Contents

4. Thermal equilibrium, temperature, and chance

Timing

This Part is the key to the whole Unit. For some students, it will also conclude the main work of the Unit, except for a glance at one or two applications from Part Six. The times suggested for earlier Parts have been limited deliberately, so as to allow breathing space for Part Four.

We think it may need about a week and a half, say 9 or 10 periods. It could be taught, without interruptions or questions, in much less, but a good deal of time is likely to be needed for discussion, in which students go back over ideas from earlier Parts, and try to see their application here.

Teachers who think their classes can cope with Part Five would do well to go a little more quickly through Part Four. Discussion which might be slow and diffuse in the context of Part Four can be held over to Part Five, when it should be easier to sharpen the issues for those who can grasp the more theoretical development in Part Five.

An improvement in our knowledge of heat, which has been attained by the use of thermometers, is the more distinct notion we have now than formerly of the distribution of heat among different bodies. Even without the help of thermometers, we can perceive a tendency of heat to diffuse itself from any hotter body to the cooler ones around it, until the heat is distributed among them in such a manner that none of them is disposed to take any more from the rest. The heat is thus brought into a state of equilibrium.

This equilibrium is somewhat curious. [Our emphasis.] We find that, when all mutual action is ended, a thermometer applied to any one of the bodies undergoes the same degree of expansion. Therefore the temperature of them all is the same. No previous acquaintance with the peculiar relation of each body to heat could have assured us of this, and we owe the discovery entirely to the thermometer. We must therefore adopt, as one of the most general laws of heat, the principle that all bodies communicating freely with one another, and exposed to no inequality of external action, acquire the same temperature, as indicated by a thermometer. All acquire the temperature of the surrounding medium.

By the use of thermometers we have learned that, if we take a thousand, or more, different kinds of matter - such as metals, stones, salts, woods, cork, feathers, wool, water, and a variety of other fluids - although they be all at first of different temperatures, and if we put them together in a room without a fire, and into which the sun does not shine, the heat will be communicated from the hotter of these bodies to the colder, during some hours perhaps, or the course of a day, at the end of which time, if we apply a thermometer to them all in succession, it will give precisely the same reading. The heat, therefore, distributes itself upon this occasion until none of these bodies has a greater demand or attraction for heat than every other of them has; in consequence, when we apply a thermometer to them all in succession, after the first to which it is applied has reduced the instrument to its own temperature, none of the rest is disposed to increase or diminish the quantity of heat which that first one left in it. This is what has been commonly called an equal heat, or the equality of heat among different bodies; I call it the equilibrium of heat.

From Joseph Black (1803), Lectures on the elements of chemistry. Reprinted in Conant, J. B. (1957) Harvard case histories in experimental science, Volume 1, Harvard University Press

Perhaps Joseph Black's discovery seems obvious or not very important. But it is important, because it is the reason why we can talk at all about the temperature of an object. First of all, in this Part, we discuss thermal equilibrium and temperature, as they appear to us on the large scale, and mention some of their practical implications. Then we shall try to get some idea of what thermal equilibrium is like, and what temperature means, when they are related to matter on the atomic scale, and to the energy shared amongst atoms.

Demonstration 9.6 A heat engine

The device shown in figure 21, a sandwich of semiconductor materials with cans on either side, produces enough electrical energy to turn a small motor if boiling water is poured into one can.

Figure 21: A thermoelectric heat engine.

More of something good may be better

still - so we try putting more hot water into the other can. The motor stops!

But if cold water is put in the other can, the motor runs well, and, if ice is used, it runs even better. Indeed it will run if the hot can is filled with cool water from the tap as long as the other can is filled with solid carbon dioxide (dry ice). It will not run if each side of the device is equally hot or cool.

Q1 If the motor is running now, will it still be running in an hour's time? Why not?

Q2 Where does the energy to run the motor come from?

Q3 Why is there a heat flow from the hot can into the semiconductor sandwich?

Q4 Why is the sandwich cooler than the hot can?

Q5 Would the motor go on running if the cold can were not there?

Q6 Why must some heat flow into the cold can? Is this energy available to run the motor?

Q7 Is all the energy going out of the hot can into the semiconductor available to run the motor?

This engine cannot help but be inefficient to some extent. We shall return to the inefficiency of such engines later on, in Part Six, and outline why steam engines, petrol engines, gas turbines, and rocket engines must inevitably all be inefficient. We note now that temperature differences are useful, for they can be used to drive engines.

In this Part, we shall not pursue these points much further, but will use the engine to illustrate Joseph Black's discoveries about thermal equilibrium.

Demonstration: 9.6 A heat engine

     1072     semiconductor thermopile unit 
     1023     solar motor 
     1004/1   voltmeter (1 V) 
      542     thermometers ( -10 to 110°C) 2 
      507     stop watch or stop clock 
      176     12 volt battery (to give 2V d.c.) 
     1001     galvanometer (internal light beam) 
     1054/2B  copper wire, 32 s.w.g., and constantan wire 32 s.w.g. 
       77     aluminium block 
     1053     slab of expanded polystyrene, Sellotape, card 
              supply of ice and of boiling water, salt, or solid C02 
     1000     leads 

The flat surfaces of the cans which will be in contact with the semiconductor thermopile should be lightly greased to obtain good thermal contact, and then held against the thermopile with a rubber band. The spindle of the motor should be passed through the centre of a small strip of card ( 40 mm × 10 mm), to act as a rotation indicator, and the motor connected directly to the thermopile unit.

Fill one can with boiling water. After a minute or two, try to start the motor by turning the spindle. When it is running, fill the other can with boiling water. The motor should stop within half a minute. Note that the thermopile unit will be permanently damaged if the temperature of the junctions is allowed to exceed 100°C.

The motor should also be run with hot water in one can and ice (or ice and salt) in the other, and with cold water in one can and ice and salt or solid carbon dioxide in the other. If the motor fails to run, a meter can be used instead.

Reversal. Replace the motor with a 2 V battery which will deliver up to 10 A, starting with cold tap water in both cans. One can warms up and the other cools down.

The temperatures are most conveniently indicated using a copper-constantan thermocouple connected to the light beam galvanometer.

The couple, made of one length of constantan wire joined to two lengths of copper, should have one junction at a steady temperature inside the hole in the aluminium block and the other in one of the two cans as in figure 22. Connect the copper wires to the galvanometer.

galvanometer thermocouple to battery (several amperes) thermopile unit aluminium block

Figure 22: Reversal of the heat engine.

(If a single junction couple is used, the galvanometer reading will drift as its terminals are warmed by the lamp in the galvanometer.) Use the x .01 position of the galvanometer initially, setting the zero to the centre of the scale. Alter the sensitivity if necessary to keep the deflections on the scale.

Follow the rise and fall of temperature of the water in the two cans by moving the couple from one to the other. See figure 23.

Thermal equilibrium

Disconnect the battery, and follow the return to thermal equilibrium. It is quicker to remove the cans from the sandwich and hold them directly in contact, using a rubber band. Stand them on an insulating surface, such as a piece of expanded polystyrene.

Since this experiment will bring out the very possibility of measuring temperature, it is best to describe what is happening without using the term. Instead, one can say that the galvanometer indication is big or small, this way, or the other way.

If the slow approach to equilibrium tries students' patience, mix the two lots of water!

Thermal equilibrium

Q8 Now imagine a film of the thermocouple device running a motor, but shown backwards. What would it seem to show? Say what energy transfers would seem to be taking place.

Interestingly, the reverse of the heat engine experiment can happen. The motor, run backwards as a dynamo, does not actually deliver enough current to produce an observable effect, so we replace it by a battery. When this is done, with equally cool water in each can, one lot of water grows warmer, and the other lot grows cooler.

This result may seem surprising, for we are all very accustomed to the inexorable one-way cooling down of hot things and warming up of cold things.

But if the battery is taken away, there are no surprises. Heat flows once again from the hot can to the cold can. A thermocouple connected to a meter, and placed alternately in each can, gives meter indications that are, at first, large but in opposite directions for the two cans. As time goes by, the two indications both become more and more nearly the same. See figure 23. Incidentally, the device is itself a thermocouple, but one between two sorts of semiconductor.

Note that so far as the conservation of energy is concerned (the basis of the First Law of Thermodynamics), energy could flow from cold to hot, as long as the same amount of energy left the cold object as reached the hot one.

Time Meter indication (thermocouple) +> 0> -> battery connected battery removed this can is warmed this can is cooled

Figure 23: Changes in temperature when the heat engine is reversed.

But, without external assistance, it never does. The approach to thermal equilibrium is one-way only. This is the theme of the Second Law of Thermodynamics, which will concern us more and more in this Unit.

Just now, we use the device to illustrate a third point, called the zeroth law of thermodynamics.

The First Law and energy conservation. See Appendix A for a discussion of the difference between the conservation of energy and the First Law.

