Thermal equilibrium

The quantity with which thermodynamics concerns itself especially and uniquelyis temperature. This quantity has meaning only for large systems, whereby “large” we mean “having a large number of degrees of freedom.” Thenumber of degrees of freedom in a system is directly related to the num-ber of fundamental constituents in the system—in general, the number ofmolecules—that make it up. We will usually be talking about systems with∼ 1023 molecules in them, which makes them truly large.

Temperature is an “average” property of a system. In a finite systemwe can reasonably expect there to be fluctuations in the temperature, justas there were fluctuations in the number of molecules in the top or bottomhalf of the box that we looked at in the previous lecture. These fluctuationswill be small for a large system, just as the fluctuations in the box were, ina percentage sense. The temperature will have fluctuations that are, in apercentage sense, on the order of 1/

√N , where N is the number of degrees

of freedom in our system. That is, if we say that the temperature of a systemwith N degrees of freedom is T0, what that means is that at any given pointin time the temperature will be T0[1± A/

√N ], where A is a fixed number.

We have an intuitive notion of what temperature is. It measures how hotsomething is: the higher the temperature, the hotter the object. Temperaturethus has a relative meaning. It also has an absolute meaning, in that we canquantify the temperature of an object. That is, we can assign a number tothe temperature, and we can use that number to characterize not only howhot that object is, but also how hot any object is that happens to be at thattemperature. Note that such a quantification is possible only if equality oftemperature has some of the properties of an equivalence relation. In fact,it has all the properties of an equivalence relation. To see that this is so,let’s devise, for our temporary use, a symbol to denote temperature equality.

Here it is:T=. So—the mathematical statement A

T= B means “system A is

at the same temperature as system B.”Temperature equality has the following three properties:

1. Reflexivity: AT= A. A system is at the same temperature as itself.

2. Symmetry: If AT= B then B

T= A. If system A is at the same tempera-

ture as system B, then system B is at the same temperature as systemA.

12

3. Transitivity: If AT= B and B

T= C, then A

T= C. If systems A and

B are at the same temperature, and systems B and C are also at thesame temperature, then systems A and C are at the same temperature.

Now, the notion of temperature is one that applies in equilibrium only.In particular, it applies when there is thermal equilibrium. An iron rodthat is red-hot at one end and ice-cold at the other is not in thermal equi-librium, and cannot be said to be at a single temperature. If left alone, andsufficiently isolated from its environment, the rod will eventually settle intoa state in which it is at a uniform temperature throughout. We think ofthermal equilibrium as a state that is spontaneously achieved by a systemleft to itself. It is also achieved by a pair of system left alone but allowed tointeract appropriately. The appropriate form of interaction occurs when thesystems are in thermal contact. Basically, this means that they are touching,or are close enough to each other that they can exchange heat.

The notion of heat plays a fundamental role in thermodynamics. In theearly development of this field, heat was thought of as some kind of a fluid(called “caloric”) which contained “hotness.” Nowadays we recognize heat asa kind of energy. What takes place between two systems in thermal contactis energy exchange. Energy flows from the hotter system—the system athigher temperature—to the colder system. This flow continues until theyare at the same temperature.

The properties of temperature and the notion of thermal equilibrium as astate in which everything is at the same temperature and as a state to whichsystems naturally tend constitutes the zeroth law of thermodynamics.

The first law of thermodynamics can be understood in two ways. First,we can think of it as the law of conservation of energy. Second, we can thinkof this law as a definition of heat. It applies to a system that has undergonesome change in which (a) some heat has flowed into or out of the system, (b)some work was performed by or on the system and (c) the system’s internalenergy has changed. If we let d-Q be the amount of heat that has flowed intothe system, d-W equal the amount of work performed by the system and dUequal the change in the internal energy of the system, the first law presentsus with the equality

dU = d-Q− d-W (41)

Note the lines throught the d’s in d-Q and d-W . This is to distinguish be-tween these two quantities and dU . Let’s take a little time to discuss thisdistinction.

13

State variables

A state variable is a quantity that characterizes some aspect of a systemin thermal equilibrium. Temperature is just such a quantity. Another oneis volume; pressure is a third. A system at thermal equilibrium is at somespecific temperature, occupies a certain volume and is subject to a well-defined pressure. These state variables take on values that are determined bythe current state of a system in thermal equilibrium. They are indepenedentof the history of the system.

The internal energy of a system is a state variable. Although energy isuniquely defined only to within an additive constant, once we have estab-lished a zero for energy a system can be said o have a perfectly well-definedenergy in thermal equilibrium. The change in energy in the first law aswritten above appeard in the form dU . This is to signify that what we arerepresenting is the difference between two values of a variable that takes on adefinite value in the equilibrium state of a system. By contrast, the two otherterms, d-Q and d-W are not representable as differences between two statevariables. A state in thermal equilibrium does not have a certain amount of“heat” or a certain amount of “work” in it. We can imagine performing thefollowing procedure on a container of liquid:

Rotating this shaft stirs theliquid, which heats up

Heat is drawn off through the walls

Figure 3: The Joule paddle experiment.

If the amount of heat energy that is drawn out of the liquid equals the amountof work performed in stirring the liquid, then we can continue allowing theprocess pictured above to operate for an arbitrarily long time, performing anarbitrarily large amount of work on the liquid and siphoning off an arbitrarilylarge amount of heat—in the end, not having changed the liquid’s internal

14

energy, or, indeed, any other aspect of the system. Thus, neither work norheat satisfies the criteria for a state variable. There is a state variable relatedto heat. We will need to go into the second law of thermodynamics beforewe can develop it. First, let’s discuss a simple thermodynamic system.

The ideal gas

Ideal gases satisfy two laws:

PV = NkBT (42)

U = Nf(T ) (43)

Here,

P is the pressure under which the gas is kept.

V is the volume it occupies.

N is the number of molecules in the gas.

T is the temperature of the gas.

U is the internal energy of the gas, and

kB is a constant (actually a conversion factor) called the Boltzmannconstant.

In the case of the monatomic ideal gas,

f(T ) =3

2kBT (44)

The ideal gas law (42) implies a temperature scale. We will adopt this scalewithout further discussion, at least for the time being. As we will see whenwe get into a deeper discussion of the second law there is a natural scale fortemperatures. This scale turns out to be the same as the scale implied by(42) and (44).

Let us see how the first law operates when applied to a simple processinvolving a cylinder of an ideal gas. Suppose the gas were to push the pistonup by a short distance ∆x. The amount of work, d-W , performed by the gas

15

Figure 4: A cylinder of gas that is compressed or allowed to expand, throughthe action of a movable piston.

equals the force it exerts on the piston times the displacement of the piston,∆x. Thus

d-W = P × A×∆x

= P × [A∆x] (45)

Now, A∆x is the amount of extra volume occupied by the gas now that thepiston has been moved outwards. Thus, we also have in this case d-W = PdV .As it turn out, this formula holds when the volume of the gas changes in morecomplicated ways. Thus, we may write for the first law as applied to a simpleprocess involving a volume change of the gas

dU = d-Q− PdV (46)

Note that in order for this expression to apply, the force exerted by the gason the piston must be the force the gas applies to its container in thermalequilibrium. In other words, the expansion has to take place reasonablyslowly. If the expansion is too rapid, there are complications.

Let’s now consider two ways in which the gas can expand from an initialvolume, Va, to a final volume, Vb. The two kinds of expansion that we willlook at are isothermal and adiabatic.

16

Isothermal expansion

If the temperature of the gas in the container is held constant and the gas isallowed to expand slowly, we have

d-W = PdV =nkBT

VdV (47)

so the total amount of work performed in the expansion is∫ Vb

Va

PdV =∫ Vb

Va

nkBT

VdV

= NkBT∫ Vb

Va

dV

V

= NkBT ln(

Vb

Va

)(48)

Because the temperature of the gas does not change, the internal energy ofthe gas remains constant, which means that, by the first law, there is a flowof heat into the cylinder that compensates for the energy loss due to workperformed by the gas on the piston. In other words

dU = d(

3

2NKBT

)= 0

= d-Q− PdV (49)

So d-Q = PdV = nkBTdV/V , and the total amount of heat that flows intothe system is equal to∫ Vb

Va

NkBT

VdV = NkBT ln

(Vb

Va

)(50)

Conversely, we can imagine an isothermal compression of the gas from avolume Vb to a volume Va in which the work done on the piston by the gas is

NkBT ln(

Va

Vb

)= −NkBT ln

(Vb

Va

)(51)

and in which the net heat flow out of the gas and into the environment isNkBT ln(Vb/Va).

A slow istothermal expansion from Va to Vb followed by a slow isother-mal compression back to Va leaves everything—the gas, its container andthe surrounding environments—exactly as it started. In other words, a slowisothermal expansion, or a slow isothermal compression, can be entirely un-done. Both are reversible processes.

17

Adiabatic expansion

An adiabatic process involves no heat transfer between the gas and its en-vironment. That is, d-Q will equal zero in this process. Once again, we willconsider only slow processes. Imagine, now, a slow expansion by a thermallyisolated cylinder of ideal gas, from a volume Va to a volume Vb. From thefirst law, we have

dU = −PdV (d-Q = 0) (52)

We also have

dU = d[3

2NkBT

]=

3

2NkBdT (53)

so,

3

2NkBdT = −PdV

= −NkBT

VdV (54)

or3

2dT = −T

VdV (55)

si thatdT

T= −2

3

dV

V(56)

If we integrate both sides of this equation, we obtain

lnT

T0

= −2

3ln

V

V0

(57)

orT

T0

=(

V

V0

)−2/3

(58)

Since V0, the initial volume, is Va, and the final volume is Vb, we have

Tb = Ta

(Va

Vb

)2/3

(59)

As the volume increases, the temperature drops, which is to be expected, asthe internal energy of the gas, which performs work on the piston, also drops.

18

The total amount of work performed by the gas in expanding is the dif-ference between the initial and the final internal energy, there being no heatexchange to contribute to the right hand side of the first law. This meansthat the total work done is given by

3

2NkBTa −

3

2NkBTb =

3

2Ta

[1−

(Va

Vb

)2/3]

(60)

The pressure at Vb is given by

PbVb = NkBTb

= NkBTa

(Va

Vb

)2/3

(61)

This means thatPb = NkBTaV

2/3a V

−5/3b (62)

Instead of falling off like 1/Vb, as it did in the case of isothermal expansion,

the pressure falls off like V−5/3b . On the basis of this consideration alone we

can infer that the gas does less work in expanding adiabatically from Va toVb than in expanding between the same two volumes isothermally.

Like the slow isothermal process, the adiabatic process discussed abovecan be performed in reverse, and, hence, undone. It is also reversible.

The two specific heats of an ideal gas, cV and cP

The specific heat of a substance tells us how much heat must be fed undercertain circumstances to raise the temperature a fixed amount of that sub-stance a given number of degrees. We’ll start with a more general quantity,the heat capacity. This quantity is defined, generically as the following ratio

C =d-Q

dT(63)

The right hand side of (63) is the ratio of the heat fed in to the temperatureincrease.

