This preview shows pages 1–10. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 41 The Brownian Movement 41—1 Equipartition of energy The Brownian movement was discovered in 1827 by Robert Brown, a botanist.
While he was studying microscopic life, he noticed little particles of plant pollens
jiggling around in the liquid he was looking at in the microscope, and he was wise
enough to realize that these were not living, but were just little pieces of dirt moving
around in the water. In fact he helped to demonstrate that this had nothing to do
with life by getting from the ground an old piece of quartz in which there was
some water trapped. It must have been trapped for millions and millions of years,
but inside he could see the same motion. What one sees is that very tiny particles
are jiggling all the time. This was later proved to be one of the eﬁ'ects of molecular motion, and we can
understand it qualitatively by thinking of a great push ball on a playing ﬁeld,
seen from a great distance, with a lot of people underneath, all pushing the ball
in various directions. We cannot see the people because we imagine that we are
too far away, but we can see the ball, and we notice that it moves around rather
irregularly. We also know, from the theorems that we have discussed in previous
chapters, that the mean kinetic energy of a small particle suspended in a liquid
or a gas will be ng even though it is very heavy compared with a molecule. If it
is very heavy, that means that the speeds are relatively slow, but it turns out,
actually, that the speed is not really so slow. In fact, we cannot see the speed of
such a particle very easily because although the mean kinetic energy is ng, which
represents a speed of a millimeter or so per second for an object a micron or two
in diameter, this is very hard to see even in a microscope, because the particle
continuously reverses its direction and does not get anywhere. How far it does get
we will discuss at the end of the present chapter. This problem was ﬁrst solved by
Einstein at the beginning of the present century. Incidentally, when we say that the mean kinetic energy of this particle is
ng, we claim to have derived this result from the kinetic theory, that is, from
Newton’s laws. We shall ﬁnd that we can derive all kinds of things—marvelous
things—‘from the kinetic theory, and it is most interesting that we can apparently
get so much from so little. Of course we do not mean that Newton’s laws are “little”
—they are enough to do it, really~what we mean is that we did not do very much.
How do we get so much out? The answer is that we have been perpetually making
a certain important assumption, which is that if a given system is in thermal
equilibrium at some temperature, it will also be in thermal equilibrium with
anything else at the same temperature. For instance, if we wanted to see how a
particle would move if it was really colliding with water, we could imagine that
there was a gas present, composed of another kind of particle, little ﬁne pellets
that (we suppose) do not interact with water, but only hit the particle with “hard”
collisions. Suppose the particle has a prong sticking out of it; all our pellets have
to do is hit the prong. We know all about this imaginary gas of pellets at tempera
ture T——it is an ideal gas. Water is complicated, but an ideal gas is simple. Now,
our particle has to be in equilibrium with the gas of pellets. Therefore, the mean
motion of the particle must be what we get for gaseous collisions, because if it
were not moving at the right speed relative to the water but, say, was moving
faster, that would mean that the pellets would pick up energy from it and get
hotter than the water. But we had started them at the same temperature, and we
assume that if a thing is once in equilibrium, it stays in equilibrium—parts of it
do not get hotter and other parts colder, spontaneously. 41—1 41—1 Equipartition of energy 41—2 Thermal equilibrium of
radiation 41—3 Equipartition and the quantum
oscillator 41—4 The random walk (a) (b) Fig. 4l—l. (a) A sensitive lightbeam
galvanometer. Light from a source L is
reﬂected from a small mirror onto a
scale. (b) A schematic record of the
reading of the scale as a function of
the time. R R (a) (b) Fig. 41—2. A highQ resonant circuit.
(0) Actual circuit, at temperature T.
