efficiently emit visible light (430–770 THz) because of this size mismatch. But it's possible with some antennas like plasmonic nanoantennas. Out of several devices that emit visible light in a controlled way, plasmonic nanoantennas are the closest to traditional radio antennas.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.