kubo solutions

8/10/2019 Kubo Solutions

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Jonathan McFaddenSpring 2010

PHYS 7400Kubo Solutions

Chapter 1 - Solutions

Problem 1-1): Explain Daltons law for an ideal gas mixture on the basis of elementary kinetictheory of gas molecules.

Solution 1-1): From Bernoullis Formula, the pressure of an ideal gas can be given as p = 23 n, where n is the

average number of molecules per unit volume and is the average kinetic energy per molecule. Now, Daltons Lawstates that PT =

ipi. Applying Bernoullis Formula to Daltons Law gives

PT =i=1

pi=i=1

2

3nii (1.1.1)

whereni is the average number of molecules of the ith kind per unit volume and i is the average kinetic energyper molecule of the ith kind. Noting these definitions for ni and i, we can express them as

ni=Ni

V

i=

TiNi

whereNi is the total number of molecules of the ith kind in the system, Vis the volume of the system, and Ti isthe total kinetic energy of all of the molecules of the ith kind. Applying these expressions for ni and i to theexpression in (1.1.1) gives

PT =i=1

2

3nii=

2

3

i=1

NiV

TiNi

=

2

3

i=1

TiV

= 2

3V

i=1

Ti (1.1.2)

Now, the total kinetic energy of all of the molecules of the ith kind, Ti, can be expressed as

Ti= j=1

p2

ij2mi

wherepij is the momentum of thej th molecule of theith kind of molecules and mi is the mass of the ith kind ofmolecule. Applying this expression for Ti to the expression in (1.1.2) gives

PT = 2

3V

i=1

Ti= 2

3V

i=1

j=1

p2ij2mi

(1.1.3)

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Problem 1-2): Imagine a fictitious surface element in an ideal gas. Supposing that momentumtransfer takes place through the surface element due to penetration by gas molecules, find theformula for calculating the pressure which both sides of the surface element exert upon each other(Lorentzs Method). Assume that the gas molecules obey a Maxwellian velocity distribution.

Solution 1-2): Using Daltons Law, the pressure exerted across the surface can be given as PT

=p1

+p2

, where p1is the pressure on one side, Side 1, of the membrane and p2 is the pressure on the other side, Side 2, of the surface.

From Bernoullis Principle, the pressure of an ideal gas can be given as p = 23 n, where n is the number density1 of

the gas in question and is the average kinetic energy per molecule. Adapting Bernoullis Principle for thissituation and applying that adaptation to Daltons Law, as it pertains to this system, gives

PT =p1+p2 =2

3(n11+ n22) (1.2.1)

wheren1 and 1, and n2 and 2 are the number densities and average kinetic energies of the gases on sides Side 1and Side 2, respectively. Choosing the surface to lie in the yz-plane, the only velocities in thex direction matter,therefore, 1 and 2 can both be rewritten as 1 =

12

m1vx21 and 2 =

12

m2vx22, respectively. This allows the

expression in (1.2.1) to be rewritten as

PT =2

3

n1

1

2m1vx

21

+ n2

1

2m2vx

22

Furthermore, this choice of surface orientation implies that there is only one degree of freedom of consequence inthis system, therefore the re-expression of (1.2.1) above can be simplified to

PT =n1m1vx21+ n2m2vx

22 (1.2.2)

Additionally, the gas is the same on both sides of the surface, thus m1 = m2 = m. Now, the distribution ofvelocities is Maxwellian on both sides so the distribution function for the velocities, f(vx, vy, vz), can be given as

f(vx, vy, vz) =

m

2kBT

32

e m2kBT(v

2x+v

2y+v

2z)

Combining the equivalence of masses with the equivalence of the velocity distributions allows it to be said thatvx

21 = vx

22 = vx

2. Using this fact, and the Maxwellian distribution of the velocities, the expression forvx2 can be

rewritten as

vx2 =

R3

d3v f(vx, vy, vz) v2x=

m

2kBT

32

v2x dvx

dvy

dvze m2kBT(v

2x+v

2y+v

2z) (1.2.3)

First, it should be noted that the interation in (1.2.3) is over all the possibleparticle velocities, not over the

velocity parameter itself. Had the integration been over the velocity parameter itself, the integral over thedistribution ofpossiblevelocities would be meaningless and would lead to a dimensionally inconsistent result whenapplied to (1.2.2) or any of its successive forms. Second, it should be noted that, technically, this should be adiscrete sum, as consideration of the number of particles or any dependent quantity (like the set of all possibleparticle velocities) is necessarily discrete. This would imply that the expression in (1.2.3)The integral in (1.2.3) is over all velocities in the x-direction because both positive and negative velocities willeffect the pressure experienced across the surface. The integrals over vy andvz are overall velocities for the same

1The phrase, number density, is equivalent to the statement, average number of molecules per unit volume. (See Problem 1 fromChapter 1)

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reason; however, their inclusion does not change the result when the integral is evaluated because they have noweighting factor in the integral. It will be shown that the results be unchanged for vx

2 given as

vx2 =

dvx f(vx)v2x=

m

2kB

T12

dvx v2x e

m2kBTv2x (1.2.4)

where the exponent of the m2kBTgoes from 3 to 1 in order to reflect the fact that the integralhere has only one

degree of freedom (compared to the three degrees of freedom in other expression for vx2). To continue,

m1 = m2 = m and vx21 = vx

22 = vx

2 are applied to the expression in (1.2.2) to yield

PT =n1m1vx21+ n2m2vx

22

=n1mvx2 + n2mvx

2

=mvx2 (n1+ n2)

Noting that n1+ n2 = n, wheren is the total number density in the system, allows this above re-expression of(1.2.2) above to be rewritten as

PT =mvx2 (n1+ n2) = mnvx

2

Finally, the expressions for vx2 from (1.2.3) and (1.2.4) are applied to this expression to give

PT =mnvx2 =mn

m

2kBT

32

v2x dvx

dvy

dvze m2kBT(v

2x+v

2y+v

2z) (1.2.5)

and

PT =mnvx2 =mn dvx f(vx)v 2x= m2kBT12

dvx v2x e

m

2kBTv2x (1.2.6)

Using Mathematica to evaluate the integrals in (1.2.5) and (1.2.6) gives the final result as

PT =nkBT (1.2.7)

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Problem 1-3): A rarefied gas is contained in a vessel of volume V at pressure p. Supposing that thedistribution of molecules of the gas is Maxwellian, calculate the rate at which the gas flows out ofthe vessel into a vacuum through a small hole (of area A). Taking the wall of the hole as the y-zplane, find the velocity distribution in the x-direction of the gas molecules moving out of the hole.

Solution 1-3): The number of particles leaving the vessel at any point, N, in time depends on the area of the holein the vessel, A, the positional distribution of the particles (over unit volume), n (x) at that point in time, and theposition of each of the particles, x, at that point in time. Thus, we sum over all of the possible positions of theparticles to obtain the expression for N as

N=

n (x) A x (1.3.1)

where the sum here is over all positions, (x,y,z), contained within the vessel. Now, the interest is in the rate atwhich particles are leaving the vessel, so an expression fordN/dt is required. This is accomplished by convertingthe positional distribution, n (x), in (1.3.1) to a velocity distribution, n (v), and the position, x, to a velocity,v, togive

dNdt =n (v) A v (1.3.2)

where the sum is over all possible positive velocities, (vx, vy, vz); the reason for omitting the negative velocities willbe discussed later. To go from (1.3.1) to (1.3.2), the expression inside the sum in (1.3.1) is differentiated withrespect to t to give

d

dt[n (x) A x] = A

x

d n (x)

dt + n (x)

dx

dt

Noting thatn (x) represents the positional distribution and is therefore time-independent, it is clear that dn(x)dt = 0.

Thus, the expression for ddt[n (x) A x] becomes

ddt

[n (x) A x] = An (x)dxdt

=A n (x) v

Since dn(x)dt = 0 and dn(v)dt = 0 (for the same reason stated above) and since the positional distribution at anytime

can be made dependent on the velocities of the particles, it can be said that n (x) = n (v). Thus

dN

dt =

d

dt

n (x) A x

=d

dt[n (x) A x]

=A

xd n (x)

dt + n (x)

dx

dt

= A

n (x) v

=An (v) v (1.3.3)The importance ofn (x) = n (v) can now be seen, as soon as the time derivative moved inside the sum, the index ofsummation changes from all possible positions in the container to all possible positive velocities. Thus, the sum in(1.3.3) is over all possible velocities, as in (1.3.2). To proceed from here, it is assumed that the number of particlesis large enough that the discrete sum in (1.3.3) can be converted into a continuous integral to give

dN

dt =

n (v) A v dN

dt =

dv n (v) A v (1.3.4)

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Taking the plane to lie in the y-z plane and the velocity distribution to be Maxwellian, the integral in (1.3.4)becomes

dN

dt = dv n (v) A v=

m

2kB

T32

0

vx dvx

dvy

dvze m2kBT(v

2x+v

2y+v

2z) (1.3.5)

or equivalently, by omitting the integrals for vy andvz as discussed in Problem 1-2,

dN

dt =

dv n (v) A v=

m

2kBT

12

0

dvx vx e m2kBTv

2x (1.3.6)

Here, the integrals for the velocity are from 0 because the presence of the vacuum outside the vessel ensuresthat the velocities of all of the particles passing through the hole will be positive. That is to say that the flux ofparticles through the hole will be directed entirely outward, as no particles will be entering the vessel from thevacuum. Finally, the integrals in (1.3.5) and (1.3.6) are evaluated using Mathematicato yield

dNdt

=AnkBT

2m (1.3.7)

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Problem 1-4): There is a furnace containing a gas at a high temperature. Through a small windowof the furnace, one observes, using a spectrometer, a spectral line of the gas molecules. The widthof one observed spectral line is broadened (this is called Dopler broadening). Show that the relationbetween the spectral line intenstity, I, and the wavelength, , is given by the following formula:

I() exp

mc2 ( 0)2220kBT

Here, T is the temperature of the furnace, c is the speed of light, m is the mass of a molecule, and 0is the wavelength of the spectral line when the molecule is at rest

Solution 1-4): The doppler redshift parameter for light can be expressed as

d= 0

0

Multiplying both sides of this expression byc yields

c d= c 0

0

Sinced is a ratio such that d [0, 1), the expression c d is defined to be c d= v, wherev is some positive velocitywhich is a fraction ofc. Specifically, this velocity v is the velocity with which the particles in the furnace aremoving. Therefore, the velocity v can be given as

v= c d= c 0

0(1.4.1)

Now, the intensity of the light is proportional to the number of light emitting particles, n, and the amount of light

each particle emits, which is in turn proportional to the velocity of each particle. For a system of many particleslike this one, the dependence of the intensity of the light on the velocity of each particle becomes a dependence onthe distribution of the velocities of all of the particles in the distribution, f(v). Again, the assumptions is madethat the number of particles is sufficient to allow for treatment as a continuous distribution. Thus

I(v) n f(v) (1.4.2)

wherev is the velocity in the direction of the observer. Due to the hight temperature environment, the particles areassumed to have a Maxwellian distribution, thus the expression in (1.4.2) becomes

I(v)

mn2

2kT1/2

e m2kTv

2

(1.4.3)

Applying the expression and the expression in (1.4.1) to this expression gives

I()

mn2

2kT

1/2e m2kT

c00

2(1.4.4)

To obtain the final answer, it is noted that

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e m2kT

c00

2=e

mc2(0)22kT20

and that, as far as functional proportionality goes, multiplicative constants are inconsequential, the expression in

(1.4.4) becomes

I()

mn2

2kT

1/2e m2kT

c00

2 e

mc2(0)2

2kT20 (1.4.5)

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Problem 1-5): A mass point with mass m moves within the range 0 x l and is reflected by wallsat x= 0 and x= l.

(i) Illustrate the trajectory of this mass point in the phase space (x, p)

(ii) Find the volume of the phase space, 0(E), with energy smaller than E

(iii) Show that 0(E) is kept constant when the wall at x= l is moved slowly (adiabatic invariance).

(iv) Going over to quantum mechanics, find the number, 0(E), of quantum states with energybelow E and compare it to 0(E).

Solution 1-5): The trajectory of the mass point in the phase space (x, p) can be illustrated as

Figure 1.5-1 : Phase space (x, p)

The system has one degree of freedom; therefore, 0(E) is just an area in the 2-dimensional phase spacerepresenting the motion of the mass. Therefore, 0(E) (or the area enclosed in Figure 1.5-1 ) can be expressed as

0(E) = l (p (p)) = l (2p) = 2lp (1.5.1)

The motion of the mass implies that its Hamiltonian is equivalent to the Hamiltonian of a free particle, therefore

H = p2

2m=E

Solving this for p gives p =

2mE. Applying this solution to the expression in (1.5.1) gives

0(E) = 2lp = 2l

2mE (1.5.2)

To show that0(E) = 0 when the wall at x = l is moved slowly, it is noted that the decrease in the momentum ofthe mass,p, during the collision with the moving wall can be expressed using the fraction of change in length ofthe masss allowed coordinate space to the original length of the masss allowed coordinate space. This can beexpressed at

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p = p ll

(1.5.3)

Noting that the expression from (1.5.2) implies 0(E) = (2lp) = 2pl+ 2lp, the expression in (1.5.3) is applied

to yield

0(E) = (2lp) = 2pl+ 2lp

= 2pl+ 2l

p l

l

= 2pl 2pl = 0

In the quantum case, this is just the infinite square well. Thus the energy for this initial system (before the wallstarts to move) can be given as

En=n222

2ml2

Solving this for n gives

n=

2Enml2

22 =

a

2Enm

(1.5.4)

The expression in Eqn (1.20a) in Kubo gives

0(E) =

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Problem 1-6): Find the number of quantum states for a particle contained in a cubic box with edgelength l, and compare it with the volume in classical phase space. Obtain also the the density of thestates.

Solution 1-6): This can be interpreted as a three dimensional infinite square well; therefore, the energy levels ofthe system can be expressed as

Enx,ny,nz = 22

2ml2

n2x+ n2y+ n

2z

(1.6.1)

From Eqn (1.20a) in Kubo, 0(E) can be expressed as

0(E) =

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must be evaluated. To evaluate this integral definitely, proper bounds must be applied. Since the particle must belocated within the box, qx, qy, and qz can all be integrated from 0 l. This allows the integral for 0(E) to berewritten as

0(E) = dpx dpydpz l

0 dqx l

0 dqy l

0 dqz

=l3

dpx dpydpz =V

dpx dpydpz (1.6.5)

by again noting thatl3 =V. To find bounds for the integration ofpx, py, and pz, it is noted that the Hamiltonianfor the motion of the particle inside the cube can be written as

H = p2

2m=

p2x2m

+p2y2m

+ p2z2m

=E0

whereE0 is some energy that satisfies the inequality 0 E0 E. This expression for the particles Hamiltonianimplies that

p2x+p2y+p

2z =p p= 2mE0

Applying the inequality 0 E0 E to this expression implies

0 p2x+p2y+p2z 2mE

This implies that the volume of interest in momentum space (which is a three dimensional subspace of the six

dimensional phase space for the system) is a volume of radius pr0 =

p2x+p2y+ p

2z =

2mE. Therefore, the

integral in (1.6.5) is converted into the spherical coordinate representation of momentum, (pr, p, p), to give

0(E) = V

dpx dpydpz = V

p2rsinpdprdpdp (1.6.6)

This conversion to the spherical momentum coordinates allows the bounds of integration for pr, p, and p to be

respectively given as 0 2mE, 0 and 0 2. Applying these integration bounds to the integral in (1.6.6)gives

0(E) = V

p2rsinpdprdpdp= V

2mE0

p2rdpr

0

sinpdp

20

dp

Evaluating this result with Mathematica gives

0(E) = V

2mE0

p2rdpr

0

sinpdp

20

dp=4

3V (2mE)

3/2(1.6.7)

To compare the number of states in the quantum case to the number of states in the classical case, it is noted that,for large E, 0(E) 0(E) /h3, where h3 is used instead ofh because the system has three degrees of freedominstead of just one. Applying the expressions from (1.6.7) and (1.6.4) to 0(E) 0(E) /h3 gives

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0(E) 0(E)h3

V

623(2mE)

3/2

1

2333 4

3V (2mE)

3/2

V

623(2mE)

3/2

V623

(2mE)3/2

which implies, as expected, that the number of states for the quantum and classical systems is equivalent in thelargeElimit. Finally, to obtain the density of states for the quantum case, the expression for 0(E) in (1.6.4) isdifferentiated with respect toE, using Mathematica, to give

(E) =d0(E)

dE =

d

dE

V

623(2mE)

3/2

=

V

23

m2E

2

as the expression for the density of states, (E).

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Problem 1-7): What does a surface of constant energy look like in the phase space of an oscillator offrequency ? Find the volume (E) in the phase space with energy below E. Then find the numberof quantum states, 0(E) with energy below E for this oscillator, and show that when Eis large wehave

0(E) 0(E)h

Solution 1-7): First, an expression for the Hamiltonian of the oscillators motion in phase space is required.Assuming a one dimensional oscillator, the Hamiltonian can be written as

H = p2

2m+

1

2m2q2

Noting thatH =Eand that = 2, this expression for the Hamiltonian of the system can be rewritten as

1 = p2

2mE

+m (2)

2

2E

q2 (1.7.1)

Noting that the equation for an ellipse in the pq-plane can be given, in general, as

p2

a2+

q2

b2 = 1

Comparing this with the expression in (1.7.1) gives a and b as

a=

2mE

and

b= 1

2

2Em

These expressions for a and b allow the surface of constant energy in the oscillators phase space to be represented as

Figure 1.7-1 : Phase space (p, q) for an oscillator of frequency

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The area of an ellipse, A, can be given as A = ab. Applying the expressions for a and b given above to thisexpression for the area gives

A= a b=

2mE 1

22E

m =

4E

2 =

E

which is also the expression for 0(E) is the phase space with an energy less than E. Thus

0(E) = A =E

(1.7.2)

In the quantum case, the energy of the oscillator can be given as

En=

n +

1

2

= 2

2

n +

1

2

= h

n +

1

2

Solving this for n0, in terms ofE, using Mathematica gives

n0 = E

h 1

2 (1.7.3)

wheren0 is the quantum number which corresponds to the energy E. Applying this expression for n0 to theexpression from Eqn (1.20a) in Kubo gives

0(E) =

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Problem 1-8):

(i) When a system ofNoscillators (N 1) with total energy E is in thermal equilibrium, find theprobability that a given oscillator among them is in a quantum state n.

(ii) When an ideal gas ofNmonoatomic molecules with total energy E is in thermal equilibrium,

show that the probability of a given particle having an energy = p2/2m is proportional toe/kBT

Solution 1-8): The expression, at the quantum level, for the energy levels of any given oscillator (the ithoscillator) can be given as

Eni =

ni+

1

2

Defining the total number of possible quantum states the system can take as M implies that

n1+ n2+

+ nN=M (1.8.1)

The total energy of the system of oscillators can be expressed as the sum of the energies of each of the Noscillators

ET =Ni=1

Eni =En1 + En2+ EnN =

n1+1

2

+

n2+

1

2

+ +

nN+

1

2

Since there are N energy terms in the above expression, there areNof the 1/2 terms in that expression as well;therefore, the expression for the total energy of the system can be rewritten as

ET = n1+

1

2+

n2+

1

2+ +

nN+

1

2=

n1+ n2+ + nN+ N1

2Now, applying the expression for the total number of possible quantum states from (1.8.1) to this expression for thetotal energy gives

ET =

n1+ n2+ + nN+ N1

2

=

M+

1

2N

There are Mpossible states that the system can take and there are N 1 ways to order theNoscillators;therefore, there are (M+ N 1)! total number of ways the system can be arranged. To find the number ofpermutations, or the thermodynamic weight, of the system ofNoscillators, the (M+ N 1)! total number of waysthe system can be arranged is divided by the M! number of ways the possible states that the system can bearranged and the (N 1)! number of ways the Noscillators can be arranged to give

WM=(M+ N 1)!

