Thermal Physics Lecture 7

7/30/2019 Thermal Physics Lecture 7

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Physics 301 27-Sep-2002 7-1

States of a Particle in a Box.

In order to count states, we will use quantum mechanics to ensure that we havediscrete states and energy levels. Lets consider a single particle which is confined to a

cubical box of dimensions L L L. You might think that this is artificial and wonderhow the physics could depend on the size of a box a particle is in? It is artificial and itsa trick to make the math easier. Once a box is big enough, the physics doesnt depend onthe size of the box, and the physics we deduce must not depend in any critical way on thebox size when we take the limit of a very big box. (Of course, the volume of a system isone of the extensive parameters that describes the system and its OK for the volume toenter in a manner like it does in the ideal gas law!) In what follows, well ignore rotationaland internal energy of the particles and drop the tran subscript.

As you probably know, particles are described by wave functionsin quantum mechanics. The de Broglie relation between wavelength

() and momentum (P) is P = h/, where h is Plancks constant.The wave function for a particle in a box must be zero at the walls ofthe box (otherwise the particle might be found right at the wall). Inone dimension, suitable wave functions are (x) sin(nxx/L) where0 x L and nx is an integer. This amounts to fitting an integernumber of half wavelengths into the box. (If you recall the Bohr modelof the atom, the idea there is to fit an integral number of wavelengths inthe electrons orbit.) The momentum is Px = nxh/2L = nxh/L.The sign on the momentum indicates that the wave function isa standing wave that is a superposition of travelling waves moving inboth directions. The first three wave functions are shown in the figure. In three dimensions,we have

(x,y,z) sin(nxx/L)sin(nyy/L)sin(nzy/L) ,which corresponds to fitting standing waves in all three directions in the box. The mo-mentum is

P =h

L(nx,ny,nz) .

The energy is

E =P2

2m=

2h2

2mL2 n2x + n

2y + n

2z

.

Now we are getting close to being able to count states. Consider a three dimensionalcoordinate system with axes nx, ny, nz (regarded as continuous variables). There is astate at each lattice point (nx, ny, nz integers) in the octant where all are non-negative.Because h is so small, and also because we usually deal with a large number of particles,we will be concerned with energies where the quantum numbers (the ns) are large. How

Copyright c 2002, Princeton University Physics Department, Edward J. Groth


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Physics 301 27-Sep-2002 7-2

many states are there with energy < E? Answer:

N(< E) =1

8

4

3

E

2h2/2mL2

3.

This is just the volume of an octant of a sphere with radius given by the square root above.Its the number of states because each state (lattice point) occupies unit volume. For alarge number of states, we dont care about the small errors made at the surface of thesphere. The number of states with energy less than E is the integral of the density ofstates, n(E),

N(< E) =

E0

n(E) dE ,

where E is the dummy variable of integration. Differentiate both sides with respect to Eusing the previous result for N(< E),

n(E) = 2

E

2mL

2h

3.

Recall that when we discussed the Maxwell distribution, we concluded that the densityof states had to be proportional to

E in order to give the Maxwell distribution. Sure

enough thats what we get. All the other factors get absorbed into the overall normalizationconstant.

It will be instructive to work out a numerical value for the number of states fora typical case. So lets suppose that E = 3kT /2 where T = 273 K and the volumeL3 = 22 103 cm3. That is, we consider an energy and volume corresponding to theaverage energy and molar volume of a gas at standard temperature and pressure. For mwell use a mass corresponding to an N2 molecule. The result is about 4 1030 states,more than a million times Avogadros number.

We have been working out the states for one particle in a box. If we have morethan one particle in the box, and they are non-interacting, then the same set of states isavailable to each particle. It the particles are weakly interacting, then these states are afirst approximation to the actual states and we will usually just ignore the interactions

when it comes to counting states. With this approximation, and with a mole of particlesin the box, weve found that less than one in a million of the available states are occupied.



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Physics 301 27-Sep-2002 7-3

Partition Function for a Single Particle in a Box

We can use the same states weve just discussed to evaluate the partition function fora single particle in a box. We have

Z() =

nx,ny,nz

exp

2h2

2mL2

n2x + n

2y + n

2z

.

