Why are longwave radio towers so tall?

They are often in the hundred meters high. See the wiki link:

Why are they so tall? Because the wavelength is long, so the antenna should also be long? But to transmit in the longwave region, we only need to make the circuit oscillate in the right frequency, right? It has nothing to do with the size of the antenna, right?

• This is a problem of impedance matching for even an infinitesimally small antenna i.e., a Hertz dipole (or monopole above a ground plane) can perfectly transmit and be perfectly matched to a pure sine wave. What it cannot do is to be matched to a signal of finite bandwidth. An AM radio transmitter needs about 5kHz effective bandwidth, and that is a lot at 500kHz carrier frequency (1%). A badly matched antenna is space heater and transmitter burner... – hyportnex Dec 23 '18 at 13:31
• @hyportnex, you should post your comment as an answer --- it is more to the point than any of the posted answers. – The Photon Dec 23 '18 at 18:35
• @ThePhoton am happy to oblige... see below. – hyportnex Dec 23 '18 at 21:13
• @hyportnex excellent answer, glad you did! – uhoh Dec 24 '18 at 1:18
• Yes. Because the wavelength is long, the antenna should be long. – phoog Dec 24 '18 at 15:49

This is a problem of impedance matching for even an infinitesimally small antenna, i.e., a Hertz dipole (or monopole above a ground plane) can perfectly transmit and be perfectly matched to a pure sine wave. What it cannot do is to be matched to a signal of finite bandwidth.

The impedance of a short dipole of length $$h$$ and radius $$a$$ is approximately $$Z_{dipole} \approx 20 (kh)^2 - \mathfrak {j} \frac{120(\textrm{ln}(h/a) -1)}{\textrm{tan}(kh)}$$ with $$k=2\pi/\lambda$$. A short dipole, (infinitesimal or otherwise), one whose length is less than $$\lambda/10$$ is, therefore, a frequency varying capacitor in series with a tiny radiation resistance, so it is almost an open circuit. To make this radiate one can just place a series inductor to resonate the capacitor "out". So far so good but the radiation resistance is still in the micro- or milli-ohms, for example, a 1m long dipole at 1MHz has radiation resistance $$800 \textrm{m}\Omega$$.... Thus this antenna as a load represents an enormous reflector on a, say, $$50\Omega$$ transmission line unless one also makes an impedance transformer to match it to the free space whose impedance is $$120\pi=377 \Omega$$.

One may try to add other reactive elements such as another inductor in parallel so that the two inductors also act as an impedance transformer but now one quickly runs into a very severe practical limitation: the larger the transformer ratio one needs the larger the circulating currents will be, and unfortunately the circuit losses are proportional to $$I^2$$. It is quite possible that the radiation resistance of a short dipole is in fact much less than the parasitic resistance of the matching network, and thus most of the transmitter's power is dissipated in the circuit (mostly in the coils) and not radiated out. Compare this case with that of a half-wave dipole whose impedance is almost real and $$\approx 73 \Omega$$.

One must also mention a further more subtle but in practice also more difficult complication. For a fixed antenna size as one increases the wavelength (reduces the frequency) the ratio of the reactance (imaginary part) to resistance (real part) of the antenna impedance also increases, in fact the increase is much faster than linear (quasi-exponential). This results in an essentially exponentially shrinking operating bandwidth because the lumped element matching circuit has an impedance that is a rational function (ratio of polynomials) of frequency. As an example, an AM radio transmitter needs about 5kHz effective bandwidth, and that is a lot at 500kHz carrier frequency (1%). A badly matched antenna is at worst a transmitter burner and at best a space heater...

My answer is completely different. Longwave antennas usually are quarter-wave antennas, also known as Marconi antennas from its inventor. As the name says, they have length about 1/4 the wavelength to be transmitted. Compare them to dipole antennas generally used for shorter wavelengths, both in transmission and in reception, which are one half-wave long. A quarter-wave antenna may be seen to be completed to a dipole by its mirror image formed by a ground plane (usually the physical ground beneath it).

The physical reason why an antenna is to be in a simple (1/4, 1/2) ratio to wavelength is not very easy to explain. Keep in mind that among e.m. devices antennas are the hardest to understand and also to design. They are at the interface between two domains: the one of electric circuits (resistors, capacitors, inductors, generators) and the one of e.m. waves. Furthermore, both in case of receiving and of transmitting anntennas it's of paramount importance their efficiency, i.e. their ability to emit the highest possible fraction of the power generated by the transmitter, or that to absorb and make usable by amplifying circuits as much power as possible from an impinging wave.