(b) Artiﬁcial circuit, with an ideal (noise
less) resistance and a "noise generator" G. This proposition is true and can be proved from the laws of mechanics, but
the proof is very complicated and can be established only by using advanced
mechanics. It is much easier to prove in quantum mechanics than it is in classical
mechanics. It was proved ﬁrst by Boltzmann, but for now we simply take it to be
true, and then we can argue that our particle has to have ng of energy if it is hit
with artiﬁcial pellets, so it also must have ng when it is being hit with water at
the same temperature and we take away the pellets; so it is ng. It is a strange
line of argument, but perfectly valid. In addition to the motion of colloidal particles for which the Brownian move
ment was ﬁrst discovered, there are a number of other phenomena, both in the
laboratory and in other situations, where one can see Brownian movement. If we
are trying to build the most delicate possible equipment, say a very small mirror
on a thin quartz ﬁber for a very sensitive ballistic galvanometer (Fig. 41—1), the
mirror does not stay put, but jiggles all the time——all the time—so that when we
shine a light on it and look at the position of the spot, we do not have a perfect
instrument because the mirror is always jiggling. Why? Because the average kinetic
energy of rotation of this mirror has to be, on the average, lkT. What is the meansquare angle over which the mirror will wobble? Suppose
we ﬁnd the natural vibration period of the mirror by tapping on one side and
seeing how long it takes to oscillate back and forth, and we also know the moment
of inertia, I. We know the formula for the kinetic energy of rotation—it is given
by Eq. (19.8): T = %Iw2. That is the kinetic energy, and the potential energy that
goes with it will be proportional to the square of the angle—it is V = %a02.
But, if we know the period to and calculate from that the natural frequency (do =
27r/to, then the potential energy is V 2 gauge? Now we know that the average
kinetic energy is ékT, but since it is a harmonic oscillator the average potential
energy is also §kT. Thus was = em
or (92) = kT/Iwg. (41.1) In this way we can calculate the oscillations of a galvanometer mirror, and thereby
ﬁnd what the limitations of our instrument will be. If we want to have smaller
oscillations, we have to cool the mirror. An interesting question is, where to
cool it. This depends upon where it is getting its “kicks” from. If it is through the
ﬁber, we cool it at the top—if the mirror is surrounded by a gas and is getting hit
mostly by collisions in the gas, it is better to cool the gas. As a matter of fact, if
we know where the damping of the oscillations comes from, it turns out that that is
always the source of the ﬂuctuations also, a point which we will come back to. The same thing works, amazingly enough, in electrical circuits. Suppose that
we are building a very sensitive, accurate ampliﬁer for a deﬁnite frequency and have
a resonant circuit (Fig. 41—2) in the input so as to make it very sensitive to this
certain frequency, like a radio receiver, but a really good one. Suppose we wish
to go down to the very lowest limit of things, so we take the voltage, say off the
inductance, and send it into the rest of the ampliﬁer. Of course, in any circuit
like this, there is a certain amount of loss. It is not a perfect resonant circuit, but
it is a very good one and there is a little resistance, say (we put the resistor in so we
can see it, but it is supposed to be small). Now we would like to ﬁnd out: How
much does the voltage across the inductance ﬂuctuate? Answer: We know that
%L12 is the “kinetic energy”——the energy associated with a coil in a resonant circuit
(Chapter 25). Therefore the mean value of %L12 is equal to §kT—that tells us what
the rms current is and we can ﬁnd out what the rms voltage is from the rms current.
For if we want the voltage across the inductance the formula is 171, = iwLI: and
the mean absolute square voltage on the inductance is (Vi) = L2w3 I2), and
putting in %L(I2) = %kT, we obtain (V%) = LwﬁkT. (41.2) So now we can design circuits and tell when we are going to get what is called
Johnson noise, the noise associated with thermal ﬂuetuations! 41—2 Where do the ﬂuctuations come from this time? They come again from the
resistor—they come from the fact that the electrons in the resistor are jiggling
around because they are in thermal equilibrium with the matter in the resistor, and
they make ﬂuctuations in the density of electrons. They thus make tiny electric
ﬁelds which drive the resonant circuit. Electrical engineers represent the answer in another way. Physically, the
resistor is effectively the source of noise. However, we may replace the real circuit
having an honest, true physical resistor which is making noise, by an artiﬁcial
circuit which contains a little generator that is going to represent the noise, and
now the resistor is otherwise ideal—no noise comes from it. All the noise is in
the artiﬁcial generator. And so if we knew the characteristics of the noise generated
by a resistor, if we had the formula for that, then we could calculate what the circuit
is going to do in response to that noise. So, we need a formula for the noise
ﬂuctuations. Now the noise that is generated by the resistor is at all frequencies,
since the resistor by itself is not resonant. Of course the resonant circuit only
“listens” to the part that is near the right frequency, but the resistor has many
different frequencies in it. We may describe how strong the generator is, as follows:
The mean power that the resistor would absorb if it were connected directly across
the noise generator would be (E2)/R, if E were the voltage from the generator.