Referring, again, to our cylinder of ideal gas, we can imagine feeding theheat into the container under one of two conditions:

1. The container is kept at constant volume.

19

2. The container is kept under a constant presssure.

In the first case, we may define a heat capacity at constant volume, CV , by

CV =

(d-Q

dT

)V

(64)

In the second case, the heat capacity at constant pressure is given by

CP =

(d-Q

dT

)P

(65)

Let us see what these ratios are equal to for the ideal gas

Constant volume

We havedU = d-Q (PdV = 0) (66)

or

d[3

2NkBT

]=

3

2NkBdT

= d-Q (67)

so (d-Q

dT

)V

=3

2NkB (68)

The heat capacity of a set of N molecules of monatomic ideal gas in a con-tainer kept at fixed volume is 3

2NkB.

Constant pressure

At constant pressure, the first law tells us that

dU = −PdV + d-Q (69)

which means that d-Q is given by

d-Q = dU + PdV (70)

20

Since PV = NkBT , we have for a system at constant pressure, P ,

PdV = NkBdT (71)

Furthermore,

dU = d[3

2NkBT

]=

3

2NkBdT (72)

so

d-Q =3

2NkBdt + NkBdT

=5

2NkBdT (73)

and

CP =

(d-Q

dT

)P

=5

2NkBT (74)

Notice that CP > CV , and that both are positive.

An irreversible expansion

Suppose we were to allow the gas in the cylinder to expand as follows: Instep a in Figure 5, a partition slides in immediately below the piston. Instep b, the piston is raised. No work is done on it because there is no gasbelow it. We are assuming that the environment is a perfect vacuum, sothere is no gas outside the container pushing the piston down. Finally, instep c, the partition is removed, and the gas fills the volume below the raisedpiston. Because the partition is infinitesimally thin (an assumption here)no work is required to push it in, nor is there any work done by the gaswhen it is removed. Finally, we assume that there is no heat exchange withthe environment during this process. Thus, both d-W and d-Q are equal tozero—which means that the internal energ, and hence, the temperature ofthe gas—is unchanged. If we want to return the volume of the gas to its

21

a

b

c

Figure 5: A three-step irreversible expansion in a container of ideal gas

original value, we can try slow compressions—either isothermal or adiabatic,and see what is involved.

Let’s assume that the original volume of the gas was Va and that it ex-panded in the above process to a volume Vb. In an adiabatic compression wewill have had to perform an amount of work given by

−3

2nkBT0 +

3

2NkBT0

(Vb

Va

)2/3

=3

2NkBT0

[(Vb

Va

)2/3

− 1

](75)

Here, T0 is the temperature of the expanded gas. Furthermore, the tempera-ture of the gas will have increased as the result of the adiabatic compression.The increase is from T0 to T0(Vb/Va)

2/3. Thus, in compressing the gas, wehave heated it up at the expense of externally supplied energy.

In an isothermal compression, the temperature of the gas has remainedconstant, which means that its internal energy has not changed. We havehad to perform work to effect the reversal of the free expansion. The netamount of this work is

−∫ Va

Vb

PdV = −NkBT0

∫ Va

Vb

dV

V

= NkBT0 ln(

Vb

Va

)(76)

22

Because dU = 0 in this process and −d-W is not, there must have been aheat flow out of the gas. From the first law (with dU = 0) we have∫

d-Q =∫

d-W

= −NkBT0 ln(

Vb

Va

)(77)

The right hand side of Eq. (77) is the amount of heat flow into the gas.Thus, an amount of heat equal to NkBT0 ln(Vb/Va) has flowed out of thegas and into its environment. In an isothermal re-compression we have hadto perform work on the gas cylinder, and the energy we have put into thesystem in this process has gone into heat that was taken up by the cylinder’senvironment.

23

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~4~

Entropy, at last

Here is where we get to the definition of entropy. We imagine a system, suchas the container of ideal gas, undergoing a set of changes. Some of thosechanges are accompanied by transfer of heat to or from thermal reservoirs.Some of them involve no heat exchange. We will number the reservoirs from1 to n. The ith reservoir is at a temperature Ti. The heat extracted by oursystem from the ith reservoir will be denoted by Qi.

Finally, we will imagine that the net effect of all these changes on oursystem is to return it to its initial state. In other words, there is no net effectof this set of changes on the state of the system. For this reason, we will callthis set of changes a cycle.

A central assertion

We now make the following assertion:The cycle described above will always have the following property:

n∑i=1

Qi

Ti

≤ 0 (104)

Eq. (104) holds as an equality for a reversible cycle. In fact, if the cycle isreversible the equality must hold.

We can show that Eq. (104) holds by assuming that it does not, andthen by showing that the second law is violated. What we do is imagine ann + 1st reservoir at a temperature T0. After completing the cycle for whichthe inequality

∑ni=1

Qi

Ti> 0 holds does not hold, we run a set of n reversible

cycles between each of the n reservoirs and the new one. In the ith of thesecycles we extract an amount of heat equal to Q(i) from the new reservoirand deposit an amount Qi in the ith one. That is, we restore to each of theoriginal n reserviors the amount of heat extracted from it. Since these cyclesare all reversible, we have

Q(i)

T0

=Qi

Ti

(105)

This means that

1

T0

n∑i=1

Q(i) =n∑

i=1

Qi

Ti

> 0 (106)

34

The net heat extracted from the new reservoir is greater than zero.The end effect of all these cycles is that

1. All working components end up in their initial state.

2. All the reservoirs except one end up in their initial states.

3. The single reservoir that is not in its initial state has had a non-zeroamount of heat extracted from it.

4. By the first law of thermodynamics this heat can only have gone intowork.

The second law of thermodynamics has thus been violated.For any cycle of the kind described here, the inequality

∑ni=1

Qi

Ti> 0 is,

thus, not allowed. In the case of a cycle that is reversible, we cannot have theopposite absolute inequality,

∑ni=1

Qi

Ti< 0, because the cycle would violate

(104) when run in reverse.To recapitulate

1. For any cycle,n∑

i=1

Qi

Ti

≤ 0 (107)

2. For a reversible cycle,n∑

i=1

Qi

Ti

= 0 (108)

On the basis of the equality (108) we can define a state variable for asystem at equilibrium. Actually, this state variable is, for the time being,defined only in terms of differences. We will establish an absolute referencevalue for it when we get to a discussion of the third law of thermodynamics.

Consider a system in a state A. This is a state of thermal equilibrium, byassumption. We assign to this system a state variable, SA, which we call itsentropy when in state A. The difference between SA and SB, the entropyof the system when in state B, is given by

SB − SA =∑

i

Qi

Ti

(109)

35

Where Qi is the heat absorbed from a reservoir at temperature Ti during oneof a set of reversible steps that takes the system from state A to state B. Wecan also write the right hand side of Eq. (109) in differential form as follows:

SB − SA =∫ B

A

dQ

T(110)

Strictly speaking, I should have put a line through the d in dQ. For conve-nience only, I will leave that line out. However, you should always keep inmind that the infinitesimal dQ is not the difference between two well-definedvariables, but rather an increment of transferred heat.

The denominator, T , in the integrand on the right hand side of (110)corresponds to the temperature of the reservoir with which the system ex-changes the infinitesimal amount of heat energy, dQ. By the Clausius versionof the second law, and the requirement that the process taking the systemfrom state A to state B be reversible, we can infer that T must also be thetemperature of the system itself at the time of the heat transfer. Otherwisethe heat transfer, and the process itself, would be irreversible.

Note that it does not matter what particular set of reversible steps wechoose to take the system from state A to state B. The integral

∫ BA dQ/T

will be independent of our choice. If it were different for the two different setof steps, then we could take the system from state A to state B along one ofthem and from state B back to state A along the other one run in reverse, andin the process construct a reversible cycle for which

∑i Qi/Ti 6= 0, which we

now know cannot be. The difference SB−SA depends on the two states A andB and not how the system is taken between those two states. The quantitiesSA and SB are, thus, true state variables—to within additive constants, atleast.

Let’s get specific and calculate the entropy difference between an idealgas at a temperature TA and a volume VA and the same set of moleculesat a temperature TB and volume VB. First, we will calculate S(TA, VB) −S(TA, VA). To take a gas from a volume VA to a volume VB at the sametemperature, we subject it to an isothermal expansion or compression. Theheat extracted from the environment by the ideal gas compensates for thework performed on it, as the internal energy of the ideal gas does not changein the process of isothermal expansion or compression. We then find

∆Q

TA

=1

TA

∫ VB

VA

PdV

36

=1

TA

∫ VB

VA

NkBTA

VdV

= NkB lnVB

VA

= S(TA, VB)− S(TA, VA) (111)

As the volume of the gas increases, so does its entropy.To take our container of gas reversibly from a volume VB and temperature

TA to the same volume and a temperature TB, we can put it into thermalcontact with a succession of different reservoirs, each at a temperature soclose to the changing temperature of the gas that the heat transfer is essen-tially reversible. If we imagine an infinite succession of these reservoirs, eachcontributing an infinitesimal amount of heat to the process, the net changein the entropy of the gas will be∫ TB

TA

dQ

T=

∫ TB

TA

1

T

dQ

dT

∣∣∣∣∣V =VB

dT

=∫ TB

TA

1

TCV dT (112)

where, recall, CV is the heat capacity of the gas at constant volume. Wehave already calculated this heat capacity at 3NkB/2. Thus

S(TB, VB)− S(TA, VA) =∫ TB

TA

3

2NkB

dT

T

=3

2NkB ln

TB

TA

(113)

What this means is that

S(TB, VB)− S(TA, VA) = [S(TB, VB)− S(TA, VB)] + [S(TA, VB)− S(TA, VA)]

=3

2NkB ln

TB

TA

+ NkB lnVB

VA

= NkB ln

[VB

VA

(TB

TA

)3/2]

(114)

Choosing our arbitrary reference system to have a temperature and a volumeequal to unity, we can write

S(TA, VA) = NkB ln[VAT

3/2A

](115)

S(TB, VB) = NkB ln[VBT

3/2B

](116)

37

That is, in general S(T, V ) = NkB ln(V T 3/2), to within additive constants.Note that the entropy is a monotonically increasing function of temperature.

Recall, now, the free expansion of the ideal gas from its initial volumeto a greater one. According to our results for entropy, the gas under freeexpansion increases its entropy by an amount NkB ln(Vfinal/Vinitial). The “freeadiabatic compression” of the ideal gas, the process that we would have hadto wait essentially forever to carry out, would have involved a net decrase inthe gas’s entropy. According to the second law of thermodynamics, this freecompression cannot occur. Thermodynamics says it is impossible. Detailedinvestigation reveals that it is possible, but only in the mathematical sense.One will never succeed in actually carrying it out.

