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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 temperatur...
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 Spring '09
 LimKong
 Physics, Energy, coo

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