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Unformatted text preview: 32 Radiation Damping. Light Scattering 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 see is the amount of energy that passes per square meter per
second through a surface normal to the direction in which the radiation is going: S = eoc(E2). (32.1) Any oscillating charge radiates energy; for instance, a driven antenna radiates
energy. If the 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 is, to the driving circuit the antenna acts like a resistance,
or a place where energy can be “lost” (the energy is not really lost, it is really radi
ated 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 where 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, if it is well built it will appear as almost a pure resistance, with very little
inductance or capacitance, because we would like to radiate as much energy as
possible out of the antenna. This resistance that an antenna shows is called the
radiation resistance. If a current I 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 re
sistance. The rate at which power is radiated by the antenna is proportional 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 (12)
is the radiation resistance. An interesting question is, what is this radiation resistance due to? Let us
take a simple example: let us say that currents are driven up and down in an an
tenna. We ﬁnd that we have to put work in, if the antenna is to radiate energy.
If we take a charged body and accelerate it up and down it radiates energy; if it
were not charged it would not radiate energy. It is one thing to calculate from the
conservation of energy that energy is lost, but another thing to answer the question,
against what force are we doing the work? That is an interestingand very difficult
question which has never been completely and satisfactorily answered for electrons,
although 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 ﬁnd
out how much work they do, and so ﬁnd the right rule for the radiation resistance.
When we say “We can calculate—” that is not quite right—we 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 us to calculate the ﬁelds inside the wave zone. Of course,
since conservation of energy is valid, we can calculate the result all right without
knowing the ﬁelds 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 32—1 32—1 Radiation resistance 32—2 The rate of radiation of energy
323 Radiation damping 32—4 Independent sources 32—5 Scattering of light Fig. 32—1. The area of a spherical
segment is 27rr sin 0 ° r d0. 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. That is, if 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 effect.
This delay in the timing makes for a lack of balance, so, as a net effect, the thing
holds itself back by its bootstraps! This model of the origin of the resistance to
acceleration, the radiation resistance of a moving charge, has run into many
diﬁiculties, because our present view of the electron is that it is not 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 mecha
nism 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. At a moment when the acceleration is, say,
vertical, we know that the electric ﬁeld 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
ﬁeld and thus the energy eocE2 leaving through a unit area per second. The quantity 50c appears quite often in expressions involving radiowave prop
agation. Its reciprocal is called the impedance of a vacuum, and it is an easy number
to remember: it has the value l/e 0c = 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 ﬁnd that _ q2a’2 sin2 6 S _ l67r360r2c3 (322) is the power per square meter radiated in the direction 0. We notice that it goes
inversely as the square of the distance, as we said before. Now suppose we wanted
the total energy radiated in all directions: then we must integrate (32.2) over all
directions. First we multiply by the area, to ﬁnd the amount that ﬂows within a
little angle d0 (Fig. 32—1). We need the area of a spherical section. The way
to think of it is this: if r is the radius, then the width of the annular segment is
r d0, and the circumference is 21rr sin 6, because r sin 6 is the radius of the circle.
So the area of the little piece of the sphere is 27rr sin 0 times r d0: (M = 27rr2 sin 0d0. (32.3) By multiplying the ﬂux [(322), the power per square meter] by the area in square
meters included in the small angle d0, we ﬁnd the amount ofenergy that is liberated in this direction between 0 and 0 + do; then we integrate that over all the angles
0 from O to 180°: 2,2 7' P=/SdA= 4"— , sin3 6 d0.
87r€()C' 0 (32.4) By writing sin3 0 = (1 — cos2 0) sin 0 it is not hard to show that f; sin3 0 d0 =
4/3. Using that fact, we ﬁnally get = I“ " (32.5) 61reoc3 32—2 This expression deserves some remarks. First of all, since the vector 21’ had a
certain direction, the a’2 in (32.5) would be the square of the vector 51’, that is,
a’  a’, the length of the vector, squared. Secondly, the ﬂux (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. This is not exactly true; it is
only an approximate idea. The exact time when the energy is liberated canhot be
deﬁned precisely. All we can really calculate precisely is what happens in a complete
motion, like an oscillation or something, where the acceleration ﬁnally ceases.
