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Lzsn2)S/9
Fig. 32-1. The area of a spherical segment is 27r sin 6 - r d6.
liberated in this direction between Ø and Ø + đđ; then we integrate that over all
the angles Ø from 0 to 180P:
q22 (*
P= Jsaa = “! sin” Ø d0. (32.4)
87coc? /ọ
By writing sin” Ø = (1— cos2 Ø) sin Ø it is not hard to show that J sin3 Ø dØ = 4/3.
Using that fact, we finally get
P=_-—.. 32.5
6zegc3 (325)
This expression deserves some remarks. Eirst of all, since the vector ø“ had a
certain đirection, the 2 in (32.5) would be the square of the vector a', that is,
d - da", the length of the vector, squared. Secondly, the ñux (32.2) was calculated
using the retarded acceleration; that is, the acceleration at the time at which the
energy now passing through the sphere was radiated. We might like to say that
this energy was in fact liberated at this earlier time. 'Phis is not exactly true; it
is only an approximate idea. The exact time when the energy is liberated cannot
be defined precisely. All we can really calculate precisely is what happens in a
complete motion, like an oscillation or something, where the acceleration ñnally
ccases. Then what we fnd is that the total energy fux per cycle is the average
of acceleration squared, for a complete cycle. 'Phis is what should really appear
in (32.5). Or, iŸit is a motion with an acceleration that is initially and fñnally
zero, then the total energy that has flown out is the time integral of (32.5).
To illustrate the consequences of formula (32.5) when we have an oscillating
system, let us see what happens if the displacement + of the charge is oscillating
so that the acceleration ø is —œ2zo€?“†, "The average of the acceleration squared
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over a cycle (remember that we have to be very careful when we square things
that are written in complex notation——it really is the cosine, and the average
Of cos2 œf is one-half) thus is
(a2) = 3 zã.
'Therefore
q2u1z?
Pp= 12cac3` (32.6)
The formulas we are now discussing are relatively advanced and more or less
modern; they date from the beginning of the twentieth century, and they are very
famous. Because of their historical value, it is important for us to be able to read
about them ¡in older books. In fact, the older books also used a system of units
diferent from our present mks system. However, all these complications can be
straightened out in the ñnal formulas dealing with electrons by the following
rule: The quantity gỆ/4zco, where q is the electronic charge (in coulombs), has,
historically, been written as e2. It is very casy to calculate that e in the mks
system is numerically equal to 1.5188 x 10~1*, because we know that, numerically,
qe = 1.60206 x 10~†12 and 1/4zco = 8.98748 x 10. Therefore we shall often use
the convenient abbreviation :
c2 = -®—, (32.7)
47m €0
TỶ we use the above numerical value of e in the older formulas and treat them as
though they were written in mks units, we will get the right numerical results.
Eor example, the older form of (32.5) is P = 3c2a'2/c. Again, the potential
energy of a proton and an electron at distance r is qg2/4zcạr or e2/r, with
e = 1.5188 x 101 (mks).
32-3 Radiation damping
Now the fact that an oscillator loses a certain energy would mean that if we
had a charge on the end oŸ a spring (or an electron in an atom) which has a
natural frequency œạọ, and we start it oscillating and let it go, it will not oscillate
forever, even ï i is in empty space millions of miles from anything. There is no
oil, no resistance, in an ordinary sense; no “viscosity.” But nevertheless it will
not oscillate, as we might once have said, “forever,” because If it is charged it is
radiating energy, and therefore the oscillation will slowly die out. How slowly?
'What is the @Q of such an oscillator, caused by the electromagnetic efects, the
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so-called radiation resistance or radiation damping of the oscillator? The @Q of
any oscillating system is the total energy content of the oscillator at any time
divided by the energy loss per radian:
Q= uy à
Or (another way to write i0), since đW/dó = (dW/đt)/(do/dt) = (dW/dL) /e,
= —: 32.8
ẹ@ dW/dt (3238)
Tf for a given @ this tells us how the energy of the oscillation dies out, đW/dt =
—(u/Q)W, which has the solution W = Wse—*“1⁄® ¡f Wg is the initial energy
(at £ = 0).
To ñnd the Q for a radiator, we go back to (32.8) and use (32.6) for đdW/di.
Now what do we use for the energy W/ of the oscillator? 'Phe kinetic energy
of the oscillator is jn2?, and the mean kinetic energy is mœ2z2/4. But we
remember that for the total energy of an oscillator, on the average half is kinetic
and half is potential energy, and so we double our result, and fñnd for the total
energy of the oscillator
W = šmuŸzã. (32.9)
'What do we use for the frequency in our formulas? We use the natural frequency œọ
because, for all practical purposes, that is the frequency at which our atom is
radiating, and for rm we use the electron mass rm=;. hen, making the necessary
divisions and cancellations, the formula comes down to
1 4me2
—= =>: 32.10
@Q_ 3Am,c2 ( )
(In order to see it better and in a more historical form we write iÈ using our
abbreviation g2/4zco = e2, and the factor œo/c which was left over has been
writben as 2/A.) Since Q is dimensionless, the combination e2/mn„c? must be
a property only of the electron charge and mass, an intrinsic property of the
electron, and i9 must be a lengfh. It has been given a name, the classical electron
radius, because the early atomic models, which were invented to explain the
radiation resistance on the basis of the force of one part oŸ the electron acting
on the other parts, all needed to have an electron whose dimensions were of this
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general order of magnitude. However, this quantity no longer has the signifcance
that we believe that the electron really has such a radius. Numerically, the
magnitude of the radius is