We can now state another version of the second law; a system inthermal isolation will never decrease its entropy. To see that this isso, imagine that we had a system in thermal isolation that we have succeededin taking from a state A to a state B with a consequent decrease in entropy.We then allow the system to establish thermal contact with various reservoirsand take it reversibly from state B to state A. In this process the entropychanges, and the change in the entropy satisfies∫ A

B

dQ

T= SA − SB > 0 (117)

The cycle consisting of the change in the system from state A to state Bwhile in thermal isolation followed by the reversible undoing of this changeviolates the rule

∮dQ/T ≤ 0. The second law is thus violated if the entropy

of a system in thermal isolation decreases.We can extend this conclusion; any change in a system in thermal

isolation that involves an increase in the system’s entropy is irre-versible. After all, if such a change were reversible, we could run it backardsand violate the second law by reducing the entropy of a system in thermalisolation.

We now have another rule; any change in a system that can bereversed leaves the entropy of the system unaltered.

In the case of the monatomic ideal gas, we have

S = NkB ln[V T 3/2

](118)

If a container of such a gas is reversibly expanded in thermal isolation, then,in order that the entropy does not change, the temperature and volumemust be related via T ∝ V −2/3. This is our previously-derived result for therelationship between T and V in adiabatic expansion or compression.

38

The first law in terms of exact differentials

Let’s return to the first law and consider small changes involving reversibleprocesses. The first law is

dU = d-Q− d-W (119)

In a reversible process d-Q/T = dS, where dS is the infinitesimal change inthe entropy of the system. Thuse, we have

dU = TdS − d-W (120)

For the ideal gas,dU = TdS − PdV (121)

Now, everything is in terms of exact differentials.What this last formula tells us is that in any reversible change of our

cylinder of gas in which the number of molecules remains constant, the changein the internal energy is expressed entirely in terms of the change in theentropy of the gas and the change in the voume of the cylinder. From this,we conclude that the temperature is not an independent state variable, nor ispressure. We can express the internal energy entirely in terms of the entropyan the volume—or U = U(S, V ). Since

dU(S, V ) =

(∂U

∂S

)V

dS +

(∂U

∂V

)S

dV (122)

we have

T =

(∂U

∂S

)V

(123)

P = −(

∂U

∂V

)S

(124)

The temperature and the pressure are (plus or minus) partial derivatives ofthe internal energy with respect to the entropy and the volume of the system.

In order to complete our specification of the independent state variableswe must also consider the effect on the internal energy of the system of achange in the number of molecules that it contains. This effect is easy toassess in a simple case. Imagine that we increase the number of molecules

39

by attaching a smaller container of the gas that is at the same temperatureand pressure. The number of molecules in the smaller container is ∆N , thevolume of this container is ∆V = (V/N)∆N , the internal energy of thiscontainer is ∆U = (U/N)∆N , and the entropy in this container is equalto ∆S = (S/N)∆N . Note that there is a strong assumption here. We areassuming that the entropy of a system increases linearly with the number ofmolecules in it. If you check back to our current expression for the entropy,you will see that this assumption is violated by that expression. We’ll returnto this point shortly.

Under the assumption that the internal energy can be written as U(S, V, N),we have the general result

∆U =U

N∆N

=

(∂U

∂S

)V,N

∆S +

(∂U

∂V

)S,N

∆V +

(∂U

∂N

)S,V

∆N (125)

From this, we infer(∂U

∂N

)S,V

=U

N−(

∂U

∂S

)V,N

S

N−(

∂U

∂V

)S,N

V

N

=U

N− T

S

N+ P

V

N(126)

We call the quantity on the left hand side of Eq. (126) µ, also known as thechemical potential. This means that we have

µ =U

N− T

S

N+ P

V

N(127)

orU = TS + µN − PV (128)

40

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+

The Helmholtz free energy of the monatomic

ideal gas

Let’s now construct the Helmholtz free energy of the monatomic ideal gas.We can accomplish this by integrating up the equations of state

PV = NkBT (179)

U =3

2NkBT (180)

The two equations of state can reframed in terms of derivatives of the Helmholtzfree energy. For example, Eq. (179) takes the form

−∂F (T, V, N)

∂V= P (T, V, N)

=NkBT

V(181)

Integrating up

F = −∫ NkBT

VdV

= −NkBT ln V + f(T,N) (182)

The function f(T, N) is an “integration constant.”In the case of the equation of state, Eq. (180), the relationship is

U(T, V, N) = F (T, V, N) + TS(T, V, N)

= F (T, V, N)− T∂F (T, V, N)

∂T

= −T 2 ∂

∂T

(F (T, V, N)

T

)(183)

Thus,

−T 2 ∂

∂T

(F (T, V, N)

T

)=

3

2NKBT (184)

Integrating this up

F (T, V, N) = −T∫ 3

2

NkB

TdT

= −3

2NkBT ln T + Tg(V, N) (185)

53

Combining (182) and (185), we find

F (T, V, N) = −NkBT ln(V T 3/2

)+ Th(N) (186)

where h(N) represents the remaining integration constant. We will shortlyargue that the free energy must be extensive. This means that

h(N) = NkB ln N + CN (187)

which means, at last, that

F (T, V, N) = −NkBT ln(

V

NT 3/2

)+ CTN (188)

The constant C in (187) and (188) cannot be determined in the absence offurther information with regard to the thermodynamic behavior of the idealgas.

That the ideal gas equations of state are obtained from the expression(188) for the Helmholtz free energy by differentiation is effectively obvious,given that we obtained that expression by integrating up those equations ofstate. Instead of taking derivatives explicitly to show that this is the case,let’s perform a Legender transformation to construct the internal energy asa function of S, V and N . We have

U(S, V, N) = F (T (S, V, N), V, N) + T (S, V, N)S (189)

where T (S, V, N) is obtained by inverting

S = −∂F (T, V, N)

∂T(190)

Making use of (188), we obtain

S = NkB ln(

V

NT 3/2

)+ (

3

2kB − C)N (191)

If you recall our previous results for the entropy of the ideal gas, you willnote that the last term on the right hand side of (191) is new. This lastterm is a new additive pieces that do not affect any of the calculations wehave already carried out, as we have been assessing entropy changes in whichN , the number of molecules in a system remains a constant. Again, we are

54

confronted with the fact that entropy is properly defined only to within anadditive constant. Let’s set the coefficient of N in the last term on the righthand side of Eq. (191) equal to kBD. Then, solving that equation for T , wefind

T =(

N

Ve

SNkB

−D)2/3

(192)

Inserting this solution into the equation (189), we obtain for the internalenergy of the ideal gas

U(S, V, N) =3

2NkB

(N

Ve

SNkB

−D)2/3

(193)

If we take the derivative of this equation with respect to S and set it equalto T , we end up with the relationship (192).

Extensive and intensive quantities

There are two kinds of variables in thermodynamics. The first kind, ex-tenstive variables grow linearly with the size of the system. Examples ofextensive variables are entropy, volume and number of molecules. When thesize of the system doubles, so does each extensive variable. The other kind ofvariable is intensive. An intensive variable stays the same as the size of thesystem increases. Examples of intensive variables are temperature, pressureand chemical potential.

Note that the internal energy is extensive, as is the Helmholtz free energy.

Conjugate variables

Thermodynamic variables come in pairs. Each pair contains one extensivevariable and one intensive variable. Each variable in the pair is a conjugateof the other one. Here is the three pairs in the case of the ideal gas: S ⇔ T ,V ⇔ P , N ⇔ µ. Recall the form that we had for the internal energy:U = TS − PV + µN . The internal energy is equal to the sum of (plus orminus) the product of conjugate variables. This is not the only relationshipinvolving conjugate variables. The partial derivative of a thermodynamicpotential (two examples of which are the internal energy and Helmholtz freeenergy) with respect to one of the appropriate independent variables yields

55

(plus or minus) its conjugate variable. That is,

∂F (T, V, N)

∂T= −S (194)

∂U(S, V, N)

∂S= T (195)

Thermodynamic potentials in general

The Legendre transformation can be utilized to construct other quantitiesderived from the internal energy. These quantities are known in general asthermodynamic potentials. Here are a few examples of thermodynamicpotentials, and the relationship between those thermodynamic potential andthe quantities with which we have already been dealing.

The Gibbs free energy

The Gibbs free energy is properly expressed in terms of the temperature,pressure and number of molecules. Its relation to the Helmholtz free energyis

G(T, P,N) = F + PV (196)

where V is expressed as a function of P , T and N via the inversion of therelationship

P = −∂F

∂V(197)

The enthalpy

The enthalpy is properly expressed in terms of the entropy, pressure andnumber of molecules. Its relation to the internal energy is

H(S, V, N) = U + PV (198)

where V is expressed as a function of S, P and N via the inversion of

P = −∂U

∂V(199)

56

The grand potential

The grand potential is properly expressed in terms of the temperature, vol-ume and chemical potential. Its relation to the Helmholtz free energy is

Ω = F − µN (200)

where N is expressed in terms of T , V and µ via the inversion of

µ =∂F

∂N(201)

The general form of thermodynamic potentials

Because U = TS − PV + µN , we also know that

F = −PV + µN (202)

G = µN (203)

H = TS + µN (204)

Ω = −PV (205)

The list above clearly does not exhaust the list of all possible thermody-namic potentials. The potential that will turn out to be most useful willclearly depend on the situation. As noted earlier, the appropriate choiceis controlled by the variables that are most “natural” in a given situation.There is one set that cannot be used, and, that is one consisting only ofintensive quantities. For the ideal gas, there is no meaningful thermody-namic potential constructed out of the temperature, pressure and chemicalpotential. For one thing, if one asks what is left after subtracting µN fromthe Gibbs thermodynamic potential to perform a Legendre transformationto a thermodynamic potential in terms of T , P and µ, we end up with zero.That there is no well-defined thermodynamic potential expressed entirely interms of intensive quantities is obvious if one realizes that in specifying thetemperature, pressure and chemical potential, one has no information withrespect to the size of a system.

57

A couple of other systems

The van der Waals gas

The van der Waals equation of state is(P + a

(V

N

)2)

(V −Nb) = NkBT (206)

What this equation of state tells us is that the effective pressure in increasedbecause of interactions between the molecules. This is the source of theterm a(N/V )2. The coefficient a is related to the strength of the attractiveinteractions. Furthermore, the effective volume decreases because of hardcore repulsion between the molecules. This is the source of the term Nb.The coefficient b is related to the volume of a single molecule. If we assumethat the heat capacity at constant volume is equal to

3

2NkB (207)

we can obtain the Helmholtz free energy by integration. This will be assignedas a homework problem.

