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- Short Answer(60 points)
(a) (6 points) If a three dimensional system has 10^6 particles, how many dimen- sions does its phase space have? You may neglect internal degrees of freedom of the particles. Answer: Number of dimensions of phase space = 6N = 6 × 106
(b) (6 points) Sketch the Fermi-Dirac distribution as a function of energy E at T = 0 and at T > 0. Solution: The Fermi-Dirac distribution is given by
f (E) =
eβ(E−μ)^ + 1
where μ is the chemical potential. At T = 0, μ(T = 0) = EF where EF is the Fermi energy. At T = 0, single particle states with E < EF are occupied while those states with E > EF are empty. So at T = 0, the Fermi-Dirac distribution function is a step function.
F
T = 0
Fermi Distribution Function
f (E)
E
T > 0
E
(c) (6 points) Consider a one dimensional line of noninteracting spins in a mag- netic field H = H ˆz. Each spin has a magnetic momentμ⃗. The energy of a spin in a magnetic field is E = −μ⃗ · H⃗. i. (3 points) What is/are the equilibrium spin configuration(s) at tempera- ture T = 0? You may make a sketch. Explain your answer. Answer: At T = 0 the system is in its ground state. This lowest energy state is where the spins point up in the same direction as the magnetic field.
H
ii. (3 points) What is/are the equilibrium configuration(s) at very high tem- peratures (μH ≪ kB T )? You may make a sketch. Explain your answer. Answer: At high temperatures, the spins point in random directions. The spins want to minimize their free energy F = E − T S and maximize their entropy S. Pointing in random directions corresponds to most of the configurational states. S = kB ln Ω where Ω is the number of states available to the system and kB is Boltzmann’s constant.
H
(f) (6 points) What is the most important fundamental assumption of statistical mechanics? Answer: An isolated system in equilibrium is equally likely to be in any of its accessible states.
(g) (6 points) Consider a system with N noninteracting four dimensional harmonic oscillators at temperature T. What is the mean potential energy P E of the system? What is the mean kinetic energy KE of the system? Answer: From the equipartition theorem
P E = 4
N kB T
= 2 N kB T KE = 4
N kB T
= 2 N kB T
P E = 2N kB T KE = 2N kB T
(h) (6 points) Deuterium is an isotope of hydrogen that is used to make so-called heavy water. Its nucleus consists of a proton and a neutron. Is a neutral deuterium atom (^2 H) a boson or a fermion? Explain your reasoning. Solution: A neutral deuterium is a fermion. It has one proton, one neutron and one electron. Adding 3 spin-1/2 particles yields a half-integer spin, so it is a fermion. (i) (6 points) In the pressure versus temperature phase diagram for 3 He, the melting curve has a negative slope (dp/dT < 0). Like most substances, solid (^3) He is denser than liquid 3 He. Suppose we have an equilibrium mixture of solid and liquid 3 He. If a small amount of heat Q is added to this mixture, will some solid converted into liquid or will some liquid be converted into solid? Or neither? Assume the mixture is still an equilibrium mixture of solid and liquid after the heat is added (though with possibly different proportions of liquid and solid). Explain your reasoning. Answer: Liquid 3 He will be converted into solid. To understand why, note that from the Clausius-Clapeyron equation:
dp dT
∆s ∆v
where s is the entropy per mole and v is the volume of a mole. Since the solid is denser than the liquid, the volume of 1 mole of the solid is less than the volume of 1 mole of the liquid. So
∆v = vsolid − vliquid < 0 (3)
Thus the Clausius-Clapeyron equation implies that
∆s = ssolid − sliquid > 0 ssolid > sliquid
So when heat Q is added, the entropy of the mixture increases by an amount
∆S =
Q
T
This added entropy goes into converting liquid into solid which is the higher entropy phase. As explained in the typed lecture notes (Lecture 7), the spin entropy in solid 3 He is much larger than the spin entropy in the liquid phase. Note that when solid and liquid coexist, the temperature T stays constant as long as P is constant.
