# Why do radio waves spread out while higher frequency waves travel in beams?

Why is it that radio waves spread out in proportion to the square of the distance, while higher frequency electromagnetic waves, like microwaves, infrared waves, light, etc are able to propagate as beams? What fundamental property allows higher energy waves to travel differently than lower energy?

Due to diffraction, wave effects become more important as the size of the wave source becomes comparable to the length of the wave. Visible light has micrometer-scale wavelengths, so a millimeter-sized light source is thousands of wavelengths across and diffraction isn't a very big deal. But radio wavelengths can be many meters, producing similar collimation for a radio "beam" would require an emitting antenna hundreds or thousands of kilometers across.

You can use the same logic to think about shadows. A hair that's less than a millimeter across can cast a well-defined shadow, while radio waves diffract around buildings. However larger objects can cast well-defined radio shadows: for instance astrophysical radio sources disappear when they are covered by the Moon or the Sun, which are both very many wavelengths across.

Note that even "collimated" light undergoes dispersion. Any sort of focusing optical system will produce a beam waist at some finite distance from the final focusing element (mirror or lens or whatever); beyond that beam waist the intensity of the light falls off like $r^2$ just as if a light source were at that location. A perfectly collimated beam of light is prohibited by the uncertainty principle, unless the beam is infinitely wide.

• +1, but I slightly disagree with the uncertainty principle comments for reasons discussed in my answer. I'm being a bit pedantic, and I think you're quite within rights to make your comments to an "advanced" audience, but it could be a little confusing to someone sorting these ideas out. – WetSavannaAnimal Oct 6 '17 at 6:14

They all spread out, but generally speaking the directivity of an antenna or emitter (assuming optimal design) is proportional to its volume, in wavelengths cubed. So a tiny high-frequency antenna with a volume of five wavelengths cubed has the same directivity as a huge low-frequency antenna also with a volume of five wavelengths cubed. Of course, other things also affect directivity, like temperature, the particular antenna design, etc. That's why I said "generally speaking".

• What do you mean by the antenna volume? Say, I use a straight wire sticking out up from my walkie-talkie. What is the volume in this case? – safesphere Oct 4 '17 at 22:58
• This is NOT a precise or scientific measurement, its a "generally speaking" thing. The volume includes a wavelength, or so, in every dimension (i.e. much of the near field). – Digiproc Oct 5 '17 at 1:46
• I see. A double linear size antenna ($2^3$ volume) would produce the same directional pattern for a doubled wavelength. Makes sense. – safesphere Oct 5 '17 at 7:06

This is simply a question of the beamwidth relative to the wavelength. Unless the transmitter is extremely big, the broadcast "beam" is less than a wavelength across. A small microwave dish is hundreds or thousands of wavelengths across; thus it acts like a huge phased array. On the other hand, the Very Large Arrays used in astronomy do the same for radiowaves; as transmitters they could form collimated beams and as receivers they are superbly directional. A quantity that yields intuition for the phenomenon is the Rayleigh length:

$$z_R = \frac{\pi\,\mu^2}{4\,\lambda}$$

and this is the axial distance a beam of width $\mu$ (as measured by the $1/e^2$ intensity diameter) must propagate before spreading out to twice its beginning diameter. You can see very easily how my comments about relative width and wavelength size are summed up in this equation. The Rayleigh length applies to Gaussian beams, which spread the least possible amount for a given RMS beamwidth. This last assertion is related to the Heisenberg uncertainty principle insofar that the mathematics that lead to it are the same as those that lead to the Heisenberg uncertainty principle, but I disagree slightly with Rob's answer that this is the uncertainty principle, because we're talking simply about a Fourier transform relationship between the superposition weights in the plane wave decomposition of a propagating field, rather than a Fourier transform between co-ordinate systems wherein conjugate quantum observables are multiplication operators (which, by dint of the Fourier transform relationship, leads to the same mathematics). Perhaps I'm being a little pedantic, but to me the HUP is explicitly about quantum measurement and observables. More on this below.

Further Details

On noting that the modes of the Helmholtz equations are plane waves, we can describe diffraction as a scrambling of the wavefront by dint of the different phase delays undergone by the plane wave components propagating in different directions, as I discuss further in my answer here or this one here. Continuing with this line of thought: Because there is a Fourier transform relationship between the field at a given transverse plane and the superposition weight function for the plane wave decomposition, there is a Heisenberg inequality between a beam's divergence angle and its beamwidth, as measured by the RMS spreads of divergence angle as I discuss here. The minimum uncertainty product is realized by a Gaussian Beam, and one should use the equations that describe this special, minimum uncertainty product case to get insight and intuition into your question. The numerical aperture (half the divergence angle) $\eta$ and beamwidth $\mu$, both measured by $1/e^2$ diameters / cones, are related by:

$$\eta = \frac{2\,\lambda}{\pi\,\mu}$$

There is another level to the question, and it is to do with the intention or purpose of why the different frequencies were put to use in the first place: but first I freely admit that a lot of the discussion in the excellent answers already given are beyond me.

The direct answer is that radio waves, of MW and HF frequencies, were often deliberately broadcast via omnidirectional antennas for specific reasons. Therefore of course the reduction in field strength with distance followed an easily understood square law. The other factor was that it is physically difficult and expensive, though always possible, to beam these lower frequency MW and HF frequencies. Indeed I worked at a complex of about 850 acres of giant rhombic antennas, necessary to make the "radio waves" into beams in the pre-satellite era, for transmission to the other side of the world, rhombic limitations and "broader" diffraction notwithstanding.

The higher frequencies which can be beamed much more cheaply, also, for practical considerations, do actually need to be beamed in order to keep the received field strength up to a reasonable level at the receiving end, many wavelengths away from the transmitter.

The other advantage of beaming VHF, UHF & microwave, etc, besides the great economy and ease of construction of physically small highly efficient beam antennas, is that diffraction at these frequencies also helps to make the beam narrow, hence a low-powered (and much cheaper) transmitter suffices.

I answer this way because implicit in the question was a suspicion (on my part) that some sorts of electromagnetic waves were somehow deemed "different" to others, which isn't really the case. The developing history of electromagnetic waves almost demanded that the lower frequencies "had" to be utilised first, with their observed omnidirectional capability in practical use - but that capability was not intrinsic, it had much to do with physical antenna sizes and mechanical limitations - and dare I suggest that that this is also a legitimate answer to the question?