Then what we ﬁnd is that the total energy ﬂux per cycle is the average of accelera
tion squared, for a complete cycle. This is what should really appear in (32.5).
Or, if it is a motion with an acceleration that is initially and ﬁnally 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 x of the charge is oscillating
so that the acceleration a is —w2xoei“‘t. The average of the acceleration squared
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 wt
is onehalf) thus is (d2) = §w4x3.
Therefore P = q2w4xg .
127reoc3 (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 different from our present mks system. However, all these complications can
be straightened out in the ﬁnal formulas dealing with electrons by the following
rule: The quantity qf/47reo, where qe is the electronic charge (in coulombs), has,
historically, been written as e2. It is very easy to calculate that e in the mks system
is numerically equal to 1.5188 X 10—”, because we know that, numerically,
q, = 1.60206 X 10—19 and 1/47reo = 8.98748 X 109. Therefore we shall often
use the convenient abbreviation (12
e2 = 47:60 (32.7) If 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.
For example, the older form of (32.5)is P = g—ezaz/c3. Again, the potential energy
of a proton and an electron at distance r is qf/47reor or ez/r, with e = 1.5188 X
10—14 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 of a spring (or an electron in an atom) which has a natural
frequency we, and we start it oscillating and let it go, it will not oscillate forever,
even if it 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 eﬁects, the
socalled 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: W
Q T HW/dqs '
32—3 Or (another way to write it), since dW/d¢ = (dW/dt)/(d¢/dt) = (dW/dt)/w, wW
Q ‘ dW/dt' (32.8) If for a given Q this tells us how the energy of the oscillation dies out, dW/dt =
—(w/Q)W, which has the solution W = Woe‘M/Q if W0 is the initial energy
(att = 0). To ﬁnd the Q for a radiator, we go back to (32.8) and use (32.6) for dW/dt. Now what do we use for the energy W of the oscillator? The kinetic energy
of the oscillator is émv 2, and the mean kinetic energy is ma) 2x3/4. 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 ﬁnd for the total energy of the
oscillator W = %mw2xg. (32.9) What do we use for the frequency in our formulas? We use the natural frequency
we because, for all practical purposes, that is the frequency at which our atom is
radiating, and for m we use the electron mass me. Then, making the necessary
divisions and cancellations, the formula comes down to 1 4 2
a = “7:62. (32.10) (In order to see it better and in a more historical form we write it using our ab
breviation qf/41re0 = e2, and the factor coo/c which was left over has been written
as 21r/>\.) Since Q is dimensionless, the combination ez/mec2 must be a property
only of the electron charge and mass, an intrinsic property of the electron, and it
must be a length. 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 of the electron acting on the other parts, all needed
to have an electron whose dimensions were of this general order of magnitude.
However, this quantity no longer has the signiﬁcance that we believe that the elec
tron really has such a radius. Numerically, the magnitude of the radius is e2 mec2 r0 = = 2.82 x 10—15m. (32.11) Now let us actually calculate the Q of an atom that is emitting light—let us
say a sodium atom. For a sodium atom, the wavelength is roughly 6000 angstroms,
in the yellow part of the visible spectrum, and this is a typical wavelength. Thus 3
Q = 47:;0 z 5 x 107, (32.12) so the Q of an atom is of the order 108. This means that an atomic oscillator will
oscillate for 108 radians or about 107 oscillations, before its energy falls by a factor
l/e. The frequency of oscillation of light corresponding to 6000 angstroms,
v = c/A, is on the order of 1015 cycles/sec, and therefore the lifetime, the time it
takes for the energy of a radiating atom to die out by a factor l/e, is on the order
of 10—8 sec. In ordinary circumstances, freely emitting atoms usually take about
this long to radiate. This is valid only for atoms which are in empty space, not
being disturbed in any way. If the electron is in a solid and it has to hit other atoms
or other electrons, then there are additional resistances and diﬁerent damping. The effective resistance term “t in the resistance law for the oscillator can be
found from the relation 1/ Q = v/wo, and we remember that the size of ’Y deter
mines how wide the resonance curve is (Fig. 23—2). Thus we have just computed
the widths of spectral lines for freely radiating atoms! Since A = 27rc/w, we ﬁnd
that AA = 271'cAat/ot2 = 27rc‘Y/wg = 27rc/Qw0
= x/Q = 41rr0/3 = 1.18 x 10‘14m. (32.13)
32—4 32—4 Independent sources In preparation for our second topic, the scattering of light, we must now discuss
a certain feature of the phenomenon of interference that we neglected to discuss