But we would like to know in more detail how much power there is at every fre
quency. There is very little power in any one frequency; it is a distribution. Let
P(w) dw be the power that the generator would deliver in the frequency range dw
into the very same resistor. Then we can prove (we shall prove it for another case,
but the mathematics is exactly the same) that the power comes out P(w) dw = (2/7r)kT dw, (41.3) and is independent of the resistance when put this way. 412 Thermal equilibrium of radiation Now we go on to consider a still more advanced and interesting proposition
that is as follows. Suppose we have a charged oscillator like those we were talking
about when we were discussing light, let us say an electron oscillating up and down
in an atom. If it oscillates up and down, it radiates light. Now suppose that this
oscillator is in a very thin gas of other. atoms, and that from time to time the atoms
collide with it. Then in equilibrium, after a long time, this oscillator will pick up
energy such that its kinetic energy of oscillation is %kT, and since it is a harmonic
oscillator, its entire energy of motion will become kT. That is, of course, a wrong
description so far, because the oscillator carries electric charge, and if it has an
energy kT it is shaking up and down and radiating light. Therefore it is impossible
to have equilibrium of real matter alone without the charges in it emitting light,
and as light is emitted, energy ﬂows away, the oscillator loses its kT as time goes
on, and thus the whole gas which is colliding with the oscillator gradually cools
off. And that is, of course, the way a hot stove cools on a cold night by radiating
the light into the sky, because the atoms are jiggling their charge and they contin
ually radiate, and slowly, because of this radiation, the jiggling motion slows down. On the other hand, if we enclose the whole thing in a box so that the light does
not go away to inﬁnity, then we can eventually get thermal equilibrium. We may
either put the gas in a box where we can say that there are other radiators in the
box walls sending light back or, to take a nicer example, we may suppose the box
has mirror walls. It is easier to think about that case. Thus we assume that all the
radiation that goes out from the oscillator keeps running around in the box.
Then, of course, it is true that the oscillator starts to radiate, but pretty soon it
can maintain its kT of kinetic energy in spite of the fact that it is radiating, because
it is being illuminated, we may say, by its own light reﬂected from the walls of the
box. That is, after a while there is a great deal of light rushing around in the box,
and although the oscillator is radiating some, the light comes back and returns
some of the energy that was radiated. 41—3 We shall now determine how much light there must be in such a box at tempera
ture T in order that the shining of the light on this oscillator will generate just
enough energy to account for the light it radiated. Let the gas atoms be very few and far between, so that we have an ideal
oscillator with no resistance except radiation resistance. Then we consider that at
thermal equilibrium the oscillator is doing two things at the same time. First, it
has a mean energy kT, and we calculate how much radiation it emits. Second, this
radiation should be exactly the amount that would result because of the fact that
the light shining on the oscillator is scattered. Since there is nowhere else the energy
can go, this effective radiation is really just scattered light from the light that is
in there. Thus we ﬁrst calculate the energy that is radiated by the oscillator per second,
if the oscillator has a certain energy. (We borrow from Chapter 32 on radiation
resistance a number of equations without going back over their derivation.) The
energy radiated per radian divided by the energy of the oscillator is called 1 / Q
(Eq. 32.8): 1 / Q = (dW/dt)/woW. Using the quantity ‘Y, the damping constant,
this can also be written as 1/ Q = W/wo, where (.00 is the natural frequency of the
oscillator—if gamma is very small, Q is very large. The energy radiated per second
is then dW _ woW _ w0W7 _
117 ‘ ? — wo _ 7W. (41.4) The energy radiated per second is thus simply gamma times the energy of the
oscillator. Now the oscillator should have an average energy kT, so we see that
gamma kT is the average amount of energy radiated per second: (dW/dt) = M. (41.5) Now we only have to know what gamma is. Gamma is easily found from Eq.
(32.12). It is — “’0 — 2 ’0‘“ (41.6) where r0 = eZ/mc2 is the classical electron radius, and we have set >\ = 27rc/wo.
Our ﬁnal result for the average rate of radiation of light near the frequency we is therefore
dW _ 2 rotong
7? _ 5 c ' (41.7) Next we ask how much light must be shining on the oscillator. It must be
enough that the energy absorbed from the light (and thereupon scattered) is just
exactly this much. In other words, the emitted light is accounted for as scattered
light from the light that is shining on the oscillator in the cavity. So we must now
calculate how much light is scattered from the oscillator if there is a certam amount
—unknown—of radiation incident on it. Let 1(w) dw be the amount of light energy
there is at the frequency to, within a certain range do.) (because there is no light at
exactly a certain frequency; it is spread all over the spectrum). So [(w) is a certain
spectral distribution which we are now going to ﬁnd—it is the color of a furnace at
temperature T that we see when we open the door and look in the hole. Now
how much light is absorbed? We worked out the amount of radiation absorbed
from a given incident light beam, and we calculated it in terms of a cross section.