- (10 points) Quantum Mechanical Harmonic Oscillator: The energy levels of a quantum mechanicial harmonic oscillator are given by:
εn = n¯hω
where n = 0, 1 , 2 , .... We have set the zero of energy to coincide with the energy of the state n = 0, i.e., we set the zero point energy to zero. What is the average energy E of this oscillator at temperature T? Answer:
Z =
X n
e−βεn^ =
X n
e−βn¯hω^ =
1 − e−β¯hω
E = −
∂ ln Z ∂β
=
∂β
ln
1 − e−β¯hω
1 − e−β¯hω
∂β
−e−β¯hω
hωe¯ −β¯hω 1 − e−β¯hω =
¯hω eβ¯hω^ − 1
E = (^) eβ¯hω¯hω (^) − 1
- (10 points) A solid is heated from an initial temperature Ti to its melting tempera- ture Tm. Over this temperature range, it has a heat capacity given by C(T ) = AT 3 where A is a constant and T is temperature. Once the solid reaches the melting temperature, it melts with a latent heat of Lm. What is the total change in entropy ∆S when this solid is heated from Ti to Tm and then melts completely at Tm? Solution: Split the problem into 2 parts: 1) Entropy change during heating from Ti to Tm, and 2) Entropy change during melting.
- To find the entropy change during heating, start with
∆S =
Z (^) Tm
Ti
C(T )
T
dT
Z (^) Tm
Ti
AT 3
T
dT
= A
Z (^) Tm
Ti
T 2 dT
A
T (^) m^3 − T (^) i^3
- The entropy change during melting is given by
∆S =
Lm Tm
∆S = A 3 (T (^) m^3 − T (^) i^3 ) + L Tmm
or
ρd^2 k =
A
(2π)^2
d^2 k (12)
where d^2 k ≡ dkxdky is an element of area in 2D “k space.” Notice that the number
of states ρ is independent of ⃗k and proportional to the area A under consideration. So the “density of states”, i.e., the number of states per unit area, lying between ⃗k and ⃗k + d⃗k is d^2 k/(2π)^2 which is a constant independent of the magnitude or shape of the area A. Note that ρ denotes the number of single particle states.
Since k–space is isotropic in this case, i.e., the same in every direction, then the number of states in a circular annulus lying between radii k and k + dk is
ρkdk = ρd^2 k =
A
(2π)^2
(2πkdk) =
A
2 π
kdk (13)
At T = 0 the Fermi distribution is a step function that tells us that the states below the Fermi energy are occupied and those above are unoccupied. The Fermi energy is the energy of the highest occupied state. The chemical potential at T = 0 is called the Fermi energy, EF = μ(T = 0). The Fermi energy depends on how many electrons there are; the more electrons, the larger the Fermi energy. More precisely, the higher the density of electrons, the higher the Fermi energy. We can calculate this dependence as follows. Since all states with k ≤ kF are occupied at T = 0, we have
N = 2
A
(2π)
Z (^) kF
0
kdk = 2
A
(2π)
k^2 F
A
(2π)
k^2 F (14)
where the factor of 2 is for the 2 possible spin states of an electron. Solving for kF yields
kF =
s 2 πN A
So the Fermi energy is
EF =
¯h^2 k^2 F 2 m
¯h^2 2 m
2 πN A
π¯h^2 m
N
A
EF = (π¯h^2 N )/(mA)
- (10 points) For blackbody radiation, the mean energy of the photon gas is E = bV T 4 where V is the volume of the cavity containing the photon gas, T is the equilibrium temperature of the photon gas, and b is a constant. The mean pressure on the walls of the cavity is p = E/ 3 V.
(a) (5 points) Suppose the cavity is in contact with a heat bath at temperature T. If the cavity expands isothermally and reversibly from volume Vi to volume Vf , how much heat Q is absorbed by the system? Solution: Denote the heat absorbed by the system by Q. From the first law of thermodynamics,
dE = dQ − pdV dQ = dE + pdV
At constant temperature
E = bV T 4 dE = bT 4 dV p =
E
3 V
bV T 4 3 V
b 3
T 4
dQ = dE + pdV
= bT 4 dV +
b 3
T 4 dV
Q =
bT 4
Z (^) Vf
Vi
dV
bT 4 (Vf − Vi)
This is the heat absorbed by the system.
Q = 43 bT 4 (Vf − Vi)