The ideal paramagnet

To study the equilibrium properties of a magnetic solid, we introduce twonew variables, the total magnetization of the sample, M , and the externallyapplied magnetic field, H. Although both of those quantities are vectors, wewill assume a direction, and deal only with amplitudes. This means that bothM and H are scalar variables. In fact, they are a conjugate pair, M beingthe extensive one and H being intensive. The magnetic Gibbs free energyof a magnetic system is written as, G(T, H, N). The total magnetization ofthis system is given by

M = −∂G(T, H, N)

∂H(208)

Given that the Gibbs free energy is, like all thermodynamic potentials, ex-tensive, and given that the only extensive variable in the set T,H, N is N ,the number of molecules in this system, we know that the magnetic Gibbsfree energy must have the form

G(T,H, N) = Ng(T, H) (209)

58

The equation of state of the ideal paramagnet is the Curie formula

M = NDH

T(210)

Here, D is a system-dependent constant. Note the appearance of the factorN . This is to ensure that the magnetization is extensive. To add to ourknowledge of the thermodynamic properties of this system, we’ll assume aheat capacity at zero magnetic field given by

CH = αNT 3 (211)

Here, α is a system-dependent coefficient. This form for the heat capacity ischaracteristic of crystalline systems in which the low-temperature propertiesare controlled by vibrations of the molecules. As it turns out, the exactform is not particularly important. Now, let’s obtain the magnetic Gibbsfree energy by integrating up (210) and (211). Starting with Eq. (210), weobtain

G = −∫

MdH

= −∫

NDH

TdH

= −1

2ND

H2

T+ Nf(T ) (212)

The “integration constant” function f(T ) is obtained by performing a doubleintegration of the heat capacity. Since S = −∂G/∂T , and CH = T∂S/∂T ,we have CH = −T∂2G/∂T 2. Performing the requisite integrations, we find

G = −Nα

12T 4 + AN + BNT +

g(H)

T(213)

Combining (212) and (213), we find for the magnetic Gibbs free energy

G = −Nα

12T 4 + AN + BNT −ND

H2

2T(214)

The constants A and B cannot be determined from the information that wehave in hand. For convenience, we will set them both equal to zero. Thishas no effect on the discussion that follows.

59

The entropy of the ideal paramagnet

To find the entropy of our magnetic system, let’s take the derivative of themagnetic Gibbs free energy with respect to temperature. We have

S(T,H, N) = −∂G(T, H, N)

∂T

= −NDH2

2T 2+ N

α

3T 3 (215)

How does this entropy behave as T → 0? First, if H 6= 0, we have S → −∞.This is because of the term going as 1/T 2 on the right hand side of Eq. (215).On the other hand, if H = 0, the expression for entropy consists entirely ofthe second term on the right hand side of (215). This term goes to zeroas T → 0. The entropy at T = 0 depends on whether there is a non-zeroapplied magnetic field. As we will see next. This implies a possible strategyfor cooling a system to the absolute zero of temperature.

60

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5

Molecules in a box

Let us consider the system shown in the figure below. We will ask pertinent

Figure 1: A set of molecules in a box with an imaginary partition separatingone half from the other.

questions concerning the distribution of the molecules. For the time being,we will concentrate on the spatial distribution of the molecules, and we willask simple questions with regard to the gross features of the distribution.Here is one question:

If there are N molecules in the box, what is the probability, given noother information about the molecules that n of them are in the top half ofthe box and N − n of them are in the bottom half?

To simplify matters, we will assume that the molecules are non-interacting,and we will also neglect the effect of gravity. Then, suppose that N − 1 ofthem are in the top half and the last molecule has to decide where to go.According to this assumption, it has no preference for either half of the box.The probability that it chooses the top half is 1/2, and the probability thatit will end up in the bottom half is also 1/2. Each molecule thus has a 50%chance of ending up in either half of the box, regardless of where the othermolecules happen to be.

Let’s warm up by asking an extreme version of the above question:

What is the probability that all N of the molecules are in the top half ofthe box?

If the probability that any one of the molecules is in the top half of thebox is 1/2, then the probability that all N of them are is (1/2)N , or 1/2N .How small is this number? In other words, how large is 2N for characteristic

1

values of N? Suppose we are thinking of a macroscopic box. A reasonablevalue for the number N would then be of the order of Avagadro’s number,6× 1023. We will take N to be 1023.

How big is 21023? It is, quite obviously, a very big number, but we willfind it edifying to investigate just how large it is. Let’s start by convertingto powers of 10. We know that 210 = 1024 ≈ 103. Then

21023 =(210)1022

=(103

)1022

= 103×1022 (1)

which means that 21023 equals 1 followed by 3 × 1022 zeros. This is clearlyan enormous number. It is, in fact, unimaginably huge. To see this, let usimagine that the molecules are moving around in the box quite at randomand estimate the amount of time that we will have to wait for all of them tobe in one half of the box. Let’s take the box to be one meter per side andhave the molecules move faster than they possibly can. We’ll assume thatthey move with the speed of light. This is 3 × 108 meters per second. Fora molecule to go from one side of the box to the other takes less than 10−8

seconds. To be precise, it takes (1/3)× 10−8 seconds. We can then imaginethat every (1/3) × 10−8 seconds there is a complete re-arrangement of themolecules in the box, so that every time one of these incredibly brief timeinterval passes the molecules have another chance to all be in the same halfof the box.

There are about 3 × 107 seconds in a year, so the number of times themolecules rearrange themselves each year is

3× 107 seconds13× 10−8 seconds/rearrangement

= 9× 1015 rearrangements (2)

The molecules rearrange themselves about 1016 times each year.We will have to wait until there are the order of 103×1022 rearrangements

of the molecules until we can reasonably expect to see all of the molecules inone half of the box. That means that we will have to wait 103×1022/1016 =103×1022−16 years. The number 3 × 1022 is so much greater than 16 that wecan forget about the 16 in the last exponent above. We will have to wait103×1022 years before there will be any reasonable likelihood of seeing all themolecules in one half of the box.

2

We can expect our solar system to “live” a few billion years longer. Let’sbe generous and give it a total life-span of 100 billion years = 1011 years.We can now ask how many solar system lifetimes we will have to wait. Thisnumber is

103×1022

1016 × 1011= 103×1022−27 (3)

Again, the 3× 1022 in the exponent swamps the number we are subtracting,and we must conclude that it will take 103×1022 solar system lifetimes for thisincredibly unlikely event to occur.

Now, imagine that I asserted that it is a law of nature that in a boxwith a mole of gas in it, you will never see all of the gas molecules happento congregate in one half. Should you take this statement seriously? Clearly,it is not rigorously true. There is a non-vanishing likelihood that such anoccurrence will come about. However, if you wait around, expecting to seeit, you will be wasting your time. Although it can happen, a violation of theabove law is so highly unlikely that it might as well never take place. Thepractical difference between this “law” and a law that is always obeyed is,for all practical purposes, nonexistent.

A bit more detail

Because the molecules are non-interacting, each of them might as well flip acoin to decide which half of the box it will occupy at some particular pointin time. We can figure out the probability that n of the molecules are inone half of the box and N − n are in the other half by performing N coinflips and asking for the probability that we have n of them turning up tailsand N − n of them heads. This is a problem in combinatorics. The way wecalculate the probability is to divide the number of ways of having n tailsand N − n heads turn up in N flips by the number of ways N flips can turnout. This latter number is 2N , corresponding to the fact that there are twopossibilities per toss. The numerator of our fraction is the number of waysof dividing a set of N objects into two subsets, the first containing n and thesecond N − n elements. This is(

nN

)=

N !

n!(N − n)!(4)

3

The probability that there are n tails and N − n heads in the N flips is

N !

n!(N − n)!

1

2N(5)

To proceed further we need a good asymptotic formula for the value of n!when n is large.

Stirling’s approximation

The number n! = n(n−1)(n−2) · · · (2)(1) can be written as an integral. Wehave

n! =∫ ∞

0tne−tdt (6)

To see that the integral on the right hand side of (6) is, indeed, a representa-tion of n!, we will integrate by parts. Starting with the integral representationfor 1!: ∫ ∞

0te−tdt = −e−tt

∣∣∣∞0

+∫ ∞

0e−tdt

= 0− e−t∣∣∣∞0

= 1 (7)

If we defineF (n) ≡

∫ ∞0

tne−tdt (8)

we haveF (1) = 1 (9)

Now, what about F (n) when n > 1? Here’s an integration by parts

F (n) =∫ ∞

0tne−tdt

= −e−ttn∣∣∣∞0

+ n∫ ∞

0tn−1e−tdt

= n∫ ∞

0tn−1e−tdt

= nF (n− 1) (10)

4

So,

F (n) = nF (n− 1)

= n(n− 1)F (n− 2)

= n(n− 1)(n− 2)(n− 3) · · · × 3× 2× F (1)

= n! (11)

where we made use of (9) to eliminate F (1) from the right hand side of Eq.(11).

What, now, when n is large?In this case, tn is a rapidly increasing function of t, while e−t is a function

that ultimately dies off faster than any power of t. The integrand will looklike the curve in Figure 2.

5 10 15 20 25 30 t

2 ·10 10

4 ·10 10

6 ·10 10

8 ·10 10

1 ·10 11

1.2 ·10 11

t n e -t

Figure 2: The function tne−t, with n = 15.

At what value of the integration variable t is the integrand a maximum?To find this out, we maximize

tne−t = en ln t−t (12)

We accomplish this by maximizing the exponent on the right hand side of(12). The extremum equation is

d

dt[n ln t− t] =

n

t− 1

= 0 (13)

The solution of this equation is, clearly, t = n.

5

The integrand is, thus, at its maximum when t = n. At this value of t,the integrand is

nne−n =(n

e

)n

(14)

For large n this number is very large. Look at the vertical axis in Figure 2,where the integrand is plotted for n = 15. Supppose we shift the origin inthe graph above to t = n. That is, we replace t by x + n, where x rangesfrom −n to ∞. The integral representation of n! is, then∫ ∞

−n(x+ n)ne−(n+x)dx =

∫ ∞−n

en ln[n+x]−n−xdx (15)

Taylor expanding the logarithm in the exponent above:

ln [n+ x] = lnn+ ln[1 +

x

n

]= lnn+

x

n− 1

2

x2

n2+O

((x

n

)2)

(16)

The integral is, then,∫ ∞−n

en lnn−nen

[xn− 1

2x2

n2 +O

(x3

n3

)]−xdx =

∫ ∞−n

en lnn−n− 1

2x2

n+O

(x3

n2

)dx (17)

The gaussian factor e−x2/2n is much less than one when x = A

√n, where

A is a sizable, but fixed, number. If n is very large, we have

x3

n2∼ A3n3/2

n2

= A3n−1/2

1 (18)

and over the width of the gaussian we can neglect the term of order x3/n2 inthe integrand (and higher order terms, as well). The integral we now haveto perform is ∫ ∞

−n

(n

e

)n

e−12x2/ndx (19)

The lower limit can be replaced by −∞ with the introduction of negligibleerror (∼ e−n). We then have, in the limit of large n,

n! ≈(n

e

)n ∫ ∞−∞

e−12

x2

n dx

=(n

e

)n√2πn (20)

6

This is Stirling’s approximation, and it is an excellent approximation forn! when n is greater than 10. To see how good it is, we compare ln 15! with

ln[(

15e

)15√2π × 15

]. The results for the two are

ln 15! = 27.8993 . . . (21)

ln

[(15

e

)15√2π × 15

]= 27.8937 . . . (22)

Back to our problem of the molecules in the box. We can apply Stirling’sformula (without the square root factors, for the time being) to the proba-bility of having n molecules in one half of the box (say, the top) and n−Nin the other half. This probability is

N !

n!(N − n)!