M! (N 1)! (1.8.2)

The probability that a given oscillator in the system is in a quantum state n can be found by dividing the numberof permutations, or the thermodynamic weight WMn, for the rest of the system while one of the oscillators is inthe quantum state n by the total number of permutations, or the thermodynamic weight, of the system. Theenergy,, of the oscillator in the state n is given by

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=

n +

1

2

This allows the energy of the rest of the system, E, to be expressed as

E = ET =

M+1

2N

n +

1

2

= (M n) +

2 (N 1) (1.8.3)

Using the same arguments that were used in constructing WM, the expression for E in (1.8.3) can be used to

expressWMn as

WM=(M n + N 1 1)!(M n)! (N 1 1)!=

(M n + N 2)!(M n)! (N 2)! (1.8.4)

The probability of one of the oscillators being in the quantum state n can therefore be given by

P r (n) =WMn

WM

Applying the expressions from (1.8.2) and (1.8.4) to this gives

P r (n) =WMn

WM=

(M+ N n 2)!(M n)! (N 2)!

M! (N 1)!(M+ N 1)!

= M(M 1) (M 2) (M n + 1)

(M+ N 1) (M+ N 2) (M+ N n 1)(N 1) (1.8.5)

Assuming that the system is sufficiently large, it can be said that M n. Using this with the fact that N 1allows the expression in (1.8.5) to be recast, approximately, as

P r (n) Mn

(M+ N)n+1 N=

N

M+ N

M

M+ N

n(1.8.6)

To find the probability that a given particle, in an ideal gas ofNmonoatomic particles with total energy E, hasthe energy = p2/2man expression for the number of states for the entire system, 0(E, N), and an expression forthe number of states of the rest of the system (the remaining N 1 particles), when one of the particles has energy, 0(E , N 1), must be found. Since this is an ideal gas, the Hamiltonian of the entire system, and thus itsenergy can be expressed as

E=Nj=1

p2xj2m

+p2yj2m

+p2zj2m

=

3Ni=1

p2i2m

(1.8.7)

To find the expression for 0(E, N), the expression from Eqn (1.17b) in Kubo can be used. Thus, the integral

0(E, N) = 0(E)

h3NN!=

1

h3NN!

d =

1

h3NN!

V

V

dq1 dqNP

P

dp1 dpN (1.8.8)

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must be evaluated. In (1.8.8), dqi= dqxidqyidqzi anddpi= dpxidpyidpzi , while V R3 and P R3 represent thethree-spaces of the allowed positions and momenta, respectively. To find integration bounds for the q integrals it isassumed that the gas is contained in a container of volume V, therefore

V

dqi= V dqxidqyidqzi =V V V dq1 dqN=VN

Applying this result to the integral in (1.8.8) gives

0(E, N) = 1

h3NN!

V

V

dq1 dqNP

P

dp1 dpN= VN

h3NN!

P

P

dp1 dpN (1.8.9)

To find integration bounds for the p integrals, the expression in (1.8.7) is considered. It implies that all of thepossible states of the system lie within a 3N-dimensional hypersphere of radius

2mE, since

E=3N

i=1p2i2m

3N

i=1p2i =p

21+p

22+ +p23N= 2mE

Thus, the p integral in (1.8.9) evaluates to the volume 3N-dimensional hypersphere of radius

2mE. Now thevolume,V, of an arbitrary, s-dimensional hypersphere can be given as

V=Csrs = s/2

s2 + 1

rsThe hypersphere here has a radius ofr =

2mEand 3Ndimensions, thus this general expression for the volume of

a hypersphere allows the p integrals in (1.8.9) to be evaluated to give

P P dp1 dpN=

3N2

3N2 + 1

2mE3N

=

3N2

3N2 + 1(2mE)3N2

Therefore, the expression for 0(E, N) in (1.8.9) becomes

0(E, N) = VN

h3NN!

P

P

dp1 dpN= VN

h3NN!

(2mE)

3N2

3N2 + 1

(1.8.10)

Using similar arguments and the fact that = p2/2m, the expression for 0(E , N 1) can be given as

0(E , N 1) = VN1

h3(N1) (N 1)!

(2m (E ))3N12

3N+1

2

(1.8.11)

Now the probability that a given particle has the energy = p2/2m is defined by

P r () =W(E , N 1)

W(E, N) (1.8.12)

WhereW(E, N) andW(E , N 1) are the thermodynamic weights of the system as a whole and the remainderof the system when one of the particles has energy = p2/2m, respectively. Noting that, in general,W(E, N) = 0(E, N) E, the expressions in (1.8.10) and (1.8.11) then imply

17


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W(E, N) =d0(E, N)

dE E=

d

dE

VN

h3NN!

(2mE)

3N2

3N2 + 1

E

= d

dE VN

h3NN! (2m)

3N2 E

3N2

3N2 + 1 E=

VN

h3NN!

(2m)

3N2 E

3N2 1

3N2 + 1

3N

2

E

=

VN

h3NN!

(2m)

3N2 E

3N2

3N2 + 1

3N

2

E

E (1.8.13)

and

W(E , N 1) = d0(E , N 1)dE

E

= ddE VN1

h3(N1) (N 1)! (2m(E ))

3(N1)2

3(N1)2 + 1

E=

d

dE

VN1

h3(N1) (N 1)!

(2m) 3(N1)2 (E ) 3(N1)2

3(N1)2 + 1

E

=

VN1

h3(N1) (N 1)!

(2m) 3(N1)2 (E ) 3(N3)2

3(N1)2 + 1

3(N 1)

2

E

=

VN1

h3(N1) (N 1)!

(2m)

3(N1)2 (E ) 3(N1)2

3(N1)

2 + 1

3(N 1)

2

E

E (1.8.14)

Now, the expressions form (1.8.13) and (1.8.14) are applied to the expression for the probability in (1.8.12) to yield

P r () =W(E , N 1)

W(E, N) =

VN1

h3(N1)(N1)!

(2m)

3(N1)2 (E)

3(N1)2

( 3(N1)2 +1)

3(N1)

2

EE

VN

h3NN!

(2m)

3N2 E

3N2

( 3N2 +1)

3N

2

EE

= h3N

V

2m

3N 1

3N

3N

2 + 1

3(N1)2 + 1

(E ) 3(N1)2E

3N2

E

E

= (3N 1) h3N

3V

2m E

E 3N2 E

(E )5N2

= (3N 1) h3N

3V

2m

E

(E ) 5N2

e3N2 log(

EE )

For a monoatomic ideal gas, E= 3N kBT /2, so the expression for P r () above becomes

P r () = (3N 1) h3N

3V

2m

E

(E ) 5N2

e

EkBT

log(EE ) (1.8.15)

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Now, logEE

can be expanded in terms of/Eas

log

E

E

=

E

2

2E2

3

2E3

For largeEandN, only the first term of this expansion matters, thus the log EE in the exponent for e in(1.8.15) is replaced with/Eto obtain

P r () = (3N 1) h3N

3V

2m

E

(E ) 5N2

e

EkBT

( E )

= (3N 1) h3N

3V

2m

E

(E ) 5N2

e kBT (1.8.16)

From the expression in (1.8.16), it is clear that

P r () e

kBT

as expected.

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Problem 1-9): A vessel of volume V contains N gas molecules. Letn be the number of molecules in apart of the vessel of volume v. Considering that the probability of finding a certain molecule in v isequal to v/V in the thermal equilibrium state of this system.

(i) Find the probability distribution f(n) of the number n.

(ii) Calculate n and (n n)2.(iii) Making use of Stirlings formula, show that when N and n are both large, f(n) is

approximately Gaussian.

(iv) Show that in the limit ofv/V 0 and V , with N/V held constant, f(n) approaches thePoisson distribution f(n) = en (n) /n!

Solution 1-9): This is clearly a Binomial Probability Distribution, since the probability of finding a certainmolecule inv, pr, is given as p = v/V, while the probability of finding the certain molecule in the rest of the volumecan be expressed as q= 1 p= 1 v/V. Applying this to the Binomial Probability Distribution gives f(n) as

f(n) =

(k+ n)!

n!k! pn

qk

=

N!

n! (N n)!pn

qN

n

(1.9.1)

= N!

n! (N n)!pn (1 p)Nn

= N!

n! (N n)! v

V

nV vV

Nn=

N!