The zeroth law. Since the First Law stole the number one, this yet more basic law was given the number zero. It says that if A is in thermal equilibrium with B, and B with C, then A will be in thermal equilibrium with C.

The zeroth law

What is thermal equilibrium? Objects in contact, like the warm and the cold cans, usually change their behaviour for a while. But soon all such changes stop, and the two are in equilibrium.

Suppose a third thing, such as a thermocouple, is put in contact with one can, and gives a certain reading. If it is put in the other, before the cans are in equilibrium, its reading changes. The couple was not in equilibrium with the second can. If it is put in the second can when the cans are in equilibrium, its reading doesn't change. The couple was already in equilibrium with the second can, just by being in equilibrium with the first.

Is this a long way of saying something very easy? Perhaps it is. But unless these things are so, measuring temperature is not even possible, indeed temperature means nothing. Suppose you used a thermometer whose indication is, say, 20°C, to find all those objects in a room which are at 20°C. That is, the thermometer indication stays put when it is placed in contact with them. Then the law we are thinking about says that none of them will change if they are put in contact with one another.

A set of objects of which we wish to say, all these are at the same temperature, must have something in common. And that something is that if they are put in contact, they will all be found already to be in thermal equilibrium.

This is called the 0th, or zeroth, law of thermodynamics. It is simple, but basic. If it were not so, a thermometer could give different readings for two things which were in equilibrium. Or perhaps one would find that of two things with the same reading, one of them was the hotter, because heat flowed from it to the other. Either event would make temperature measurement as we know it a useless procedure, because it would tell us nothing about the future behaviour of the bodies involved.

An analogy: getting on with someone

If A gets on well with B, and B gets on well with C, does A get on well with C? We could not be sure, so it would be impractical to think of measuring people so as to put each onto a scale of getting on well with others, and using the scale to predict who would get on with whom.

Q9 What does it mean, to say that two things are at the same temperature?

Q10 If two things are at the same temperature, do they contain equal amounts of energy?

Scales of temperature

Temperature exists is the message of the zeroth law. Making a scale of temperature, that is, putting a number to it, is a different matter.

Many scales have been invented, and many substances have been used in thermometers. Each scale, be it Celsius, Fahrenheit, Reaumur, or Kelvin, depends on deciding what number to assign to the temperatures of some things, such as ice, boiling water, or the lowest temperature possible, absolute zero. Each substance marks off by its changes, whether of volume, pressure, resistance, or some other property, the temperature between these fixed points. Any one thermometer can now be used to tell whether one object is hotter than another, according to how the physical property used to indicate temperature differs when the thermometer is put in contact with each in turn.

There remains, though, the question of assigning values of the temperature to the values of the physical property being used. If mercury in a thermometer placed in warm water can be seen to have completed half its expansion from ice temperature (0°C) to steam temperature (100°C}, it may seem sensible to assign the temperature 50°C on the mercury scale to that amount of expansion. But doing so is a matter of arbitrary decision, convenient but not absolutely necessary.

Furthermore. there is no guarantee that a coil of resistance wire, placed in the same warm water, will be found to have completed just half its increase of resistance from 0°C to 100°C. If the increase is less than half complete, the resistance coil temperature of the water will be (by the same decision) less than 50°C, disagreeing with the mercury expansion temperature.

These disagreements are usually not large, and their details are mainly important only in accurate work, or to thermometer designers. The existence of such disagreements indicates that we do not have here a fundamental basis for setting up a temperature scale. So we say no more about such temperature scales, though much of the rest of the Unit (especially Parts Four and Five) will pursue the problem of finding a fundamental basis for the idea of temperature measurement. This inquiry will lead us to the absolute, or Kelvin scale of temperature.

The large scale and the small scale

In the rest of this Part, we shall ask and begin to answer such questions as What is going on when things reach thermal equilibrium?, How do atoms come to share out energy so that the temperature is uniform?, and What is the difference between hot and cold so far as atoms in a body are concerned?

We shall be thinking about the large-scale (macroscopic) events discussed above and trying to form a small-scale (atomic, microscopic) picture of what is going on.

Einstein's model solid

To think about the questions raised above, we need to fix attention on something definite. It is no use trying to think about gases, solids, and liquids all at once, for each has its own rules for the motion and energy of its atoms. We need to find one definite picture of how atoms share energy in one kind of material. And it had better be as simple a picture as possible, to start with.

The best theoretical physicists often succeed because they can invent a really simple model which nevertheless gives useful ideas about complicated reality. It was Einstein who proposed the simple model we shall talk about. It is a model of a solid.

Q11 In a solid crystal such as sodium chloride, the ions are held in place by balanced forces acting on each ion. If one ion is displaced and released, how will it move?

Q12 Would you expect the oscillations of one ion to affect others nearby?

Q13 If all the ions in a row were oscillating in unison except one, what would you expect to happen to that one?

Einstein suggested thinking of atoms or ions as being fixed in place in a solid and vibrating as harmonic oscillators (remember Unit 4). In general they do not but it is much simpler to discuss their energy if we imagine a model in which they do. This is because such an oscillating atom can only have one of a series of definite energies, and these energy levels are equally spaced, like the rungs of a ladder. In Unit 2 you saw that there was experimental evidence that the electrons within atoms could be in only one out of a series of energy levels at a time. Those levels were not equally spaced. We are now talking about the energy of the atom when the whole atom is oscillating back and forth inside a crystal, not about the energy of electrons within the atom. These oscillation levels are equally spaced, if the oscillator is harmonic. The next experiment offers some experimental evidence to support this assertion. A theoretical reason to support the assertion can be found in Unit 10, Waves, particles, and atoms.

Einstein then introduced a further simplification that was very helpful, even though it was obviously wrong. It takes real cleverness to dare to do that, and to choose the right way to make the model wrong. The trick is to leave out those complications which won't affect the answers to the questions one is interested in (without knowing what those answers are!). Einstein's model achieves this for the answers we shall obtain.

a b

Figure 24: Models of a solid

a A balls and springs model of a solid (later developed theoretically by Born and von Karman, in an improvement upon Einstein's arguments).

b Einstein's model, with nearly rigid prison cells around each oscillating atom.

Figure 24 illustrates this simplification. As the answer to question 13 suggests, the oscillations of one atom in a regular array, all linked by springy bonds as in figure 24 a, will affect the oscillations of others. In fact, figure 24 a is like a wave model of trolleys joined by springs (as used in Unit 4) but extended to two or to three dimensions. Indeed, those who developed Einstein's simpler arguments had to argue in terms of wave energy travelling all through the crystal. The trouble is that the energy levels of such linked oscillators affect each other so much that one cannot speak of the energy on one atom at all. Einstein's idea was to imagine the atom-oscillators not as if they were linked together but as if each were enclosed in a rigid prison cell, as in figure 24 b. But this was going too far, for then energy Would not be transferred from one atom to another at all, and there certainly would not be any heat flow to be discussed. So Einstein made the model a bit fuzzy at the edges, and allowed himself to imagine the walls of the prison cells as being just a little flexible, but not so much that the motion of one atom governed that of another nearby. Then energy could go from atom to atom, but the energy levels of each would be unaffected by the motions of any of the rest of the atoms.

Please don't think that Einstein proposed the model as a good way of imagining the arrangement of atoms in a solid. He proposed it as the simplest model he could think of which would still allow him to argue about heat flow and thermal equilibrium. As our present interest is in these matters, and not in the nature of solids, it is sensible to adopt the simplest possible model. But remember, although we shall talk of the energy of one atom, the model was invented to make that possible. For a real solid, it is usually not possible to think like that at all, and one must imagine compression waves travelling through all the solid and think of all the atoms sharing the energy cooperatively. (Real solids are like communist societies - all for one and one for all - while Einstein's model is like that simplified economic theory which supposes that there are many independent small manufacturers and shopkeepers, each acting independently of the rest, except for some flow of money between them. Their wealth can be counted up for each individual. In the ideal communist society the wealth belongs to everybody.) Some of the results of thinking about the model are true only for the model, but many are true of much more complex models, and just happen to emerge rather easily from the simple model. We shall try to make it clear which results are of each kind. Analogous arguments using models have appeared elsewhere in other Units.

See Appendix B for reasons for choosing the Einstein model, and for more about how the theory can be developed.

Demonstration 9.7 A ladder of equal energy levels

We said that an atom vibrating like a harmonic oscillator has an energy limited to one of a series of equally spaced values. The energy level spacing is actually hf, where h is Planck's constant and f is the frequency the oscillator has in the ordinary theory of oscillators (2πf= √(k/m)) where k is the spring constant and m the mass).

Imagine the levels to be like the rungs of a ladder, as in figure 25. There is a lowest rung, rung zero, and others above it. If the atom passes from rung one to rung two it has to acquire one lump of energy, magnitude ε. But on another occasion it may pass from rung five to rung four: it then gives up the same amount of energy ε.

