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Solutions to sample quiz problems and assigned problems Sample Quiz Problems Quiz Problem 1. Prove the expression for the Carnot efficiency for a perfectly reversible Carnot cycle using an ideal gas. Solution: The ideal Carnot cycle consists of four segments as follows (1) An isothermal expansion during which heat QH is added to the system at temperature TH ; (2) an adiabatic expansion during which the gas cools from temperature TH to TC ; (3) An isothermal compression during which heat QC is extracted from the system to a cold reservoir at temperature TC ; (4) Adiabatic compression during which the temperature of the gas rises from TC to TH. The efficiency of the engine is given by η = W/QH , where W is the useful work done by the system and QH is the heat added to the system. This is a thermodynamic cycle provided it is carried out reversibly (i.e. kept very close to equilibrium), and then the internal energy is the same at the beginning and the end of the cycle. In that case, dU = 0, and Wtotal = QH − QL, so we can also write the efficiency as
η = 1 −
QL
QH
TL∆S
TH ∆S
TL
TH
The introduction of Entropy by Clausius greatly simplified understanding of engines.
Quiz Problem 2. Show that in the eigenfunction basis, the von Neuman entropy S = −kB tr(ˆρln(ˆρ)) reduces to the Gibbs form S = −kB
i piln(pi) where the sum is over all eigenstates. Solution. See Eq. (12) of Lecture notes.
Quiz Problem 3. Use Stirling’s approximation to show that,
ln(
N
n
) ≈ −N [pln(p) + (1 − p)ln(1 − p)], where p = n/N (2)
Solution. Using Stirling’s approximation we have,
ln(
N
n
) = ln(N !) − ln(n!) − ln(N − n)! = N ln(N ) − N − nln(n) + n − (N − n)ln(N − n) + (N − n) (3)
which reduces to
ln(
N
n
) = N ln(N ) − nln(n) − (N − n)ln(N − n). (4)
Defining p = n/N gives,
ln(
N
n
) = N ln(N ) − pN ln(pN ) − N (1 − p)ln(N (1 − p)) = −N [pln(p) + (1 − p)ln(1 − p)] (5)
Quiz Problem 4. If S = −kB N [pln(p) + (1 − p)ln(1 − p)], by doing a variation with respect to p find the value of p that gives the maximum entropy. Demonstrate that it is a maximum by showing that the second derivative with respect to p is negative.
Solution. Taking a derivative with respect to p gives, ∂S ∂p = −kB N [ln(p) − ln(1 − p)] (6)
The extrema occur at values of p = p∗ where this expression is zero, which yields p∗ = 1/2. To determine whether this is a minimum of maximum, we take the second derivative to find that,
∂^2 S ∂p^2 = −kB N [
p
1 − p ]p=p∗ = −kB N (7)
Since the second derivative is negative the extremum at p∗ = 1/2 is a maximum. We have thus found the value of the probability at which the entropy is a maximum.
Quiz Problem 5. Use the Master equation to prove the second law of thermodynamics, i.e. in a closed system dS/dt ≥ 0.
Solution. see lecture notes page 8.
Quiz Problem 6. Give three examples of systems where the ergodic hypothesis fails. Explain why it fails in these cases.
Solution. Non-interacting classical gas, as in the absence of interactions the systems is non-chaotic. Coupled harmonic oscillators with no other interactions, as the normal modes do not share energy. Any system with spon- taneous symmetry breaking at low temperature, e.g. magnetic systems. Once in the up magnetized state at low enough temperature the system stays in that state. Similarly once a crystal has formed from the gas phase at low temperature it almost never transforms to another crystal phase.
—————–
Quiz Problem 7. Why is the ideal gas law still a good starting point for high temperature gases, even though a non-interacting gas is non-ergodic?
Solution. Even very small interactions can make the system chaotic. These interactions have little effect on the equilibrium thermodynamic behavior though they are critical to the transport properties.
—————–
Quiz Problem 8. Using pi = e−βEi^ /Z in the Gibbs form for the entropy, show that F = −kB T ln(Z), where F = U − T S is the Helmholtz free energy. Here Z =
i e −βEi (^) is the canonical partition function.