It is just as though we said that all of the light that falls on a certain cross section
is absorbed. So the total amount that is reradiated (scattered) is the incident
intensity 1(w) dw multiplied by the cross section a. The formula for the cross section which we derived (Eq. 31.19) did not have
the damping included. It is not hard to go through the derivation again and put in
the resistance term which we neglected. If we do that, and calculate the cross
section the same way, we get 81rr3 w4 414 Now, as a function of frequency, a, is of signiﬁcant size only for a) very near
to the natural frequency (.00. (Remember that the Q for a radiating oscillator is
about 108.) The oscillator scatters very strongly when w is equal to wo, and very
weakly for other values of (.0. Therefore we can replace w by too and w2 — cog
by 2w0(w — coo), and we get 21rr3w3 =aw~mﬁ¥wm‘ Now the whole curve is localized near 0) = am. (We do not really have to make
any approximations, but it is much easier to do the integrals if we simplify the
equation a bit.) Now we multiply the intensity in a given frequency range by the
cross section of scattering, to get the amount of energy scattered in the range dw.
The total energy scattered is then the integral of this for all w. Thus 0'3 (41.9) dag/8 = /0 1(w)a,(w)dw (41 . 10)
2717'?)ng (to) do.) =ﬂMw—MVIMM' Now we set dWs/dt = 37kT. Why three? Because when we made our
analysis of the cross section in Chapter 32, we assumed that the polarization was
such that the light could drive the oscillator. If we had used an oscillator which
could move only in one direction, and the light, say, was polarized in the wrong
way, it would not give any scattering. So we must either average the cross section
of an oscillator which can go only in one direction, over all directions of incidence
and polarization of the light or, more easily, we can imagine an oscillator which
will follow the ﬁeld no matter which way the ﬁeld is pointing. Such an oscillator,
which can oscillate equally in three directions, would have 3kT average energy
because there are 3 degrees of freedom in that oscillator. So we should use 37kT
because of the 3 degrees of freedom. Now we have to do the integral. Let us suppose that the unknown spectral
distribution I(w) of the light is a smooth curve and does not vary very much across
the very narrow frequency region where a, is peaked (Fig. 41—3). Then the only
signiﬁcant contribution comes when w is very close to coo, within an amount
gamma, which is very small. So therefore, although 1(a)) may be an unknown and
complicated function, the only place where it is important is near to = too, and
there we may replace the smooth curve by a ﬂat one—a “constant”——at the same
height. In other words, we simply take [(00) outside the integral sign and call it
[(wo). We may also take the rest of the constants out in front of the integral, and
what we have left is w dw
0 (w — 600V ‘1‘ 72/4 Now, the integral should go from 0 to 00, but 0 is so far from wo that the curve is
all ﬁnished by that time, so we go instead to minus oo—it makes no difference and
it is much easier to do the integral. The integral is an inverse tangent function of
the form fdx/(x2 + a2). If we look it up in a book we see that it is equal to
1r/a. So what it comes to for our case is 27/7. Therefore we get, with some re
arranging, : 37kT. (41.11) 2 2
%7r0w01(w0) 972kT _ 41r2r3w3 [(090) = (41.12) Then we substitute the formula (41.6) for gamma (do not worry about writing
coo; since it is true of any coo, we may just call it w) and the formula for 1(a)) then
comes out (0sz .