An oscillating atom may absorb the energy needed to jump from one level to another, from some light. hf (again) gives the energy absorbed, the frequency f being that of the light. A spectrum of light, that has passed through some oscillating atoms that have absorbed energy from it shows dark lines at particular frequencies f, and these frequencies yield evidence of the energies hf absorbed from the light. (In Unit 4 we argued about this for the absorption of infra-red radiation by sodium chloride.)

5 4 3 2 1 0 lowest rung ε ε

Figure 25: A ladder of levels

A solid, even a transparent solid, is quite unsuitable for showing the equally spaced ladder of levels. As we explained above, real solids are not much like the Einstein model made of many independent oscillators. For a demonstration of the ladder, the best thing to use is a gas of two-atom molecules, like iodine, I2. Being in a gas, the molecules are nicely separated and don't affect each other's vibrations. Being made of two atoms, each molecule can vibrate in and out like two masses joined by a spring.

Figure 26: Part of the absorption spectrum of iodine vapour.

From Gaydon, A. G. (1968) Dissociation energies and spectra of diatomic molecules, Chapman & Hall.

Figure 27: Observing the iodine absorption spectrum.

Figure 26 is a photograph of the iodine vapour absorption spectrum. You can see the spectrum for yourself by looking at a filament lamp through a diffraction grating, and putting a tube full of iodine vapour in the way as in figure 27. The vapour is produced by warming the tube gently, there being one small iodine crystal in the bottom of the tube. The red, yellow, and part of the green regions of the spectrum are crossed by a ladder of many dark lines.

Figure 28 suggests how the ladder of lines arises.

low level vibrational levels

Figure 28

In the warm iodine vapour, nearly all molecules are in the lowest state of electron energy, and many have the least possible vibration energy. A molecule can absorb energy by jumping to a higher state of electron energy, but the excited molecule can also vibrate in one of a ladder of vibrational energy levels. So there is a whole ladder of absorption lines corresponding to an electron jump, combined with a jump to one or other of the levels in the ladder of vibration energies that go with an excited state of the electrons in the 12 molecule. Ladders of levels give ladders of absorption lines.

Demonstration 9.7 A ladder of equally spaced energy levels

      94A      lamp, holder, and stand 
       27      transformer 
    191/2      fine grating (about 300 lines per millimetre) 
     1055      test-tube, hard glass, 150 × 25 mm, with cork 
     1056      iodine crystals 
      508      Bunsen burner 

The test-tube with one or two small iodine crystals in it should be warmed over most of its length, using a low burner flame. The tube should be lightly corked. Warming of the lower part of the tube should be continued until the iodine vaporizes. The tube may now be supported in front of the line filament lamp and the spectrum of the emerging light examined by holding the grating close to the eye. It is necessary to warm the sides of the tube to prevent condensation of the iodine vapour there.

If the vapour is too dense, too little light comes through for much to be seen. If it is not dense enough, the contrast between the absorption bands and the continuous background is poor. It is best to warm the tube until a strongly coloured vapour fills it, and then to watch the lamp through grating and tube as the vapour cools. Make sure the tube is clean to start with, and remove iodine condensed on the sides by warming the sides before repeating an observation.

Four students can look at each lamp and tube, if they each have a grating. Four or more sets of apparatus may therefore be needed, with sixteen gratings.

Over-simplifications in the vibrational energy level story

There is not one single level from which the 12 molecules jump to higher levels, but the ground state is split into vibrational levels too. At moderate temperatures, the population of vibrational levels above the lowest is not very big, but careful inspection of the spectrum reveals at least two ladders of lines corresponding to jumps from the two lowest vibrational levels. See figure 29 a.

one ladder second ladder a b 0 1 2 3 not many molecules up here in a cool gas

Figure 29

In addition, the 12 molecules can also rotate. This splits each vibrational level into many finely spaced rotational levels (figure 29 b). As a result, each absorption line is really many closely spaced lines. Through a grating the lines appear broader than, for example, the absorption lines in sodium vapour, but the rotational lines are not resolved.

Zero-point energy

As stated above, the energy level spacing of a quantum harmonic oscillator with a classical frequency f is hf. The lowest level, however, has energy not equal to zero but to ½hf. The other levels have energy (1 +½)hf, (2+½)hf, and so on. Actually, the ladder in figure 25 has been drawn with its lowest rung half a rung spacing above the ground.

A game of chess with quanta for chessmen

The chess board (though it has only 36 squares) in figure 30 represents the array of atoms in a solid, each square, like each atom, having a fixed site within the whole.

Figure 30: A chess board of atomic sites in a solid. Photographs Michael Plomer

Each square represents a site where there is an atom. The energy of an atom at one site will be represented by piling counters on that site. No counters on a site means that the atom at that site is in its lowest energy level. One counter is added to represent lifting it to the first level, another to represent lifting it to the next, and so on. Figure 30 a shows the 36-atom solid with each atom in its lowest level.

Figure 30 b shows it with each atom in the next level up, with one counter on each site.

Figure 31: Transferring one quantum of energy. Photographs, Michael Plamer.

How shall we represent energy passing from atom to atom? The energy levels are equally spaced. Therefore, if a counter is picked up from an atom in the third level (with three counters, so that it then has two and has dropped one level), the same counter can be put down anywhere else and will represent the atom it now joins jumping up one level. Figures 31 a and b show one atom going from level 3 to level 2, and another going from the lowest level to the first level above that, all by moving one counter.

If a counter, representing one quantum of energy, were not put down once it had been picked up, the total energy would have decreased. It must go somewhere; to begin with we shall investigate what happens when the solid is well insulated so that no energy leaks in or out. So a counter picked up from one site must be put down on some site or other.

The fact that the counters representing quanta (which are real bits of plastic or metal) are distinct from one another, is not true of quanta. But the moves could just as well be made by keeping a paper and pencil tally of the state of each atom, remembering to move one atom down an energy level for every other atom one moved up one level. The plastic counters just save one having to remember or write down the state of each atom.

Game: 9.8 Moving quanta at random

How shall we choose which atoms gain energy and which lose energy? The clue to Understanding diffusion in Part Three was that there were no rules other than those of chance: if atoms move about quite by chance, then diffusion will inexorably happen if there are many atoms involved. A predictable, one-way diffusion process emerges from total atomic chaos.

We shall try a similar idea with quanta of energy moving about amongst fixed atoms. The questions we shall try to answer are, Does a tidy pattern of behaviour emerge when many quanta move at random among many atoms? and Is there anything like a one-way process going on?

Here are the rules of the game. Put down some counters anywhere you like. One counter on each site, as in figure 30 b, makes a good starting point. Throw a pair of six-sided dice to get two random numbers which pick out one site for you at random. If, for instance, the dice show 4 and 6, go to the site four across and six down. Pick up one counter from that site. Cast the dice again, and put the counter down on the site they now indicate. Then do the whole move, picking up and putting down a counter, again and again.

What should you do when you hit upon a site with no counters, wanting to pick up a counter from it? Since there is a lowest possible energy level, and the atom on that site is already in this level, it can go no further down. You will have to try again for another site, perhaps several times, until you find one with a counter to pick up.

The game does have definite rules. They are meant to achieve chaos. There are no rules saying where a quantum (counter) must go from each particular place: it may go anywhere. In that sense, there are no rules, or if you prefer, only that of chance.

Game: 9.8 Moving quanta at random

    1053 counter (tiddlywinks are suitable) 36 
    1055 dice 2 
    1054 graph paper 2 sheets 

Have a six-by-six grid ruled on one sheet of graph paper for each player, the squares being at least 30 mm on a side. It will be convenient to have the results of at least eight games available for averaging, so up to three hundred counters are needed.

The other sheet of graph paper is needed for recording a bar chart of the number of sites having no counters, one counter, two, and so on. It is convenient to plot such charts after twenty, sixty, and a hundred moves.

The rules of the game are given above, in the text. The game, and others like it, will shortly be shown on film, played by a computer using 900 sites. The purpose of playing the game with a 36-site board is that students should be clear about what the computer is doing, and see for themselves at first hand some of the expected and the unexpected results. If they use small numbers themselves, they will more easily appreciate the need for the large numbers used by the computer.

It is particularly important that playing the game should teach them what the bar chart of numbers of atoms having each energy represents. This is the beginning of an understanding of the idea of a distribution.

Null moves. A detailed analysis of the game shows that the null moves, when a site is picked on to have a quantum removed, but happens to have no quanta to donate, must be counted when the average behaviour of the system is calculated. See Appendix C.

More realistic neighbour to neighbour moves?

Students may want to exchange energy only between adjacent atoms, feeling that this is more realistic. This produces a problem at the edges of the board, and is better avoided. Say that one of the jumps of a quantum in the game might be the result of several steps in a real crystal, and that our aim is to use as simple a rule as possible. The diffusion game was not realistic either, but it predicted the average behaviour quite well even so.

