Homework 1David Sirajuddin

Department of Nuclear Engineering and Engineering PhysicsPhysics 715 - Statistical Mechanics

February 8, 2010

1. Consider the Schrodinger equation:

i~dψ

dt= Hψ

where H is time-independent.

(a) What happens to this equation under the time-reversal transformation: t → −tand all quantities are complex conjugated? What are the physical implications ofyour result?

Under the time-reversal transformation and complex conjugation, Schrodinger’sequation takes the form:

−i~ d

d(−t)ψ∗(~r,−t) = H†ψ∗(~r,−t)

+i~d

dtψ∗(~r,−t) = H†ψ∗(~r,−t)

where the superscripts ∗ and † denote the complex conjugate and the Hermitianadjoint respectively. Given that the Hamiltonian is independent of time, theHermitian adjoint operation returns the same operator since the Hamiltonian isthen self-adjoint (H† = H). Thus,

+i~d

dtψ∗(~r,−t) = Hψ∗(~r,−t)

This is of the same form as the original Schrodinger’s equation. The equation canbe recasted identically if there exists a time-reveral operator L that is invertible,such that,

+Li~d

dtψ∗(~r,−t) = LHψ∗(~r,−t)

i~d

dt(Lψ∗(~r,−t)) = H[Lψ∗(~r,−t)]

i~d

dtψ∗(~r, t) = Hψ∗(~r, t)

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Provided that the Hamiltonian and the time-reversal operator commute. Thus,the procedure explicitly shows that the complex conjugation and time-reversal ofSchrodinger’s equation produces another solution in a special case. The physicalimplications of taking t→ −t suggest there exists no preferred direction in time.That is to say, the procedure informs that Schrodinger’s equation is reversible.

(b) How does your result to part (a) appear to contradict the Second Law of Thermody-namics?

The conclusion of part (a) admits the consequence of reversibility, i.e. a deter-ministic evolution is predictable in both directions of time. The contradiction ofthe Second Law involves the tenet that the entropy S of a system must alwaysincrease. Solutions involving time propogating in the opposite direction providesimplies it is possible for the entropy to decrease, or that extrapolation backwardsin time could admit instances of higher entropy than at later times.

(c) How can you resolve the apparent contradiction of part (b)?

The contradiction is resolved by noting the criterion for its applicability, i.e thecommutation between the Hamiltonian and a time-reversal operator. It is sug-gested that this special case is never realized in reality for macroscopic systems.The Schrodinger equation is an isolated system with its outside interactions mod-eled through the Hamiltonian (in a potential field term). The averaging procedureof the equation to compute tractable quantities ruins the reversability.

2. (Huang 6.3) Consider a system of N free particles in which the energy of each particlecan assume two and only two distinct values, 0 and E(E > 0). Denote by n0 and n1

the occupation numbers of the energy level 0 and E, respectively. The total energy ofthe system is U .

(a) Find the entropy of such a system.

The entropy S is calculated as per the “recipe” provided by Huang, using the definitionS = k ln Γ, where k is Boltzmann’s constant, and Γ is the total number of microstatesthe system may achieve. The quantity Γ is calculated by noting the N free particles mayonly take on two distinct values. The number of states may be modeled as a functionof the ratio U/E (i.e. Γ = Γ(U/E)). This allows a direct computation of the number ofmicrostates by way of the binomial coefficient,

(Nr

)= N !

r!(N−r)! , for the ratio r = U/E,without repeat counting.The choice of counter r = U/E is furnished by use of the total energy U . For example,suppose N = 4 providing for the total number of possible states to be 16. If n0 = 3particles are at zero energy, and n1 = 1 particles are at energy E, then r = U/E =(0+0+0+E)/E = 1, such that

(N1

)= 4 possible states are calculated. This computation

predicts the correct number of states. If N = 4, n0 = n1 = 2, then r = U/E =(0 + 0 + E + E)/E = 2E/E = 2, so that

(N2

)= 6 microstates.