7r2c2 1(a)) = (41.13) And that gives us the distribution of light in a hot furnace. It is called the black
41—5 Fig. 41—3. The factors in the integrand
(41.10). The peak is the resonance curve
l/(w — mg)2 + 72/4. To a good ap
proximation the factor I(w) can be
replaced by l(wo). Ital) Fig. 41—4. The blackbody intensity
distribution at two temperatures, accord
ing to classical physics (solid curves). The
dashed curves show the actual distribution. body radiation. Black, because the hole in the furnace that we look at is black when
the temperature is zero. Inside a closed box at temperature T, (41.13) is the distribution of energy of
the radiation, according to classical theory. First, let us notice a remarkable feature
of that expression. The charge of the oscillator, the mass of the oscillator, all
properties speciﬁc to the oscillator, cancel out, because once we have reached equilib
rium with one oscillator, we must be at equilibrium with any other oscillator
of a different mass, or we will be in trouble. So this is an important kind of check
on the proposition that equilibrium does not depend on what we are in equilibrium
with, but only on the temperature. Now let us draw a picture of the 1(w) curve
(Fig. 41—4). It tells us how much light we have at different frequencies. The amount of intensity that there is in our box, per unit frequency range,
goes, as we see, as the square of the frequency, which means that if we have a box
at any temperature at all, and if we look at the xrays that are coming out, there
will be a lot of them! Of course we know this is false. When we open the furnace and take a look
at it, we do not burn our eyes out from xrays at all. It is completely false. Further
more, the total energy in the box, the total of all this intensity summed over all
frequencies, would be the area under this inﬁnite curve. Therefore, something is
fundamentally, powerfully, and absolutely wrong. Thus was the classical theory absolutely incapable of correctly describing the
distribution of light from a blackbody, just as it was incapable of correctly de
scribing the speciﬁc heats of gases. Physicists went back and forth over this deriva
tion from many different points of view, and there is no escape. This is the pre
diction of classical physics. Equation (41.13) is called Rayleigh’s law, and it is the
prediction of classical physics, and is obviously absurd. 41—3 Equipartition and the quantum oscillator The difﬁculty above was another part of the continual problem of classical
physics, which started with the difﬁculty of the speciﬁc heat of gases, and now has
been focused on the distribution of light in a blackbody. Now, of course, at the
time that theoreticians studied this thing, there were also many measurements
of the actual curve. And it turned out that the correct curve looked like the dashed
curves in Fig. 41—4. That is, the xrays were not there. If we lower the temperature,
the whole curve goes down in proportion to T, according to the classical theory,
but the observed curve also cuts off sooner at a lower temperature. Thus the low
frequency end of the curve is right, but the highfrequency end is wrong. Why?
When Sir James Jeans was worrying about the speciﬁc heats of gases, he noted
that motions which have high frequency are “frozen out” as the temperature goes
too low. That is, if the temperature is too low, if the frequency is too high, the
oscillators do not have kT of energy on the average. Now recall how our derivation
of (41.13) worked: It all depends on the energy of an oscillator at thermal equilib
rium. What the kT of (41.5) was, and what the same kT in (41.13) is, is the mean
energy of a harmonic oscillator of frequency to at temperature T. Classically, this
is kT, but experimentally, nol—not when the temperature is too low or the oscillator
frequency is too high. And so the reason that the curve falls off is the same reason
that the speciﬁc heats of gases fail. It is easier to study the blackbody curve than it
is the speciﬁc heats of gases, which are so complicated, therefore our attention is
focused on determining the true blackbody curve, because this curve is a curve
which correctly tells us, at every frequency/what the average energy of harmonic
oscillators actually is as a function of temperature. Planck studied this curve. He ﬁrst determined the answer empirically, by
ﬁtting the observed curve with a nice function that ﬁtted very well. Thus he had
an empirical formula for the average energy of a harmonic oscillator as a function
of frequency. In other words, he had the right formula instead of kT, and then by
ﬁddling around he found a simple derivation for it which involved a very peculiar
assumption. That assumption was that the harmonic oscillator can take up energies
only fit.) at a time. The idea that they can have any energy at all is false. Of course,
that was the beginning of the end of classical mechanics. 41—6 The very ﬁrst correctly determined quantummechanical formula will now be
derived. Suppose that the permitted energy levels of a harmonic oscillator were
equally spaced at hwo apart, so that the oscillator could take on only these different
energies (Fig. 41—5). Planck made a somewhat more complicated argument than
the one that is being given here, because that was the very beginning of quantum
mechanics and he had to prove some things. But we are going to take it as a fact
(which he demonstrated in this case) that the probability of occupying a level of
energy E is P(E) = axe—EH”. If we go along with that, we will obtain the right
result. Suppose now that we have a lot of oscillators, and each is a vibrator of fre
quency we. Some of these vibrators will be in the bottom quantum state, some will
be in the next one, and so forth. What we would like to know is the average energy
of all these oscillators. To ﬁnd out, let us calculate the total energy of all the oscilla
tors and divide by the number of oscillators. That will be the average energy per
oscillator in thermal equilibrium, and will also be the energy that is in equilibrium
with the blackbody radiation and that should go in Eq. (41.13) in place of kT.