Other random number generators

Dice serve well for a six-by-six board. A box containing balls numbered from one up to the total number of sites is also good, if the balls are jumbled between moves and are returned to the box after each draw. Numbered cards in a box are simpler, but experience suggests that they are not usually easy to shuffle well. Tables of random numbers are best avoided, for students are still learning what random means. It must be clear to them that nobody could have predicted beforehand how their own game would go.

How many squares?

The smaller the number of squares the quicker the game but the less reliable the results. 36 is the minimum. 64 is quite nice, but 100 is too many, for the game has then to go on for several hundred moves before it reaches statistical equilibrium. As the games will, in any case, be shown on a computer-made film with 900 sites it is probably best to keep to 36 sites now.

Atoms and oscillators

As indicated at the end of this Part, each atom can oscillate independently in each of three perpendicular directions. Thus, one site, corresponding to one oscillator, represents only one-third of an atom. Strict usage would require oscillator to be substituted for atom in all of this Part.

What happens in the game?

When you play the game you may notice a number of things. As you ought to have expected, the nice tidy one-counter-per-site pattern soon vanishes. And it never returns: the game leads, one-way only, away from such tidy patterns.

After about a hundred moves it is interesting to ask whether another hundred moves will make any more change. In one sense, obviously the answer would be yes. If you go on and on, some, at least, of the atoms which now have, say, two counters will have some other number of counters. The counters keep on moving. Yet, in another sense, not much more change happens, however long you go on after the first long run of moves.

Q14 Suppose you made the following records after 100 and 200 moves:

A the locations of sites with, say, two counters. 
B the number of sites with, say, two counters. 

Which record would be about the same each time, do you think?

The distribution of quanta among atoms

To start with, you may have had 36 sites each with one counter. No sites had no counters, and no sites had two, or any larger number of counters.

After a while, some sites have no counters, fewer than 36 have one, some have two, and so on. A good way to follow such changes is to plot bar charts with columns whose heights represent the numbers of atoms having no quanta, one quantum, two quanta, and so on. This is called the distribution of the atoms among the possible energy levels.

Number of sites 36 0 0 1 2 3 4 Number of quanta a Number of sites 20 10 0 0 1 2 3 4 5 b 0 1 2 3 4 0 1 2 3 4 5 0 1 2 3 4 Number of quanta Number of sites 20 10 0 0 1 2 3 4 0 1 2 3 4 5 0 1 2 3 0 1 2 3 4 5 Number of quanta Number of sites 20 10 0 0 1 2 3 4 0 1 2 3 4 0 1 2 3 0 1 2 3 Number of quanta

Figure 32: Twelve sample games (36 sites, 36 quanta, 100 moves) played by computer. a Initial distribution. b Distributions after 100 moves.

Figure 32 a shows what the bar chart looks like if you start with one counter on each atom - a single column 36 units high. Figure 32 b shows what happened to the distribution after 100 moves in each of twelve different trials of the game. Figure 33 shows the average of these twelve games.

Number of sites 200 100 0 0 1 2 3 4 5 Number of quanta ratio 1.72 ratio 1.36 ratio 2.24 ratio 3.4 ratio 2.0

Figure 33: Total of 12 computer samples (36 sites, 36 quanta, 100 moves in each sample).

Q15 Look at figure 33, or at the average of the bar charts obtained in your class. Suggest a rule about whether the numbers of atoms with n quanta will be more than, fewer than, or the same as with n + 1 quanta.

Q16 Is the difference between the number of atoms having four quanta and the number having three quanta the same as the difference between the number of atoms having one quantum and the number having none?

Q17 Speculate: what might be true of the ratios between the numbers of atoms in adjacent columns of the chart, but for chance variations because of the small total numbers involved?

Q18 Just before question 15, we introduced the term distribution. Figure 33 shows a distribution. To which aspect of figure 33 does this term refer?

  • a The number of atoms with no quanta.
  • b The collection of numbers of atoms with no quanta, one quantum, two quanta, and so on.
  • c The total number of atoms.
  • d The average number of quanta per atom.

Q19 Suppose one of next year's students were just starting to play the game.

  • a Could you predict in advance which atoms would give or receive quanta?
  • b Could you predict in advance anything about how the game would go?

The quantum shuffling game played with large numbers of atoms by a computer

Questions 15 to 19 may have brought out some of the results of the game of randomly moving quanta in an Einstein model of a solid. But because chance is at work and the numbers involved are small, the results vary quite a lot. So you ought to be doubtful about whether what you have seen is to be expected or is an accident. Ought the distribution to slope downwards towards higher energy states of atoms? Does it have a definite shape or not? If it does, what is the shape?

Such questions can only be answered by a trial with larger numbers of atoms and quanta (or by a theoretical discussion). Chance is like that: only trials with large numbers give reliable answers. A poll of ten people is a bad guide to how the country will vote in an election, but a poll of a thousand can be more reliable.

Sometimes it is possible to calculate in advance what to expect. This is possible for the number of times zero will turn up at roulette, but not for the number of votes in an election. It is possible to make such a calculation for the quantum shuffling game, but it is not easy. Part Five indicates how it may be done, and the calculation is given in Appendix B. For many problems, calculations are too difficult or impossible, and they are often tackled by making many random trials to see what happens. The problem of the most likely length occupied by a polymer molecule that can coil and uncoil at random, has been tackled in this way, for instance. This way of going about a problem is called, for obvious reasons, the Monte Carlo method. The advent of computers which can rapidly make many random trials according to certain rules has greatly increased the popularity of Monte Carlo methods.

We arranged for a computer to play the game you have just tried, but with 900 sites and 900 counters. The computer simultaneously made a film of what was happening in its head, so that you can see how the quanta move about and what happens to the bar chart showing the distribution.

computer films

The six short films on one reel are:

  • 1 Equilibrium of randomly shuffled quanta
  • 2 Different initial arrangement of quanta
  • 3 Fewer atoms and fewer quanta
  • 4 The biography of one atom
  • 5 A colder solid
  • 6 Heat goes from hot to cold

The series is entitled Change and chance: a model of thermal equilibrium in a solid.

The first film starts by showing the game played very slowly for two moves, so that students may see that the rules are exactly the same as those they have used. Then the 900 quanta are shuffled more and more rapidly for 10 000 moves, by which time the bar chart has changed to the expected exponential form. Then a further 10 000 moves follow, showing that statistical equilibrium has been reached, for the distribution (bar chart) merely fluctuates a little around a steady exponential shape. The average of the last 10 000 moves is then shown to be rather accurately exponential in form.

The second film shows that 900 quanta and 900 atoms reach the same final distribution even though they are differently arranged to begin with.

The third film shows that the final distribution has the same form if only 625 quanta and 625 atoms are used. The steepness of the distribution thus depends, in this model, on the number of quanta per atom.

The fourth film shows that the biography of one atom followed for 800 000 moves has much the same pattern as the distribution of numbers of atoms having various energies. Over a long period, one atom has zero quanta on as many occasions as it has one quantum, as, at an instant, there are atoms with no quanta rather than one quantum. This film is of importance mainly in Part Five. It can be passed over lightly now.

The fifth film shows the more steeply sloping distribution reached by 900 atoms sharing only 300 quanta.

The sixth film proves that 900 atoms sharing 900 quanta is hotter than 900 atoms sharing 300 quanta, for when these two are put together, energy flows spontaneously from the former to the latter, until they reach a common equilibrium.

Still pictures from the computer films of randomly shuffled quanta

In all, the computer made six short films, the series being called Change and chance: a model of thermal equilibrium in a solid. Figures 34 to 65 are still frames from the films, for you to inspect at leisure. (The frames are not prints from the film negative, but were made at the same time as the film, using the same data. The computer printed the pictures directly onto sensitized paper as well as onto film.)

Equilibrium of randomly shuffled quanta (figures 34 to 42)

In this film, the first, 900 atoms share 900 quanta, starting with one quantum on each atom. After making two moves in slow motion, to show that the rule used is the same as the rule we suggested that you should use in the game (9.8), moves are made at high speed until 10,000 moves have been made. The distribution has changed to a downward-sloping shape. Then a further 10,000 moves are made, during which the distribution fluctuates a little, but does not change its overall shape. Finally, the average distribution for the last 10,000 moves is calculated and displayed.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY

Figure 34: 900 atoms each with one quantum.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 1 AN ATOM IS CHOSEN AT RANDOM

Figure 35: One atom is chosen as the first step in moving a quanta.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 1 THIS ATOM LOSES A QUANTUM

Figure 36: The atom chosen in Figure 35 loses a quanta.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 1 ANOTHER ATOM IS CHOSEN AT RANDOM

Figure 37: An atom is randomly chosen to receive the quantum removed in figure 36.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 1 THIS ATOM GAINS A QUANTUM

Figure 38: The quantum removed from an atom in figure 36 is given to another atom. One move is completed.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 100

Figure 39: 100 moves like those in figures 36 to 38 have been completed. Much of the original pattern still persists but several atoms have no quanta or two quanta.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 5000

Figure 40: The system of 900 atoms and 900 quanta after 5000 moves.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 10000

Figure 41: The effect on figure 40 of 5000 more moves. The distribution has not changed essentially, though individual atoms have different number of quanta.