We will make a negligibly small error by converting the sums to integrals,

Z() =

0

dnx

0

dny

0

dnz exp

2h2

2mL2

n2x + n

2y + n

2z

,

=

2m

h 3

L3

0

dx

0

dy

0

dz exp(x2 y2 z2) ,

(rescaling variables)

=1

84

2m

h

3L3

0

dr r2 er2 ,

(changing to spherical coordinates and integrating over angles)

=1

84

2m

h

3V

1

2

3

2

,

=V

(2h2/m)3/2((3/2) =

/2) ,

= nQV .

The volume of the system is V = L3 and the quantity that occurs in the last line, nQhas the dimensions of an inverse volume or a number per unit volume. m is the squareof a typical momentum. So h2/m 2, and the volume associated with nQ is roughlya cube of the de Broglie wavelength. This is roughly the smallest volume in which youcan confine the particle (given its energy) and still be consistent with the uncertaintyprinciple. K&K call nQ the quantum concentration. A concentration is just a number perunit volume, and nQ can be thought of as the concentration that separates the classical

(lower concentrations) and quantum (higher concentrations) domains. For a typical gasat STP, the actual concentration n = N/V is much less than the quantum concentration(by the same factor as the ratio of the number of states to the number of molecules wecalculated earlier), so the gas can be treated classically.



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Physics 301 27-Sep-2002 7-4

Partition Function for N Particles in a Box

If we have N non-interacting particles in our box, all with the same mass, then (seethe homework) the partition function for the composite system is just the product of the

partition functions for the individual systems,ZN() = Z1()

N wrong!

where ZN is the N-particle partition function and Z1 is the 1-particle partition functioncalculated in the previous section.

Why is this wrong? Recall that the partition function is the sum of Boltzmannfactors over all the states of the composite system. Writing ZN as a product includesterms corresponding to molecule A with energy Ea and molecule B with energy Eb andvice versa: molecule A with Eb and molecule B with Ea. However, these are not differentcomposite states if molecules A and B are indistinguishable! The product overcounts the

composite states. Any given Boltzmann factor appears in the sum roughly N! times morethan it should because there are roughly N! permutations of the molecules among thesingle particle states that give the same composite state.

Why roughly? Answer, if there are two or more particles in the same single particlestate, then the correction for indistinguishability (what a word!) is not required. However,weve already seen that for typical low density gasses, less than one in a million singleparticle states will be occupied, so its quite safe to ignore occupancies greater than 1. (Ifthis becomes a bad approximation, other quantum effects enter as well, so we need to dosomething different, anyway!)

To correct the product, we just divide by N!,

ZN() =1

N!ZN1 =

1

N!(nQV)

N .

To find the energy, we use

U = 2

log ZN ,

= 2

( log N! + Nlog nQ + Nlog V) ,

= 2

(Nlog nQ) ,

(derivatives of N! and log V give 0)

= 2

Nlog

m

2h2

3/2,

=3

2N ,



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Physics 301 27-Sep-2002 7-5

which expresses the energy of an ideal gas in terms of the temperature. Weve obtained thisresult before, using the Maxwell distribution. Note that our correction for overcounting ofthe microstates does not appear in the result.

In lecture 6, we noted that

p =

U

V

,N

,

and we remarked that keeping the entropy constant while changing the volume of a systemmeans keeping the probability of each microstate constant. The average energy is

E =s

Es P(Es) ,

and keeping the probabilities of the microstates constant means that P(Es) doesnt change.Thus, changing the volume at constant entropy changes the energy through changes inenergies of the individual states. For each single particle state,

Es 1L2

V2/3 ,

which meansdU

U=

dEsEs

= 23

dV

V,

all at constant . Then the pressure is

p =

U

V

,N

=2

3

U

V.

Again, this is a result weve seen before.



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Physics 301 27-Sep-2002 7-6

Helmholtz Free Energy

Recall the expression for the conservation of energy,

dU = d p dV ,where we have omitted the chemical potential term, since we wont be contemplatingchanging the number of particles at the moment.