As far as pure theory is concerned it wouldn't be necessary a special size for an antenna having to transmit a given wavelength (i.e. a given frequency). But on the practical side things are very different. Consider a Marconi antenna of the wrong size. In general it would be seen by transmitting equipment as a reactive load, i.e. a resistor in series with an inductor or a capacitor. An inductor if too long, a capacitor if too short. This can be cured by impedance matching: by adding in series to the antenna a capacitor in the former case, an inductor in the latter.

But an antenna of the wrong length suffers another drawback: its radiation resistance (RR) drops substantially. RR multipled by current squared gives radiated power. So a low RR requires a higher current to give the required power, and this in turn requires a step-down transformer to match the high antenna current to the output characteristics of transmitter. The high antenna current moreover increases power dissipated in the unavoidable ohmic resistance, so that radiation efficiency lowers.

... to transmit in the longwave region, we only need to make the circuit oscillate in the right frequency

That is the point. For lower frequency the electrons - which are accelerated inside the antenna rod -, need a longer distance inside the rod. Would the disturbance of these electrons reach the end of the rod too fast, the power of the antenna generator would not do work, but only heat the system. The frequency, one want to transmit and the length of the rod are not independent regarding the energy efficiency and the quality of the sine form of the transmitted energy.

To give an example of a sharp pulsed wave signal try to imagine what happens with the electrons inside a superconducting rod. The generator nearly at once pushes all available electrons to the end of the rod and after until the end of the first half cycle runs again a huge ohmic resistance (we are tolking about an open circuit). The accelerated electrons emit at one moment a huge number of photons - I suppose in the range of X-Rays - and after nothing and only with the beginning of the second half cycle of the generator the same emission happens again, this time in the opposite direction of course.

It has nothing to do with the size of the antenna

For very low frequencies it is impossible to built such long antennas. But it’s helpful to add to the end of the rod additional metal which works like a capacitor Dachkapazität (available only in the German wiki). So the acceleration of photons lasts longer and the efficiency of the generator and the quality of the wave is improved.

• IMHO there is no meaningful physics in your answer. For instance, apparently you have no idea of how electrons move in a conducting wire. At frequencies of many kHz electrons oscillate in a very very short distance. I made no estimate, but if you wish I'll do. – Elio Fabri Dec 23 '18 at 20:03
• @ElioFabri I’m aware of the drift velocity and I wrote „Would the disturbance of these electrons reach the end of the rod too fast...“. Did you agree, that then lower the frequency - for rods of the same length-, then more electrons “fill” the end of the rod? – HolgerFiedler Dec 23 '18 at 21:00
• @The_Sympathizer A potential difference between the generator and the end of the rod travels with maximum the drift velocity of the conductor. The heigher the generators frequency the faster the drift changes its sign. In the case of low frequencies the drift finishes before the generator changes its potential and instead of a sine signal one get peaks. To last the drift longer, the roof capacity was invented. And about an electron, “moving all the way”, please change the text, if you think that this was written in it. It was not my intention, to be understood like you interpret it. – HolgerFiedler Dec 24 '18 at 6:30
• Dachkapazität in English is called a capacitance or capacity hat, or a top load. – Phil Frost Dec 26 '18 at 13:27
• "...the electrons - which are accelerated inside the antenna rod -, need a longer distance inside the rod." Sorry, they are not inside, because of the Skin Effect. – Mike Waters Dec 28 '18 at 2:46

Antenna performance is strongly affected by the presence of the ground nearby. The standard rule of thumb is to raise the antenna to a height above ground of about one half of the wavelength it will operate at, in order to minimize power loss in the ground and radiation directionality effects. At low frequencies- say, ~1 MHz- the wavelength is 300 meters which would put the antenna wire 150 meters above ground. It is not uncommon to take some effectiveness loss and raise the antenna up 1/4 wavelength instead.

• I think you meant 1 MHz ($=10^6$ Hz) not 1 mHz ($=10^{-3}$ Hz). – Gary Godfrey Dec 23 '18 at 9:55

Most antennas are "line of sight". That is, you want the EM wave to pass through as little obstructions as possible, as the the photons will have a higher likelihood of scattering in random directions or being absorbed. So to maximize coherence of the signal we try to limit the amount of obstructions it comes in contact with.

Since they are radio waves they can pass through most things, but you don't want them to pass through very thick media like mountains etc as the will make the signal weaker.