0 443 1 223 2 121 3 59 4 31 5 13 6 5 7 2 8 2 9 1 50 100 150 200 250 300 350 400 450 ENERGY

Figure 42: The average distribution over the last 10,000 moves out of 20,000. The numbers over each column are the numbers of atoms having each number of quanta. Each is nearly twice as big as the next.

Different initial arrangement of quanta (figures 43 to 47)

In the second film there are still 900 atoms and 900 quanta, as in the first film, but the quanta are differently arranged at the start, to see if the starting point has any influence on the distribution reached after many moves.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 0

Figure 43: There are 900 quanta and 900 atoms, but 300 atoms have no quanta each, 300 have one quantum each, and 300 have two each. The initial distribution records this new starting point.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 100

Figure 44: The system shown in figure 43 after 100 moves. The initial arrangement still largely persists, though some atoms have acquired three quanta each.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 5100

Figure 45: After 5100 moves the initial arrangement seen in figure 43 has quite vanished. The distribution is like that obtained in the first film, figure 40.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 10000

Figure 46: Approximately 5000 moves after figure 45 there is no essential change in the distribution, though individual atoms have different energies from those they had before.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY

Figure 47: The bar chart shows the average distribution over the last 10 000 moves out of 20 000. This distribution is very much the same as that obtained in the first film.

Fewer atoms and fewer quanta (figures 48 to 50)

The third film shows fewer atoms and fewer quanta, but as there are 625 of each, there is still one quantum per atom as in the first two films. The film shows that the final distribution is also the same in form as that obtained before. The shape of the distribution thus depends on the ratio of atoms to quanta, not on their absolute numbers.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 ENERGY 0

Figure 48: 625 atoms each have one quantum.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 ENERGY 10000

Figure 49: The distribution produced by random shuffling of quanta in the system shown in figure 48 is now changing no further.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 ENERGY 10000

Figure 50: The distribution shown is the average over the last 10 000 moves out of 20 000 for 625 atoms sharing 625 quanta. It has the same form as those shown in figures 47 and 42. Each column is twice as tall as the one next to it.

The biography of one atom (figures 51 to 56)

In the fourth film the fortunes of just one atom are followed, counting the number of times it has no quanta, the number of times it has one quantum, two quanta, and so on.

In the first three films, twice as many atoms had no quanta at any instant, as had one quantum, twice as many had one as had two, and so on. The idea is to see whether, as time goes by, the energy of one atom fits this pattern. That is, will it have no quanta twice as often as one, one quantum twice as often as two, and so on?

Figure 51: One atom is selected for attention.

TOTAL MOVES 0 MOVES WITH 0 0 0 1 0 2 0 3 0 4 0 5 0 6

Figure 52: The selected atom has three quanta. No moves have yet been followed.

TOTAL MOVES 1 MOVES WITH 0 0 0 1 0 2 1 3 0 4 0 5 0 6

Figure 53: On the first move followed, the selected atom happens to retain its three quanta (the move affects other atoms). One count appears in the register for moves with three quanta on the selected atom.

TOTAL MOVES 100000 MOVES WITH 0 0 2328 1 2615 2 2417 3 1289 4 1121 5 230 6

Figure 54: After 10000 moves, the atom has not yet had any moves with no quanta, though it has had one quantum as at the 10 000th move, on 2328 occasions. In the short run, chance can play many tricks. But as there are nearly 1000 other atoms, this one has probably only had about ten changes in its state so far.

TOTAL MOVES 100000 MOVES WITH 49437 0 29126 1 15189 2 3608 3 1289 4 1121 5 230 6

Figure 55: After 100 000 moves the registers show that the selected atom has had no quanta more often than one, one quantum more often than two, and so on. But the number in each register is only very roughly twice that in the next one down.

TOTAL MOVES 800000 MOVES WITH 331780 0 206516 1 143247 2 67666 3 28511 4 10358 5 4787 6

Figure 56: After 800,000 moves the ratios of numbers of moves with one, two, etc. quanta are rather nearer to two for more of the registers. But it would not be easy to be confident that this atom has no quanta twice as often as it has one, and so on for all numbers of quanta on the atom.

The film had to stop after 800,000 moves (figure 56), for reasons of cost. But the computer continued to follow the fate of the selected atom up to ten million moves. Table 4 shows the results. By ten million moves it is fairly clear that in the long run one atom is twice as often to be found with any particular number of quanta, as it is to be found with one more than that number.

total moves            1 000 000   5 000 000   10 000 000   
moves with 0 quanta      437 597   2 427 674    4 936 888   
moves with 1 quantum     260 630   1 241 675    2 501 041   
moves with 2 quanta      165 721     668 649    1 274 341   
moves with 3 quanta       80 911     327 425      611 711   
moves with 4 quanta       31 678     166 232      326 452   
moves with 5 quanta       11 541      80 714      164 727   
moves with 6 quanta        4 787      39 602       88 933   

Table 4: Continuation of from four to ten million moves.

A colder solid (figures 57 to 60)

In the fifth film, only 300 quanta are shared among 900 atoms. (Perhaps you would guess that this smaller energy would produce a model of a solid that is colder than the one shown before, which had more energy. Whether this is so or not must wait to be demonstrated until the sixth film.) The fifth film shows that the distribution is now steeper than before, with a larger ratio between numbers of atoms with a certain number of quanta and those with one more than that number of quanta.

However, the distribution has the same basic shape, in that this ratio is constant right across the distribution.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 0

Figure 57: 900 atoms share only 300 quanta.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 100

Figure 58: After 100 moves the original arrangement persists.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 ENERGY 10000

Figure 59: After 10 000 moves the distribution has reached a steady form.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 ENERGY

Figure 60: The average distribution, taken over the last 10 000 moves out of 20 000. It is steeper than that in figure 42, for 900 atoms and 900 quanta. That is, fewer atoms have large energies. There are four times as many with no quanta as with one, and so on.

Heat goes from hot to cold (figures 61 to 65)

In the sixth film, the model solid with 900 quanta and 900 atoms from the first film is allowed to exchange energy with the model from the fifth film, which has only 300 quanta among 900 atoms. One might expect the model with less energy to behave as if it were the colder of the two: this film shows that it does.

The two solids are first allowed to reach equilibrium within themselves, quanta being shuffled within each but not being allowed to pass from one to the other. It is as if there were an insulating wall of, say, expanded polystyrene between them.

Then the insulating wall is removed, and the computer is instructed to treat the two solids together as one large solid, shuffling quanta amongst all the sites in the combined system. Soon the quanta are evenly spread and the effect of this spreading on the two distributions of atoms among energy levels can be seen.

Quite by chance, energy goes from one to the other. The one which loses energy must previously have been the hotter of the two; the one which gains energy, the colder.

Figure 61: On the left, 900 quanta among 900 atoms. On the right, 300 quanta among 900 atoms. The two are not allowed to exchange energy, but quanta are being shuffled about within each one.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 ENERGY 0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 ENERGY

Figure 62: The two distributions for the two groups of atoms shown in figure 61. The lefthand distribution slopes less steeply than the righthand one.

Figure 63: The two groups of atoms just as they are put 'in contact', but before any quanta have been moved about at random following this change.

Figure 64: ... some time later! Quanta in the system shown in figure 63 have been allowed to move about at random, favouring moves from one group of atoms no more or less than moves from the other. But the quanta are now evenly spread between the two. After this, quanta still shift about, but there is no more overall change.

0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 ENERGY 0 1 2 3 4 5 6 7 8 9 50 100 150 200 250 300 350 400 450 500 550 600 650 700 ENERGY

Figure 65: The distributions of atoms among energy levels for the system shown in figure 64, after equilibrium has been reached. The two distributions are essentially similar, sloping downwards equally. They both slope more steeply than the lefthand distribution in figure 62, but less steeply than the righthand distribution in figure 62.

What can be learned about thermal equilibrium from the computer films?

Randomness

Q20 Was the computer instructed to try to achieve the steady distribution shown in figures 40, 41, 42, 47, and elsewhere? What was it told to do?

Q21 Would it have been possible in the first film, figures 34 to 42, to predict in advance the locations of atoms which would have no quanta at the 10 000th move?

Q22 In the sixth film, figures 61 to 65, was the computer instructed to move quanta from left to right rather than from right to left? What was it told to do?

Q23 In the 5000 moves from figure 40 to figure 41, is an atom with as many as seven quanta especially likely to lose one? Is an atom with no quanta especially likely to acquire one?

Q24 Look at figure 39.

