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diference we have taken to be zero) may be comparable to or larger than the
small Tung term, and our approximation will no longer be valid.
--- Trang 554 ---
Miqcli(frore I)crrtppirntgg. Lígphhí Secrffor-rrtg/
32-1 Radiation resistance
In the last chapter we learned that when a system is oscillating, energy is
carried away, and we deduced a formula for the energy which is radiated by an
oscillating system. If we know the electric field, then the average of the square
of the field times cọc is the amount of energy that passes Der square meter Der
second through a surface normal to the direction in which the radiation is goïng:
8 = (ạc(E®*). (32.1)
Any oscillating charge radiates energy; for instance, a driven antenna radiates
energy. Ifthe system radiates energy, then in order to account for the conservation
of energy we must find that power is being delivered along the wires which lead
into the antenna. 'That ¡s, to the driving circuit the antenna acts like a resisfance,
or a place where energy can be “lost” (the energy is not really lost, it is really
radiated out, but so far as the circuit is concerned, the energy is lost). In an
ordinary resistance, the energy which is “lost” passes into heat; in this case the
energy which is “lost” goes out into space. But from the standpoint of circuit
theory, without considering +0here the energy goes, the net effect on the circuit is
the same——energy is “lost” from that circuit. Therefore the antenna appears to
the generator as having a resistance, even though it may be made with perfectly
good copper. In fact, 1ƒ it is well built ít will appear as almost a pure resistance,
with very little inductanee or capacitance, because we would like to radiate as
much energy as possible out of the antenna. 'Phis resistance that an antenna
shows is called the radiation resistance.
TÍ a current ƒ is going to the antenna, then the average rate at which power
is delivered to the antenna is the average of the square of the current times the
resistance. The rate at which power is rød¿atcd by the antenna is proportional
--- Trang 555 ---
to the square of the current in the antenna, of course, because all the fields are
proportional to the currents, and the energy liberated is proportional to the
square of the field. The coefficient of proportionality between radiated power
and (T2?) is the radiation resistance.
An interesting question is, what is this radiation resistance due to? Leb us
take a simple example: let us say that currents are driven up and down in an
antenna. W© find that we have to put work in, iŸ the antenna is to radiate energy.
TỶ we take a charged body and accelerate it up and down it radiates energy; 1Í
1E were not charged it would not radiate energy. Ïlt is one thing to calculate
from the conservation of energy that energy is lost, but another thing to answer
the question, øgœ#ns‡ t0uhøt ƒorce are we doïing the work? 'Phat is an interesting
and very dificult question which has never been completely and satisfactorily
answered for electrons, althouph it has been for antennas. What happens is this:
in an antenna, the fields produced by the moving charges in one part of the
antenna react on the moving charges in another part of the antenna. We can
calculate these forces and fnd out how much work they do, and so ñnd the right
rule for the radiation resistance. When we say “We can calculate—” that is not
quite right— cannot, because we have not yet studied the laws of electricity at
short distances; only at large distances do we know what the electric field is. We
saw the formula (28.3), but at present it is too complicated for ws to calculate
the fñelds inside the wave zone. Of course, since conservation of energy is valid,
we can calculate the result all right without knowing the fñelds at short distances.
(As a matter of fact, by using this argument backwards it turns out that one can
find the formula for the forces at short distances only by knowing the field at
very large distances, by using the laws of conservation of energy, but we shall not
go into that here.)
The problem in the case of a single electron is this: if there is only one charge,
what can the force act on? It has been proposed, in the old classical theory, that
the charge was a little ball, and that one part of the charge acted on the other
part. Because of the delay in the action across the tiny electron, the force is not
exactly in phase with the motion. 'Phat is, IÝ we have the electron standing still,
we know that “action equals reaction.” So the various internal forces are equal,
and there is no net force. But if the electron is accelerating, then because of the
time delay across it, the force which is acting on the front from the back is not
exactly the same as the force on the back from the front, because of the delay in
the efect. This delay in the timing makes for a lack of balance, so, as a net efect,
the thing holds itself back by its bootstrapsl 'This model of the origin of the
--- Trang 556 ---
resistance to acceleration, the radiation resistance of a moving charge, has run
into many difculties, because our present view of the electron 1s that it is nof a
“little ball”; this problem has never been solved. Nevertheless we can calculate
exactly, of course, what the net radiation resistance force must be, i.e., how much
loss there must be when we accelerate a charge, in spite of not knowing directly
the mechanism of how that force works.
32-2 The rate of radiation of energy
Now we shall calculate the total energy radiated by an accelerating charge.
'To keep the discussion general, we shall take the case of a charge accelerating any
which way, but nonrelativistically. A% a moment when the acceleration is, say,
vertical, we know that the electric ñeld that is generated is the charge multiplied
by the projection of the retarded acceleration, divided by the distance. So we
know the electric field at any point, and we therefore know the square of the
electric ñeld and thus the energy cocF2 leaving through a unit area per second.
The quantity cọc appears quite often in expressions involving radiowave
propagation. Its reciprocal is called the #npedønce oƒ a 0uacuwm, and 1t is an easy
number to remember: it has the value 1/eoc = 377 ohms. So the power in watts
per square meter ¡is equal to the average of the field squared, divided by 377.
Using our expression (29.1) for the electric field, we fnd that
g_— 04s” 6 (33.2)
16r2cgr2c3
1s the power per square meter radiated in the direction Ø. We notice that it goes
inversely as the square of the distance, as we said before. NÑow suppose we wanted
the total energy radiated in all directions: then we must integrate (32.2) over all
directions. Eirst we multiply by the area, to ñnd the amount that flows within a
little angle đØ (Eig. 32-1). We need the area of a spherical section. The way to
think of it is this: 1Ý r is the radius, then the width of the annular segment is z đÓ,
and the cireumference is 277 sin Ø, because 7 sin Ø is the radius of the circle. So
the area of the little piece of the sphere is 2zz sin Ø times r đ6:
dA = 2m sin 0 d0. (32.3)
By multiplying the ñux [(32.2), the power per square meter| by the area in square
meters included in the small angle đØ, we fnd the amount of energy that is