If we have a system whose temperature is fixed by thermal contact with a heat bath,it is convenient to eliminate the differential in the entropy in favor of a differential in thetemperature. For this we use a Legendre transformationexactly the same kind of trans-formation used in classical mechanics to go from the Lagrangian, a function of coordinatesand velocities, to the Hamiltonian, a function of coordinates and momenta.

Define the Helmholtz free energy by

F = U .

Not all authors use the symbol F for this quantityI believe some use A and there maybe others. In any case,

dF = dU d d = d p dV d d = dp dV .

If a system is placed in contact with a heat bath and its volume is fixed, then its freeenergy is an extremum. As it turns out, the extremum is a minimum. To show this, weshow that when the entropy of the system plus reservoir is a maximum (so equilibrium isestablished), the free energy is a minimum.

= r(U Us) + s(Us) ,= r(U) Us(r/U) + + s(Us) ,= r(U) Us/ + s(Us) ,= r(U) (Us s(Us)) / ,= r(U) Fs/ .

In the above, the subscripts r and s refer to the reservoir and system and U is the fixedtotal energy shared by the reservoir and system. Note that unlike our derivation of theBoltzmann factor, the system here need not be so small that it can be considered to bein a single stateit can be a macroscopic composite system. However, it should be muchsmaller than the reservoir so that Us U. Also, the partial derivatives above occur atfixed volume and particle number. Since r(U) is just a number, and is fixed, maximizing requires minimizing F.



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Physics 301 27-Sep-2002 7-7

From dF = dp dV, we see

=

F

V

and p =

F

V

.

Substituting F = U in the right equation above,

p =

U

V

+

V

.

This shows that at fixed temperature, if a system can lower its energy by expanding, thenit generates a force, pressure, that will create an expansion. This is probably intuitive,since we are used to the idea that the equilibrium state is a minimum energy state. If thesystem can increase its entropy (at fixed temperature) by expanding, this too, generates aforce to create an expansion.

Note that

V

= 2F

V =

2F

V=

p

V

.

The outer equality in this line is called a Maxwell relation. These occur often in thermody-namics and result from the fact that many thermodynamic parameters are first derivativesof the same thermodynamic potential, such as the free energy in this case.

The Free Energy and the Partition Function

Consider F = U and = (F/)V. Putting these together, we have

U = F + ,

= F

F

V

,

= 2 (F/)

.

Recall the expression for energy in terms of the partition function

U = 2 log Z

.

Comparing with the above, we see

F

= log Z+ C ,



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Physics 301 27-Sep-2002 7-8

where C is a constant independent of. In fact, the constant must be zero in order to givethe correct entropy as 0. If is sufficiently small, only the lowest energy (E0) stateenters the partition function. If it occurs g0 different ways, then log Z log g0 E0/and = F/ (log g0 E0 C)/ = log g0 C. So C = 0 in order that theentropy have the correct zero point. Then

F = log Z or Z = eF/ .

Remembering that the Boltzmann factor is normalized by the partition function to yielda probability, we have

P(Es) =eEs/

Z= e(F Es)/ .

Just for fun, lets apply some of these results using the partition function for the ideal

gas we derived earlier.

F = log Z ,= log((nQV)N/N!) ,= (Nlog nQ + Nlog V Nlog N + N) (Stirlings approx.) ,= Nlog

(m /2h2)3/2(V /N)

N .

With p = F/V, we havep = N/V ,

the ideal gas law again. For the entropy, = F/, = Nlog(nQV /N) + (3/2)N + N ,

= N

log

nQn

+5

2

,

where n = N/V is the concentration. This called the Sackur-Tetrode formula. Note thatif one considers the change in entropy between two states of an ideal gas,

f

i =

3

2

Nlogf

i

+ NlogVf

Vi

,

a classical result which doesnt contain Plancks constant. However, to set the zero pointand get an absolute entropy, Plancks constant does enter since it determines the spac-ing between states and their total number. The overcounting correction does not makeany difference in the pressure above, but it does enter the entropyas might have beenexpected. A final note is that these expressions for an ideal gas do not apply in the limit 0. (Why?)


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Thermal Physics Lecture 7