  • a Is one particular atom more likely to be chosen in the hundred and first move than any other atom?
  • b Is it likely that the atom picked upon to lose a quantum will have just one quantum?
  • c Is it likely that the atom chosen to gain one quantum will have just one already?
  • d Is it likely that the hundred and first move will end up with one atom more with no quanta, one atom more with two quanta, and one atom fewer with one quantum?
  • e Do the impartial random moves produce a definite trend in overall behaviour at this stage?

Statistical equilibrium

Q25 Are there about as many atoms with one quantum in figure 40 as in figure 41? Compare figures 45 and 46 too.

Q26 Have the same atoms got one quantum in figures 40 and 41? Compare figures 45 and 46 too.

Q27 Say in what sense there has been a change from figure 40 to figure 41.

Q28 Say in what sense there has been no change from figure 40 to figure 41.

Q29 Say in what sense the outcome of the different changes to the different starting points shown in figures 43 and 34 is, in the end, essentially the same. (See figures 41 and 46.)

Q30 In the computer film, did the distribution change in any way between the moments shown in figures 40 and 41 ? Did its general shape change?

Q31 Hard. Suppose that another time, a set of moves led back from figure 40 to figure 39 (the computer could be run backwards, perhaps). If the moves after that were random, what would you expect to happen? Why do you think no such backward run occurred in any of the films?

The shape of the equilibrium distribution

Questions 25 to 31 were about the fact that one definite distribution is ultimately reached by the game working itself out, wherever it starts off from. The following questions are about the shape of this distribution.

Q32 Look at figure 42.

  • a What, approximately, is the ratio of the number of atoms with no quanta to the number with one quantum?
  • b What, approximately, is the ratio of the number of atoms with one quantum to the number with two?
  • c What, approximately, is the ratio of the number of atoms with two quanta to the number with three?
  • d Give a general rule for the shape of the distribution.
  • e In radioactive decay, if half the nuclei decay in time T, a further half will decay in a further time T. What is the.rule for the shape of the distribution of numbers decaying in successive equal time intervals? What is the name of the mathematical form of the process of radioactive decay?
  • f What is the mathematical form of the distribution of atoms among energy levels; thatis. the number of atoms with particular energies plotted against those energies?

Q33 Look at figure 60, the distribution for the colder model solid.

  • a Is the ratio of the number of atoms with a certain energy, to the number with one more energy quantum, a constant for this distribution?
  • b Is this ratio the same as for figure 42, the hotter model solid? What is the ratio?
  • c If there are N atoms sharing a total of q quanta, there are theoretical arguments which say that the ratios discussed in question 32 and in this question should be equal to (1+N/q). Test this prediction for figure 42 (N = 900, q = 900), for figure 60 (N = 900, q = 300), for figure 65 (N = 1800, q = 1200), and for figure 50 (N = 625, q = 625).

The theoretical reasons are outlined in Part Five, and given in more detail in Appendix B. Here you are merely making an experimental test of the answer they give. We shall use this answer at the end of Part Four to make a prediction about the heat capacity of solids.

Q34 In what sense is the distribution in figure 50 the same as that in figure 42? Why is it the same?

Heat going from hot to cold

Q35a Starting at figure 63, was the computer instructed to try to reach something like figure 64? What was it told to do?

Q35b In figure 63 the top lefthand atom has one quantum; in figure 64 it still has one quantum. In figure 63, the bottom lefthand atom has one quantum; in figure 64 it has no quanta. Could these results have been predicted in advance?

Q35c In figure 63 the atoms on the left share more quanta than those on the right. In figure 64, they share more nearly equal numbers. Could this change have been predicted in advance?

Q36a When the two halves of figure 63 are allowed to exchange quanta quite at random, there is a net flow of energy from left to right. Does such a flow occur from the part with a steeply sloping distribution to the part with a shallow or gently sloping distribution, or the other way round?

Q36b When a saucepan of soup is put on the plate of an electric cooker, does heat flow from high to low temperature, or the other way around?

Q36c After the stage shown in figure 64, quanta travel in equal numbers either way. Will the distributions shown in figure 65 change any more? How do they compare in terms of steepness?

Q36d If a casserole is put in an oven, when will its temperature stop rising and stay steady?

Q36e Which half of figure 63 is hotter? What about figure 64?

Q36f In questions 32 and 33 you thought about the ratio of the number of atoms with any particular energy to the number with one more quantum of energy. This ratio, being the same for all such pairs, describes the whole of a distribution by just one number. Steeply sloping distributions have a large ratio, shallow ones have a small ratio. Which go together, large ratio and high temperature (and small ratio and low temperature) or the other way around?

Part Four will conclude by showing how to fix upon a fundamental measurement of temperature, related to the distribution ratio.

The likely thing happens

Q37a Why, in a large school, is the number of people absent with colds almost constant from day to day (except during epidemics)?

Q37b Did the computer know that the equilibrium distributions should be exponential?

Q37c Is it fair to say that in each film, anything could happen, so what did happen was what was likely to happen?

Summary: chance and the flow of heat

If you have watched the computer films and thought about the questions related to them, you now have some reason to believe the following things.

Two lumps of solid of the same material (not in contact) which have the same proportion of internal energy to number of atoms will sooner or later, probably already, have the same distribution of atoms amongst possible energy levels, and the same temperature. (They had better not be compressed by different amounts - that would alter the energy levels.) It doesn't matter how they start off; they end up the same.

This will have come about by chance. In either lump, different atoms will be at different energies from time to time, but the relative numbers of atoms with different energies will stay remarkably steady, just because they are not made to do so, but are allowed to share energy at random.

Equilibrium is not static, in one sense, because the atoms continually exchange energy. However, it is steady, in that the equilibrium distribution, once reached, is permanent until outside conditions alter. But fluctuations around the average steady equilibrium distribution do occur, as chance tries out the possible variations on the arrangement of quanta among atoms.

The approach to equilibrium is definitely a one-way-only process. It comes about inexorably, but quite by chance, because what happens is what is likely to happen. If there are many atoms and quanta, what is likely to happen is so very, very likely that it always does happen in the end.

For the simple Einstein model solid, the equilibrium distribution is exponential in form. The distribution has a constant ratio of the number of atoms with one particular energy to the number of atoms with one quantum more energy. So atoms with no quanta are more common than atoms with many quanta. The distribution ratio is related to the number of atoms sharing each quantum. N atoms sharing q quanta have a distribution ratio of (1+N/q). (See question 33, or Part Five, or Appendix B.)

The distribution ratio is related to the temperature. Small ratios go with high temperatures; large ratios with low temperatures. The ratio and the temperature both become equal when two objects come together in thermal equilibrium. Temperature and distribution ratio both describe the object as a whole, summarizing its average behaviour. Each is a quantity that decides how likely it is that heat will flow one way or the other.

Heat goes from hot to cold in one way only, inexorably, but not because it is pushed. Heat flow occurs by chance, and happens because it is likely to happen. It happens inexorably because, with many atoms and quanta, it is very, very likely to happen.

The absolute Kelvin temperature scale

Note to teachers

It is suggested in the Introduction that some students might stop at the end of Part Four, though others may go on a little further. This brief piece about Kelvin temperature and the piece about the heat capacity of a solid which follows it are meant to round off Part Four for those who stop there. Part Five develops the same ideas in more detail, with less assertion. Those who are going on could pass straight from here to Part Five.

We decided that a high temperature goes with a low value of the ratio that describes the equilibrium distribution of the Einstein model solid. Now temperature ought to have a numerical value; we expect to be able to say that the temperature of something is, say, 300 K. How might distribution ratios be linked up with temperatures measured in kelvins, so as to go further than to be able vaguely to say that the temperature is high or low? This is what we now consider.

The temperature is related to the steepness of the distribution. Fixing the relation between them is mainly a matter of deciding how to relate them, not of discovering anything new. For we already know what hot means for the model: a small slope or ratio. It is just a matter of giving the right kind of number to the slope.

ln (number of atoms 14 13 12 11 10 9 8 7 6 0 1 2 3 4 5 6 7 8 Quanta of energy on an atom energy difference ε constant difference ln 2 slope (ln 2) / ε

Figure 66: Logarithmic plot of the distribution of atoms among energy levels for an Einstein solid with 106 atoms sharing 106 quanta, each of size ε.

Figure 66 illustrates a slightly different way of showing the slope of a distribution. It is drawn for an Einstein model with N = 106 atoms and q = 106 quanta, so that the ratio (1 + N/q) is equal to 2. (See question 33)

Unlike earlier distribution diagrams, the quantity plotted against the energy of an atom is the logarithm (base e) of the number of atoms with that energy, not simply the number itself.

A detailed calculation shows that we should expect 5 × 105 atoms with no quanta. ln(5 × 105) = 13.122. The number with one quantum will be half as many, 2.5 × 105. But ln(2.5 × 105) must simply be less than ln(55 × 105) by the amount ln 2. So the logarithmic distribution plot drops like a straight staircase, each step being equal to ln 2.