previously. This is the question of when interference does not occur. If we have
two sources S1 and S 2, with amplitudes A 1 and A2, and we make an observation
in a certain direction in which the phases of arrival of the two signals are 45 1 and m
(a combination of the actual time of oscillation and the delayed time, depending on
the position of observation), then the energy that we receive can be found by
compounding the two complex number vectors A1 and A2, one at angle ¢1 and
the other at angle m (as we did in Chapter 30) and we ﬁnd that the resultant energy
is proportional to Air: = A? + A3 + 2A1A2 COS (¢>1  ¢2) (3214) Now if the cross term 2A1A 2 cos (¢1 — (1)2) were not there, then the total energy
that would be received in a given direction would simply be the sum of the energies,
A? + A3, that would be liberated by each source separately, which is what we
usually expect. That is, the combined intensity of light shining on something
from two sources is the sum of the intensities of the two lights. On the other hand,
if we have things set just right and we have a cross term, it is not such a sum, because
there is also some interference. If there are circumstances in which this term is of
no importance, then we would say the interference is apparently lost. Of course,
in nature it is always there, but we may not be able to detect it. Let us consider some examples. Suppose, ﬁrst, that the two sources are
7,000,000,000 wavelengths apart, not an impossible arrangement. Then in a given
direction it is true that there is a very deﬁnite value of these phase differences. But,
on the other hand, if we move just a hair in one direction, a few wavelengths, which
is no distance at all (our eye already has a hole in it that is so large that we are
averaging the effects over a range very wide compared with one wavelength) then
we change the relative phase, and the cosine changes very rapidly. If we take the
average of the intensity over a little region, then the cosine, which goes plus, minus,
plus, minus, as we move around, averages to zero. So if we average over regions where the phase varies very rapidly with position,
we get no interference. Another example. Suppose that the two sources are two independent radio
oscillators—not a single oscillator being fed by two wires, which guarantees that
the phases are kept together, but two independent sources—and that they are not
precisely tuned at the same frequency (it is very hard to make them at exactly the
same frequency without actually wiring them together). In this case we have what
we call two independent sources. Of course, since the frequencies are not exactly
equal, although they started in phase, one of them begins to get a little ahead of
the other, and pretty soon they are out of phase, and then it gets still further ahead,
and pretty soon they are in phase again. So the phase difference between the two
is gradually drifting with time, but if our observation is so crude that we cannot
see that little time, if we average over a much longer time, then although the
intensity swells and falls like what we call “beats” in sound, if these swellings and
fallings are too rapid for our equipment to follow, then again this term averages out. In other words, in any circumstance in which the phase shift averages out, we
get no interference! One ﬁnds many books which say that two distinct light sources never interfere.
This is not a statement of physics, but is merely a statement of the degree of sensi
tivity of the technique of the experiments at the time the book was written. What
happens in a light source is that ﬁrst one atom radiates, then another atom radiates,
and so forth, and we have just seen that atoms radiate a train of waves only for
about 10—8 sec; after 10‘8 sec, some atom has probably taken over, then another
atom takes over, and so on. So the phases can really only stay the same for about
10’8 sec. Therefore, if we average for very much more than 10"8 see, we do not
see an interference from two different sources, because they cannot hold their
phases steady for longer than 10‘8 sec. With photocells, very highspeed detection 325 / +At<*
Scattered radiation Y Incident beam
(unpolarized) Fig. 32—2. A beam of radiation falls
on an atom and causes the charges
(electrons) in the atom to move. The
moving electrons in turn radiate in
various directions. is possible, and one can show that there is an interference which varies with time,
up and down, in about 10—8 sec. But most detection equipment, of course, does
not look at such ﬁne time intervals, and thus sees no interference. Certainly with
the eye, which has a tenthofasecond averaging time, there is no chance whatever
of seeing an interference between two different ordinary sources. Recently it has become possible to make light sources which get around this
effect by making all the atoms emit together in time. The device which does this is
a very complicated thing, and has to be understood in a quantummechanical way.