Thus, given this formulation, the entropy S may be expressed as,

S = k ln(

N !r!(N − r)!

)= k [lnN !− ln r!− ln(N − r)!] (1)

2

Each term is of the form lnx!, where x is an integer. An approximation for such termsmay be obtained by approximating the sum as a continuous function,

lnx! = ln [x(x− 1)(x− 1) . . . 2 · 1]= lnx+ ln(x− 1) + . . .+ ln 2 + ln 1

=x∑

m=1

lnm

Multiplying by a form of unity: 1 = (m + 1) - m,

lnx! =x∑

m=1

lnm [(m+ 1)−m]︸ ︷︷ ︸∆m

=x∑

m=1

lnm∆m

'∫ x

1lnmdm

⇒ lnx! ' x lnx− x (2)

This is, of course, Stirling’s approximation. Applying Eqn. (2) with each term in Eqn.(1),

lnN ! ' N lnN −Nln r! ' r ln r − r

ln(N − r)! ' (N − r) ln(N − r)− (N − r)

⇒ ln(

N !r!(N − r)!

)= N lnN −N − r ln r + r − (N − r) ln(N − r) + (N − r)

= N lnN − r ln r −����(N − r)−N ln(N − r) + r ln(N − r) +����(N − r)

= N [lnN − ln(N − r)] + r [ln(N − r)− ln r]

= N ln(

N

N − r

)+ r ln

(N − rr

)= −N ln

(N

N − r

)+ r ln

(N − rr

)ln(

N !r!(N − r)!

)= −N ln

(1− r

N

)+ r ln

(N

r− 1)

Inputting this result into Eqn. (1):

S = k ln Γ(r)

' kr ln(N

r− 1)− kN ln

(1− r

N

)3

Recalling that r = U/E gives a final expression for the entropy S,

S = k

(U

E

)ln

[N

(U

E

)−1

− 1

]− kN ln

[1− 1

N

(U

E

)](3)

(b) Find the most probable values of n0 and n1, and find the mean square fluctuations ofthese quantities.Given that the particles may only take on an energy of 0 or E, the occupation numbersn0 and n1 follow a binomial distribution. The most probable values of n0 and n1

correspond to the modes of the distribution. The proceeding work derives the mode nfor n = {n0, n1} follows that of Ross [2].Given the discrete binomial probability (P ) mass function for N particles

P (i) =(N

i

)pi(1− p)n−i

,where i = 0, 1, . . . , N , and p = 1/2 is the probability of taking on an energy 0 or E, themode is proposed to be the largest integer less than or equal to n = (N + 1)p = b(N +1)/2c. This is to say, the cumulative probability P{N = i} increases monotonically toa peak value (mode) and decreases for a values larger than i corresponding to n. Theproof proceeds by determining the ratio P{n = i}/P{n = i− 1}.

P{n = i}P{n = i− 1}

=N !

i!(N−i)!pi(1− p)N−i

N !(N−i+1)!(i−1)!p

i−1(1− p)N−i+1

=��N !

i���(i−1)!���(N−i)!���pi−1p���

���(1− p)N−i

��N !(N−i+1)���(N−i)!���(i−1)!�

��pi−1���

���(1− p)N−i(1− p)(Sorry for the mess above, I will clean it up if I have enough time)

=(N − i+ 1)pi(1− p)

Thus, an index i corresponds to greater frequency if P{n = i} ≥ P{n = i − 1}, i.e., ifand only if

(N − i+ 1)p ≥ i(1− p)(N + 1)p−��ip ≥ i−��ip

⇒ i ≤ b(N + 1)pc

Identifying the particular value i as the mode n, and the probability of taking on anenergy as p = 1/2 completes the proof,

Most probable occupation number n = b(N + 1)pc (4)4

where n = {n0, n1} is constrained by the criterion n0 + n1 = N

The mean square fluctuation 〈(δn)2〉 is given by,

〈(δn)2〉 =〈n2〉 − 〈n〉2

〈n〉2(5)