If we write ε for the energy of one quantum, the energy difference between levels, the slope of the logarithmic plot is simply (ln 2)/ε. In general, for N atoms and q quanta, the slope will be [ln(1 + N/q)]/ε.

Now the temperature T is high when the slope is small. So a reasonable decision would be to have T inversely proportional to the slope. With a constant k to turn the proportionality into an equality, we have:

kT = ε/ln(1 + N/q).

This is just a decision. We could have had T directly proportional to the slope, but then hot things would have low numerical values of 'temperature, which happens not to be normal practice. We could have had T inversely proportional to the square of the slope. That would merely be unnecessarily complicated.

Does this temperature T behave sensibly? If the solid acquires very many quanta, q becomes big and (1 + N/q) comes nearer and nearer to 1. So ln (1 + N/q) becomes nearly zero, and T goes off towards infinity. More and more energy makes the solid hotter and hotter.

Previously, we chose to make kT = ε/ln(1 + N/q). If we had used (1 + N/q) in place of its logarithm, which might seem simpler, the temperature would have had an upper limit. The sum of (1 + N/q) can never be less than one, so kT would have a largest value equal to ε. There would be an absolute hottest on such a temperature scale. That is why we did use the logarithm.

What about less energy? The solid cannot have fewer than no quanta, q = 0. Then both (1 + N/q) and ln(1 + N/q) become as big as one pleases, and the temperature T becomes zero. So this temperature T has the value zero at the point where the Einstein solid is as cold as it could ever be.

So T behaves sensibly, and has its zero at the absolute zero, that is, at the coldest imaginable condition.

A temperature T whose meaning is restricted to one very special model may not seem to be much use. But it happens that the definition can be generalized so as to cover any model involving particles distributed over energy levels, even if the levels are not equally spaced. So you may regard the explanation of how T can be interpreted in terms of a model of atoms and quanta as a simple example of a more general, but more complicated, interpretation. The essence of the idea is correct, but the details need attention in a fuller treatment.

Note to teachers kT= ε/ln(1 + N/q)

Appendix B, The Einstein model and the Kelvin temperature, discusses more fully the introduction of a temperature based on statistical ideas.

The equation given serves only for an Einstein solid. But it is a special case of an equation of quite general application. As shown in Part Five, ln(1 + N/q) is the change in the logarithm of the number of ways of arranging q quanta among N oscillators, when one more quantum of energy ε is added to the system.

Thus the equation can be written kT = ε/ Δ ln W, where W is the number of arrangements of quanta, and Δ ln W is the change corresponding to the addition of energy ε. This result is general, and may be applied to any material whatever. Usually it is hard to get a value for W or Δ ln W; the Einstein solid is of interest because calculating the value of Δ ln W is not too difficult.

The reason for introducing the energy ε into the equation is passed over in the treatment in Part Four, but a reason is given in Part Five and in Appendix B.

same exponential curve (in equilibrium) a Level spacing ε ε b Level spacing 2ε c Arbitrary levels

Figure 67: Equilibrium of materials with different level spacings.

The reason is illustrated in figures 67 and 68. Figure 67a shows the distribution for an Einstein solid of level spacing ε. Figure 67b shows such a solid with level spacing 2ε. Figure 67c shows the distribution for a material with arbitrary level spacings. All have been drawn in the form which deeper theory predicts they will have if they are mutually in equilibrium. The quantity which is the same for all, and which so relates to their mutual temperature, is not the ratio between numbers of atoms in adjacent levels. The numbers of atoms in levels fit onto the same exponential curve if it is plotted on the same energy scale.

Figure 68 shows the corresponding logarithmic plots. The quantity that is the same for all three is the slope, which is the difference in the logarithms of numbers of atoms in adjacent levels divided by the level difference.

ln n same straight line, slope = (Δ ln n)/ε a Level spacing ε ε ln n b Level spacing 2ε ln n c Arbitrary levels

Figure 68: Logarithmic plots of the distributions shown in figure 67.

The heat capacity of an Einstein solid

In the equation kT = ε/ln(1 + N/q), the quantity k was put in as a constant of proportionality. We shall now think about its meaning, and estimate its value by experiment.

We shall need one mathematical trick. Notice that for hot enough objects, the number of quanta is big compared to the number of atoms, so N/q is small. Now ln(1 + x) is nearly equal to x itself when x is small. Table 5 gives some values:

x     1+x   ln (1+x)   
1.0   2.0   0.693   
0.5   1.5   0.405   
0.25  1.25  0.223   
0.1   1.1   0.095   
0.01  1.01  0.00995 

Table 5

If the Einstein solid is hot enough, we can replace ln(1 + N/q) by N/q. Then the expression for T is very simple:

kT ≈ ε / (N/q) = qε/N

But qε is the number of quanta multiplied by the energy of a quantum, and is thus the total energy shared by the atoms, which we shall call U.

Now we are talking about real solids, and we have to be a bit careful about N. It is not the number of atoms, but the number of single separate oscillators in the model. Each atom in its prison cell (figure 24 b) can oscillate in three directions at right angles, as the real solid is three-dimensional. So each atom is like three independent oscillators. If there is one mole of material, there are L = 6 × 1023 atoms and 3L oscillators. So to talk about atoms in solids, N must be replaced by three times the number of atoms, in a solid element made of single atoms.

For one mole of atoms, putting N = 3L (where L is the Avogadro constant) and the total energy U = qε, we have

kT ≈ U/3L

or

3kT ≈ U/L

Notice that U/L is the mean energy per atom. It is roughly equal to 3kT.

Constant thermal capacity for many elements

The above result has a startling consequence. Suppose one mole of atoms of a solid element is warmed up, rising in temperature by ΔT, its energy U rising by ΔU, then

3k ΔT ≈ ΔU/L

or

3kL ≈ ΔU/ΔT

But ΔU/ΔT is the heat capacity of one mole of the element; the energy delivered to raise the temperature one degree (at fixed volume). The argument above says that the molar heat capacities of all solid, monatomic elements at high enough temperatures will all be the same, always supposing that the Einstein model is good enough to describe them. For the quantity 3kL is the same for all elements, L being the Avogadro constant and k the constant of proportionality we put into the equation defining temperature.

Table 6 gives some heat capacities for one mole of some elements.

Iodine is included to remind you that molecules arranged in a solid will take up more energy, because (among other things) the molecule can vibrate within itself as well as vibrating as a whole. Of the monatomic solid elements, a great number have very similar molar heat capacities, running at about 25 J mol-1 K-1, which is striking confirmation of the general correctness of the prediction.

Element   Cp/J mol-1 K-1   Element   Cp/J mol-1 K-1
carbon:                    gold        25.2   
1 diamond    6.06   
2 graphite   8.65          nickel      26.0   
boron       12.0           cobalt      25.6   
beryllium   17.8           manganese   26.3   
silicon     19.8           chromium    23.3   
aluminium   24.8           molybdenum  23.5   
copper      24:5           tungsten    25.0   
iron        25.2           vanadium    24.5   
lead        26.8           titanium    25.1   
zinc        25.0           magnesium   23.9   
tin (grey)  25.8           calcium     26.3   
bismuth     25.5           strontium   25.1   
antimony    25.4           barium      26.4   
arsenic     25.0           lithium     23.6   
selenium    24.9           sodium      28.4   
germanium   26.1           potassium   29.2   
cadmium     25.9           caesium     31.0   
silver      25.5           iodine (12) 55.0   

(Cp is the heat capacity per mole at constant pressure. The value at constant volume, which should have been used, is too little different from Cp to matter. All values are given for temperature 298 K and pressure one atmosphere.)

Table 6 Molar heat capacities.

Dulong, Petit, Einstein, and heat capacities

This rule of constant heat capacities is not a new one. Dulong and Petit discovered it as a rough rule by studying heat capacity data a very long time ago. It was used by chemists to help settle atomic masses.

It is not a completely general rule. At low temperatures, q is not large and N/q is not small, so that the approximation to the logarithm of 1 + N/q fails. The effect is that the heat capacity drops to lower values. Just how hot an element must be for q to be much larger than N depends on the energy level spacing ε. (A few large-sized quanta represent as much energy as many smaller ones.) In table 6, the elements like carbon, for which the heat capacity is less than about 25 J mol-1 K-1, are those for which the quanta are large, and relatively few in number. They do not really spoil the story, for it is found that at higher temperatures, their heat capacities do rise to the value given by Dulong's and Petit's rule.

Einstein actually invented his model to try to explain how and why the heat capacities do fall off at low temperatures. To do so is not very hard: one simply does the (harder) calculation of ΔU/ ΔT without making the approximation ln(1 + N/q) ≈ N/q. The agreement with experimental data is then good, though not perfect, over a wide range of temperatures.

Note to teachers

It would be unfair to claim the prediction of Dulonq's and Petit's rule as a great success for the Einstein model, pleasing as it is. For it is very insensitive to the assumptions that go into the model used to predict it. It is given by a classical argument which assigns average equal energy ½kT to each degree of freedom. Each oscillator has two degrees of freedom - kinetic and potential energy terms - and each atom is equivalent to three oscillators.