1

2N= exp [lnN !− lnn!− ln(N − n)!−N ln 2]

≈ exp[N ln

(N

e

)− n ln

(n

e

)− (N − n) ln

(N − ne

)−N ln 2

]= exp [N lnN − n lnn− (N − n) ln(N − n)−N ln 2] (23)

Let’s write n = (N/2) + ∆. We already know that the most probable distri-bution will have half of the molecules in each side of the box. Our exponentis, now

N lnN −(N

2+ ∆

)ln(N

2+ ∆

)−(N

2−∆

)ln(N

2−∆

)−N ln 2 (24)

We can Taylor expand the second and third terms in (24) with respect to ∆.Given that

d

dxx lnx = ln x+ 1 (25)

d2

dx2(x lnx) =

d

dx(lnx+ 1)

=1

x(26)

we have(N

2+ ∆

)ln(N

2+ ∆

)=N

2lnN

2+ ∆

[N

2+ 1

]+

1

2∆2 × 2

N(27)

7

and(N

2−∆

)ln(N

2−∆

)=N

2lnN

2−∆

[N

2+ 1

]+

1

2(−∆)2 × 2

N(28)

This means that our exponent is

N lnN − 2[N

2lnN

2

]− 2∆2

N−N ln 2 = −2∆2

N(29)

Forgetting about the square roots in Stirling’s formula, which is what we’vebeen doing up to now, we now have for the probability that there are N/2+∆molecules in one half of the box and N/2−∆ molecules in the other

e2∆2/N (30)

Notice that this is a gaussian distribution. If we had put the square rootsin and carried them along in our calculation, the result for this probabilitywould have been √

2

πNe−2∆2/N (31)

The width of the probability distribution in (31) is proportional to√N .

That is, the gaussian starts to fall off dramatically when ∆>∼√N . If there

are ∼ 1023 molecules in the box, the gaussian e−2∆2/N starts to fall off when∆>∼√

1022 = 1011, or when there is an excess of ∼ 1011 molecules in one halfof the box.

Now, imagine that I told you that it is a law of nature that, given a boxcontaining some molecules flying around at random, one half of the box willalways contain half of the molecules. How seriously should you take me?Clearly, what I said is not always true. In fact, it is in general not true at all.According to our results, th probability that exactly half of the molecules arein one half of the box is √

2

πNe−2(∆=0)2/N =

√2

πN(32)

When N ∼ 1023, this probability is vanishingly small (actually it is ∼ 10−11).On the other hand, if I were to say that half of the molecules, more or less,are in the bottom (or top) half of the box, I would be making an essentiallycorrect statement. The percentage difference between N/2 and N/2±

√N is

8

√N/(N/2) ∼ 10−11 (if N ∼ 1023), so half of the molecules are in the bottom

half of the box, to within errors of a few parts per trillion.Let’s be a little more quantitative and ask what the mean square deviation

is from the predicted—or average—configuration. We want to calculate

〈∆2〉 =∑∆

√2

πNe−2∆2/N∆2

≈∫ ∞−∞

√2

πNe−2∆2/N∆2d∆ (33)

Gaussian integrals: a review

We want to be able to evaluate∫ ∞−∞

x2ne−ax2

dx (34)

When n = 0, we have ∫ ∞−∞

e−ax2

dx =

√π

a

= π1/2a−1/2 (35)

If we note that

x2n = (−1)ndn

dane−ax

2

(36)

we have, straightforwardly,∫ ∞−∞

x2ne−ax2

dx =∫ ∞−∞

(−1)ndn

dxne−ax

2

dx

= (−1)ndn

dan

[π1/2a−1/2

](37)

so, for instance, ∫ ∞−∞

x2e−ax2

dx = − d

daπ1/2a−1/2

= π1/2 1

2a−3/2 (38)

9

Back to the quantity of interest

Our integral is√2

πN

∫ ∞−∞

e−2∆2/N∆2d∆ =

√2

πN

∫ ∞−∞

e−a∆2

d∆ a =2

N

=

√2

πNπ1/2

(2

N

)−3/2

=N

4(39)

So, ∆2 = N/4. The mean square deviation in the number of molecules inone half of the box is of order N . On the average, there are, then,

√〈∆2〉 =

√N

2(40)

extra molecules on one side of the box or the other.

10

Principles of statistical mechanics

We will begin our discussion of statistical mechanics with a definition ofentropy that is suitable for use in the consideration of the behavior of largesystems. For the time being, the definition will be simply stated. The validityof this definition will be establshed through its application to a few systems.

Statistical definition of entropy

The entropy, S, of a system is defined as follows:

S = kB ln number of states available to the system (1)

Here, kB is Boltzmann’s constant.

An example of the application of this definition of en-tropy: a spin 1/2 system

Let’s assume that we have N ∼ 1023 spin 1/2 particles. The z-component ofthe spin of each points either up or down. The net magnetic moment in thez-direction of this collection of particles is given by

Mz = µ

[N∑i=1

si

](2)

where si = ±1 for each individual spin, and µ is the product of the Bohrmagneton and the gyromagnetic ratio.

Suppose we do not specify Mz. Then, each spin can be in one of twostates, and the system as a whole can be in one of 2N states. According toour definition of entropy,

S = kB ln 2N

= NkB ln 2 (3)

Notice that according to this definition, the entropy is extensive.Now, suppose we specify Mz. That means that we must have N+ spins

with Si = +1 and N− spins with Si = −1, where

N+ +N− = N (4)

µ (N+ −N−) = Mz (5)

1

In other words,

N+ =1

2

(N +

Mz

µ

)(6)

N− =1

2

(N − Mz

µ

)(7)

We have already worked out the number of ways of separating N objects intotwo piles, one containing N+ objects and the other with N− objects. Thecombinatorial factor is

N !

N+!N−!(8)

Thus,

S = kB lnN !

N+!N−!

= kB lnN !− kB lnN+!− kB lnN−! (9)

Making use of Stirling’s formula:

lnn! ≈ n lnn

e(10)

we have

S

kB≈ N ln

N

e−N+ ln

N+

e−N− ln

N−

e= N lnN −N+ lnN+ −N− lnN−

= N lnN − 1

2

(N +

Mz

µ

)ln

[1

2

(N +

Mz

µ

)]

−1

2

(N − Mz

µ

)ln

[1

2

(N − Mz

µ

)](11)

Using

N =

(1

2

(N +

Mz

µ

)+

1

2

(N − Mz

µ

))(12)

we have

S

kB=

1

2

(N +

Mz

µ

)lnN +

1

2

(N − Mz

µ

)lnN

2

−1

2

(N +

Mz

µ

)ln

[1

2

(N +

Mz

µ

)]− 1

2

(N − Mz

µ

)ln

[1

2

(N − Mz

µ

)]

= −1

2

(N +

Mz

µ

)ln

[1

2

(1 +

Mz

µN

)]− 1

2

(N − Mz

µ

)ln

[1

2

(1− Mz

µN

)]

= −1

2

(N +

Mz

µ

)ln

1

2− 1

2

(N − Mz

µ

)ln

1

2

−1

2

(N +

Mz

µ

)ln

[1 +

Mz

µN

]− 1

2

(N − Mz

µ

)ln

[1− Mz

µN

]

= N ln 2− 1

2

(N +

Mz

µ

)ln

[1 +

Mz

µN

]− 1

2

(N − Mz

µ

)ln

[1− Mz

µN

](13)

Notice that the entropy if this system is extensive, in that if N → 2N andMz → 2Mz, then S(N,Mz)→ S(2N, 2Mz) = 2S(N,Mz).

Entropy versus energy for this model

If there is an external field on the syhstem of spin 1/2 magnetic moments,then the energy of interaction of the moments with the magnetic field, alsoknown as the Zeeman energy, is given by

E = − ~M · ~H→ −MzH (14)

That is, we assume that the magnetic field points in the zdirection.The free energy of this system is

G = U − TS= −MzH

−kBTN ln 2− 1

2

(N +

Mz

µ

)ln

[1 +

Mz

µN

]− 1

2

(N − Mz

µ

)ln

[1− Mz

µN

](15)

If this free energy is to be minimized with respect to the net magnetization,Mz, then

∂G

∂Mz

=∂

∂Mz

[−HMz − kBT

N ln 2− 1

2

(N +

Mz

µ

)ln

[1 +

Mz

µN

]

3

−1

2

(N − Mz

µ

)ln

[1− Mz

µN

]]= 0 (16)

or

0 = −H − kBT

− 1

µln

(1 +

Mz

µN

)− 1

2µN

N + Mz

µ

1 + Mz

Nµ

+1

µln

(1− Mz

µN

)+

1

2µN

N − Mz

µ

1− Mz

Nµ

(17)

orµH

KBT= ln

1 + Mz

µN

1− Mz

µN

(18)

or

eµH/kBT =1 + Mz

µN

1− Mz

µN

(19)

Solving for Mz/µN ,

Mz

µN=

eµH/kBT − 1

eµH/kBT + 1

= tanh(

1

2

µH

kBT

)(20)

Thus,

Mz = µN tanh(

1

2

µH

kBT

)(21)

When H is small,

tanh(

1

2

µH

kBT

)≈ µH

2kBT(22)

and

Mz ≈µ2NH

2kBT

=µ2

2kB

NH

T(23)

4

The total magnetic moment in the z-direction obeys the Curie law. On theother hand, when H is large,

tanhµH

2kBT≈ 1 (24)

andMz ≈ µN (25)

The magnetization is saturated, as all the moments are essentially perfectlylined up.

If we plug our results back into the expression for the free energy, we find

G = −µNH tanhµH

2kBT−NkBT ln 2

+N

2kBT

(1 +H tanh

µH

2kBT

)ln[1 +H tanh

µH

2kBT

]+N

2kBT

(1−H tanh

µH

2kBT

)ln[1−H tanh

µH

2kBT

](26)

Given our statistical definition of entropy and thermodynamic approachesbased on combinations of energy and entropy, we have ended up with an ex-pression for the magnetic Gibbs free energy of a set of magnetic momentscarried on spin-1/2 particles that interact with an externally imposed mag-netic field, but not with each other.