Thus we let N 0 be the number of oscillators that are in the ground state (the
lowest energy state); N1 the number of oscillators in the state E1; N 2 the number
that are in state E2; and so on. According to the hypothesis (which we have not
proved) that in quantum mechanics the law that replaced the probability e_P‘E'/kT
or e_K'E"kT in classical mechanics is that the probability goes down as e‘”’”,
where AE is the excess energy, we shall assume that the number N1 that are in the
ﬁrst state will be the number N 0 that are in the ground state, times e‘M/kT. Simi
larly, N 2, the number of oscillators in the second state, is N 2 = Noe—2W”. To
simplify the algebra, let us call e“'l"’”‘T = x. Then we simply have N1 = N 0x,
N2 = Noxz, . . . ,Nn = Nox”. The total energy of all the oscillators must ﬁrst be worked out. If an oscillator
is in the ground state, there is no energy. If it is in the ﬁrst state, the energy is hwo,
and there are N1 of them. So N lhw, or thox is how much energy we get from
those. Those that are in the second state have Zhwo, and there are N 2 of them,
so N 2  2hw = Zthox2 is how much energy we get, and so on. Then we add
it all together to get Em = Nohw(0 + x + 2x2 + 3x3 + . . .). And now, how many oscillators are there? Of course, N 0 is the number that
are in the ground state, N1 in the ﬁrst state, and so on, and we add them together: N101 = No(1 + x + x2 + x3 + . . .). Thus the average energy is _ Etot =1_Vohw(0+x+2x2+3x3+...).
Ntot N0(l +x+x2+...) Now the two sums which appear here we shall leave for the reader to play with and
have some fun with. When we are all ﬁnished summing and substituting for x in
the sum, we should get—if we make no mistakes in the sum— hw eﬁu/kT _ 1 (E) (41.14) (E) = (41.15)
This, then, was the ﬁrst quantummechanical formula ever known, or ever discussed,
and it was the beautiful culmination of decades of puzzlement. Maxwell knew
that there was something wrong, and the problem was, what was right? Here
is the quantitative answer of what is right instead of kT. This expression should,
of course, approach kT as w —> 0 or as T —> 00. See if you can prove that it does—
learn how to do the mathematics. This is the famous cutoff factor that Jeans was looking for, and if we use it
instead of kT in (41.13), we obtain for the distribution of light in a black box hw3 dco
7“_2c2(e7im/ICT _ We see that for a large to, even though we have (.03 in the numerator, there is an e
raised to a tremendous power in the denominator, so the curve comes down again and does not “blow up”—we do not get ultraviolet light and xrays where we do
not expect them! 1(a)) do) = (41.16) 41—7 —‘— E4  41w P.  onp(4fm/kT) 95 AexMSM/k'l'l E2  211w Pz  AcxuZMIKT)
N
—l— E,  11w P,  Aupt‘ﬁw/kT)
No
Eu 0 Po A
Fig. 41—5. The energy levels of a harmonic oscillator are equally spaced:
5,, = nhw. 36 Fig. 41—6. A random walk of 36
steps of length I. How far is $36 from B?
Ans: about 6! on the average. One might complain that in our derivation of (41.16) we used the quantum
theory for the energy levels of the harmonic oscillator, but the classical theory in
determining the cross section 08. But the quantum theory of light interacting with a
harmonic oscillator gives exactly the same result as that given by the classical theory.
That, in fact, is why we were justiﬁed in spending so much time on our analysis of
the index of refraction and the scattering of light, using a model of atoms like
little oscillators—~the quantum formulas are substantially the same. Now let us return to the Johnson noise in a resistor. We have already re
marked that the theory of this noise power is really the same theory as that of the
classical blackbody distribution. In fact, rather amusingly, we have already
said that if the resistance in a circuit were not a real resistance, but were an antenna
(an antenna acts like a resistance because it radiates energy), a radiation resistance,
it would be easy for us to calculate what the power would be. It would be just the
power that runs into the antenna from the light that is all around, and we would
get the same distribution, changed by only one or two factors. We can suppose
that the resistor is a generator with an unknown power spectrum P(w). The spec
trum is determined by the fact that this same generator, connected to a resonant
circuit of any frequency, as in Fig. 41—2(b), generates in the inductance a voltage
of the magnitude given in Eq. (41.2). One is thus led to the same integral as in
(41.10), and the same method works to give Eq. (41.3). For low temperatures the
kT in (41.3) must of course be replaced by (41.15). The two theories (blackbody
radiation and Johnson noise) are also closely related physically, for we may of
course connect a resonant circuit to an antenna, so the resistance R is a pure
radiation resistance. Since (41.2) does not depend on the physical origin of the
resistance, we know the generator G for a real resistance and for radiation resist
ance is the same. What is the origin of the generated power P(w) if the resistance R
is only an ideal antenna in equilibrium with its environment at temperature T?