Solution. The Gibbs formula for the entropy is
S = −kB
i
piln(pi). (8)
Using the Boltzmann probability in the canonical ensemble pi = exp(−βEi)/Z, we have,
S = −kB
i
pi[ −Ei kB T −^ ln(Z)] =^
U
T +^ kB^ ln(Z)^ so^ U^ −^ T S^ =^ −kB^ T ln(Z),^ (9)
where we used
i pi^ = 1, U^ =^
i piEi —————–
Quiz Problem 9. If x = e−β(Ei−Ej^ ), so that wji → f (x). Show that the Metropolis algorithm is given by f (x) = min(x, 1).
Solution With this notation, the Metropolis algorithm is: If x > 1, wji = 1, and if x < 1, wji = x. This can be put in one equation as wji = f (x) = min(1, x).
Quiz Problem 16. In a MC calculation of the ferromagnetic Ising model on a square lattice, you calculate the magnetization as a function of temperature as accurately as you can. However your simulations remain rounded at the critical point, instead of showing the behavior m ∼ |Tc − T |β^ expected in the thermodynamic limit. Explain the physical origins of the rounding that you observe. Your experimental colleague does a measurement of a model Ising magnet that consists of quantum dots with around 1000 spins in each quantum dot. She also observes a rounding of the behavior of the magnetization. Provide some reasons why her magnetization experiment also gives a rounded behavior near the critical point.
Solution. Three reasons for rounding of the transition are finites size effects, sample inhomogeneity and non- equilibrium effects. Non-equilibrium effects occur when the system is not fully equilibrated. In experiments all three effects occur, while in simulations the sample inhomogeneity effect is avoided.
Problem 17. Derive the relation,
P V = N kB T +
internal∑
k
~rk · F~k (11)
Explain the physical origin of each of the three terms. This relation is not always true for periodic boundary condi- tions, though rather fortuitiously it is true for periodic boundaries provided the interaction potential is a pair potential.
Solution. See Eqs. (98-101) of the notes.
—————– Problem 18. Show that (δE)^2 = kB T 2 CV (12) Solution. See Eq. (129-132) of the notes.
Solutions to assigned problems Problem 1. Find the relation between pressure and volume for an ideal gas under going an adiabatic process.
Solution:. In an adiabatic process no heat is added to the system, so dU +dW = 0, where dW = P dV. Combining the ideal gas law and the equipartion theorem, we have
dU = αN kB dT = αd(P V ) = α(V dP + P dV ) (13)
where α = DOF/2. Using the first law, we then have,
−P dV = α(V dP + P dV ) or α dP P = −(α + 1) dV V
Integrating gives,
P P 0
V 0
V
)1+1/α^ (15)
which may be written in its most common form,
P V γ^ = constant where γ = α + 1 α (16) For a mono-atomic ideal gas in three dimensions there are three translational degrees of freedom, so DOF = 3, α = 3/2, and γ = 5/3.
Problem 2: Derive the ideal monatomic gas law using kinetic theory. You may use the equipartition result U = 32 N kB T.
Solution: The internal energy of an ideal gas is purely kinetic energy, so that,
U =
N kB T =
i
m < [(vix)^2 + (viy )^2 + (viz )^2 ] >=
N m < ~v^2 > (17)
The pressure is calculated by considering a particle incident normally on a perfectly reflecting wall,
Fx = max = m ∆px δt
2 mvx δt
The time taken for the particle to strike the wall again is δt = 2L/vx, so that
Fwall = 2 mv^2 x 2 L so that P =
i F^ xi A
m
i(v xi^2 ) AL
Noting that V = AL, with
i(v xi)^2 =^1 3 < ~v (^2) > and combining (12) and (14) yields,
P V = m
i
(vix)^2 = N kB T (20)
Problem 3. Find the degeneracy Ω(E, N ) for N a spin 1/2 Ising system with Hamiltonian H = −h
i Si. Solution: The energy of a spin state depends only on the number of up spins Nu, and the number of down spins Nd, but not their arrangement. We then have,
E = − h 2 (Nu − Nd), with degeneracy Ω(E) =
N!
Nu!Nd!
