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8.08 Problem Set # 1 Feb. 3, 2010 Due Feb. 10, 2010 Problems: 1. (10 pts.) (Problem 8.1 in K. Huang’s book) A perfect crystal has N lattice sites and M interstitial locations. A energy ∆ is needed to remove an atom from a site and place it in an interstitial, when the number n of displaced atoms is much less than N and M . (a) How many ways are there to remove n atoms from N lattice sites? (b) How many ways are there to place n atoms on M interstitials? (c) Use the microcanonical ensemble to calculate the entropy S as a function of total energy E. Find the temperature T as a function of total energy E. (d) Show that average number n of displaced atoms at temperature T is given by n2 = e−∆/kB T (N − n)(M − n) Obtain n for ∆ ≫ kB T and for ∆ ≪ kB T . (e) Use this model for defects in a solid. Let N = M and ∆ = 1eV. Find the concentration n/N of the defects for temperature T = 300K and T = 1000K. Solution: (a) The number of ways to choose n atoms to remove from N sites is CnN =

N! n!(N − n)!

(b) The number of ways to choose n interstitials out of M is CnM =

M! n!(M − n)!

(c) The number of states at total energy E = n∆ is the number of ways to choose n atoms from N sites and place them in M interstitials, that is Γ(n) = CnN × CnM =

N !M ! n!(N − n)!n!(M − n)!

Using Stirling approximation ln N ! ≈ N ln N − N when N ≫ 1, we obtain the entropy S(E) = ≈ = = =

kB ln Γ(n) [ ] kB N ln N + M ln M − 2n ln n − (N − n) ln(N − n) − (M − n) ln(M − n) [ M N −n M − n] N − M ln + n ln + n ln kB − N ln N −n M −n n n [ M N −n M −n NM ] N kB − N ln − M ln + n ln + n ln + n ln 2 N −n M −n N M n [ n n ] NM ) kB n ln 2 − (N − n) ln(1 − ) − (M − n) ln(1 − n N M

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where Boltzmann constant kB = 1.3806503 × 10−23 m2 kg s−2 K−1 . The temperature is defined through 1 ∂S ∂(kB ln Γ(n)) kB ∂ ln Γ(n) = = = T ∂E ∂(n∆) ∆∂n which gives

∆ ∂ ln Γ(n) (N − n)(M − n) = = ln kB T ∂n n2

(d) The previous equation can be rewritten as n2 = e−∆/kB T (N − n)(M − n) The low-temperature and high-temperature limits are { √ N M e−∆/2kB T n≈ NM N +M

(kB T ≪ ∆) (kB T ≫ ∆)

(e) The conditions satisfy the low-temperature limit: the defect concentration { −20 n e ≈ 2 × 10−9 (T = 300K) −∆/2kB T ≈e ≈ −6 e ≈ 2.5 × 10−3 (T = 1000K) N

2. (10 pts.) (Problem 8.4 in K. Huang’s book) The unwinding of double-stranded DNA is like unzipping a zipper. The DNA has N links, each of them can be in one of two states: a closed state with energy 0 and an open state with energy ∆. A link can be in the open state only when all the links to its left are open (see the figure below).

(a) Show that the partition function of the DNA chain has a form QN =

1 − e−(N +1)∆/kB T 1 − e−∆/kB T

(b) Find the average number of open links in kB T ≪ ∆ limit. Solution: (a) The possible states are labeled by the number of open links n = 0, 1, 2, . . . , N . The state labeled by n has n open links and energy of En = n∆. The partition function is QN =

N ∑

e−βn∆ =

n=0

1 − e−β(N +1)∆ 1 − e−β∆

(b) The average number of open links is n=−

1 ∂ ln QN e−β∆ (N + 1)e−β(N +1)∆ E =− = − ∆ ∆ ∂β 1 − e−β∆ 1 − e−β(N +1)∆

The second term in Eq. (2) is negligible for large N . Therefore at low temperature β∆ ≫ 1, we have n ≈ e−β∆ .