It is called a laser, and it is possible to produce from a laser a source in which the
interference frequency, the time at which the phase is kept constant, is very much
longer than 10‘8 see. It can be of the order of a hundredth, a tenth, or even one
second, and so, with ordinary photocells, one can pick up the frequency between
two different lasers. One can easily detect the pulsing of the beats between two
laser sources. Soon, no doubt, someone will be able to demonstrate two sources
shining on a wall, in which the beats are so slow that one can see the wall get bright
and dark! Another case in which the interference averages out is that in which, instead of
having only two sources, we have many. In this case, we would write the expression
for Ai as the sum of a whole lot of amplitudes, complex numbers, squared, and
we would get the square of each one, all added together, plus cross terms between
every pair, and if the circumstances are such that the latter average out, then there
will be no effects of interference. It may be that the various sources are located in
such random positions that, although the phase difference between A2 and A3 is
also deﬁnite, it is very different from that between A 1 and A2, etc. So we would
get a whole lot of cosines, many plus, many minus, all averaging out. So it is that in many circumstances we do not see the effects of interference,
but see only a collective, total intensity equal to the sum of all the intensities. 32—5 Scattering of light The above leads us to an effect which occurs in air as a consequence of the
irregular positions of the atoms. When we were discussing the index of refraction,
we saw that an incoming beam of light will make the atoms radiate again. The
electric ﬁeld of the incoming beam drives the electrons up and down, and they
radiate because of their acceleration. This scattered radiation combines to give a
beam in the same direction as the incoming beam, but of somewhat different
phase, and this is the origin of the index of refraction. But what can we say about the amount of reradiated light in some other direc
tion? Ordinarily, if the atoms are very beautifully located in a nice pattern, it is
easy to show that we get nothing in other directions, because we are adding a lot
of vectors with their phases always changing, and the result comes to zero. But
if the objects are randomly located, then the total intensity in any direction is the
sum of the intensities that are scattered by each atom, as we have just discussed.
Furthermore, the atoms in a gas are in actual motion, so that although the relative
phase of two atoms is a deﬁnite amount now, later the phase would be quite differ
ent, and therefore each cosine term will average out. Therefore, to ﬁnd out how
much light is scattered in a given direction by a gas, we merely study the effects of
one atom and multiply the intensity it radiates by the number of atoms. Earlier, we remarked that the phenomenon of scattering of light of this nature
is the origin of the blue of the sky. The sunlight goes through the air, and when
we look to one side of the sun—say at 90° to the beam—we see blue light; what
we now have to calculate is how much light we see and why it is blue. If the incident beam has the electric ﬁeld E = Eoef‘” at the point where the
atom is located, we know that an electron in the atom will vibrate up and down in
response to this E (Fig. 32—2). From Eq. (23.8), the amplitude will be A quO = . 32.15
X m(w3 — (92 + iw’Y) ( ) 32—6 We could include the damping and the possibility that the atom acts like several
oscillators of different frequency and sum over the various frequencies, but for simplicity let us just take one oscillator and neglect the damping. Then the response
to the external electric ﬁeld, which we have already used in the calculation of the
index of refraction, is simply ﬂ = quO (32.16) m(w(2) — wz) We could now easily calculate the intensity of light that is emitted in various
directions, using formula (32.2) and the acceleration corresponding to the above it. Rather than do this, however, we shall simply calculate the total amount of
light scattered in all directions, just to save time. The total amount of light energy
per second, scattered in all directions by the single atom, is of course given by
Eq. (32.7) So, putting together the various pieces and regrouping them, we get P = [(q3w4/12r6003)q3E3/m3(w2 — w3)2]
= (%606E3)(87F/3)(q3/16W2€3m364)[w4/(w2 — 60(2))2]
(%606E3)(87"(2)/3)[w4/(w2 — 003?] (3217) for the total scattered power, radiated in all directions.