VNn! (N n)! vn (V v)Nn (1.9.2)

wheref(n) is the probability of finding n molecules in the volumev andk = N n molecules in the rest of thevolume, V v. To find expression for n, it is simplest to proceed with the form of the binomial distribution givenin (1.9.1) and the apply the values for p and qonce an expression for n has been found in terms of them. For thissystemp = v/V and q= 1

v/V. Therefore, to find n, the expression

n=Nn=1

n f(n) =Nn=1

n

N!

n! (N n)!pnqNn

(1.9.3)

must be evaluated. Noting that

(p + q)N =Nn=1

N!

n! (N n)!pnqNn

both sides of this are differentiated with respect to p to give

p

(p + q)N

=

p

Nn=1

N!

n! (N n)!pnqNn

N(p + q)N1

=

Nn=1

N!

n! (N n)! n pn1qNn

Multiplying both sides of this expression byp gives

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p

N(p + q)N1

= p

Nn=1

N!

n! (N n)! n pn1qNn

pN(p + q)N

1

=

Nn=1

N!n! (N n)! n pnqNn (1.9.4)The right side of the expression in (1.9.4) is that same as the right side of the expression in (1.9.3), thus

n=Nn=1

n f(n) =Nn=1

n

N!

n! (N n)!pnqNn

= pN(p + q)

N1

Noting that, by the definition of the Binomial Probability Distribution, p + q= 1, the expression for n becomes

n= pN(p + q)N1

=N p

Finally, the fact that p = v/V is applied to express n as

n= N p= Nv

V (1.9.5)

To calculate (n n)2, it is noted that (n n)2 can be alternatively expressed as

(n n)2 =n2 n2 (1.9.6)

The expression for n2 can be found by squaring the result obtained in (1.9.5), so all that remains to do is find anexpression for n2. As when calculating n, the form of the binomial distribution given in (1.9.1) will be used withthe values p = v/V andq= 1

v/V substituted in the final step. Thus, the expression

n2 =

Nn=1

n2 f(n) =

Nn=1

n2

N!

n! (N n)!pnqNn

(1.9.7)

must be evaluated. Now, the expression in (1.9.4) differentiated with respect to p to obtain

p

pN(p + q)

N1

=

p

Nn=1

N!

n! (N n)! n pnqNn

N(p + q)N1 +pN(N

1) (p + q)N2 =

N

n=1 N!

n! (N n)!n2pn1qNn ,

which is then simplified by noting, by the definition of the Binomial Probability Distribution, that p + q= 1. Thus

N+ pN(N 1) =Nn=1

N!

n! (N n)! n2pn1qNn

Multiplying both sides of this expression byp yields

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p (N+ pN(N 1)) =p Nn=1

N!

n! (N n)! n2pn1qNn

pN+ p2

N2

p2

N=

Nn=1

N!n! (N n)! n2pnqNnwhich can be applied to the expression for n2 in (1.9.7) since is clear that this result and the result in (1.9.7) haveequivalent right hand sides. Thus, the expression in (1.9.7) becomes

n2 =Nn=1

n2

N!

n! (N n)!pnqNn

= pN+ p2N2 p2N

Applyingp = v/V to this expression allows n2 to be expressed as

n2 =pN+ p2N2 p2N=NvV + N2v2

V2 Nv2

V2

which, along with the result from (1.9.5), can be applied to the expression for (n n)2 in (1.9.6) to obtain

(n n)2 =n2 n2 =

Nv

V + N2

v2

V2 Nv

2

V2

Nv

V

2=N

v

V

1 v

V

(1.9.8)

Stirlings formula is given, in general, as lnp! p lnp p, for p R p 0. This is applied to the expression forf(n) in (1.9.2) by first taking the natural log of both sides to obtain

ln [f(n)] = ln N!VNn! (N n)! vn (V v)Nn= ln [N!] + ln [vn] + ln

(V v)Nn

ln VN ln [n!] ln[(N n)!]

= ln [N!] ln [n!] ln[(N n)!] + (N n) l n [V v] Nln [V] + n ln [v] (1.9.9)

Applying Sterlings formula to the ln [N!], ln[n!], and ln [(N n)!] terms in (1.9.9) gives

ln [N!] Nln N Nln [n!] n ln n n

ln[(N n)!] (N n) l n [N n] + n N

Thus, the expression in (1.9.9) becomes

ln [f(n)] (Nln N N) (n ln n n) ((N n) l n [N n] + n N)

Using Mathematica to simplify this expression gives

f(n) = n ln v

n

+ (n N) ln

n Nv V

+ Nln

N

V

(1.9.10)

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SinceN is large, Mathematicais used to expand this expression about n = Nv/V gives

ln [f(n)]

V2

2N v (v V)

n N vV

2(1.9.11)

Noting that the expression for a Gaussian distribution, g (n) is proportional to

g (n) e (n)22 (1.9.12)For this system, and are given as

= n= Nv

V

and

= (n n)2 =n2 n2 =NvV 1

v

Vso the expression in (1.9.12) becomes

g (n) e (NvnV)2

2Nv(vV)

Since this is to be compared to the expression for ln [f(n)], the natural log ofg (n) must be taken. Thus

ln [g (n)] (N v nV)2

2N v (v V) (1.9.13)

Using Mathematica, the expression in (1.9.13) is compared with the expression in (1.9.12) to yield

(N v nV)22N v (v V)

V2

2N v (v V)

n N vV

2= 0

which implies that, for large N, the distribution given by f(n) becomes Gaussian.

To show thatf(n) becomes a Poisson distribution as v/V 0 andV , while N/V remains constant, theexpression for f(n) in (1.9.1) is examined, noting that p = v/V andq= 1 p= 1 v/V. Thus

f(n) = N!

n! (N n)!pnqNn =

N!

n! (N n)! v

V

n 1 v

V

NnApplying n= Nv/V v= nV/Nto this expression for f(n) using Mathematicayields

f(n) =NN (N n)Nn nn N!

n! (N n)!Before proceeding, this new expression for f(n) is rewritten in the more useful and suggestive form

f(n) = (N n)Nn NNN!

(N n)!

nn

n!

(1.9.14)

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Clearly, the expression in (1.9.14) is close to what would be expected for a Poisson Distribution, in that the nn and1/n! terms are already present, since a Poisson Distribution, h (n) can be given as

h (n) = en

nn

n! (1.9.15)

Therefore, all that remains to do is show that the terms not in parentheses in (1.9.14) are, at least, proportional toen. That is to say, it remains to be shown that

en NN (N n)Nn N!

(N n)!

To accomplish this, (N n)Nn is expanded, using the Binomial Theorem, to yield

(N n)Nn =Nn

i=1

NNni ()i

(N n)!i! (N n i)!

= (N n)!

Nn

i=1

NNni ()ii! (N n i)!

Now, this to binomial expansion for (N n)Nn is multiplied by the expression

NNN!(N n)!

to obtain the result

(N n)Nn NNN!

(N n)! = NNN!(N n)!

(N n)!

Nni=1

NNni ()ii! (N n i)!

=NNN!

Nni=1

NNni ()ii! (N n i)!=Nni=1

NNN! NNni

(N n i)!

()ii!

=Nni=1

N! Nni

(N n i)!

()ii!

Now, the approximation

N! Nni

(N

n

i)!

1

can be made for large N; however, since N/V is constant as V , it is guaranteed that Nwill be sufficientlylarge. Thus, the expression in (1.9.16) becomes

(N n)Nn NNN!

(N n)! =Nni=1

N! Nni

(N n i)!

()ii!

Nni=1

ii!

Nn1i=0

ii!

(1.9.16)

Noting that the Taylor series expansion for e can be given as

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e =j=0

ii!

Kj=0

ii!

for a sufficiently large K R+

, the expression in (1.9.16) clearly implies

(N n)Nn NNN!

(N n)! =e

Applying this expression to the expression for f(n) in (1.9.14) yields

f(n) = (N n)Nn NNN!

(N n)!

nn

n!

= e

nn

n!

(1.9.17)

Thus, for v/V 0 and V , whileN/V remains constant, f(n) does indeed become a Poisson distribution.

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Problem 1-10): As illustrated in Figure P1.10, a string with a lead ball of mass m is slowly pulledupward through a small hole. Consider the work done on the system during this process and findthe change in the energy and frequency of this pendulum during this adiabatic process, assumingthe amplitude of the pendulum to be small.

Figure P1.10

Solution 1-10): First, expressions for the energy of the system and the tension in the pendulum string arerequired. To create the desired expression for the energy of the system and the tension in the pendulum string, thevariables , l, Tp, and Tare defined, respectively, to be the angle the pendulum makes with the y-axis, the lengthof the pendulum, the tension in the pendulum string, and the force pulling the pendulum up. These quantities areillustrated below in Figure 1.10-1

Figure 1.10-1

Using the variables illustrated above, the energy of the system can be expressed as

26


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E= Ep mgl=

1

2ml22 mgl+ mg (l l cos )

=1

2 ml22 mgl+ mgl (1 cos ) (1.10.1)

whereEp= E+ mgl is the energy associated with the motion of the pendulum and as such, dependent on .Furthermore, the tension in the pendulum string can be given using the expression

T=ml2 + mg cos (1.10.2)

To simplify the expression for the energy of the system in (1.10.1), the trig identity cos ( ) cos( + ) == 2sin sin is employed by setting = = /2 to obtain

1

cos = cos 0

cos = cos

2

2 + cos

2

2= 2sin

2 sin

2Thus, the expression for E in (1.10.1) becomes

E=1

2ml22 mgl+ mg (1 cos ) = 1

2ml22 mgl+ mgl

2sin

2

sin

2

Since is said to be small, the small angle approximation for sine can be used. Thus, the expression for Ecan begiven, approximately, as

E=1

2ml22 mgl+ mgl

2sin

2

sin

2

=12

ml22 mgl+ mgl

2

2

2=

1

2ml22 mgl+1

2mgl2 (1.10.3)

Similarly, to simplify the expression for the tension in the pendulum string in (1.10.2), the same trig identity usedabove is used to obtain

cos = 1 + 2 sin

2

sin

2

Thus, the expression for T in (1.10.2) becomes

T=ml2 + mg

1 2sin

2

sin

2

Again, the fact that is considered to be small is used. Thus, the small angle, approximation for sine, allows theexpression for Tto be rewritten, approximately, as

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T=ml2 + mg

1 2sin

2

sin

2

=ml2 + mg 1 2

22

=ml2 + mg 1

2mg2 (1.10.4)

Now, the expressions for the energy of the system and the tension in the pendulum string are time averaged overthe period of one oscillation. Since the pendulum is executing simple harmonic motion, the average kinetic andpotential energies resulting for the motion of the pendulum are equal, thus