It is the radiation [(02) in the space at temperature Twhich impinges on the antenna
and, as “received signals,” makes an effective generator. Therefore one can deduce
a direct relation of P(w) and 1(a)), leading then from (41.13) to (41.3). All the things we have been talking about—the socalled Johnson noise and
Planck’s distribution, and the correct theory of the Brownian movement which we
are about to describe—are developments of the ﬁrst decade or so of this century.
Now with those points and that history in mind, we return to the Brownian move
ment. 41—4 The random walk Let us consider how the position of a jiggling particle should change with time,
for very long times compared with the time between “kicks.” Consider a little
Brownian movement particle which is jiggling about because it is bombarded on
all sides by irregularly jiggling water molecules. Query: After a given length of
time, how far away is it likely to be from where it began? This problem was solved
by Einstein and Smoluchowski. If we imagine that we divide the time into little
intervals, let us say a hundredth of a second or so, then after the ﬁrst hundredth of
a second it moves here, and in the next hundredth it moves some more, in the
next hundredth of a second it moves somewhere else, and so on. In terms of the
rate of bombardment, a hundredth of a second is a very long time. The reader
may easily verify that the number of collisions a single molecule of water receives
in a second is about 10 14, so in a hundredth of a second it has 1012 collisions, which
is a lot! Therefore, after a hundredth of a second it is not going to remember what
happened before. In other words, the collisions are all random, so that one “step”
is not related to the previous “step.” It is like the famous drunken sailor problem:
the sailor comes out of the bar and takes a sequence of steps, but each step is
chosen at an arbitrary angle, at random (Fig. 41—6). The question is: After a long
time, where is the sailor? Of course we do not know! It is impossible to say.
What do we mean—he is just somewhere more or less random. Well then, on the
average, where is he? 0n the average, how far away from the bar has he gone?
We have already answered this question, because once we were discussing the 41—8 superposition of light from a whole lot of different sources at different phases,
and that meant adding a lot of arrows at different angles (Chapter 32). There we
discovered that the mean square of the distance from one end to the other of the
chain of random steps, which was the intensity of the light, is the sum of the
intensities of the separate pieces. And so, by the same kind of mathematics, we
can prove immediately that if RN is the vector distance from the origin after N steps,
the mean square of the distance from the origin is proportional to the number N
of steps. That is, (Riv) = NL2, where L is the length of each step. Since the num
ber of steps is proportional to the time in our present problem, the mean square
distance is proportional to the time: (R2) = at. (41.17) This does not mean that the mean distance is proportional to the time. If the mean
distance were proportional to the time it would mean that the drifting is at a nice
uniform velocity. The sailor is making some relatively sensible headway, but only
such that his mean square distance is proportional to time. That is the character
istic of a random walk. We may show very easily that in each successive step the square of the distance
increases, on the average, by L2. For if we write RN = RN_1 + L, we ﬁnd that
Rir is RN RN = R13 = R134 + 2RN_1  L + L2, and averaging over many trials, we have (Rgzv) = (R12V_1> + L2, since (RN_1 L)
= 0. Thus. by induction, R12v = NL2. (41.18) Now we would like to calculate the coefﬁcient a in Eq. (41.17), and to do so
we must add a feature. We are going to suppose that if we were to put a force
on this particle (having nothing to do with the Brownian movement—we are
taking a side issue for the moment), then it would react in the following way against
the force. First, there would be inertia. Let m be the coefﬁc1ent of inertia, the
eﬂective mass of the object (not necessarily the same as the real mass of the real
particle, because the water has to move around the particle if we pull on it). Thus
if we talk about motion in one direction, there is a term like m (d2x/dt2) on one side.