• Most antennas are "line of sight". - This is not true at longwave frequencies. The BBC longwave transmitter at Droitwich (UK) is intended cover the whole of England and Wales, but in practice it can be easily received by British ex-pats in the south of France, and even as far away as Italy. – alephzero Dec 23 '18 at 14:54
• Moreover the ionosphere will function like a mirror and will bounce the signal back towards earth. – ZeroTheHero Dec 23 '18 at 19:28

$$\let\d=\delta \let\lam=\lambda \let\om=\omega \let\s=\sigma \let\Om=\Omega \def\qy#1#2{#1\,\mathrm{#2}} \def\Rr{R_{\mathrm r}} \def\half{{1 \over 2}} \def\10#1#2{#1\cdot10^{#2}} \def\PD#1#2{{\partial#1 \over \partial#2}}$$ I want to add another answer rather than editing the former, since what follows is strictly connected to @The_Sympathizer's answer. On the other hand a comment is too much limited in size for my reasoning.

Let me first correct a minor point. The_Sympathizer writes

To have such a resonant pattern, of course, we need an antenna that is a full wavelength long, just as the case for the organ tubes and sound.

In fact the shortest resonant length is $$\lam/2$$ ($$\lam/4$$ for the "virtual" one). The same holds for organ pipes and other wind instruments too).

But my central argument is the following. The_Sympathizer writes

you get a transient, extremely rapidly alternating (with the frequency of the radio transmission) charge imbalance where that one half of the antenna is developing a net positive charge and the other half developing a net negative charge, and that reverses once per wave cycle.

This is true but raises a doubt. How can such imbalance arise if not with electrons migrating from one side ot the antenna to the opposite? Strange as it may appear, the only way to solve the issue is in examining some numbers.

Let's begin by choosing a reasonable size for the antenna and its power. There is a considerable range of possible values, from high frequency low power amateur transmitters to low frequency extra-high power broadcasters. I'll take values rather near the former limit, but any reader will be able to change numbers and follow how that modifies my conclusions - if it does. Then

• $$\nu = \qy{100}{MHz} \quad \lam = \qy 3 m$$ (frequency, wavelength)
• $$l = \lam/4 = \qy{0.75}m$$ (1/2 antenna length)
• $$d = \qy1{cm} = \qy{0.01}m$$ (diameter of antenna conductor)
• $$I_0 = \qy 2 A$$ (peak antenna current).

Given radiation resistance $$\Rr = \qy{73}\Om$$ radiated power is $$\half\,\Rr\,I_0^2 = \qy{146}W$$.

We'll also need the effective conductor cross section, taking into account skin effect. From https://chemandy.com/calculators/skin-effect-calculator.htm we get

• skin effect depth $$\d = \qy{6.5}{\mu m}$$

and cross section is $$\s = \pi d\d = \qy{\10{2.0}{-7}}{m^2}$$.

Another datum we need is electron number density $$n$$ (number of free electrons per unit volume). If the antenna is made of copper, assuming one free electron per atom, from known density of copper and its atomic mass we have

• $$n = \qy{\10{8.5}{28}}{m^{-3}}\$$ (electron number density).

We're now ready to begin reasoning. In our antenna a standing wave is present, of wavelength $$\lam$$ and frequency $$\nu$$. All interesting quantities must have an expression consistent with the standing wave and relevant boundary conditions. For instance the current will be $$I(x,t) = I_0 \cos kx \,\cos \om t \qquad \left(\!k = {2\pi \over \lam},\quad \om = 2\pi\nu = c\,k\!\right)$$ (We must choose $$\cos kx$$ in order that $$I$$ vanishes at antenna's ends ($$x=l$$). The choice $$\cos\om t$$ for time dependence is arbitrary, only amounting to fix the wave's phase at $$t=0$$.)

We have in general $$I = -n\,\s\,e\,v$$ ($$v$$ average electrons speed in point $$x$$ at time $$t$$). If we assume a collective average motion of electrons, described by a displacement $$D(x,t)$$, we'll write $$v = \PD Dt.$$ Then $$\PD Dt = -{I \over n\,\s\,e}$$ immediately integrable to $$D = -{I_0 \over n\,\s\,e\,\om}\,\cos kx\,\sin\om t.$$ So the maximum displacement occurs at antenna centre ($$x=0$$) and is $$|D_0| = {I_0 \over n\,\s\,e\,\om} = \qy{\10{1.2}{-12}}m = \qy{1.2}{pm}.$$

When a physicist encounters such an unbelievable result he has two moral duties:

• to check his reasoning and calculations as accurately as he can
• to find an alternative way to reach, at least grossly, the same result.