KE= P E

1

2ml22 =

1

2mgl2

l2 =g 2

Applying this to the time average of the expression for T gives

T =ml2 + mg 12

mg2

=ml2 + mg 12

mg2

=mg 2 + mg 12

mg2

=mg+1

2mg2

Since the amount of work done on the system when the string is pulled up by l is given asT l, the averageamount of work done on the system, and thus the change in the energy of the system, can be expressed as

E= T l = mg l 12

mg2 l (1.10.5)

by using the result for Tabove. Thus, using the expression for E= T l in (1.10.5) and the fact thatEp= E+ mgl, the change in the energy of the pendulum is simply

Ep= E+ mgl= mgl 12

mg2 l+ mgl= 12

mg2 l (1.10.6)

Now, combining the expression for Egiven in (1.10.3) with Ep= E+ mgl allows Ep to be expressed as

Ep= E+ mgl = 12

ml22 mgl+12

mgl2 + mgl = 12

ml22 +12

mgl2

ThisEp can be given as

Ep=1

2ml22 +

1

2mgl2

=1

2ml22 +

1

2mgl2 =mgl2 (1.10.7)

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Comparing the expressions in (1.10.6) and (1.10.7) gives

Ep

Ep=

12 mg2 lmgl2

= l2l

Ep= l2l

Ep (1.10.8)

Since the energy of a harmonic oscillator is proportional to is frequency, , the expression in (1.10.8) implies thatthe change is the frequency of oscillation can be expressed as

= l2l

=|l|

2l

since l 0, as l is decreasing.

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Problem 1-11): Prove, on the base of the principle of entropy increase, that when a coordinate xvaries extremely slowly through a quasi-static adiabatic process (1.35), the entropy,S(x), does notchange.

Solution 1-11): It is to be shown that dSdt = 0, whereS(x (t)) is a function ofx (t), which is a function oft;

therefore, using the chain rule, d

Sdt can be rewritten as

dSdt

=dx

dt

dSdx

(1.11.1)

Nowx, varies adiabatically, so dxdt 0. Applying this, to the expression in (1.11.1) implies that

dSdt

=dx

dt

dSdx

= 0 dSdx

= 0

QED.

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Problem 1-12): As in figure P12.1, a system under consideration is contained in a box made ofinsulating walls and a moveable piston.

Figure P1.12

A weight w is placed on the piston. Regarding the total system, including the weight, as an isolatedsystem derive, from the micro-canonical ensemble, the relation

p= EVSApply this formula to an ideal gas of monoatomic molecules and prove the equation of state p= 32

EV.

Solution 1-12): To begin, the volume enclosed, V, is defined as V =xA where A is the area of the piston and x isthe potion of the piston above the bottom of the cylinder. Clearly, the pressure exerted by the piston is then givenbyp = wA , so a differential expression for

wA is required. Since the system will come to equilibrium when the number

of available states is the greatest, when entropy is the greatest, an expression for the entropy of the system isrequired. From eqn (1.18) in Kubo, entropy can be given as

S(x) = kBln [ (E, V)] = kBln[ (x)] (1.12.1)

since both E, the total energy of the gas, and V , the volume of the container and thus the gas, both depend on x.Thus, the point at which the entropy is greatest can be found using

dSdx

= 0

Applying the expression for (1.12.1) to this and using the chain rule gives

dSdx

=kB

E

x

ln[(E, V)]

E +

V

x

kBln[ (E, V)]

V

= 0 (1.12.2)

An expression forV in terms ofx already exists, so all that remains is to find an expression for Ein terms ofx;

however, the total energy of the system ETis the sum of energy of the gas and the potential energy of the weight.Thus,

ET =E+ wx E= ET wx

Applying this expression for EandV =Ax to the expression in (1.12.2) yields

31


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0 = w

ln[(E, V)]

E

V

+ A

ln[(E, V)]

V

E

=

w

S

EV + AS

VEwhich implies that

w

A=

SV

E

SE

V

1(1.12.3)

since

E

x =

(ET wx)x

= wV

x =

(Ax)

x =A

Using this expression, the chain rule, and p = w/A, the expression for the pressure in the container becomes

p=w

A=

SV

E

SE

V

1=

E

V

S

(1.12.4)

Finally, the entropy of an ideal gas can be given by

S= N kB

lnV

N +

3

2ln

2E

3N + ln

m3/2 e3/2

2

3

Applying the expression forSto the expression for p in (1.12.3), using Mathematica, and noting that p = w/A,yields

p=w

A=

SV

E

SE

V

1=

kBN

V

3kBN

2E

1=

2E

3V

Noting that for a monatomic ideal gas, E= 32 N kBTverifies this expression for the pressure, since

p=2E

3V =

2

3V

3

2N kBT

=

kBT N

V

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Problem 1-13): There is a one-dimensional chain consisting ofn (n 1) elements, as seen in figureP1.13. Let the length of each element be a and the distance between the end points x, as shown.

Figure P1.13

Find the entropy of this chain as function ofx and obtain the relation between the temperature, T,of the chain and the force (tension) which is necessary to maintain the distance x, assuming the

joints to turn freely.

Solution 1-13): First, it is noted that the total distance, x, can be expressed in terms of the total number ofsegments going to the right, n+; the total number of segments going to the left, n; and the length of eachindividual segment, a. Clearly, this can be accomplished using the relation

x= (n+ n) a

Similarly, the total number of segments, n, can be expressed in terms ofn+ andn by

n= n++ n

Solving these expressions for n+ andn in terms ofx, a and n gives n+ andn as

n+ =1

2

n +

x

a

(1.13.1)

n =1

2

n x

a

(1.13.2)

Since the entropy of the chain is sought, the expression for eqn (1.18) in Kubo is useful as it expresses entropy interms of the number of system states,

S= kbln [W(x)] (1.13.3)

This is useful, since the number of states for this system is easily obtained using

W(x) = n!

n+!n!

Applying this to the expression for the entropy of the system in (1.13.3) yields

S= kbln [W(x)] = kbln

n!

n+!n!

= kB(ln n! ln n+! ln n!)

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Now, n, n+ andn are all considered to be large, thus Stirlings Approximation can be used to simplify theexpression for the entropy, yielding

S= kB[n ln n n (n+ln n+ n+) (n ln n n)]

Noting that n = n++ n, it is clear that this expression for the entropy further simplifies to

S= kB(n ln n n+ln n+ n ln n)

Now, using the expressions for n+ andn in (1.13.1) and (1.13.2) are applied, using Mathematica, to yield

S= kB

an ln 4 + 2an ln [an] (x an) l n [an x] (x + an) l n [an + x]

2a

(1.13.4)

Now, the tension in the system, T, can be expressed using the Helmholtz Free Energy

T=F

x

T

=(U T S)

x

T

However, due to the fact that the joints of the system are free to rotate, the internal energy of the system, U, inindependent ofx. Thus, the expression for T simplifies to

T =

F

x

T

=

(U T S)

x

T

=

(T S)

x

T

= T

S

x

T

Applying the expression for entropy given in (1.13.4) to this expression for T, and using Mathematica to simplify,yields

T= T

S

x

T

= kBT2a

ln

2an

an + x 1

kBTa2n

x + kBT

3a4n2x3 +

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Problem 1-14): There is a system consisting ofNoscillators of frequency . Discussing the systemclassically,

(i) Find the number of states.

(ii) Using the results for the number of states, derive the relation between the energy and

temperature of the system

Solution 1-14): The Hamiltonian of this system of oscillators can be expressed as

H =Ni=1

p2i2m

+k

2q2i

by assuming that all of the oscillators have identical mass and frequency of oscillation. If the angular frequency ofthe oscillations is defined to be =

k/m= 2, then this expression for the Hamiltonian can be simplified to

H=

N

i=1

p2i

2m+

k

2q2

i= 12N

i=1

p2i + 2q2i (1.14.1)Then, the number of possible system states, 0(E) can be found by evaluating the volume integral

0(E) =0(E)

N =

1

N

Ni=1

HE

dqi dpi

=

1

N

Ni=1

PNi=1

p2i

2m+k2 q

2i

ffE

dqi dpi

(1.14.2)

whereE Ni=1 p2i2m+ k2 q2i 0, serves as the bounds of integration, sinceH =E. To simplify the Hamiltonian,and thus obtain an simpler integral than the one in (1.14.2),H will be expressed in terms of the normal coordinate R2N. Thus, the pis and qis in (1.14.1) are replaced according to

pi = i

2

qi =N+i

2

Applying these expressions for the pis andqis to the expression forH in (1.14.1) yields the simplified Hamlitonian

H = 12

Ni=1

p2i +

2q2i

=1

2

Ni=1

i

22

+ 2

N+i

2

2

=1

2

N

i=1 22i + 2

2N+i=

N

i=1 2i +

2N+i=

2N

i=1 2i (1.14.3)

The desire is to apply this simplification of the Hamiltonian to normal coordinates, given in (1.14.3), to the integralfor 0(E), given in (1.14.2); however, to accomplish this, the differential relations

dpi = di

2

dqi =dN+i

2

35


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are required. Applying these differential relations and the expression for the Hamiltonian in normal coordinates tothe integral for 0(E) in (1.14.2) yields

0(E) =0(E)

N =

1

N

N

i=1HEdqi dpi=

1

N

Ni=1

HE

di

2dN+i

2

=

2

N Ni=1

P

2Ni=1

2iE

di dN+i

=

2

N 2Ni=1

P2Ni=1

2iE

di

(1.14.4)

Thus, the integral here is over a 2N-dimensional sphere of radius

E, as opposed to over an ellipsoid as in (1.14.2).Using the expression for the volume of a sphere in an arbitrary number of dimensions, the integral in (1.14.4)

evaluates to

0(E) =

2

N 2Ni=1

P

2Ni=1

2iE

di

=

2

N2N/2

[2N/2 + 1]

E2N

=

2

NN

[N+ 1]EN

Noting that [N+ 1] =N! allows this expression for 0(E) to be simplified to its final form, yielding

0(E) = 2

N

N

[N+ 1]EN =

1

N! 2E

N

(1.14.5)

or, by noting that = 2

0(E) 1

N!