We also have N = Nu + Nd, so that
E = − h 2 (2Nu^ −^ N^ )^ or^ Nu^ =^
N
2 −^
E
h , Nd^ =^
E
h −^
N
Since entropy is the kB ln(Ω(E)), we use Stirling’s approximation to write,
ln(Ω(E)) = N lnN − N − NulnNu + Nu − NdlnNd + Nd = N lnN − NulnNu − NdlnNd (23)
It is useful to introduce the fraction f = Nu/N , so that Nd/N = 1 − f and
ln(Ω) = N lnN − f N ln(f N ) − (1 − f )N ln[(1 − f )N ] = N [f ln(f ) + (1 − f )ln(1 − f )] (24)
which shows that the entropy in this model is extensive. In terms of f , E = (N h/2)(2f − 1). We define = E/N , so that f = 1/ 2 − /h. Finally the entropy is,
S = −kB ln(Ω(E)) = −kB N [(1/ 2 − /h)ln(1/ 2 − /h) + (/h − 1 /2)ln(/h − 1 /2)] (25)
Problem 5. By taking a time derivative of the definition of the density operator in terms of a set of wavefunctions drawn from an ensemble, ˆρ(t) =
j |ψj^ (t)^ >< ψj^ (t)|, derive the von Neumann time evolution equation (12). Solution. Take a time derivative of the definition to find, ∂ ρˆ(t) ∂t
j
∂|ψj (t) > ∂t < ψj (t)| + |ψj (t) > ∂ < ψj (t)| ∂t
Using the Schr¨odinger equation i¯h∂|ψ(t) > /∂t = H|ψ(t) > and its adjoint, yields,
∂ ρˆ(t) ∂t
i¯h
[
j
H^ ˆ|ψj (t) >< ψj (t)| − |ψj (t) >< ψj (t)| Hˆ] or i¯h ∂^ ρˆ(t) ∂t = [ H,ˆ ρˆ] (37)
—————– Problem 6. Prove the central limit theorem for a set of N independent subsystems that each have an energy Ei drawn from a Gaussian distribution with standard deviation σ. You will need to use the integral, ∫ (^) ∞
−∞
e−ax (^2) +bx dx = ( π a )^1 /^2 e b 42 a (38)
Solution: Consider any intensive thermodynamic variable Xi (e.g. E/N ) that has a different value in each of N subsystems. Consider each subsystem to have the same volume and that each subsystem is statistically equivalent. Each subsystem is assumed to have a statistical behavior that is Gaussian distributed, so that,
P 1 (Xi) =
2 πσ^2
e−^
X i^2 2 σ^2 (39)
where we assumed that each subsystem has the same statistical variations, quantified by σ. We would like to calculate the statistical behavior of the quantity
XN = 1 N
∑^ N
i=
Xi (40)
The statistical behavior of XN is given by,
PN (XN ) =
2 πσ^2
)−N/ 2 ∫^ ∞
inf ty
i
dXie−^
X^2 i 2 σ^2 δ(
i
Xi − N XN ) (41)
Using the delta function representation,
δ(α) =
2 π
−∞
eiyαdy (42)
we find,
PN (XN ) =
2 πσ^2
)−N/ 2 ∫^ ∞
inf ty
2 π
−∞
dy
i
dXie−^
X i^2 2 σ^2 eiy(
i Xi−N XN^ )^ (43)
which is equivalent to
PN (XN ) =
2 πσ^2
)−N/ 2 ∫^ ∞
−∞
dy 2 π e−iyN XN^ [I]N^ (44)
where
I =
−∞
dXe−^ 2 Xσ^22 eiyX^ =
2 πσ^2
e−^ σ^22 y 2 (45)
Finally,
PN (XN ) =
−∞
dy 2 π e−iyN XN^ e−^ N σ 22 y^2 = 1 (2πN )^1 /^2 σ e−X
(^2) N / 2 σ (^2) N (46)
where σN = σ/N 1 /^2
Problem 7. Find the entropy of mixing for a system consisting of m different atom types, with ni of each atom type and with the total of atoms, N =
∑m i ni. Solution: The number of ways of arranging the system is given by the multinomial,
Ω({ni}) = N^! n 1 !n 2 !...nm!
Using Stirling’s approximation we find,
ln(Ω) = N lnN − N − [
i
(niln(ni) − ni)] = N lnN − N − [
i
(N piln(N pi) − N pi)] (48)
where pi = ni/N. Using
i pi^ = 1 this simplifies to, S = kB ln(Ω) = −kB
i
piln(pi) (49)
which is the same as the Gibbs entropy.
Problem 8. Derive Stirling’s approximation ln(N !) = N ln(N ) − N.
Solution. Expanding the logarithm of N! as a sum we write,
ln(N !) =
∑^ N
i=
ln(i) ≈
∫ N
1
ln(x)dx = [xln(x) − x]|N 1 = N ln(N ) − N + 1 (50)
In the large N limit the constant “one” is negligible.