The Einstein model also gives the rule correctly. Its success was rather in predicting, for the first time, the form of departures from the rule. Other models, in which the energy is carried about by elastic waves, also yield the rule, and are better than Einstein's at getting the form of departures from it correct. Current research aims to improve the agreement between theory and experiment by making more detailed and realistic calculations based on the motions of interacting sets of atoms, and by allowing, for example, for the effects of non-linearity in the displacement-force law, and so on.

Value of the Boltzmann constant k and mean energy per atom

It seems reasonable that the roughly constant value 25 J mol-1 K-1 is a fair estimate of the quantity 3kL predicted for the molar heat capacity of one-atom solid elements by the Einstein model.

3kL = 25 J mol-1 K-1

but L = 6 × 1023 mol-1

so k=1.39 × 1023 JK-1

The value given in data books is

k = 1.380 × 10-23 JK-1

k is called the Boltzmann constant. We have presented it as a scale constant. introduced so that temperature T is measured on a particular scale.

In measuring the heat capacities, the temperatures were measured on the usual scale with about 273 one-degree intervals between absolute zero and the temperature of melting ice. The measurement of k must be in joules per kelvin units if T is in kelvins, so that kT is in joules, although k could be chosen to be any size whatever. But the value of k settles the scale of T. If k has the value above, T is in kelvins, and the scale of T is called the absolute Kelvin scale.

Since U/L ≈ 3kT (at high enough temperatures) and U/L is simply the mean energy per atom, we can say that the temperature T is roughly proportional to the mean energy per atom. This is not an exact result, but it is a useful one.

Also, kT has the order of magnitude of the mean energy per atom. At room temperature, 300 K, the value is of the order of 4 × 10-21 J. It reaches one electronvolt, 1.6 × 10-19 J, at about 10 000 K.

Physicists like to remember that kT is of the order of the mean energy per atom. They like it so much that they sometimes forget that it is only an approximate result, and that the temperature is not exactly proportional to the mean energy per atom.

You should also be able to remember a similar result in the kinetic theory of gases. There, too, the mean energy per molecule is proportional to the temperature, and is roughly equal to kT.

Note to teachers

Mean energy per atom

Appendix B, The Einstein model and the Kelvin temperature shows that of two materials, it is quite possible for the one with the lower mean energy per atom to be the hotter. However, for one material, it will be true that if the temperature rises, so does the mean energy per atom.

Kinetic theory of gases

See the introduction to this Unit, and the Teachers' handbook. The kinetic theory ought to be shown to all students at some stage. As it is fully discussed in the Nuffield O-level programme, we arrive at similar ideas at Advanced level by a different route, using solids.

But if students took a different O-level course, and missed the theory, it must be fitted in now, and an appropriate sacrifice of other material must be made. The rest of this Unit could be one such sacrifice.

See, for textbook material for students:

Rogers, Physics for the inquiring mind, Chapters 25 and 30.

PSSC. College physics, Chapter 20; Physics, Chapter 9.

Nuffield Physics Special, Molecules and motion.

Experiment: 9.9 Measuring the Boltzmann constant k

This hardly counts as an experiment: you may have done it before, you know what will happen, and you know the answer. But it may help to remind you how the ideas fit together.

You just have to measure how much energy ΔU is needed to warm an aluminium block by a measured temperature rise ΔT. Then k is given by

k ΔT ≈ U/3 × (number of atoms).

Find the mass of the block and calculate how many atoms of aluminium it contains. The energy can be delivered and measured electrically.

Calculate ΔU divided by three times the number of atoms; the rise in mean energy per oscillator in the Einstein model. As this rises, more oscillators fill high energy levels and fewer occupy low levels. The distribution becomes less steep, which is to say that the aluminium is hotter. Indeed, a little energy goes at once to the thermometer, just as energy flowed from the hotter, less steeply sloped distribution to the cooler, more steeply sloped distribution in the sixth part (Heat goes from hot to cold) of the Change and chance film. The thermometer rises because the aluminium's distribution has shifted.

The thermometer is marked in Kelvin-sized degrees (though the markings are probably Celsius degrees.) These degrees have an arbitrarily decided size: 273 of them from absolute zero to the ice point, and 100 of them from ice to boiling water.

The constant k links arbitrarily-sized Kelvin degrees with shifts in distributions of oscillating atoms in energy levels, which is the basis of our fundamental idea of temperature. We understand by k the mean rise in energy per oscillator divided by the change in Kelvin temperature (as long as the temperature is high enough).

Experiment: 9.9 Measuring the Boltzmann constant k

     77   aluminium block   
     75   immersion heater   
    542   thermometer ( -10 to 110°C)   
    507   stop watch or stop clock   
 1003/5    ammeter (10 A)   
 1004/3    voltmeter (100 V)   
     or   
   1005   multi-range meter (25 V)   
     42   lever arm balance   
     59   l.t. variable voltage supply (d.c.)   
    and   
   1064   low voltage smoothing unit   
    or   
    176   12 volt battery   
   1000   leads   

The aluminium block has a large hole drilled in it to take the heater, and a smaller hole tor the thermometer. A drop of oil in the thermometer hole helps thermal transfer. The ammeter and voltmeter are used to record the power delivered to the heater. The temperature rise is noted at its maximum, after a few minutes heating. In essence, the experiment is an electrical measurement of the heat capacity of the aluminium block.

The Boltzmann factor

We have introduced the temperature as a measure of how gently an exponential distribution slopes. Thus, at high temperatures, there will be proportionately more atoms in high energy levels, having many quanta, than at low temperatures. This fact can now be expressed quantitatively.

The factor (1 + N/q) is the constant ratio between numbers of atoms in adjacent energy levels. If nlow is the number in the lower level, nhigh is the number in the higher level, and the levels differ by energy ε,

nlow/nhigh = (1 + N/q)

But from the previous way in which temperature was introduced,

ln(1 +N/q) = ε/kT.

Thus ln(nlow/nhigh) = ε/kT.

so that (nlow/nhigh) = eε/kT

When T is small, ε/kT is large, and eε/kT is yet larger, so that there are many more atoms in the low level compared to those in the higher level.

The ratios of numbers of atoms in any pair of levels is also obtainable, since the ratio is the same for each adjacent pair of levels. The factor eε/kT just becomes eε/kT × eε/kT × eε/kT.... = eε+ε+ε.../KT, each level having eε/kT times more atoms than the next. Thus, for any pair of levels different by energy E,

nlow/nhigh = eE/kT

or, since it is often more interesting to know how many high energy atoms there are,

nhigh/nlow = 1/eE/kT

which is

nhigh/nlow = e-E/kT

The factor e-E/kT is called the Boltzmann factor. Its importance is that its use is not restricted to simplified situations like Einstein solids, but is valid for most sorts of materials. In Part Six, it will be used to study processes that depend on those atoms or molecules which have particularly high energies, such as those which evaporate from wet clothes on a washing line, or those which are able to take part in chemical reactions.

Summary of Part Four

In the Summary: chance and the flow of heat, we set out the ideas to be learned about thermal equilibrium from the study of the model of a solid, using the random games and computer-made films. You should look at this again.

Since then we have introduced temperature into the argument. Before, we saw that heat goes from hot to cold because that is the likely thing to happen. To say that one thing is hotter than another is just to say that energy is likely to flow from the hotter one to the colder one. It is so very likely that it happens as inexorably as gases diffuse and mix (or as tidy desks become untidy).

Temperature is a number related to how likely it is that energy will flow from one thing to another. Here this is partly concealed, but Part Five brings it out into the open. If you stop at Part Four, just remember that the temperature describes the distribution of atoms among energy levels. If there are a high proportion of atoms with large energy, the temperature is high. And an object with such a gradually sloping distribution readily loses energy to another object with a steeper distribution slope, which has fewer atoms in high energy levels.

Sooner or later, objects in contact come to equilibrium. When they have, their temperatures are equal. They are equally likely to pass energy one way as the other. Their distributions have equalized, so that the relative numbers of atoms in levels different by the same energy are identical.

A high temperature does not push heat from hot to cold. The flow happens by the chance, random, sharing out of energy among atoms. The temperature is an indicator of how likely it is that a body will, on balance, lose or acquire energy in this random shuffling process.

A place to stop

You could stop here, if you have had enough. Part Five goes on to develop the idea of temperature and the chance of heat flow more quantitatively. Part Six contains a number of important applications of the ideas, to engines, to chemistry, to pressures of vapours and electric currents in some materials, and to other matters. It reaches the great Second Law of Thermodynamics, and the idea of entropy.

But what you have done may seem enough. Even the Second Law is there, for it may be expressed in the form, heat flow, left to itself, always goes from hot to cold. You already know why this is so, and that is no mean thing.