E

N(1.14.5 a)

To find the relation between the energy and the temperature of the system, the expression

S

E =

1

T (1.14.6)

from eqn (1.28) in Kubo is noted. Thus, an expression for the entropy of the system is required; however, noting

eqn (1.18) in Kubo and the expression for 0(E) given in (1.14.5), the entropy of the system can be given as

S= kBln[0(E, N)] = kBln[0(E)]

=kBln

1

N!

2E

N

=kB

Nln

2E

ln N!

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Applying this expression for S to the expression in (1.14.6) yields

1

T =

S

E =

E

kB

Nln

2E

ln N!

=kBN

Eln2E = kBN 2 1E= 2kBNE (1.14.7)

Finally, this expression is solved for Ein terms ofT and = 2is applied to the result in (1.14.7) to yield thefinal results

E=2kBN

T =

kBN

T

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Problem 1-15): For an oscillator with mass m and angular frequency ,

(i) Calculate the partition function classically.

(ii) Calculate the partition function quantally (quantum mechanically).

(iii) Find the internal energy, entropy, and the heat capacity of a system consisting ofN suchoscillators in terms of temperature.

Solution 1-15): The Hamiltonian of a single oscillator can be given as

H = p2

2m+

k

2q2 (1.15.1)

where the angular frequency, , is defined =

k/m. Using the expression given in eqn (1.71 a) in Kubo and theexpression for the Hamiltonian of a single oscillator given in (1.15.2) allows the classical partition function for thesingle oscillator to be given as

Z0C = 1

eH/kBTdpdq= 1

e p2

2mkBT+ kq2

2kBTdpdq

Carrying out this integration using Mathematicaand noting that =

k/m gives

Z0C = 1

e p22mkBT

kq2

2kBT dpdq=2kBT

m

k =

2kBT

(1.15.2)

For the quantum case, the energy of a single oscillator can be given as

En= n +

1

2 (1.15.3)

Using the expression given in (1.71b) in Kubo and the expression for the nth energy state of the oscillator given in(1.15.3), the quantum partition function for the single oscillator to be expressed as

Z0q =

n=0

eEn =n=0

e kBT(n+

12 )

Using Mathematicato evaluate this sum gives

Z0Q =

n=0e

kBT

(n+ 12 ) =1

2csch

2kBT

(1.15.4)

To find the internal energy of a system ofNoscillators (of the kind described above), the expression from eqn (1.80a) in Kubo is used

U=

ln ZQ (1.15.5)

Since the Noscillators are identical the partition function of the system, ZQ can be given using the partitionfunction of one of the oscillators. Thus, using the expression in (1.15.4) and = 1/kBT, ZQ can be given as

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ZQ= ZNQ0 =

1

2csch

2kBT

N=

1

2csch

2

N(1.15.6)

This expression for ZQ allows the expression for Uin (1.15.5) to become

U=

ln ZQ=

ln

1

2csch

2

N

which, using Mathematica, evaluates to

U=

ln ZQ=N

2 coth

2

(1.15.7)

Now, the heat capacity of a system can be expressed using the relation

C=U

T

N,

(1.15.8)

Before applying the expression for Ugiven in (1.15.7) to this relation for C, the substitution = 1/kBT is made toobtain

U=N

2 coth

2

=

N

2 coth

2kBT

(1.15.7 a)

Applying this expression for U to the relation in (1.15.8), usingMathematica , yields

C= UTN, = T N

2 coth 2kBTN, = N

2

2

4kBT2 csch2 2kBT (1.15.9)

Finally, to find the entropy, it is noted that

S=U F

T (1.15.10)

whereF is the Helmholtz Free Energy. Using the expression in eqn (1.81) in Kubo, the Helmholtz Free Energy canbe expressed in terms of the partition function, ZQ, as

F= kBTln ZQApplying the expression for ZQ from (1.15.6) allows Fto be given as

F= kBTln ZQ= kBTln

1

2csch

2

N

Using this expression for Fand the expression for U, given in (1.15.7 a), in the expression for Sin (1.15.10) allowsSto be given as

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S=U F

T =

1

T

N

2 coth

2kBT

+ kBTln

1

2csch

2

N

Using Mathematica to simplify this result, gives

S=N

2T coth

2kBT

kBln

2N sinhN

2kBT

(1.15.11)

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Problem 1-16): Show that, the temperature is uniform, the pressure of a gas in a uniform

gravitational field decreases with height according to p (z) = p (0) emgz/kBT, where m is the mass ofa molecule.

Solution 1-16): The pressure at a given point z1 above z = 0 depends on the number of particles above that

point,n (z) and the weight, mg , of each of the particles above the heightz0. Thus,

p (z) = mg n (z) (1.16.1)

Since the Hamiltonian for the ith particle of this gas can be given as

Hi = p2

2m+ mgz

the number of particles above the point z0 can be given as

n (z) = N

d3p

Lx

0

dx Ly

0

dy

z0

dz P(z)

=N

d3p

Lx0

dx

Ly0

dy

z0

dze(p

2/2m+mgz)

Z (1.16.2)

whereZis the partition function of the system. This partition function can be given as

Z=

d3p

Lx0

dx

Ly0

dy

0

dz P(z)

=

d3p

Lx0

dx

Ly0

dy

0

dz e(p2/2m+mgz)

=

LxLzg

8m35 =

A

g8m3

5

using Mathematica to evaluate the integrals. Applying this expression for Zto the expression for n (z) in (1.16.2)gives

n (z) =N g

A

5

8m3

d3p

Lx0

dx

Ly0

dy

z0

dz e(p2/2m+mgz)

=N g

A

5

8m3

egmz1

A

g

8m3

5

= N egmz1

Applying this to the expression in (1.16.1) gives p (z) as

p (z) = mg n (z) = mg Negmz1 (1.16.3)

To complete the proof, an expression for p (z) atz = 0, p (0), is required. Setting z = 0 in (1.16.3) gives p (0) asp (0) =mgN, thus the expression for p (z) in (1.16.3) becomes

p (z) = mg Negmz1 =p (0) egmz1

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Problem 1-17): Consider an ideal gas consisting ofNparticles obeying classical statistics. Supposethat the energy of one particle, , is proportional to the magnitude of momentum, p, such that = cpwhere c R. Find the thermodynamic functions of this gas without considering the internalstructure of the particles.

Solution 1-17): The desire is to calculate, in terms ofV, N, andT, the Helmholtz Free Energy, F; the pressure,p; the internal energy, U; the entropy, S; the enthalpy, H; the Gibbs Free Energy, G; the heat capacity at constantvolume, CV ; and the heat capacity at constant pressure, Cp; therefore, an expression for the systems partitionfunction is required. Using the expression in eqn (1.71 a) in Kubo, with HN= = pc, allows the partition functionof the system, Z, to be given as

Z= 1

N!

1

3

dqx dqydqz

dpx dpydpze

EN/kBTN

Noting that theqintegrals evaluate to

dqx dqydqz = Vand that EN= = cp allows the expression for Z tobe rewritten as

Z= 1N!

V3

dpx dpydpzepc/kBT

NConverting the remaining integral, in p, in the expression for Z to spherical coordinates, in p, yeilds

Z= 1

N!

V

3

20

0

0

epc/kBTp2 sinpdpdpdp

NEvaluating this integral using Mathematicagives the final expression for Z as

Z= VN! kBTc 3

N

(1.17.1)

Using this expression for Z inF= kBTln Z gives

F= kBTln Z= kBTlnV

N!

kBT

c

3N= kBT Nln

V

N!

kBT

c

3= kBT N

ln [V] + 3ln

kBT

c

ln N!

SinceN is considered to be large, Stirlings formula applies; therefore, the expression for F becomes

F = kBT N

ln [V] + 3ln

kBT

c

ln N!

= kBT N

ln [V] + 3ln

kBT

c

Nln N N

(1.17.2)

Now, it is noted that p and Ucan be expressed, using F, as

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p=

F

V

T,N

U=

T2

T F

TV,NApplying the expression for F in (1.17.3) to these expressions, using Mathematica, gives p and U as

p=kBN T

V (1.17.3)

U= 3kBN T (1.17.4)

Now, the enthaply of the system, H, can be given as H=U+pVand the entropy of the system, S, can be given asS= (U F)/T. Therefore, using the expressions for p and U from (1.17.3) and (1.17.3) and Mathematica, H canbe expressed as

H=U+ pV = 4kBN T (1.17.5)

and using the expressions forF andU from (1.17.2) and (1.17.4), Scan be expressed as

S=U F

T =kBN

3 N Nln N+ ln [V] + 2ln

kBT

c

(1.17.6)

Finally, the heat capacity at constant volume, Cv, can be expressed as Cv = UT ; the heat capacity at constant

pressure,Cp, can be expressed as Cp= HT; and the Gibbs Free Energy, G, can be expressed as G = F+pV, thus

Cv, Cp andG can be given as

Cv =

T [3kBT N] = 3kBN

Cp=

T [4kBT N] = 4kBN

G= kBT N

1 + N+ Nln N ln [V] 3 ln

kBT

c

using Mathematica .

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Problem 1-18): We are given an ideal gas consisting ofNmonoatomic molecules and a systemconsisting ofNoscillators. Supposing that they have a canonical distribution at temperatureT, findthe most probable value, E, of the total energy Eof the respective systems and confirm that itagrees with the mean value Ein the canonical distribution.

