NEC-LIST: Whence does an antenna radiate?

Jack Belrose <john.belrose_at_crc.ca> wrote:

> . . . Radiation results from the acceleration of charge. Where
> does this happen? Near the ends of the antenna, since here the
> velocity of a current element comes to zero, and accelerates very
> quickly again to its velocity on the center part of the antenna. So
> the ends of a dipole are important. The expectation that "maximum
> radiation results from the parts of the dipole where current is a
> maximum" is not true....

I disagree -- reluctantly, because I have such high regard for Jack
Belrose, who has taught me very much and has surely forgotten more
about antennas than I will ever learn. Right or wrong, I hope by
these comments to evoke educational comments from the many NEC-LIST
subscribers who understand radiation better than I.

Here are four reasons/arguments why, e.g., for a uniform, straight,
thin, half-wavelength-long, wire "dipole", most of the radiation comes
from the middle, not the ends of the wire. The radiation originates
in proportion to the current density. On a uniform, straight, thin,
half-wavelength-long wire, the current density varies cosinusoidally
with a maximum at the midpoint of the wire and with nodes (zeroes) at
the ends.

1. The argument that "near the ends of the antenna,...the velocity of
a current element comes to zero, and accelerates very quickly" seems
plausible by analogy with the simple pendulum or mass-spring
oscillator. Another supportive analog is water sloshing back and
forth between the ends of atrough or bathtub. Such analogies may fit
another kind of electromagnetic oscillator-radiator, in which a
charged particle such as an electron bounces back and forth, or
oscillates between, the walls of a potential well; but such analogies
are not appropriate for the conducting wire dipole. A better analogy
in this case is that of small waves on the surface of a deep ocean.
My point is that, in a conducting wire, there is a virtually infinite
reservoir of mobile charge, and that the oscillations represent
infinitesimal perturbations of this charge distribution. In argument
#4 below I attempt to show the absurdity of the first-mentioned class
of analogies.

2. Anyone having real-world, practical experience with thin wire
dipole antennas knows that the radiation field is affected very little
by bending the ends, i.e., the last (say) 10% of the wire. This fact
is a simple proof that the ends do not contribute much to the
radiation; and they most certainly do not dominate it.

3. Anyone having access to NEC can confirm argument #2 with a
simulation. If you've never done so, do it now. NEC is not the real
world, but NEC does such a good job of simulating the real world that,
IMO, a NEC experiment is meaningful.

4. Now let's see what the charge-sloshing-back-and forth analogy
implies. In this view, the current, I, as a function of the distance,
x, from the center of the wire, and of the time, t, is related to the
charge density Q and to the charge velocity V (not to be confused with
Voltage) by

     I(x,t) = Q(x,t) * V(x,t) Eq. [1]
or
     V(x,t) = I(x,t) / Q(x,t). Eq. [2]

The analogy with fluid flow is clear: mass flow is the product of mass
density and velocity. If we're talking about water in a trough, the
mass density (per unit length) is the water depth.

Now the acceleration, A = A(x,t), is the partial derivative of the
velocity V with respect to time t, at fixed x. Thus

     A = Idot/Q - I*Qdot/Q^2, [3]

where function arguments have been suppressed for simplicity; a "dot"
suffix denotes the partial derivative with respect to t at fixed x;
and ^2 denotes an exponent 2, i.e., the square. The
charge-acceleration density, QA, to which the radiation field is
proportional, is thus

     QA = Idot - I*(Qdot/Q) . [4]

For an infinitely thin wire the current, I, the charge density, Q, and
their "dot" time derivatives have the functional forms (as a NEC-4
simulation that I just did confirms):

     I = Io * cos(kx) * cos(omega*t) [5]

     Q = Qo * sin(kx) * sin(omega*t) [6]

     Idot = - Io * omega * cos(kx) * sin(omega*t) [7]

     Qdot = Qo * omega * sin(kx) * cos(omega*t) [8]

where Io, Qo, k, and omega are (related) constants; k = 2 * pi /
lambda; and without loss of generality the phase of the current has
been set to zero ("cosine") here.

Substituting Eqs. [5] through [8] into [4] now yields

     QA = - Io * omega * cos(kx) * sin(omega*t)

                 [ Io * omega * cos(kx) * cos^2(omega*t) ]
              - [---------------------------------------] [9]
                 [ sin(omega*t) ] .

Substituting 1 - sin^2(omega*t) for cos^2(omega*t) and canceling then
yields

     QA = Io*omega * cos(kx) / sin(omega*t), [10]

which is alarming because the denominator sin(omega*t) goes through
zero twice per cycle, when the charge-density goes through zero. Of
course, in the sinusoidal steady state the charge-density _must_ go
through zero twice per cycle, and unless the current vanishes
simultaneously, the velocity and the acceleration must blow up (i.e.,
go to infinity). But the current does not vanish; in fact the current
has a maximum when the charge-density goes through zero.

If QA _did_ blow up, there would be an impulse of radiation, having a
broad spectrum (kinda like synchrotron radiation), even though the
current and charge oscillated harmonically. Clearly no such
phenomenon occurs.

On the other hand, looking back at Eq. [1] and recognizing that Q(x,t)
is the sum of a small spatially and temporally varying perturbation,
and a very large constant, leads to the recognition that, to first
order,

     QA = Idot = - Io * omega * cos(kx) * sin(omega*t). [11]

Thus, QA has the same spatial variation, i.e., as cos(kx), as the
current I.

Which is what we knew had to be true. After all, you get the
radiation field by integrating the _current_ over the wire.

73 de Chuck W1HIS
Received on Thu Feb 17 2000 - 14:48:26 EST