Again, owing to both occupation numbers having the same distribution, n = {n0, n1}.Ross indicates the expectation value of the first and second moments of the occupationnumbers are given by [2]:

〈n〉 = Np

〈n2〉 = Np[(N − 1)p+ 1]

Proof of these propositions are not provided, but may be located in Ross [2], or anystandard probability text. For the probability p = 1/2, this admits 〈n〉2 = N2/4, and

〈n2〉 = Np[(N − 1)p+ 1]

=N

2

[N − 1

2+ 1]

=N

2

[N

2− 1

2+ 1]

=N

2

[N

2+

12

]⇒ 〈n2〉 =

N2

4− N

4

Inserting these expressions into Eqn. (5) provides,

〈(δn)2〉 = ��N2

4 −N4 −��

N2

4

N2/4

=−N/4N2/4

=−���N/4N����(N/4)

⇒ = − 1N

Taking the absolute value, the final result may be quoted,

Mean square fluctuation of occupancy number 〈(δn)2〉 =1N

(6)

Which, reassuringly, tends to zero as N → ∞, i.e. the result obeys the strong/weaklaws of large numbers.

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(c) Find the temperatue as a function of U , and show that it can be negative.The temperature T may be obtained from the entropy S calculated in part (a),

T =(∂U

∂S

)V

⇒ 1T

=(∂S

∂U

)V

Where U is the total internal energy of the system. Noting that U = U(S), the partialderivative may translated to an ordinary derivative:

1T

=(∂S

∂U

)V

⇒ 1T

=dS

dU

and the constant volume condition is understood. The differentiation is more conve-niently facilitated by using the form of entropy given above Eqn. (3), where S = S(r),r = U/E. Translating differetial operators, it is written,

d

dU=

d

dr

dr

dU=

1E

d

dr

so that dSdU = 1

EdSdr . Calculating the derivative then admits T = T (r),

1T

=dS

dU

=1E

dS

dr

=1E

d

dr

[kr ln

(N

r− 1)− kN ln

(1− r

N

)]=

k

E

[ln(N

r− 1)− �r

r−�2

Nr − 1

+��N ��1N

1− rN

]

=k

E

[ln(N

r− 1)−����1

N − r+����1

N − r

]=

k

E

[ln(N

r− 1)]

⇒ T (r) =E/k

ln(Nr − 1

)Recalling r = U/E yields the temperature T as a function of total energy U

T (U) =E/k

ln(

NU/E − 1

) (7)

To show that the system may take on negative temperatures, the logarithm argumentmust be between 0 and 1. Explicitly, a condition can be found,

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ln(

N

U/E− 1)

< 0

N

U/E− 1 < 1

N

U/E< 2

⇒ U/E > N/2

A similar calculation for the lower bound, ln[N(U/E)− 1] > 0 gives an upper limit forU/E < N . Thus, the system may have negative temperatures if the ratio N/2 < U/E < N .This statement equivalently implies that more than half the free particles must take onan energy E. For if the total energy U =

∑i ni = n0(0) +n1(E) = n1(E), this indicates

that n1 > N/2 so that U/E > (N/2)(E/E) > N/2.

(d) What happens when a system of negative temperature is allowed to exchange heat witha system of positive temperature?So as to not contradict the Clausius statement, the hotter temperature (T > 0) musttransfer heat to the lower temperature (T < 0) until an equilibrium temperature isreached. Curiously, the result of part (c) shows T (U) to admit a trend that binds onlythe occupancy number n1 (population of particles with energy E). The function T (U)shows that in order to gain higher temperature, n1 must decrease so that n0 increases(concentration of particles with zero energy). The process is counterintuitive in therespect that to increase the temperature of the system, the total energy U must belowered.

3. Find the equations governing an adiabatic transformation of an ideal gas.

A set of equations relating the pressure (P ), temperature (T ), and the volume (V )may be obtained in terms of the ratio of specific heats γ = Cp/CV .