And next, we want also to assume that if we kept a steady pull on the object, there
would be a drag on it from the ﬂuid, proportional to its velocity. Besides the inertia
of the ﬂuid, there is a resistance to ﬂow due to the viscosity and the complexity of
the ﬂuid. It is absolutely essential that there be some irreversible losses, something
like resistance, in order that there be ﬂuctuations. There is no way to produce the
kT unless there are also losses. The source of the ﬂuctuations is very closely related
to these losses. What the mechanism of this drag is, we will discuss soon—we shall
talk about forces that are proportional to the velocity and where they come from.
But let us suppose for now that there is such a resistance. Then the formula for
the motion under an external force. when we are pulling on it in a normal manner, is dzx dx The quantity n can be determined directly from experiment. For example, we can
watch the drop fall under gravity. Then we know that the force is mg, and p is mg
divided by the speed of fall the drop ultimately acquires. Or we could put the
drop in a centrifuge and see how fast it sediments. Or if it is charged, we can put
an electric ﬁeld on it. So a is a measurable thing, not an artiﬁcial thing, and it is
knowu for many types of colloidal particles, etc. Now let us use the same formula in the case where the force is not external,
but is equal to the irregular forces of the Brownian movement. We shall then try
to determine the mean square distance that the object goes. Instead of taking the
distances in three dimensions, let us take just one dimension, and ﬁnd the mean
of x2, just to prepare ourselves. (Obviously the mean of x2 is the same as the mean
of y2 is the same as the mean of 22, and therefore the mean square of the distance 41—9 is just 3 times what we are going to calculate.) The xcomponent of the irregular
forces is, of course, just as irregular as any other component. What is the rate of
change of x2? It is d(x2)/dt = 2x(dx/d1), so what we have to ﬁnd is the average
of the position times the velocity. We shall show that this is a constant, and that
therefore the mean square radius will increase proportionally to the time, and at
what rate. Now if we multiply Eq. (41.19) by x, mx(d2x/dt2) + ux(dx/ dz) = sz.
We want the time average of x(dx/dt), so let us take the average of the whole
equation, and study the three terms. Now what about x times the force? If the
particle happens to have gone a certain distance x, then, since the irregular force is
completely irregular and does not know where the particle started from, the next
impulse can be in any direction relative to x. If x is positive, there is no reason why
the average force should also be in that direction. It is just as likely to be one
way as the other. The bombardment forces are not driving it in a deﬁnite direction.
So the average value of x times F is zero. On the other hand, for the term
mx(d2x/dt2) we will have to be a little fancy, and write this as d2x _ m d[x(dx/dt)] _ m (dx)2 "MHz—2 _ dt d7 Thus we put in these two terms and take the average of both. So let us see how
much x times the velocity should be. Now x times the velocity has a mean that
does not change with time, because when it gets to some position it has no re
membrance of where it was before, so things are no longer changing with time.
So this quantity, on the average, is zero. We have left the quantity mvz, and that
is the only thing we know: m02/2 has a mean value ﬁkT. Therefore we ﬁnd that d2 d
<mxaﬁx + ,u<xd—:C = (xF,) implies
_ 2 £1 2 =
(mv>+2dt(x> 0,
01‘
2
‘12:) = 2 (41.20) Therefore the object has a mean square distance (R2), at the end of a certain amount
of t, equal to (R2) = 6kTi. (41.21)
And so we can actually determine how far the particles go! We ﬁrst must determine
how they react to a steady force, how fast they drift under a known force (to
ﬁnd it), and then we can determine how far they go in their random motions.
This equation was of considerable importance historically, because it was one of
the ﬁrst ways by which the constant k was determined. After all, we can measure
it, the time, how far the particles go, and we can take an average. The reason that
the determination of k was important is that in the law PV = RT for a mole,
we know that R, which can also be measured, is equal to the number of atoms in a
mole times k. A mole was originally deﬁned as so and so many grams of oxygen—
16 (now carbon is used), so the number of atoms in a mole was not known, orig
inally. It is, of course, a very interesting and important problem. How big are
atoms? How many are there? So one of the earliest determinations of the number
of atoms was by the the determination of how far a dirty little particle would move
if we watched it patiently under a microscope for a certain length of time. And
thus Boltzmann’s constant k and the Avogadro number N 0 were determined be
cause R had already been measured. 4110 ...
View
Full
Document
This note was uploaded on 06/18/2009 for the course PHYSICS Physics taught by Professor Limkong during the Spring '09 term at Uni. Nottingham  Malaysia.
 Spring '09
 LimKong
 Physics

Click to edit the document details