More on entropy

Recall our definition of entropy, (1). Suppose the system of interest consistsof n1 “type 1” components that can be in one of f1 states, n2 “type 2”components that can be in one of f2 states, . . .nm “type m” componentsthat can be in one of fm states. The total number of states available to thesystem as a whole is then fn1

1 fn22 · · · fnmm , which means that the entropy of

this system is

kB ln fn11 fn2

2 · · · fnmm = kBm∑i=1

ni ln fi (27)

In the case originally considered, there was only one type of component, amagnetic moment with spin 1/2. This moment could exist in one of twostates, so the entropy of a system consisting of N of these moments is

kB ln 2N = NkB ln 2 (28)

5

Additivity of the statistically defined entropy

Suppose we split our system into two subsystems as follows: We place n1a

of the n1 type 1 components into system a and the remaining n1b of thosecomponents into system b, and so on. This means that the total number ofstates available to system a is fn1a

1 fn2a2 · · · fnmam , while the number of states

available to system b is fn1b1 fn2b

2 · · · fnmbm . The entropies of the two systemsare given by

Sa = kBm∑i=1

nia ln fi (29)

Sb = kBm∑i=1

nib ln fi (30)

Notice that the total entropy of the combined system is

S = kBm∑i=1

ni ln fi

= kBm∑i=1

(nia + nib) ln fi

= Sa + Sb (31)

The entropy of the combined system is equal to the sum of the entropies ofthe two subsystems. In this case, at least, entropy is additive.

The effects of a constraint on the entropy

Suppose there is a constraint on the states that the system can occupy. Sup-pose, for example, that ths system, because it is energetically isolated fromthe rest of the universe, can only exist in states that have a total energy E.Suppose, furthermore, that the system is made up of N identical components(i.e. n1 = N and there is no other kind of component). Furthermore, letus assume that each component can exist in one of two states, and that theenergy of one of those states is zero, while the energy of the other state is ε.Thus, we are considering a system made up of a set of two-level components.

The total number of states available to this system is just equal to thenumber of ways of having m components in the energy equal ε level andN − m components in the energy equal to zero level, where mε = E. We

6

have solved this sort of problem before—in fact just a few pages ago—usingcombinatorial factors and Stirling’s approximation. We will now solve it in adifferent way. . . one that will allow us to look at much more complex systems.We start by noting that the number of states that the system can occupysubject to the above condition is just equal to the coefficient of emε in

[1 + eε] [1 + eε] · · · [1 + eε] = [1 + eε]N (32)

That is, the number of states that are available to the system is equal to thecoefficient of eE (since E = mε) in [1 + eε]N .

There are a number of tricks for finding that coefficient. That one thatwe will use starts with the replacement of eε with eiω. We are then interestedin the coefficient of eimω = eiEω/ε in [1 + eiω]N . We are going to obtain thatcoefficient by multiplying [1 + eiω]N by e−iE/ε, and then integrating over ωfrom −π to π. This picks out the coefficient of eimω, since, when k is aninteger ∫ π

−πeikωdω = 0 (33)

unless k = 0, when we have∫ π

−πeikωdω =

∫ π

−πdω

= 2π (34)

The quantity of interest is

1

2π

∫ π

−π

[1 + eiω

]Ne−Eω/εdω ==

1

2π

∫ π

−πeN ln[1+eiω]e−iEω/εdω (35)

Now, we will be interested in the case in which E/ε = O(N). If this is true,then both the real and the imaginary parts of the exponent will be veryrapidly varying as a function of the integration variable ω. The rapid phasevariations in the integrand will give rise to essentially complete cancellation,except where the phase is at an extremum, or, when

0 =d

dω

[N ln

[1 + eiω

]− iE

εω]

=Nieiω

1 + eiω− iE

ε(36)

7

or, whereeiω

1 + eiω=

E

Nε(37)

This extremum equation is satisfied only when ω is imaginary. Let ω = iλ.Then,

e−λ

1 + e−λ=

E

Nε(38)

Solving for λ:

e−λ =E/Nε

1− E/Nε(39)

or

λ = ln

[1− E/NεE/Nε

](40)

The integral is essentially equal to the integrand at this extremum point.Recall our discussion of Stirling’s approimation, as derived from the integralrepresentation of the factorial.

The entropy is, then, given by

S = kB ln[

1

2π

∫ π

−πeN ln[1+eiω]−iEω/εdω

]≈ kB ln

1

2π+ kB ln

[eN ln[1+e−λ]+Eλ/ε

]= kB ln

1

2π+ kB

[ln[1 + e−λ

]+E

ελ]

= kB ln1

2π+NkB ln

[1

1− E/Nε

]+ kB

E

εln

[1− E/NεE/Nε

](41)

If we are to carry through on our identification of this quantity withentropy we ought to identify the temperature, T , with ∂E/∂S, or

1

T=

∂S

∂E

=∂

∂E

[NkB ln

(1

1− E/Nε

)+ kB

E

εln

(1− E/NεE/Nε

)](42)

Carrying out the algebra, we end up with the result

1

T=kBε

ln

[1− E/NεE/Nε

](43)

8

This means that

eε/kBT =1− E/NεE/Nε

(44)

or, solving for E,

E =Nε

1 + eε/kBT(45)

Let’s use this result to construct the free energy of our system

F = E − TS (46)

we have

S = NkB ln

[1

1− E/Nε

]+ kB

E

εln

[1− E/NεE/Nε

]+ kB ln

1

2π(47)

The last term on the right hand side of Eq. (47) is negligible compared tothe first two and will be dropped from now on.

Substituting for E, we find

E − TS = −NkBT ln1 + eε/kBT

eε/kBT

= −NkBT ln[1 + e−ε/kBT

](48)

We have thus derived from our statistical definition of entropy and a math-ematical trick a quantity that contains all the thermodynamic informationfor a set of N two-level components at a temperature, T .

The next question is, of course, why we ought to believe the physicalpremises on which the above calculation was based.

9

The two-level system and the canonical ensem-

ble

Before we talk about the foundations of statistical mechanics—which will benecessary, just to establish a bit of confidence in the approach that we willbe taking—let us work a little on what we have obtained, to extract somenew results.

Extensivity of the entropy

In our calculation based on combinatorial factors and the use of Stirling’sformula we found for the entropy of a two-level system

S = NkB

ln

[1

1− E/Nε

]+

E

Nεln

[1− E/NεE/Nε

]

= Ns(E

N

)(49)

In the above expression, terms that do not grow as fast as N have beenneglected. This means that the number of configurations available to our setof N two-level components, subject to the restriction that the total energyis E, is given by

N (E) = eNs(E/N)/kB (50)

We expect this to he a general property of large systems. It is essential tothe development that follows.

Now, let’s use this result to find the probability that a given two-levelcomponent is in the excited state. If the component is excited, the amountof energy available to the remaining N−1 components is E− ε. The numberof states available to them is

e(N−1)s((E−ε)/(N−1))/kB (51)

We can write (N − 1)s((E − ε)/(N − 1)) as

(N − 1)s(

E

N − 1− ε

N − 1

)= (N − 1)s

(E

N − 1

)+ εs′

(E

N − 1

)+

ε2

N − 1s′′(

E

N − 1

)+ · · · (52)

10

The first term on the right hand side of (52) is O(N), the second term isO(1), and the remaining terms are O(1/N) and smaller. We will neglectthose remaining terms. Now,

S ′(

E

N − 1

)=

∂

∂E

[(N − 1)s

(E

N − 1

)]=

∂

∂ES(E,N − 1)

=1

TN−1

(53)

Assume that we can neglect the difference between the temperature, sodefined, for an N − 1 component system and the temperature for an Ncomponent system. Then, if the component of interest is in the excitedstate, there are

e(N−1)s(E/(N−1))/kBe−ε/kBT (54)

configurations available to the other N − 1 components. If the component ofinterest is in the ground state, there are clearly

e(N−1)s(E/(N−1))/kB (55)

configurations available.We assume that the system visits indiscriminately and with equal

weight all the configurations available to it. This means that the probabilityof finding the component of interest in its excited state is equal to the ratio

[# of configurations available when the component is in its excited state]

[# of excited state configurations]× [# of ground state configurations]

=e(N−1)s(E/(N−1))/kB−ε/kBT

e(N−1)s(E/(N−1))/kB−ε/kBT + e(N−1)s(E/(N−1))/kB

=e−ε/kBT

e−ε/kBT + 1(56)

Alternatively, the probability that the component is in its ground state is

1

e−ε/kBT + 1(57)

11

How the energy is shared in general

Let us consider two systems that share a total amount of energy E. Let’ssuppose that the first system has an energy E1 and the second system has anenergy E2. The total number of configurations available to the two systemswith this division of energy is

eS1(E1)/kBeS2(E−E1)/kB (58)

Here the first term in the product is the number of configurations availableto the first system, while the second term in the product is the number ofconfigurations available to the second. If we want to sum over all possibleways of sharing the energy, we have for the total number of configurations∫

eS1(E1)/kBeS2(E−E1)/kBdE1 (59)

Now, both exponents in the above expressions are very large. We will beable to show that the integral above is dominated by contributions fromthe region in the immediate vicinity of the value of E1 that maximizes theintegrand—or, equivalently, that maximizes

S1(E1) + S2(E − E1) (60)

To find the energy, E1, that does this, take the derivative

d

dE1

(S1(E1) + S2(E − E1)) (61)

and set it equal to zero. That is,

0 =d

dE1

(S1(E1) + S2(E − E1))

= S ′1(E1)− S ′2(E − E1)

=1

T1

− 1

T2

(62)

The dominant contributions to the configurations is from the immediatevicinity of the energies for which the temperatures of the two subsystemsare equal.

12

Let’s expand with respect to E1 in this region. Set E1 = E01 +∆E1. Then,

S1(E01 + ∆E1) + S2(E − E0

1 −∆E1)

= S1(E01) + S2(E − E0

1) + ∆E1

(S ′1(E0

1) + S2(E − E01))

+1

2

[S ′′1 (E0

1) + S ′′2 (E − E01)]

∆E21 + · · ·

= S1(E01) + S2(E − E0

1) +1

2

[S ′′1 (E0

1) + S ′′2 (E − E01)]

∆E21 + · · ·

The term proportional to ∆E1 vanishes because of the equality of the tem-peratures of the two systems. Now, by assumption

S1(E1) = N1s1

(E1

N1

)(63)

which means that

S ′1(E1) = s′1

(E1

N1

)(64)

S ′′1 (E1) =1

N1

s′′1

(E1

N1

)(65)

and so on for higher derivatives of the entropy of system 1 with respect tothe energy of that system. The same considerations apply to system 2. Thismeans that the higher one goes in the Taylor expansion of the entropy of thetwo systems sharing energy, the smaller the terms are, when N1 and N2 arelarge.

