Why do artificial transmissions of coherent EM waves tend to be isotropic at radio frequencies but directional at visible frequencies?

Question

This is a somewhat soft question, and I'm far from an expert on the subject, so it's possible that my premise is factually incorrect.

The practical generation mechanisms for useful real-world coherent EM signals (e.g. for wireless communication, radar/lidar, etc.) seem to be very different for radio vs. visible signals.

From what I can tell, radio transmissions tend to be generated by AC-driven antennas, which are naturally isotropic in at least one plane through the transmitter. While it's possible to generate a directional beam at radio frequencies, it's much harder to do and requires designing an antenna array in which each element is individually driven with a separate carefully controlled phase offset.

On the other hand, coherent visible light seems to be typically generated by lasers, which are naturally highly directional and narrow-beam. While there may be a way to generate an isotropic coherent visible-frequency signal, it seems much harder. (It is of course easy to generate an isotropic incoherent visible signal; just turn a light bulb on and off.)

It seems to me that purely mathematically, it's just as easy to construct a directional coherent RF signal or an isotropic coherent visible signal as the opposites. The difference between radio and visible frequencies seems to be only quantitative in theory, but qualitative in practice.

1. Am I correct that it's practically much easier to produce an isotropic coherent radio-frequency signal than a directional one, but conversely it's much easier to produce a directional coherent visible-frequency signal than an isotropic one?

2. If this is correct, what is the fundamental reason for this opposite behavior at different frequencies? Is it simply due to the nature of the currently practical generation mechanisms, or is there some fundamental reason intrinsic to the EM waves themselves at different frequencies?

3. If the former, is there a fundamental reason for the huge difference between the practical transmission mechanisms?

Answer 1

You wrote "While it's possible to generate a directional beam at radio frequencies, it's much harder to do and requires designing an antenna array in which each element is individually driven with a separate carefully controlled phase offset".

This is not quite true, for example, in 1955 (!) the US Army introduced the AN/FPS-16 instrumentation radar at C-band (~5.5GHz) that has antenna gain 44dB and a corresponding beamwidth $1.1^{\circ}$. Note though that the antenna to achieve this has a diameter 12ft.

In general, theoretically, any beam shape that is possible at optics could be achieved at RF (3kHz - 300GHz), theoretically. The practice is different for several reasons but the main reason is always that the beamwidth of an antenna is proportional to its wavelength and inversely proportional to its characteristic dimension (ie., "coherently illuminated size"):$\Delta \theta_{3dB} \approx k_0\frac{\lambda}{D}$, where $k_0 = O(1)$. All macroscopic optical stuff is thousands of wavelengths in size therefore the natural coherent beamwidth when used around $500nm - 10\mu m$ will be $\frac{10}{1000} \rm{rad}< 1^{\circ}$ without lifting a finger, so to speak.

Dipole antennas and their direct relatives have circular and planar symmetry and so do their radiation pattern (very "unoptical") but other radiators such as a waveguide open end or a horn radiator behaves just like a funnel, a torchlight, so to speak. There are parabolic dish antennas all the way down to lower UHF. The largest that has recently got decomissioned is the 300m (!) dish https://en.wikipedia.org/wiki/Arecibo_Telescope working between 300MHz to 10GHz. It is not just a "dish", rather it is a Gregorian telescope with an ellipsoid primary reflector and a parabolic secondary. The radiator illuminating the primary is coma-corrected (!) as any self-respecting telescope would be.

Regarding the transmission mechanism note that the spread of the wave by Friis's formula is https://en.wikipedia.org/wiki/Friis_transmission_equation $$\frac{P_r}{P_t}=\frac{A_r A_t}{\lambda^2 d^2}$$ from which you can see that for a given distance the ratio of the received to transmitted power depends of the product of the effective radiating areas of the antennas divided by the square of the wavelength. But there is an other issue that cannot be seen Friis's formula, namely the actual propagation of EM waves in a real scattering and absorbing environment. That effect is strongly frequency dependent, for example, below HF the earth is a conductor and can be used as an effective ground plane so that there is waveguide propagation between the top of the atmosphere and the surface of the earth, a wavelength size dish would be larger than the earth but a loaded monopole is practical, see https://en.wikipedia.org/wiki/VLF_Transmitter_Cutler.

Answer 2

... what is the fundamental reason for this opposite behavior at different frequencies? Is it simply due to the nature of the currently practical generation mechanisms, or is there some fundamental reason intrinsic to the EM waves themselves at different frequencies?

The cause of EM radiation in thermal sources, in lasers and in radio waves is the same. All EM radiation begins with the disturbance of charges. This is trivial, but it is helpful to keep it in mind in any discussion of EM waves.

EM radiation

The wavelengths at which the accelerated charges radiate depend on their kinetic energy.

• For a light bulb, it is mainly in the infrared range. The bandwidth does not have a clear peack, but is rather a bell-shaped curve. This has to do with the chaotic movement of the electrons in the resistance wire and thus strongly varying kinetic energies.
• Ordinary lasers are special cases of artfully designed materials and geometries where the charges are only disturbed to a very limited extent and these have sharp peaks in the emitted wavelengths.

In both cases, the disturbed electrons emit EM radiation in the form of photons. In sum, these photons do not have the properties of a wave with its oscillating energy content. Nevertheless, it is possible to generate modulated EM radiation from such sources. If you switch them on and off, you get packets of photons and, with a little effort, such modulated radiation could have a sinusoidal function.
When switching an incandescent lamp on and off with 1 Hz, we are now dealing with two bandwidths. One is the (mostly) IR radiation and the second is a modulation of this IR radiation with 1Hz.

EM waves

Radio waves have a very specific property, which even a modulated light bulb radiation do not have. The photons, emitted from accelerated charged in an antenna rod, are polarized. The surface electrons get accelerated synchronious and all in the same direction. The common electric field turns into a common magnetic field from which do to an energy excess photons are formed.

How many photons are emitted during a half period? The electrons feel the electric potential difference from the antenna generator and start moving, get accelerate, get decccalerated and get stopped. During this period they emit a lot of photons. Otherwise we would see only sharp peacks ot some moments and no radiation in between.

The crutial question is, which wavelengths the emitted photons have. Radars for example are modulated EM radiaon with frequencues from MHz to GHz. But among others they emit X-rays which is in of range 30 petahertz to 30 exahertz. Driving radio emitters with to short antennas and with pwoerful generators allows to modulate the same frequencies as well tuned emitters but makes the EM radiation harder.

Showing pictures with EM radiation from Gamma to radio waves is good for technical purposes but not instructive for a physics degree. Unfortunately, this is forgotten nowadays.