(a) TV γ−1 = constant

Beginning with the first law in differential form,

dU = dQ︸︷︷︸=0, adiabatic

−dW

dU = −dW= −PdV

dU = −PdV (8)

To obtain an expression for the internal energy dU , the definition of specific heatat constant volume, CV is used [1],

CV =

(∂U

∂T

)V

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For an ideal gas, U = U(T ) ⇒ CV =(∂U∂T

)V→ dU

dT. Thus, the differential is

separable and the internal energy may be written as dU = CvdT . Consolidatingthis result with Eqn. (8) gives,

CvdT = −PdV

CvdT = −NkT dVV

(9)

where in the last step, the ideal gas law (PV = NkT ) was used. A substitutionfor the quantity Nk may be extracted as per Huang, by examining the definitionof the specific heats at constant pressure and volume if the enforcement is madethat CV 6= CV (T ) [1].

dU = CV dT

→ U(T ) = CV T + (((((constant︸ ︷︷ ︸

may choose to be zero

U(T ) = CV T

Inserting this result into U in the definition of enthalthy H = U + PV andreplacing P with the ideal gas law (P = NkT ) yields,

H = (CV +Nk)T ⇒(∂H

∂T

)V

=dH

dT= CV +Nk = Cp

as according to the definition of specific heat at constant pressure. This impliesthat

Nk = Cp − CV

Inputting this result into Eqn. (9):

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CvdT = −(Cp − CV )TdV

V

CVdT

T= −(Cp − CV )

dV

VdT

T= −Cp − CV

CV

dV

V

dT

T= −

(CpCV− 1

)︸ ︷︷ ︸

=(γ−1)

dV

V

dT

T= −(γ − 1)

dV

VdT

T+ (γ − 1)

dV

V= 0

lnT + (γ − 1) lnV = constant

ln(TV γ−1) = constant

And, the proposition is proved,

TV γ−1 = constant (10)

(b) PV γ = constant

Beginning with the First Law (dQ = 0), the definition of specific heat at constantvolume, and the ideal gas law result from above,

dU = −PdV First Law

dU = CV dT Definition of specific heat at constant volume

PV = (Cp − CV )T Ideal gas law withNk = Cp − CV

Differentiating the ideal gas law gives,

(Cp − CV )dT = PdV + V dP

→ dT =1

Cp − CV(PdV + V dP )

Inserting this result in for dT in the second equation above,

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dU =CV

Cp − CV(PdV + V dP )

=1

Cp/Cv︸ ︷︷ ︸=γ

−1(PdV + V dP )

→ dU =1

γ − 1(PdV + V dP )

Finally, substituting this form into the First law yields,

PdV = V dP = −(γ − 1)PdV

���PdV + V dP = −γPdV +���PdV

V dP = −γPdVdP

P= −γ dV

VdP

P+ γ

dV

V=

lnP + γ lnV = constant

ln(PV γ) = constant

Which implies,

PV γ = constant (11)

(c) P γ−1T−γ = constant

The final equation may be obtained from a proper combination of the previousresults (Eqns. (10) and (11)) to eliminate the volume dependence.

Considering two states (1 and 2), Eqns (10) and (11) may be written as

T1Vγ−1

1 = T2Vγ−1

2 ⇒ T1

T2

=

(V2

V1

)γ (V2

V1

)−1

and

P1Vγ

1 = P2Vγ

2 ⇒ P1

P2

=

(V2

V1

)γSubstituting the right hand equation of the second into the right hand side of thefirst gives,

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T1

T2

=

(P1

P2

)(P1

P2

)−1/γ

T1

T2

=

(P1

P2

)1−1/γ

T1P−1+1/γ1 = T2P

−1+1/γ2

T1P− γ−1

γ

1 = T2P− γ−1

γ

2

T−γ1 P γ−11 = T−γ2 P γ−1

2

Which shows,

T−γP γ−1 = constant (12)

4. An automobile tire containing N molecules of air at pressure α times atmosphericpressure, ptire = αpair, and at an atmospheric temperature Tair is punctured so thatall the air settles down to a uniform temperature and pressure. What is the resultingchange in entropy? Use the ideal gas law, p = nkT where n is the number density.