You have only my word on the first point - about the second here's my proposal.

We have a current of $$\qy2A$$ peak at $$\qy{100}{MHz}$$ on our antenna. So it alternates 200 million times a second and in 1/(200 millionth) of a second it transfers a charge of 2/(200 million) = $$\qy{10^{-8}}C$$ from - say - left side to right side of the antenna (actually less, since current is not constant at its peak value, but too bad - it means that we are overextimating the transferred charge).

This charge amounts to a number of electrons: $${10^{-8} \over \10{1.6}{-19}} = \10 6{10}.$$

If there is one free electron per Cu atom, the interested volume will be $$\10 6{10}$$ times the volume occupied by an atom. How can we estimate such volume?

I know by heart some data:

• Cu molar mass (atomic weight): about $$\qy{60}{g/mol}$$
• Avogadro's constant: $$\qy{\10 6{23}}{mol^{-1}}.$$

These give me the mass of one Cu atom: $${\qy{60}{g/mol} \over \qy{\10 6{23}}{mol^{-1}}} = \qy{10^{-22}}g.$$ Now using

• Cu density: about $$\qy9{g/cm^3}$$

I find the volume occupied by one atom:

$${\qy{10^{-22}}g \over \qy9{g/cm^3}} = \qy{10^{-23}}{cm^3} = \qy{10^{-29}}{m^3}.$$

Now the total volume of copper interested by charge displacement is $$\10 6{10} \times \qy{10^{-29}}{m^3} = \qy{\10 6{-19}}{m^3}.$$ I had already computed (and re-checked) the cross section: $$\qy{\10 2{-7}}{m^2}$$ and the above volume corresponds to a displacement $${\qy{\10 6{-19}}{m^3} \over \qy{\10 2{-7}}{m^2}} = \qy{\10 3{-12}}m$$

Conclusion. You may change the initial data as you like. The result is so small that you will never be able to turn it into a significant one. This solves our doubt: there are so many electrons in the antenna that in order to produce an important charge imbalance (and a relevant radiation power) only a very very small fraction of them are to be moved.

A final comment. I spent a not negligible part of my time in writing the above as I found it a very useful lesson - an impressive example of a general truth. You can't understand physics, also in its seemingly intuitive aspects, without grounding your reasoning on a sound quantitative basis. Physics without numbers is just chatter.

An antenna basically works by generating an alternating electromagnetic field, which then per Maxwell's equations will propagate spatially like a wave. It in turn does this by having passed through itself an extremely high frequency, modulated alternating current (radio frequency alternating current or RF AC), just like the AC power that comes out of the mains and that runs your house, but with more signal complexity (the modulation to carry the message).

Thus it is basically just a circuit element, i.e. a conductor, but with a catch. In a normal circuit, with low- to no-frequency changes in the currents/voltages, and which is of a suitably small size, one is in a regime in which the behavior of circuits can be analyzed in terms of what are known as Kirchhoff's laws and the lumped element model. In layman's terms, this means "circuit diagrams and component rules" like $$V = IR$$ for resistors, $$C = \frac{Q}{V}$$ for capacitors, and so forth with one component on the diagram for each real component on the circuit. And the reason this can be done is that in such circuits a couple of facts about the electromagnetic situations in them hold to at least an approximate degree of truth: the fields in most components are localized to within them, and that the charge distributions across the circuit are effectively uniform.

But an antenna, in fact, is in a way a circuit which is expressly designed to violate these assumptions. It is a conductor operated at a high enough frequency and which is large enough that the rules are violated dramatically and thus its analysis requires considerably more sophistication to do in a quantitative, mathematical manner and thus to design such antennas with attention to making them possess efficiency of operation and quality and fidelity of generated signal, is also a fraught matter requiring level-99 Electrical Engineering (EE) skills.

In particular, when you have alternating currents in a circuit, what you in a sense "really" have going through the circuit is electromagnetic waves. In a way, a circuit is kind of like a "fiber optic cable for RF waves", almost. These waves are what carry the energy through the circuit and moreover this is for exactly the same reason that when you flip your light switch the light turns on effectively "instantly" despite that the electrons in a circuit move only with much lower speed, almost a crawl (at least from a classical naive point of view - actually according to QM they do indeed move pretty quick in most metallic conductors, about 1000 km/s [to the extent that speeds can be assigned a well-defined value thanks to quantum fuzziness] so enough to reach your house light in a microsec essentially literally, but still the speed the energy reaches your light is higher than even this as I believe at least can be measured with more sophisticated experiments).