We have written the result in the above form because it is then easy to re member: First, the total energy that is scattered is proportional to the square of
the incident ﬁeld. What does that mean? Obviously, the square of the incident
ﬁeld is proportional to the energy which is coming in per second. In fact, the energy
incident per square meter per second is 606 times the average (E2) of the square of
the electric ﬁeld, and if E0 is the maximum value of E, then (E2) = %E§. In
other words, the total energy scattered is proportional to the energy per square
meter that comes in; the brighter the sunlight that is shining in the sky, the brighter
the sky is going to look. Next, what fraction of the incoming light is scattered? Let us imagine a “tar
get” with a certain area, let us say a, in the beam (not a real, material target, be
cause this would diffract light, and so on; we mean an imaginary area drawn in
space). The total amount of energy that would pass through this surface 0' in a given circumstance is proportional both to the incoming intensity and to a,
and would be P : (%eocE(2))0'. (32.18) Now we invent an idea: we say that the atom scatters a total amount of in
tensity which is the amount which would fall on a certain geometrical area, and
we give the answer by giving that area. That answer, then, is independent of the
incident intensity; it gives the ratio of the energy scattered to the energy incident
per square meter. In other words, the ratio total energy scattered per second _—T'—_—_—————“ is an area.
energy 1nc1dent per square meter per second
The signiﬁcance of this area is that, if all the energy that impinged on that area were to be spewed in all directions, then that is the amount of energy that would
be scattered by the atom. This area is called a cross section for scattering; the idea of cross section is
used constantly, whenever some phenomenon occurs in proportion to the intensity
of a beam. In such cases one always describes the amount of the phenomenon
by saying what the effective area would have to be to pick up that much of the beam.
It does not mean in any way that this oscillator actually has such an area. If there
were nothing present but a free electron shaking up and down there would be no
area directly associated with it, physically. It is merely a way of expressing the
answer to a certain kind of problem; it tells us what area the incident beam would 32—7 have to hit in order to account for that much energy coming off. Thus, for our case, 2 4
a, = 8W0 . ——“’— (32.19) 3 (w2 — at? (the subscripts is for “scattering”). Let us look at some examples. First, if we go to a very low natural frequency
too, or to completely unbound electrons, for which we = 0, then the frequency w
cancels out and the cross section is a constant. This lowfrequency limit, or the
free electron cross section, is known as the Thompson scattering cross section.
It is an area whose dimensions are approximately 10"15 meter, more or less, on a
side, i.e., 10—30 square meter, which is rather small! On the other hand, if we take the case of light in the air, we remember that for
air the natural frequencies of the oscillators are higher than the frequency of the
light that we use. This means that, to a ﬁrst approximation, we can disregard w”
in the denominator, and we ﬁnd that the scattering is proportional to the fourth
power of the frequency. That is to say, light which is of higher frequency by, say,
a factor of two, is sixteen times more intensely scattered, which is a quite sizable
difference. This means that blue light, which has about twice the frequency of the
reddish end of the spectrum, is scattered to a far greater extent than red light. Thus
when we look at the sky it looks that glorious blue that we see all the time! There are several points to be made about the above results. One interesting
question is, why do we ever see the clouds? Where do the clouds come from?
Everybody knows it is the condensation of water vapor. But, of course, the
water vapor is already in the atmosphere before it condenses, so why don’t we see
it then? After it condenses it is perfectly obvious. It wasn’t there, now it is there.