This means that our integral over energy in (59) is well-approximated by

E(S1(E01)+S2(E−E0

1))/kB∫ ∞−∞

e[S′′1 (E0

1)+S′′2 (E−E0

1)]∆E21/2kBd∆E1 (66)

The integral is well-approximated by a gaussian integral. To see what theintegrand looks like in a bit more detail, we’ll consider the coefficient of ∆E2

1 .Since

S ′1(E1) =1

T1

(67)

we have

S ′′1 (E1) =d

dE1

1

T1

= − 1

T 21

dT1/dE1

= − 1

T 21

1

dE1/dT1

(68)

13

If we make the connection E = U , with U the system’s internal energy, wehave

dE

dT=

dU

dT= C (69)

where C is the appropriately defined heat capacity. Thus,

S ′′1 (E1) = − 1

T 21

1

C1

(70)

The extremum is a maximum only if the above term is negative—so werequire C1 > 0, in line with previous arguments based on thermodynamicstability. Note that we can write C1 = N1c1, with c1 an intensive quantity.We have now

e[S1(E01)+S2(E−E0

1]/kB∫ ∞−∞

e− 1

2kBT2

[1

N1c1+ 1

N2c2

]∆E2

1d∆E1

= e[S1(E01)+S2(E−E0

1]/kB√√√√ 2kBT 2π[

1N1c1

+ 1N2c2

] (71)

The entropy of this combined system is

S1(E01) + S2(E − E0

1) +kB2

ln

[2πkBT

2

1N1c1

+ 1N2c2

](72)

The first two terms dominate the expression in (72), since S1 ∝ N1 andS2 ∝ N2. The third terms goes like ln (N1, N2). The entropy of the combinedsystem is essentially the sum of the entropies of the individual componentswith the energy shared between them in such a way tha the combined entropyis maximum.

We can see what the expected fluctuations are in there energy distribu-tion. We do this by calculating

∫∞−∞ e

− 12kBT2

[1

N1c1+ 1

N2c2

]∆E2

1∆E21d∆E1∫∞

−∞ e− 1

2kBT2

[1

N1c1+ 1

N2c2

]∆E2

1d∆E1

= kBT2

[1

1N1c1

+ 1N2c2

]

14

=N1N2kBT

2

N2

c1+ N1

c2

= O(N1, N2) (73)

The root mean square deviation in energy on either side from the maximizingamount is √

〈∆E21〉 = O(N

1/21 , N

1/22 ) (74)

This quantity is small compared to the total energy in either component,which is of order N1, N2. Thus, the fluctuations in the amount of energy ineither system are small compared to the total energy, by are not absolutelysmall. Furthermore, the extent of the fluctuations is controlled by the heatcapacities of the two systems.

Two systems sharing energy, one of which is much largerthan the other

Here, we are going to look at the energy-sharing that takes place betweentwo systems, one of which will play the role of a thermal reservoir. We willutilize the subscript B to label quantities associated with the large system.Quantities associated with the smaller system will be labeled with the sub-script S. The key relationship between the two systems is NB NS. Wewill assume for the entropy of the thermal bath

SB(EB, NB) = NBsB

(EBNB

)(75)

and similarly for the system S. Then, we have for the total entropy of thecombined system

eS/kB =∫eSS(ε)/kBeSB(E−ε)/kBdε (76)

The arguments of the previous section carry through in this case. Here, theintegral is dominated by the maximum, which occurs when

S ′S(ε) = S ′B(E − ε) (77)

or, whenTS = TB (78)

15

Because the bath has so many more degrees of freedom than the system,the width of the energy distribution is dominated by the heat capacity ofthe smaller system, S. This is because the second derivative with respect toenergy of the quantity Ns(E/N) goes as 1/N , and 1/NS 1/NB.

So far, this is just a relatively simple variation on the case discussedin more generality in the preceding sections. Let’s expand SB(E − ε) withrespect to ε. Then, the integral over all possible partitions of energy betweenthe system, S, and the bath, B, takes the form

eSB(E)/kB

∫eSS(ε)/kB−ε/kBTdε (79)

where we have removed the subscript from the temperature of the bath, TB,in the integral. Let’s remove the multiplicative factor eSB(E)/kB from (79).The integral is now∫

eSS(ε)/kB−ε/kBTdε→ eSS(ε0)/kB−ε0/kBT (80)

where ε0 is the energy at which the integrand is a maximum. When logs aretaken of both sides of the expression (80), one recovers an equality, to withinterms that are small compared to the leading ones by the ratio lnN/N . Now,ε0 is equal to the energy of the system, S, with the greatest likelihood. Thatis, the result of the integration over ε of

eSS(ε)/kB−ε/kBT (81)

is, to within multiplicative corrections that are negligible after the taking oflogarithms,

eSS(US)/kB−US/kBT (82)

That is, the result of the summation over all partitions of the total energybetween the system and the bath with which it is in contact, is e−F/kBT ,where F = U − TS is the Helmholtz free energy of this system. Let’s doa little checking. We’ll start with the derivative of this candidate for theHelmholtz free energy with respect to temperature:

∂

∂T(ε0 − TS(ε0)) = −S(ε0) +

(1− T ∂S

∂ε0

)= −S(ε0) (83)

The last equality follows from the fact that ∂S/∂ε|ε=ε0 = 1/T .

16

Now, the quantity eSS(ε)/kB is the sum over all states of the system subjectto the requirement that the total energy in each of those states is equal to ε.That is

eSS(ε)/kB =∑

states with energy ε(84)

Multiplying the expression on the left hand side of (84) by e−ε/kBT and thesumming (or integrating) over all values of ε is the same as multiplying thesum on the right hand side by e−ε/kBT and then performing an unrestrictedsum over all states of the system.

Putting all this together, we end up with the result

e−F/kBT =∑

all states of the systeme−Estate/kBT (85)

The sum on the right hand side of Eq. (85) is called the partition function.

17

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)

Modifications of the ideal gas

Suppose the atoms or molecules that make up the ideal gas are in an externalpotential. The, the total energy of the system of N of them is

N∑i=1

p212m

+N∑i=1

U(~ri) (113)

The partition function is given by

Z =∑states

e−Estate/kBT

=1

N !h3N

∫· · ·

∫d3Npd3Nre

− 1kBT

∑i

p2i

2m− 1

kBT

∑iU(~ri)

=1

N !h3N

[∫ ∞−∞

e− 1

kBTp2

2m

]3N [∫e−U(~r)

kBT d3r]N

(114)

and we have for the Helmholtz free energy

F = −kBT lnZ

= −kBT ln

1

h3N

(e

N

)N[∫ ∞−∞

e− 1

kBTp2

2m

]3N [∫e−U(~r)

kBT d3r]N

= −NkBT ln[h−3

e

N(2πmkBT )3/2

∫e−U(~r)

kBT

](115)

The effect of interactions between the molecules

Suppose, now, that we introduce pairwise interactions into the total energy.Dropping the external potential from the energy, we have

E =∑i

p2i2m

+∑i>j

u(~ri − ~rj)

≡∑i

p2i2m

+∑i>j

uij (116)

The partition function is now

Z =1

N !

1

h3N

∫· · ·

∫d3Nrd3npe

−∑

i

p2i

2m−∑

i>j

uijkBT (117)

26

A little inspection reveals that this partition function is far from trivial toevaluate. In fact, for general interaction potentials, u(~r), the exact evaluationof the 6N -fold integration is impossible, at least at present.

So, what can one do?Here is one trick that works if the interaction energies are small compared

to other energies. First, we write e−uij/kBT as follows:

e−uij/kBT = 1 +[e−uij/kBT − 1

]≡ 1 + fij (118)

Then,

e−∑

i>juij =

∏i>j

e−uij/kBT

=∏i>j

[1 + fij] (119)

If iij kBT , then e−uij/kBT ≈ 1, and fij is small. We can then try to expandthe product in (119) with respect to the small quantities fij:∏

i>j

[1 + fij] = 1 +∑i>j

fij + · · · (120)

Then, the partition function is given by

Z =1

h3N1

N !

∫· · ·

∫d3Npd3Nre

−∑

i

p2i

2mkBT∏i>j

[1 + fij]

≈ 1

h3N1

N !

∫· · ·

∫d3Npd3Nre

−∑

i

p2i

2mkBT

1 +∑i>j

fij

(121)

The “zeroth order” term in the expansion above is

1

h3N1

N !

∫· · ·

∫d3Npd3Nre

−∑

i

p2i

2mkBT =1

hNV N

(N/e)N[2πmkBT ]3N/2 (122)

which is the partition function of the noninteracting gas. The “first order”term is ∑

i>j

1

h3N

(e

N

)N[∫ ∞−∞

dpe− p2

2mkBT

]3N ∫d3Nrfij (123)

27

The result of the integration above will be the same for every index pair,i, j. This means that the first order term will be equal to the result of theintegrations for a given i and j, (say, i = 2, j = 1) multiplied by the numberof distinct pairs. This number, the number of ways of making unorderedpairs out of a set of N objects, is N(N − 1)/2. The first order term is, thus,

N(N − 1)

2

1

h3N

(e

N

)N[∫ ∞−∞

e− p2

2mkBT dp

]3N ∫f12d

3r1d3r2 ×

∫d3(N−2)r

=N(N − 1)

2

1

h3N

(e

N

)N

[2πmkBT ]V N−2∫f12d

3r1d3r2 (124)

Now, because f12 = e−u12/kBT − 1 = e−u(~r1−~r2) − 1 depends only on thedifference ~r1 − ~r2, we can write the integral

∫f12d

3r1d3r2 as follows:∫

d3r1

∫d3(r1 − r2)f12 = V

∫ [e− u(~r)

kBT − 1]d3r

≡ V α1(T ) (125)

Our first order term is equal to

N2

2

(e

h3N

)N

[2πmkBT ]3N/2 V N−1α1(T )

=(eV

h3N

)N

[2πmkBT ]3N/2N

2

N

Vα1(T )

(126)

Adding the zeroth and first order terms in our new expansion, we obtain

Z =(eV

h3N

)N

[2πmkBT ]3N/2

1 +N

2

N

Vα1(T )

(127)

This means that the free energy, to first order in our expansion, is given by

F = −kBT lnZ

= −kBT ln

[2πmkBT

h2

]3N/2 (eV

N

)N 1 +

N

2

N

Vα1(T )

= −NkBT ln

[2πmkBT

h2

]3N/2 (eV

N

)N

−kBT ln

1 +N

2

N

Vα1(T )

(128)

28

The next step is to perform an entirely unjustified expansion. We write

ln

1 +N

2

N

Vα1(T )

≈ N

2

N

Vα1(T ) (129)

This expansion is valid if (N/2)(N/V )α1(T ) 1. Clearly, as the systembecomes arbitrarily large (N →∞, V →∞, N/V → a constant) the secondterm in the logarithm that we have expanded overwhelms the first one. Let’signore that problem for the time being and continue. We now have

F = −NkBT ln

[2πmkBT

h2

]3/2eV

N

−NkBT N

2Vα1(T ) (130)

We now have in hand the first two terms in the virial expansion of theHelmholtz free energy. This expansion produces corrections to the idealgas laws. The corrections are small as long as the gas is dilute and thetemperature is not too low.

