The simplest antenna is a short
(total length $l$ much smaller than one
wavelength $\lambda$) dipole
antenna, which is shown above as two colinear conductors
(e.g., wires or conducting rods). Since they are driven at
the small gap between
them by a current source
(a transmitter), the current in the bottom conductor is 180 deg
out of
phase with the current in the top conductor. The radiation
from a
dipole depends on frequency, so we consider a driving current $I$
varying sinusoidally
with
angular frequency $\omega = 2 \pi \nu$: $$I = I_0 \cos(\omega
t)~,$$
where $I_0$ is the peak current going into each half of the
dipole. It is
computationally convenient to replace the trigonometric function
$\cos(\omega t)$ with its exponential equivalent, the real part of

$$e^{-i \omega t} = \cos(\omega t) - i \sin(\omega t)$$ so the
driving
current can be rewritten as $$I = I_0 e^{-i \omega t}$$ with the
implicit understanding that only the real part of $I$
represents this current. The driving current accelerates charges
in the
antenna conductors, so we can use
Larmor's formula to calculate the radiation from the antenna by
converting from the language of charges and accelerations to
time-varying currents.

It is a
common misconception to believe that the velocities $v$ of
individual
electrons in a wire are comparable with the speed of light $c$
because
electrical
signals do travel down
wires
at
nearly the speed of light. A wire filled with electrons is
like a
garden hose already filled with an
incompressible fluid—water. When the faucet is turned on,
water
flows from the other end of a full hose almost immediately, even
though
individual water molecules are moving slowly along the hose.
As
a specific example, consider a current of 1 ampere flowing through
a
copper
wire of cross section $\sigma = 1$ mm$^2 = 10^{-6}$ m$^2$.
The
number density
of free electrons is about equal to the number density of
copper
atoms in the wire, $n \approx 10^{29}$ m$^{-3}$. In mks
units,
the charge of
an electron is

$$-e \approx 4.80 \times 10^{-12} {\rm ~statcoul}

\times {1 {\rm ~coul} \over 3 \times 10^9 {\rm ~statcoul}}

\approx 1.60 \times 10^{-19} {\rm ~coul}$$ One Ampere is one
Coulomb
per second, so the number $N$ of electrons
flowing past any point along the wire in one second is

$$N = {I \over \vert e \vert} \approx

{1 {\rm ~coul~s}^{-1} \over 1.60 \times 10^{-19}{\rm ~coul}}

\approx 6.25 \times 10^{18}{\rm ~s}^{-1}$$ The average electron
velocity is only

$$ v \approx {N \over \sigma n}

\approx { 6.25 \times 10^{18}{\rm ~s}^{-1} \over 10^{-6}{\rm ~m}^2

\times 10^{29}{\rm ~m}^{-3}} \approx 6 \times 10^{-5} {\rm
~m~s}^{-1}
\ll c$$ Thus the nonrelativistic Larmor equation may be used
directly
to calculate the
radiation from a wire.

Most practical dipoles are
half-wave
dipoles ($l \approx \lambda/2$) because half-wave dipoles are
resonant,
meaning that they provide a nearly resistive load to the
transmitter. When each half of the dipole is $\lambda/4$
long, the standing-wave current is highest at the center and
naturally
falls as $\cos(2 \pi z / \lambda)$ to almost zero at the ends of
the
conductors.