Solution 1-18): First, partition function for the ideal gas system is required. This partition function is given as

ZG= 1

N!3N

Ni=1

V3

dqi

dpi eHi

where V3 represents the three dimensional volume where the gas is contained. Using this (information about V3)and the Hamiltonian of an ideal gas,H =p2/2m= p2x+p2y+p2z /2m, the expression for ZG can be simplified to

ZG = 1

N!3N

Ni=1

V dqi

dpi e p

2i

2m

Using Mathematicato evaluate the integral over p allows the expression for ZG to finally be given as

ZG = VN

N!3N

2m

3N/2=

VN

N!3N (2mkBT)

3N/2=

VN23N

N!3N

Em

3T

3N/2(1.18.1)

since= 1/kBT and E= 3N kBT /2. Using the partition function ZG, as given in (1.18.1), the probability of theideal gas having an energy level, E, is

P rG(E) = 1

N!3N

eE

ZG

=

eE/kBT

23NVN

3T

mE

3N/2Finding the most probable value for E is equivalent to finding the maximum of this expression for the probability

of various energy values. Thus, the first derivative, with respect to E, of this expression is required. UsingMathematica , this derivative can be given as

dP rG(E)

dE =

(2E+ 3kBN T) eE/kBT

23N+1VNEkBT

3T

mE

3N/2(1.18.2)

Setting the expression in (1.18.2) equal to zero and solving for E, using Mathematica, give E as

E =

32 kBT N =32 kBN T

which is identical to the value from the ideal gas law. Now, for the system of oscillators, the energy of the system

depends on the number of states the system can take. As discussed in Problem 1-8, the total number of states thesystem of oscillators can take, M, implies the requirement that M=

Ni ni= n1+ n2+ n3+ + nN. Thus,

removing the dependence of this system on Mand replacing it with a dependence on T will yield an expression forthe most probable energy at a particular temperature. To accomplish this, an expression for the thermodynamicweight of the entire system is required. First, the maximum possible energy of this system, can be expressed as

E=

M+

1

2N

(1.18.3)

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Thus, as in Problem 1-8, the thermodynamic weight of the system can be given as

W =(M+ N 1)!

M! (N 1)!

ConsideringM andNto be sufficiently large (and sufficiently greater than 1), Stirlings Approximation is appliedto yield

ln [W] (M+ N) l n [M+ N] (M+ N) (Mln [M] M) (Nln [N] N) (M+ N) l n [M+ N] Mln M Nln N

Now, the entropy of the system, S, can be given in terms of the thermodynamic weight of the system asS= kBln [W]. Combining this expression for the entropy with the above expression for yields

S= kBln [W] = kB((M+ N) l n [M+ N] Mln M Nln N)

Noting that 1/T =S/Eand the expression for the entropy above, and applying the chain rule implies

1

T =

S

E =

M

E

S

M =

M

E

kB

M [((M+ N) l n [M+ N] Mln M Nln N)]

(1.18.4)

To proceed with the expression in (1.18.4) an expression forMin terms ofEis required. Noting the expression in(1.18.3), we clearly have

M= E

N

2 (1.18.5)

using Mathematicato solve for M. Now, Mathematica is used to evaluate the partial derivative ofSwith respect

to M andMwith respect to Eand the results are applied to the expression in (1.18.4) in order to obtain

1

T =

S

E =

M

E

S

M =

1

kBln

M+ N

M

Now, the aim is to obtain an expression for the energy of the system which depends on the temperature of thesystem and is independent ofM. Thus, the expression for Min terms ofEin (1.18.5) is applied to the aboveexpression, with the help ofMathematica, to obtain

1

T =

kB

ln

4E

2E N

=kB

ln

2E+ N

2E N

=kB

ln

E+ 12 N

E 12 N

Finally, the expression no longer depends onM, so the final step of solving for Ein terms ofT is undertaken, usingMathematica, to give

E=1

2N coth

2kBT

as the most probable value ofE, which compares well with the expression for the average value ofE for this systemgiven in expression (1.15.7 a) in Problem 1-15.

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Problem 1-19): Show that the grand canonical distribution function of a classical ideal gas ofmonatomic molecules is given by

= ef

What is the significance of and f?

Solution 1-19): In the classical case, can be found using

=

NA=0

NB=0

Ni=NA,NB,...

1

Ni!3Ni

eHNA,NB,...

PNA,NB,...j=0 Nii

d

Since this is an ideal gas, with one type of particle, this expression for simplifies to

=

N=0

1

N!3N

N

i=0 eHie d

=

N=0

1

N! 1

3 eH d

N

e

N

For a monatomic ideal gas,H can be expressed asH = p22m . Further, using eqn (1.74 c) in Kubo, e =;therefore, the expression for becomes

=N=0

1

N!

1

3

e

p2

2m d

NN

=

N=0

1

N!

3

e

p2

2m d

N (1.19.1)

Evaluating the integral in (1.19.1) gives

e

p2

2m d =

L0

dqx dqydqz

0

dpx dpydpze p22m

=L3

0

dpx dpydpze(p2x+p2y+p2z)/2m =V

m2

3/2Applying this to the expression in (1.19.1) gives

=

N=0

1N!

V

3

m

2

3/2N =

N=0

N

N!

V

3

m

2

3/2N (1.19.2)

Iffis now defined as

f= 1

3 e p2

2m d = NN! ZN=Zi (1.19.3 a)=

V

3

m

2

3/2(1.19.3)

Then, the expression for , in (1.19.2) becomes

=N=0

N

N!fN =

N=0

(f )N

N!

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which is just the series expansion for = ef. Thus,

=

N=0

N

N!fN =

N=0

(f)N

N! =ef (1.19.4)

where = e is the absolute activity and f, given in (1.19.3) or (1.19.3 a), is the partition function of the ithparticle.

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Problem 1-20): Consider a monatomic ideal gas ofNmolecules in a volume V. Show, with the helpof the T- distribution, that the number ofn of molecules contained in a small element of volume vis given by the Poisson distribution

Pn= enn

n

n!

Solution 1-20): Using the expression in Eqn (1.75) in Kubo, the distribution of particle numbers ( n/v) can begiven as

Pn(NA, NB, . . . ) =e(NAA+NBB+ )ZN

=

NA

NB

ZN

Since this is an ideal gas, there is only one type of particle. Thus, the expression forPn(NA, NB , . . . ) simplifies to

Pn

(N) = NZN

(1.20.1)

Using the the expression in (1.19.3.a) in Problem 19, ZN =fN/N! and the expression in (1.19.4), = ef, the

expression in (1.20.1) becomes

Pn(N) = NZN

=N

fN

N!

1

ef

= N

fN

N!ef =

( f)N

N! ef (1.20.2)

Finally, is must be shown that N=f. Again using the expressions from Eqn (1.75) and Eqn (1.74 c) in Kuboallows Nto be written as

N=

1

Ni=0

ii

Zi =

Ni=0

ii

1

Zi =

Ni=0

Zii

=

Ni=0

iZi

=

Ni=0

iZi

Noting that =N

i=0 iZi, allows the expression for Nto be given as

N=

Ni=0

iZi

=

=

ln

Finally, the expression from (1.19.4) in Problem 19, = ef is applied to this expression for N to give

N=ln

=

f

=f (1.20.3)

Applying this expression for Nto the expression for Pn(N) in (1.20.2) gives

Pn(N) =( f)N

N! ef =

NN

N!eN

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Problem 1-21): Show that the T-p distribution the quantity

G (T, p, N) = kBTln Y

is equal to the Gibbs free energy, where Y is the partition function.

Solution 1-21): Noting thatYcan be given as

Y =

0

ZN(V) epV dV

This expression for Y is applied to G (T, p, N) = kBTln Y to obtain

G (T, p, N) = kBTln

0

ZN(V) epV dV

(1.21.1)

The ZN(V) term in the integral is now rewritten as ZN(V) = eln[ZN(V)]; however, ln [Z(V)] =

F( , V , N ).

Thus,Z(N) can be rewritten as ZN(V) = eF(,V,N), which allows the expression in (1.21.1) to be rewritten as

G (T , p , N ) = kBTln

0

eF(,V,N)epV dV

(1.21.2)

Now, it is noted that

eG(,p) =

0

eF(,V,N)epV dV

since theVterms are be integrated to zero due to the negative exponent and the infinite integration limits, so theexpression in (1.21.2) becomes

G (T, p, N) = kBTln

eG(,p)

= kBT(G (, p)) = G (, p) = G (T, p, N)

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Problem 1-22): Classify the following particles according to Fermi or Bose statistics :

3 particle 3He

H2 molecule Positron 6Li + ion 7Li + ion

Solution 1-22): Whether a particle will obey Fermi or Bose statistics depends on the number of Fermi particlespresent. An even number of Fermi particles implies Bose statistics, while an odd number of Fermi particles impliesFermi statistics. Noting that protons, neutrons, electrons and positrons are all Fermi particles gives

3 particle - 2 protons, 2 neutrons - Bose

3He - 2 Protons, 1 neutrons, 2 electrons - Fermi

H2 molecule - 2 Protons, 2 electrons - Bose Positron - e+ - Fermi 6Li + ion - 3 Protons, 3 Neutrons, 2 electrons - Bose 7Li + ion - 3 Protons, 4 Neutrons, 2 electrons - Fermi

50


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Problem 1-23): Show that, when the density of a gas consisting of particles with mass m issufficiently low and its temperature is sufficiently high so that the condition,

mean de Broglie wavelengthmean distance between particlesis satisfied, one can use Boltzman statistics as a good approximation irrespective of whether the

particles obey Fermi of Bose statistics.

