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Feynman Physics Lectures V1 Ch41 1962-04-20 The Brownian Movement

Feynman Physics Lectures V1 Ch41 1962-04-20 The Brownian Movement

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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 light-beam 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 high-Q 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 mean-square 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. 41-2 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 re-radiated (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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