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3. Heat capacity of a spin-1/2 in magnetic field: (10 pts.) Consider a spin-1/2 in a magnetic field B. The Sz = 1/2 state has an energy gµB B/2, and the Sz = −1/2 state has an energy −gµB B/2. Assume the spin is in contact with a heat bath of temperature T . (a) Find the partition function Q and the free energy F of the spin-1/2 system. (b) Find the average energy E of the spin as a function of temperature T . (c) Find the heat capacity of the spin: C = ∂E/∂T . Solution: (a) The partition function is Q(T ) =

∑

e−βH = e−βgµB B/2 + eβgµB B/2 .

where β = 1/kB T . The free energy is F = −kB T ln Q = −kB T ln(e−βgµB B/2 + eβgµB B/2 ) (b) The average energy E is E=−

1 ∂Q ∂ ln Q gµB B gµB B =− = tan( ) Q ∂β ∂β 2 2kB T

(c) The heat capacity is C=

( gµ B )2 ∂E 1 B = kB 2 gµB B ∂T 2kB T ch ( 2k T ) B

4. (10 pts.) Consider a two dimensional ideal gas of N particles. The gas is confined within a square wall of size L. Assume that the mass of the particle is m and the temperature is T . (a) Find the free energy A of the gas. (b) Find the force F exerted on one side of the square wall by the gas. Solution: (a) The partition function Z= where λ =

√

L2N N!

∫ ∏ N d2 p⃗i i=1

h2

e−β

∑ i

p ⃗2i /2m

=

L2N N !λ2N

2π~2 /mkB T . Using ln N ! ≈ N ln N − N , we find A = −kB T ln Z = N kB T [ln(N λ2 /L2 ) − 1]

(b) The pressure is ∂A = N kB T /L2 ∂L2 In 2D the pressure is Force/Length. So the force is P =−

F = P L = N kB T /L

5. Cooling by adiabatic demagnetization: (10 pts.) (a) Consider N spin-1/2 spins in a magnetic field B. Initially, the system has a temperature T . If we slowly reduce the magnetic field to B/2, what becomes the temperature of the system? If we slowly reduce the magnetic field to zero, what becomes the temperature of the system? (Hint: the entropy remains unchanged in the above adiabatic process.) 3

(b) Consider N spin-1/2 spins in a magnetic field B. The spin system is in thermal contact with an ideal gas of N particles in a volume V . Initially, the two systems have a temperature T . Assume gµB B ≫ kB T . If we slowly reduce the magnetic field to zero, what becomes the temperature of the gas? Solution: (a) At temperature T , magnetic field B, using canonical ensemble statistics, we find the partition function to be ∑ Q(T, N ) = e−β(S1 +S2 +...+SN )gµB B all states f or N particles

∑

N ∏

all states f or N particles

i=1

=

=

N ∏

e−βSi gµB B

∑

i=1 all states f or particle i

( =

e−βgµB B/2 + eβgµB B/2

e−βSi gµB B )N

The partition function can also be calculated by noting that the spins are non-interacting (the total energy is simply the sum of the energies of each spin). So the partition function of N spins is given by Q(T, N ) = [Q(T, 1)]N , where Q(1) = e−βgµB B/2 + eβgµB B/2 is the partition function of one spin. Thus the free energy A = −kB T ln Q(T, N ) = −N kB T ln(e−βgµB B/2 + eβgµB B/2 ) = −N kB T ln(2 cosh x) where x =

gµB B 2kB T ,

and entropy S=−

∂A N gµB B = N kB ln(2 cosh x) − tanh x ∂T 2T

In an adiabatic process, entropy remains unchanged. We note from the above expression that the entropy only depends on B/T . Thus B/T remains unchanged in an adiabatic process, which gives Tfinal =

Bfinal Tinitial Binitial

When magnetic field reduces to B/2, temperature reduces to T /2. When magnetic field reduces to zero, temperature also reduces to zero. In this way the system can be “cooled” by adiabatic demagnetization. N gµB B BB (b) For the spin system, we already got S = − ∂A tanh x, where x = gµ ∂T = N kB ln(2 cosh x)− 2T 2kB T . We find that when gµB B ≫ kB T , S → 0 and when B = 0, S = N kB ln 2. Therefore the entropy difference for the spin system is ∆S = N kB ln 2.

Since the spin system and the ideal gas system form an isolated system, whose total entropy remains unchanged in an adiabatic process. ∆SV = SV (Tf inal ) − SV (T ) = −N kB ln 2. Now let us calculate the entropy of the ideal gas system SV . Consider the ideal gas in volume V . The free energy [ ] AV = N kB T ln(nλ3 ) − 1 , √

where λ=

4

2π~2 . mkB T

And entropy

( ) ∂AV 5 v SV = − + ln 3 , = N kB ∂T V 2 λ

where v=

1 V = . n N

Now the condition ∆SV = SV (Tf inal ) − SV (T ) = −N kB ln 2 we obtained above gives us N kB

(5 2

+ ln

) (5 v v ) + ln − N k = −N kB ln 2 B λ3 Tf inal 2 λ3 T ( )2/3

After some algebra, the final temperature of the ideal gas system is Tf inal =

5

1 2

T.