Explain why one might have worried about how to model the atmosphere. Should onehold the atmospheric pressure constant? or the volume? Explain why, for a tire ofrealistic size, such fears are groundless.

The process is irreversible, but the change in entropy may notwithstanding be obtainedby a difference between state values when obtained from a reversible path. The air inthe tire begins at a pressure ptire = αpair and at a temperature Tair. After expandinginto the atmosphere (reservoir), the gas must settle at an equilibrium temperature andpressure to that of the atmosphere (Tair and pair)). Labeling the two states: beforepuncture (state 1) and after equilibrium has been reached (state 2), it is evident theprocess is isothermal (T1 = T2 = Tair). Invoking the First Law (dU = 0, U = U(T )),

dQ = dW = pdV

The work pdV may be expressed in terms of other parameters by differentiating theideal gas law,

PV = NkT ⇒ PdV = NkdT − V dP = NkdT −NkT dPP

where the ideal gas law was used again in the final step. The First Law above, thenadmits

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dQ = NkdT −NkT dPP

dQ

T= Nk

dT

T︸ ︷︷ ︸=0,T=constant

−NkdPP

Integrating both sides with the identification of the entropy change ∆S =∫

dQT

,

∫dQ

T≡ ∆S = −Nk ln

(P2

P1

)= −Nk ln

(pair

ptire

)= −Nk ln

(��pair

�pair

)Implying,

∆S = Nk lnα (13)

5. (Huang 1.2)

(a) An engine is represented by the cyclic transformation shown in the accompanyingT − S diagram, where A denotes the area of the shaded region and B the area ofthe region below it. Show that this engine is not as efficient sa a Carnot engineoperating between the highest and the lowest available temperatures.

According to the figure, the efficiency η of the engine may be immediately writtendown,

η =W

Qin12

Sirajuddin, David Homework 1 – Physics 715, Spring 2010

where Qin is the heat input, and the word W = Qin −Qout as usual. Inspectingthe figure, these heats correspond to according to Q =

∫TdS. Thus, with the

identification Qin = A + B, and Qout = B, then W = A + B − B = A. Theefficiency is then seen to be,

η =A

A+B=

1

1 +B/A

For a Carnot engine, the engine operates at the highest and lowest temperaturereservoirs available, and is the shape of a rectangle. The only change is that thearea Acarnot > A. The efficiency of this engine is given as,

ηcarnot =1

1 +B/Acarnot

Comparing the two engines directly may be done with a ratio,

ηcarnotη

=1 +B/A

1 +B/Acarnot

It is immediately seen that since Acarnot > A, the ratio is greater than unity(ηcarnot/η > 1) revealing that the efficiency of the Carnot engine is greater.

(b) Show that an arbitrary reversible engine cannot be more efficient than a Carnotengine operating between the highest and the lowest available temperatures.

This part is an extension of the previous. Using the same figure, suppose anengine is operating with any shaped cycle between the two temperature limitsgiven. A Carnot cycle is a rectangle, while an arbitrary engine is of any shape.Again, the only quantity that changes is the area A on the T-S diagram. Thisarea changes with the particular shape of the cycle on the diagram. Given therelation from before,

η =1

1 +B/A

With B always constant. The efficieny increases with A. However, it is noted thatthe maximum area possible is obtained from a rectangular cycle, which is the sameas a Carnot engine. Thus, Carnot’s theorem is proven: No reversible engine can bemore efficient than a Carnot engine operating between the highest and the lowestavailable temperature.

References

[1] Huang, Kerson Statistical Mechanics 1987. John Wiley & Sons, Inc. Canada.

[2] Ross, Sheldon A First Course in Probability, 6th Ed., Prentice Hall, Inc. , 2002.

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