$\let\d=\delta \let\lam=\lambda \let\om=\omega \let\s=\sigma
\let\Om=\Omega \def\qy#1#2{#1\,\mathrm{#2}} \def\Rr{R_{\mathrm r}}
\def\half{{1 \over 2}} \def\10#1#2{#1\cdot10^{#2}}
\def\PD#1#2{{\partial#1 \over \partial#2}}$
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.