Problem 9. Which of the following is an exact differential? a) dU (x, y) = 2xdx + 3xydy; b) dU (x, y) = (− 1 /y)dx + (x/y^2 )dy; c) dU (x, y) = −ydx + xdy? For these three examples, set dU = 0 and solve the differential equations to find y(x). In cases where the differential is exact, find U (x, y).
Solution: a) We have ∂U ∂x = 2x; and
∂U
∂y = 3xy (51)
so that
∂^2 U ∂x∂y = 1/y^2 6 = ∂
2 U
∂y∂x = 3y (52)
Therefore dU IS NOT an exact differential. b) We have ∂U ∂x
= −^1
y ; and ∂U ∂y = x y^2
so that
∂^2 U ∂x∂y
y^2
∂^2 U
∂y∂x
y^2
Problem 12. From the fundamental thermodynamic relation show that, ( ∂μ ∂P
S,N
∂V
∂N
S,P
Solution: Since the independent variables are S, P, N , we choose the enthalpy and write,
dH = T dS + V dP + μdN =
∂H
∂S
P,N
dS +
∂H
∂P
S,N
dP +
∂H
∂N
S,P
dN (65)
hence, ( ∂H ∂P
S,N
= V ;
∂H
∂N
S,P
= μ. (66)
By doing a derivative with respect to N of the first equation in (65) we find, ( ∂ ∂N
∂H
∂P
S,N
S,P
∂V
∂N
S,P
while a derivative with respect to P of the second equation in (65) yields ( ∂ ∂P
∂H
∂N
S,P
S,N
∂μ ∂P
S,N
Because dH is an exact differential, the order of the derivatives of H in Eq. (66) and (67) does not matter so these expressions are equivalent hence proving the Maxwell relation (65).
—————– Problem 13. From the fundamental thermodynamic relation show that, ( ∂CP ∂P
T,N
= −T
∂^2 V
∂T 2
P,N
Solution: Since the independent variables in the expression are T, P , we consider the Gibb’s free energy G(T, P, N ), so that,
dG = −SdT + V dP + μdN =
∂G
∂T
P,N
dT +
∂G
∂P
T,N
dP +
∂G
∂N
T,P
dN (70)
hence, ( ∂G ∂T
P,N
= −S;
∂G
∂P
T,N
= V. (71)
This leads to the Maxwell relation,
−
∂S
∂P
T,N
∂V
∂T
P,N
Now take a temperature derivative at constant pressure of this equation, to find,
−
∂T
∂S
∂P
T,N
P,N
∂^2 V
∂T 2
P,N
We can change the order of the derivatives of the expression on the left hand side, and use the fact that CP = T (∂S/∂T )P to write,
−
∂(CP /T )
∂P
T,N
∂^2 V
∂T 2
P,N
; or
∂CP
∂P
T,N
= −T
∂^2 V
∂T 2
P,N
which solves the problem.
Problem 14. From the fundamental thermodynamic relation show that (here we use β = 1/T )
T
∂N
∂T
V,βμ
∂E
∂μ
T,V
Solution: Since the independent variables are β, V, βμ, we start with the entropy form of the fundamental relation,
dS = βdU + βP dV − βμdN (76)
where β = 1/T. Since the independent variables in the desired expression are β, V, βμ, we define, Y = S −βU +(βμ)N , so that,
dY (β, V, βμ) = −U dβ + βP dV + N d(βμ) =
∂Y
∂β
V,βμ
dβ +
∂Y
∂V
β,βμ
dV +
∂Y
∂βμ
β,V
d(βμ) (77)
hence, ( ∂Y ∂β
V,βμ
= −U ;
∂Y
∂(βμ)
β,V
= N, (78)
that leads to the Maxwell relation,
−
∂U
∂(βμ)
β,V
∂N
∂β
V,βμ
Since β is constant for the left hand side derivative, we can take it outside. For the right hand derivative, we use the chain rule to find
− 1 β
∂U
∂μ
β,V
∂N
∂T
βμ,V
∂T
∂β
βμ,V
which reduces to the equation posed in the problem.
Problem 15. From the fundmental thermodynamic relation show that 1 T 2
∂T
∂βμ
E,V
∂N
∂E
βμ,V
and −ρ^2
∂(S/N )
∂ρ
T,N
∂P
∂T
ρ,N
where ρ = N/V is the number density.