When you flip the switch, a sharp electromagnetic wave front is created, behind which is the energy that you associate with "electric [mains] power", and as it passes them, at nearly the speed of light, the electrons are set in motion. Unlike what one may be thinking, electrons don't actually need to reach the bulb from elsewhere for it to have power, only the wave does.

And when there's such a wave in a conductor, specifically Kirchhoff's junction rule is violated - meaning that there are, for a time, net accumulations of charge at some points along it, or places where the sum of currents going in does not equal the sum of currents going out, leading to unbalanced charges. This is because (at least from classical EM, my QED ain't good enough to give a really detailed quantum account) as the wave passes the electrons are not all moving uniformly but rather some bunch together more tightly at areas of stronger field gradient (higher wave slope, like at the middles between the peaks and troughs) than at areas of weaker gradient (lower wave slope, like at the peaks and troughs themselves). (This is also analogous, by the way, to how that gravitational forces cause stretching/squeezing when you have a gradient as in tidal forces or the passing of a gravitational wave past a point, though with the caveat that gravitational waves are tensor waves, not vector waves, so they affect things somewhat differently in terms of details.)

And the simplest, ideal antenna is then one which is exactly one half or one wavelength of such a wave, long. As when you do that, you get an electromagnetic wave - what is really the radio wave you want to transmit, confined in a bottle like light in a fiber optic cable - across the whole length that is alternating and moreover forms a standing-wave pattern because with this length of antenna, the far end and near end can both serve as nodes (the electromagnetic wave - not electrons as in another answer that was posted here - effectively shoots through and then "bounces back" against the antenna's end and thus establishes the SW pattern similar to a sound wave of the right frequency in a hollow tube like those used on a pipe organ to produce the notes, which actually can thus be thought of as a sort of analogue model for the electromagnetic situation). With that standing wave, due to the variable compression and rarefaction of electrons within the metallic conductor, you get a transient, extremely rapidly alternating (with the frequency of the radio transmission) charge imbalance where that one half of the antenna is developing a net positive charge and the other half developing a net negative charge, and that reverses once per wave cycle. This means that what you have is then an alternating dipole and thus more specifically an alternating dipole moment from the antenna, and according to Maxwell's equations an alternating dipole moment releases electromagnetic radiation. Alternatively, from a more conceptual or fundamentally theoretical point of view, you can say that a distant observer holding a charge will feel from the field of this alternating dipole in hir test charge an alternation in likewise manner. Which, effectively, is what the opposite of a transmitter - a receiver - does. And that means that information - the message being transmitted by the transmitter - is going from one point to another in space. And when that happens, since the speed of propagation of information is limited thanks to relativistic constraint (Lorentzian geometry of spacetime), then there must be something traversing the void between hir and the tower at that limited speed and thus that means in this situation radiation is present (and moreover, a though experiment of this type is an easy way to determine if in a given problem that radiation will be present or not. For example, a perfectly symmetric charged cylinder rotating around its axis - feel no change in EM field -> no information -> no radiation present. Charged cylinder tumbling end over end -> if you're a distance from the "ends" you feel it tugging periodically -> information is moving -> radiation is present.).

To have such a resonant pattern, of course, we need an antenna that is a full wavelength long, just as the case for the organ tubes and sound. And it turns out thanks to a mathematical property of the laws of electromagnetism we can do that in two ways: one is "real", to make a real antenna really one full wavelength long, and the other is "virtual", to make a half-antenna a half wavelength and to use the reflection of the waves from the ground to form a sort of "image" (this is first introduced in an EM course as the "method of image charges" for static charges and is the same basic principle) of the antenna that acts like the opposite, negatively-charged part. Antennae of other lengths will still have varying charge distributions through them, but won't be as efficient as it's off-resonance and so the amplitude of the wave in the antenna doesn't get as high and thus neither does the amplitude of what is sent out.

And to get the final answer, the wavelength of an electromagnetic wave is related to its frequency by

$$\lambda = \frac{c}{f}$$

and of course, $$c \approx 300\ 000\ \mathrm{km/s}$$, which can be written perhaps more usefully as $$300\ \mathrm{m \cdot MHz}$$. Thus a 1 MHz broadcast frequency requires a full-wave resonant antenna 300 m long, or a half-wave mirror antenna 150 m long. You can see that would make for quite a tower and indeed "quite a tower" is exactly what we have in real life!