Modification of the equation of state

Let’s see how our result for the Helmholtz free energy alters the equation ofstate that expresses the pressure, P , as a function of N , V and T . We have

P = −(∂F

∂V

)T,N

=∂

∂V

NkBT ln

[2πmkBT

h2

]3/2eV

N

+NkBTN

2Vα1(T )

=

nkBT

V−(N

V

)2 kBT

2α1(T )

=N

VkBT

[1− N

V

α1(T )

2

](131)

Notice that as N/V goes to zero, the second term in brackets on the last lineof (131) will become negligible compared to the first one. Furthermore, ifkBT is sufficiently large and u(~r) is well-behaved,

α1(T ) =∫ [

e−u(~r)/kBT − 1]d3r

→ −∫ u(~r)

kBTd3r (132)

29

so defining a as follows:

a ≡ −1

2

∫u(~r)d3r (133)

we have

P =NkBT

V− a

(N

V

)2

(134)

If the molecules attract each other, then the interaction potential u(~r) willbe negative and a as defined in (133) will be positive. Rearranging the termsin (134) we end up with the equation of state(

P + a(N

V

)2)V = NkBT (135)

This is almost the van der Waals equation of state. If we were to invokethe notion of impenetrable hard cores, and reduce the effective volume, Vin (135) by V − Nb, where b is the volume displaced by each molecule, wewould have the van der Waals equation of state in its entirety.

30

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Quantum statistical mechanics

Symmetric and antisymmetric wave functions

An important tenet of quantum mechanics is that the wave function of Nidentical particles will be either symmetric or antisymmetric with respect tointerchange of two of them. For reasons having to do with relativity andlocality in quantum field theory the parity (symmetry or antisymmetry) ofthe wave functionis tied to the intrinsic spin of the particles that it controls.Particles with integral spin have symmetric wave functions—and obey Bose-Einstein statistics. Particles with half-integral spin have wave functions thatare antisymmetric with respect to particle interchange. Such particles obeyFermi-Dirac statistics.

If identical particles can occupy single-particle energy levels, then thestate of an N -particle system is completely specified by the number of parti-cles that occupy each level. The issue of which particle occupies which level ismeaningless in the context of a symmetric or anti-symmetric wave function.In this the key difference between Fermi-Dirac and Bose-Einstein particles isthat any number of Bose-Einstein particles can occupy a given energy level,while at most one Fermi-Dirac particle can occupy a single energy level. Thisis because interchange of the two particles occupying the same energy levelyields the same wave function, which is impossible if the wave function isantisymmetrized.

To see how statistics work out, let’s calculate the grand partition functionfor each type of particle

Bose-Einstein

Let the energy of the ith level be εi. The occupation number of the ith levelwill be denoted ni. The total energy of the system is∑

i

niεi (172)

and the total number of particles is∑i

ni (173)

Then,

e−βE+βµN = e−β∑

iniεi+βµ

∑ini

40

=∏i

e−βniεi+βµni

=∏i

e−βni(εi−µ) (174)

The sum over states is a sum over all possible values of the occupationsnumbers, ni. Since each occupation number can take any value from 0 to ∞,we have ∑

states

e−βEstate+βµNstate =∞∑

n1=0

∞∑n2=0

· · ·∏i

e−βni(εi−µ)

=∏i

∞∑ni=0

e−βni(εi−µ)

=∏i

1

1− e−β(εi−µ)(175)

Thus, the grand partition function, Ω, is given by

Ω = −kBT ln Q

= −kBT ln

[∏i

1

1− e−β(εi−µ)

]= kBT

∑i

ln(1− e−β(εi−µ)

)(176)

Now,

N = −∂Ω

∂µ

= − ∂

∂µkBT

∑i

ln ln(1− e−β(εi−µ)

)

=∑

i

e−β(εi−µ)

1− e−β(εi−µ)

=∑

i

1

eβ(εi−µ)(177)

This seems to indicate that the average occupation number per energy levelwill be given by

〈ni〉 =1

eβ(εi−µ) − 1(178)

which is, in fact, the case.

41

Fermi-Dirac

Here, the grand partition function is exactly as it is in the case of Bose-Einstein particles, except that the n′is take on the values of 0 and 1 only.The grand partition function is

Q =∑

n1=0,1

e−βn1(ε1−µ)∑

n2=0,1

e−βn2(ε2−µ) · · ·

=(1 + e−β(ε1−µ)

) (1 + e−β(ε2−µ)

)=

∏i

(1 + e−β(εi−µ)

)(179)

and

Ω = −kBT ln Q

= −kBT∑

i

ln(1 + e−β(εi−µ)

)(180)

and

N = −∂Ω

∂µ

=∑

i

e−β(εi−µ)

1 + e−β(εi−µ)

=∑

i

1

eβ(εi−µ) + 1(181)

which seems to indicate that

〈ni〉 =1

eβ(εi−µ) + 1(182)

The Fermi gas

Suppose that the fermions are electrons. In particular, assume that they arethe conduction electrons in a metal. According to the currently-acceptedversion of simple metals, the conduction electrons—electrons in the outer or-bitals of the constituent atoms—are “released” and propagate freely throughthe metal. The ionized atoms that they leave provide a positively charged,effectively inert, background. In the following, we will neglect the charges

42

carried by the electrons, which means that we will ignore the electromag-netic interactions between them, and between the electrons and the ionicbackground. In light of the importance of electromagnetic interactions atthe atomic level, it may seem a little unreasonable to expect our results tobe very useful, but they turn out to correspond remarkably well with ex-perimental observations in the case of a number of real metals. One of theimportant results of mid-twentieth-century solid state physics is a series ofcalculations that show why it is permissible to neglect electrostatic inter-actions for some purposes. Actually, what those calculations show is howsome of the essential features of a non-interacting collections of electrons arepreserved when the interactions are included.

It also turns out that we are provided by nature with another system towhich the calculations are directly relevant. That system is a fluid consistingof 3He, the two-proton-one-electron isotope of Helium. Because 3He containsan odd number of Fermions, its total “spin” (internal angular momentum)is half-integral. 3He is a fluid at very low temperatures. At extremely lowtemperatures, <∼ 10−3K, it is a superfluid. The calculations below do notapply to that exotic state of matter. However, when 3He is a normal fluid itdoes behave, in many respects, the way a non-interacting gas of Fermi-Diracparticles is predicted to behave. This behavior and more complex behaviorthat can be explained in terms of interactions, is described theoretically bya model developed by the celebrated Russian physicist Lev Landau. Themodel is known as the theory of normal Fermi liquids.

Fermions are assumed to be confined to a rectangular box. The eigenstateof a single particle in such a box is a standing wave. If the dimensions of thebox are (Lx, Ly, Lz) then the wave-vector of the standing wave will be of theform

(kx, ky, kz) =

(nxπ

Lx

,nyπ

Ly

,nzπ

Lz

)(183)

The energy of this eigenstate is

E(~k)

=h2

2m

(k2

x + k2y + k2

z

)=

h2

2m

(nxπ

Lx

)2

+

(nyπ

Ly

)2

+(

nzπ

Lz

)2 (184)

The sum over eigenstates is accomplished as follows:

43

We know that a unit change in one of the quantum numbers, say, nx,leads to an incremental change in the corresponding component of ~k, kx,equal to π/Lx. That is,

∆kx =π

Lx

(185)

Furthermore, ∆ky = π/Ly, ∆kz = π/Lz. Thus, because each of the factorsin brackets below is equjal to one, the product

LxLyLz

π3(186)

is equal to one. We can turn the sum over a sufficiently smooth function of~k into an integral with the use of the above identity

∑f(~k)

=∑

f(~k) LxLyLz

π3∆kx∆ky∆kz

=V

π3

∑f(~k)

∆kx, ∆ky∆kz (187)

Since each ∆ki becomes infinitesimal as Lx, Ly, Lz → ∞ we can replace thesum above by an integral, i. e.

∑f(~k)→ V

π3

∫f(~k)dkxdkydkz (188)

The integrals are over positive kx, ky, kz. If the function has the propertyf(kx, ky, kz) = f(−kx, ky, kz) = · · ·, then we can replace the integral aboveby an integral over all k’s, so that

∑f(~k)

→ V

(2π)3

∫f(~k)d3k (189)

Finally, if we replace kx by px/h, and similarly for ky and kz,∑f(~k)

=v

(2π)3

∫f(~k)dpxdpydpz

=V

(2π)3

∫f

(~p

h

)dpxdpydpz (190)

In the case of Fermions,

Ω = −kBT ln Q

44

= −kBT∑

i

ln(1 + e−β(εi−µ)

)= −kBT

∑~k

ln(1 + e−h2k2/2mkBT+µ/kBT

)

= −kBTV

(2πh)3

∫ln(1 + e−p2/2mkBT+µ/kBT

)d3p

= −kBTV

(2πh)34π∫ ∞

0ln

(1 + e

− p2

2mkBT+ µ

kBT

)p2dp (191)

The number of particles is given by

N = −∂Ω

∂µ

=∂

∂µ

[kBT

4πV

(2π)3

∫ ∞

0ln

(1 + e

− p2

2mkBT+ µ

kBT

)p2dp

]

=V

2π2h3

∫ ∞

0

e− p2

2mkBT+ µ

kBT

1 + e− p2

2mkBT+ µ

kBT

p2dp

=V

2π2h3

=V

2π2h3

∫ ∞

0

p2dp

eβ[ε(p)−µ] + 1(192)

Here, ε(p) = p2/2m, and, as always, β = 1/kBT .

Occupation of a single energy level

Using an argument like the one that led to a result for the probability that agiven two-state atom is in the excited state, we arrive at the following resultfor the expectation, or average, value of the occupation number of the ith

single particle state—in the case of Fermions.

〈ni〉 =1

eβ(εi−µ) + 1(193)

This result arises naturally when we consider the way in which averages aretaken in the grand canonical ensemble.

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The extra factor of two

We have neglected an important property of electrons; they have two spin po-larization states—spin up and spin down. That is, each spatial eigenstate—each ~k—corresponds to two electron eigenstates, one for each spin polariza-tion. That means that we really ought to have for the formula for the numberof particles

N = 2

[V

2π2h3

∫ ∞

0

p2dp

eβ[ε(p)−µ] + 1

]

=V

π2h3

∫ ∞

0

p2dp

eβ[ε(p)−µ] + 1(194)

(195)

The limit T = 0

If T = 0, then β = ∞, and the function eβ(ε−µ) takes on one of two values: 0when ε(p) < µ and ∞ when ε(p) > µ. This means that, in the limit of zerotemperature,

1

eβ[ε(p)−µ] + 1=

1 ε(p) < µ0 ε(p) > µ

(196)

and our expression for N has the form

N =V

π2h3

∫ε(p)<µ

p2dp

=V

π2h3

∫ √2mµ

0p2dp

=V

3π2h3 [2mµ]3/2 (197)

The integral over p is confined to the interior of a sphere in p-space ofradius [2mµ]1/2. This radius is called the Fermi momentum, pF . Theenergy, p2

F /2m = µ, is called the Fermi energy, εF .

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