Solution: For the first equation, the variables held constant are βμ and E, so we use the entropy form of the fundamental relation, dS = βdU + βP dV − βμdN. We also define Y = S + (βμ)N to find a relation that has βμ as an independent variable. We then have,
dY (U, V, βμ) = βdU + βP dV + N d(βμ) =
∂Y
∂U
V,βμ
dV +
∂Y
∂V
U,βμ
dV +
∂Y
∂βμ
U,V
d(βμ) (82)
which leads to the Maxwell relation, ( ∂β ∂(βμ)
U,V
∂N
∂U
V,βμ
Using the chain rule on the left hand side yields, ( ∂β ∂(βμ)
U,V
∂β ∂T
U,V
∂T
∂(βμ)
U,V
T 2
∂T
∂(βμ)
U,V
Solution: Since the variables that are kept fixed are T, μ, V , we use the grand potential ΦG = U − T S − μN , so that,
dΦG(T, V, μ) = −SdT + P dV − N dμ = −d(P V ) (94)
Now we take a derivative of this expression with respect to x at fixed T, V, μ. Since the entropy is maximized, the Grand potential is minimized, so that ( ∂ΦG ∂x
T,V,μ
= 0 = −V
∂P
∂x
T,V,μ
Problem 18. Consider a system where the thermodynamics are a function of temperature and another variable x which can be varied by an experimentalist. Assuming that ( ∂E ∂S
x
= T ; and using F = E − T S (96)
show that, ( ∂E ∂x
S
∂F
∂x
T
Solution: We take a derivative of the equation F = E − T S with respect to x at constant T , to find, ( ∂F ∂x
T
∂E
∂x
T
− T
∂S
∂x
T
Using the formula for a non-natural derivative, we have, ( ∂E ∂x
T
∂E
∂S
x
∂S
∂x
T
∂E
∂x
S
Using this result, along with the first of Eq. (96), in Eq. (98) proves Eq. (97).
—————– Problem 19. Show that
CV = −T
∂^2 F
∂T 2
N,V
Solution: The specific heat at constant volume is given by,
CV = T
∂S
∂T
V,N
The Helmholtz free energy is given by,
F = U − T S; so dF = −SdT − P dV + μdN =
∂F
∂T
V,N
dT +
∂F
∂V
T,N
dV +
∂F
∂N
T,V
dN (102)
so that,
S = −
∂F
∂T
V,N
; and
∂S
∂T
V,N
∂^2 F
∂T 2
V,N
Multiplying by T and using Eq. (101) proves the relation (100).
Problem 20. Show that
U =
∂(F/T )
∂(1/T )
N,V
Solution: Defining β = 1/T , we have, ( ∂(βF ) ∂β
N,V
= F + β
∂F
∂β
N,V
= F − T
∂F
∂T
N,V
where the last expression on the right hand side is found using the chain rule. We also have,
dF = −SdT − P dV + μdN =
∂F
∂T
V,N
dT +
∂F
∂V
T,N
dV +
∂F
∂N
T,V
dN (106)
so that ( ∂F ∂T
V,N
= −S. (107)
Therefore, ( ∂(βF ) ∂β
N,V
= F + T S = U (108)
as posed in Eq. (104).
—————– Problem 21. Show that
V
∂P
∂T
μ,N
= S; and V
∂P
∂μ
T,N
= N (109)
Now express the pressure of the ideal classical gas in terms of the variables μ and T and using this expression verify the equations above.
Solution: We use the fact that the Gibbs free energy is μN to write ( ∂G ∂P
T,N
= N
∂μ ∂P
T,N
We also have,
dG =
∂G
∂T
P,N
dT +
∂G
∂P
T,N
dP +
∂G
∂N
T,P
dN = −SdT + V dP + μdN (111)
so that ( ∂G ∂P
T,N
= V. (112)
Using this relation in Eq. (110) proves the second identity in (109). To prove the first relation in (109), we use the triple product identity, ( ∂P ∂T
μ,N
∂μ ∂P
T,N
∂T
∂μ
P,N
From Eq. (114) and G = μN , we have,
−S =
∂G
∂T
P,N
= N
∂μ ∂T
P,N
Using the second relation in Eq. (109) for the second term in the product (113) and using (114) leads to the first relation in (109).