Solution 1-23): First, an expression for the de Broglie wavelength is required. Since , the de Broglie wavelength,

can be expressed as = /pand the p can be given as p 0 ep2/2m, can be expressed as=

p =

0

ep2/2m dp

1=

22

m (1.23.1)

using Mathematicato evaluate the integral for p. For high temperature, e()/kBT 1, thus the expression inboth Eqn (1.90) in Kubo and in Eqn (1.91) in Kubo reduce to

n =e()/kBT

which can be applied, along with e()/kBT 1, to the classical equivalent of the expression from Eqn (1.93) inKubo to obtain

N

e()/kBT d =eV

3

ep

2/2m dp= V

m

22

3/2

This reduces to

N

V m

223/2

(1.23.2)

which is equivalent to the condition given in Eqn (1.95) in Kubo for high temperature, low density systems.Additionally, the expression for in (1.23.1) can be rewritten as

=

22

m 1

3 =

m

22

3/2Applying the above expression, derived from the expression for , to the expression in (1.23.2) gives

N

V 1

3 3 V

N

V

N1/3

(1.23.3)

Now, the average distance between each particle can be found by dividing the volume occupied by the entiresystem by the average number of particles in the system and taking the cube root (to convert it from averagevolume per particle to average distance per particle). Thus, the expression in (1.23.3) implies that the de Brogliewavelength is much less than the average distance between particles as desired.

Alternatively, the problem can be solved by starting with the condition mean de Broglie wavelength meandistance between particles in the form of (1.23.3). Then the expression for the de Broglie wavelength from (1.23.1)could have been applied to give

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V

N

1/3(1.23.3)

22

m V

N1/3

22

m

3/2 V

N

N

V

m

22

3/2(1.23.2)

which is to the condition given in Eqn (1.95) in Kubo for high temperature, low density systems and to the resultfrom (1.23.2).

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Problem 1-24): Let ps be the probability that a system is in a state s with energy Es. Show that ifentropy is defined by

S= kBs

pslnps

the values ps which make Sa maximum under the condition that the mean energy of the system isE, follows the canonical distribution.

Solution 1-24): Sinceps is the probability that the system is in a given state with energy Es, by the definition ofprobabilities

s

ps = 1 (1.24.1)

must hold. Furthermore, since the average energy of the system is E, from the definition of the first moment(average),

s

Esps = E= constant (1.24.2)

must hold as well. Since the desire is to maximize S,

S= kB

s

pslnps

= kB

s

(pslnps)

ps

ps= kB

s

{lnps+ 1} ps = 0 (1.24.3)

must hold, while subject to the constraints given in (1.24.1) and (1.24.2). Thus, using the method of LagrangeMultipliers, expressions for the multipliers 1 and2, such that

0 = S+ 1

s

Esps

+ 2

s

ps

=S+ 1s

Es

psps

ps+ 2

s

psps

ps

=S+ 1s

{Es+ 1} ps

holds, are sought. This can be rewritten by applying the expression for S from (1.24.3) to give

0 = kBs

{lnps+ 1} ps+ 1s

{Es} ps+ 2s

{1} ps

0 =s

{kBlnps kB+ 1Es+ 2} ps

For the expression in (1.24.4) to hold in general,kBlnps kB+ 1Es+ 2 = 0 must also hold in general. Solvingthis forps gives

ps= exp

1Es

kB+

2kB

1

(1.24.4)

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Now, 1 and2 must be such that (1.24.1) and (1.24.2) still hold when ps is replaced by the expression in (1.24.4).Thus, from (1.24.1)

1 = sps= s

exp 1Es

kB+

2kB

1 (1.24.1 a)and from (1.24.2)

E=s

Esps=s

Esexp

1Es

kB+

2kB

1

(1.24.2 a)

SinceEs is, in general, unbounded1/kB must be less than zero for (1.24.1 a) and (1.24.2 a) to both converge.Thus,1 can be defined as 1 = kB, where R+. This allows (1.24.1 a) to be rewritten as

1 =

s

exp

Es+ 2

kB 1

= e2/kB1

seEs =e2/kB1 Z() (1.24.1 b)

by using the definition of the canonical partition function. This implies thatZ() can be given as

Z() = e12/kB

By using the definition1 = kB, R+ and this expression for the partition function,Z, the expression in(1.24.2 a) to be rewritten as

E=s

Esexp

Es+ 2

kB 1

= e2/kB1

s

Ese

Es= 1Z() s

Ese

Es (1.24.2 b)By comparing (1.24.1) with (1.24.1 b) and (1.24.2) with (1.24.2 b) and using the expression for the partitionfunction,Z() = e12/kB , it is clear that

ps= eEs e2/kB1 =eEs

1

Z() (1.24.5)

since,

s

ps= e2/kB1

s

eEs =e2/kB1 Z() = 1Z()Z() = 1

and

s

Esps= e2/kB1

s

EseEs =

1

Z()s

EseEs =

s Ese

Ess e

Es =E

Sinceps can be given as in (1.24.5), ps = eEs 1Z() , the system does indeed follow the canonical distribution.

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Problem 1-25): Derive, from an elementary molecular kinetic theory, Poissons equation,pV =constant, for a quasi-static process of an ideal gas.

Solution 1-25): First an expression for the change in the energy of the system, E, is required. SinceE= F d,Ecan be expressed as

E= F d + F d

Now, F = (Fx, Fy, Fz) soF= (Fx, Fy, Fz). Similarly, d= (x, y, z) so d= (x,y,z). These expressions allowthe expression for Eto be rewritten as

E= F d + F d

= (Fx, Fy, Fz) (x, y, z) ++ (Fx, Fy, Fz) (x,y,z)

=xFx+ yFy+ zFz+

+ Fxx + Fyy+ Fzz

The forces Fx, Fy, and Fz can be expressed as Fx= Axp, Fy = Ayp, and Fz = Azp, respectively; therefore,Fx, Fy, and Fz can be expressed as Fx= Axp + Axp, Fy = Ayp + Ayp, and Fz = Azp + Azp,respectively. This allows the expression for Eto be further refined to give

E= xFx+ yFy+ zFz+

+ Fxx + Fyy+ Fzz

=xAxp + xAxp + yAyp++ yAyp + zAzp + zAzp+

+ Axpx + Aypy+ Azpz

The areas, Ax, Ay, and Az can be expressed as Ax= yz, Ay =xz, andAz = xy, respectively; therefore, Ax, Ay,and Az can be expressed as Ax= yx + xy, Ay = zx + xz, and Az = zx + xz, respectively. Theseexpressions allow the expression for Eto be rewritten as

E= xAxp + xAxp + yAyp+

+ yAyp + zAzp + zAzp+

+ Axpx + Aypy+ Azpz

=xpyz+ xpzy+ ypzx+

+ ypxz+ zpyx + zpxy+

+ xyxp + yxzp + zxyp+

+ yzpx + xzpy+ xyp

= 3p (yzx + xzy+ xyz) + 3xyzp

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If it is noted that the volume occupied by the system, V , can be expressed as V =xyz , then Vcan be expressedas V =yzx + xzy+ xyz. Thus, the expression for Ecan be simplified to give E= 3pV + 3V p. This, inturn, can be further simplified usingpV = 23 Eto give

E= 3pV + 3V p

E3pV

= V

V +

p

p

E2E

= V

V +

p

p

Without loss of generality, the system is considered to be contained within a box. For the one dimensional case thebox is considered to be expanding in only one direction; again, without loss of generality (for the case of onedimensional expansion of the system), the expansion of the volume is said to be in the x-direction. That is to say,that one of the walls perpendicular to the x-direction is moving such that the volume contained within the systemchanges with respect to the change in x, V x. This is illustrated below in Figure 1.25-1.

Figure 1.25-1 : Adiabatic expansion of the volume occupied by the system

Therefore,, the change in the energy of the system, E, can also be given by the change in the kinetic energy of theparticles colliding with the moving wall during some period of time t. Thus, Ecan be given as

E= n t (1.25.1)

whereA is the area of the moving wall, n is the number of particles impacting the wall during the time interval t,and is the change in the average kinetic energy per particle for the particles impacting the wall. Now, can bewritten as = p2x/2m so can be given as

=

px

m

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Defining the speed with which the wall is moving as vx implies that the change in the momentum of a particlestriking the wall is2mvx. This allows to be given as

= 1

2m(px 2mvx)2 p

2x

2m=

4vxpxm + 4m2v2x2m

2vxpx+

02mv2x= 2vxpx

Thev 2x term in above is neglected due to the adiabatic nature of the expansion. That is to say, for the expansion ofthe system to be adiabatic, vx must be small; therefore v

2x must be infinitesimal and can therefore be neglected.

Applying this expression for to the expression for above allows to be expressed as

=

px

m

=

px(2vxpx)

m

=

2vxp2xm

Applying this expression for to the expression for Eand using the definition of average value allows theexpression for Eto be expressed as

E= n2vxp2xm t = n t Ly

0dy Lz

0dz

0

2vxp2xm f(p)d3p= n A vx t

0

p2xm

f(p) d3p= n V

0

p2xm

f(p) d2p (1.25.2)

wheref(p) is the distribution of the particles. The distributionf(p) depends only on pfor the same reason thatthe spatial integration in (1.25.3) is only over y and z , but not x; the region of space of interest is the moving wall.In going ton V fromn A vx t in the expressions in (1.25.3) it is noted that vx t = x, which impliesn A vx t = n A x= n V. Now, noting that

2

0

p2xm

f(p) d2p= 4 p2x2m

= 4

allows the expression for Egiven in (1.25.3) to be rewritten as

E= 2 n V

0

p2xm

f(p) d2p= 4 n V

Usingp = 23 n = 2 n= 3p and pV = 23 E = p= 23 1VE, this expression for Ebecomes

E= 4 n V = 2 2 n V = 2 V p= 43

V

V E (1.25.3)

Combining this expression for Ewith the expression for E in (1.25.1) gives

V

V +

p

p =

E

2E =

1

2E

4

3

V

V E

= 2

3

V

V V

V +

p

p = 2

3

V

V

This expression implies

V

V +

p

p = 2

3

V

V 5

3

V

V +

p

p = 0 5

3 ( ln V) + lnp= ln A (1.25.4)

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