So the mystery of where the clouds come from is not really such a childish mystery
as “Where does the water come from, Daddy?,” but has to be explained. We have just explained that every atom scatters light, and of course the water
vapor will scatter light, too. The mystery is why, when the water is condensed into
clouds, does it scatter such a tremendously greater amount of light? Consider what would happen if, instead of a single atom, we had an agglom
erate of atoms, say two, very close together compared with the wavelength of the
light. Remember, atoms are only an angstrom or so across, while the wavelength
of light is some 5000 angstroms, so when they form a clump, a few atoms together,
they can be very close together compared with the wavelength of light. Then
when the electric ﬁeld acts, both of the atoms will move together. The electric ﬁeld
that is scattered will then be the sum of the two electric ﬁelds in phase, i.e., double
the amplitude that there was with a single atom, and the energy which is scattered
is therefore four times what it is with a single atom, not twice! So lumps of atoms
radiate or scatter more energy than they do as single atoms. Our argument that
the phases are independent is based on the assumption that there is a real and large
difference in phase between any two atoms, which is true only if they are several
wavelengths apart and randomly spaced, or moving. But if they are right next to
each other, they necessarily scatter in phase, and they have a coherent interference
which produces an increase in the scattering. If we have N atoms in a lump, which is a tiny droplet of water, then each one
will be driven by the electric ﬁeld in about the same way as before (the effect of
one atom on the other is not important; it is just to get the idea anyway) and the 3
amplitude of scattering from each one is the same, so the total ﬁeld which is1
scattered is N—fold increased. The intensity of the light which is scattered is then
the square, or N 2fold, increased. We would have expected, if the atoms were
spread out in space, only N times as much as 1, whereas we get N 2 times as much as
1! That is to say, the scattering of water in lumps of N molecules each is N times
more intense than the scattering of the single atoms. So as the water agglomerates
the scattering increases. Does it increase ad inﬁnitum? No! When does this
analysis begin to fail? How many atoms can we put together before we cannot
drive this argument any further? Answer: If the water drop gets so big that from
one end to the other is a wavelength or so, then the atoms are no longer all in 32—8 phase because they are too far apart. So as we keep increasing the size of the drop
lets we get more and more scattering, until such a time that a drop gets about the
size of a wavelength, and then the scattering does not increase anywhere nearly as
rapidly as the drop gets bigger. Furthermore, the blue disappears, because for long
wavelengths the drops can be bigger, before this limit is reached, than they can
be for short wavelengths. Although the short waves scatter more per atom than
the long waves, there is a bigger enhancement for the red end of the spectrum than
for the blue end when all the drops are bigger than the wavelength, so the color is
shifted from the blue toward the red. Now we can make an experiment that demonstrates this. We can make
particles that are very small at ﬁrst, and then gradually grow in size. We use a
solution of sodium thiosulfate (hypo) with sulphuric acid, which precipitates very
ﬁne grains of sulphur. As the sulphur precipitates, the grains ﬁrst start very small,
and the scattering is a little bluish. As it precipitates more it gets more intense, and
then it will get whitish as the particles get bigger. In addition, the light which goes
straight through will have the blue taken out. That is why the sunset is red, of
course, because the light that comes} through a lot of air to the eye has had a lot
of blue light scattered out, so it is yellow—red. Finally, there is one other important feature which really belongs in the next
chapter, on polarization, but it is so interesting that we point it out now. This is
that the electric ﬁeld of the scattered light tends to vibrate in a particular direction.
The electric ﬁeld in the incoming light is oscillating in some way, and the driven
oscillator goes in this same direction, and if we are situated about at right angles
to the beam, we will see polarized light, that is to say, light in which the electric
ﬁeld is going only one way. In general, the atoms can vibrate in any direction at
right angles to the beam, but if they are driven directly toward or away from us, we
do not see it. So if the incoming light has an electric ﬁeld which changes and os
cillates in any direction, which we call unpolarized light, then the light which is
coming out at 90° to the beam vibrates in only one direction! (See Fig. 32—3.) There is a substance called polaroid which has the property that when light
goes through it, only the piece of the electric ﬁeld which is along one particular
axis can get through. We can use this to test for polarization, and indeed we
ﬁnd the light scattered by the hypo solution to be strongly polarized. 32—9 Electron
moves in plane .L'k. \Fx Atom
Incident beam + (unpolarized) Radiation scattered
.L k is plane polarized Fig. 32—3.
the polarization of radiation scattered
at right angles to the incident beam. IllustratiOn of the origin of ...
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This note was uploaded on 06/18/2009 for the course PHYSICS none taught by Professor Leekinohara during the Spring '09 term at Uni. Nottingham  Malaysia.
 Spring '09
 LeeKinohara
 Physics, Light, Radiation

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