PHGN341: Thermal Physics
Final Exam
May 6, 2009
NAME: KEY
For each problem, show all work, explain your reasoning, and state any assumptions.
1. (25) (a) Consider the entropy to be a function of the energy (U ), volume (V ), and number of particles (N ), i.e.
S(U, V, N ). Expand the differential, dS, in partial derivatives of the independent variables, {U, V, N } and solve for dU .
Then using the thermodynamic relation: dU = T dS − P dV + µdN , find {T, P, µ} in terms of the partial derivatives of
the entropy.
Solution: Rearrange the thermodynamic identity: T dS = dU + P dV − µdN . Since the entropy can be thought of as
depending on the independent variables, {U, V, N }, then
dS
=
∂S
∂S
∂S
dU +
dV +
dN
∂U
∂V
∂N
Comparing with the thermodynamic identity, one has:
1
T
=
∂S
,
∂U
P
∂S
=
,
T
∂V
µ
∂S
=−
.
T
∂N
(b) From the Sakur-Tetrode expression for the multiplicity of an ideal monotonic gas (found on the equation sheet),
derive an expression for the temperature.
Solution: The multiplicity is:
3 #N
V 4πmU 2 5
e2
,
Ω(U, V, N ) =
N 3h2 N
Therefore, the entropy is S = k log Ω(U, V, N ) = kN log U 3/2 + log{everything else} . Using the entropic definition of
temperature derived in part (a), one finds
∂S
3 kN
1
=
=
,
T
∂U
2 U
or, in the more familiar equipartition form, U = 32 N kT .
"
(c) From the Sakur-Tetrode expression for the multiplicity of an ideal monotonic gas (found on the equation sheet),
derive an expression for the pressure.
Solution: Using the multiplicity from part (b), we have S = k log Ω(U, V, N ) = kN [log V + log{everything else}]. Using
the entropic definition of pressure derived in part (a), one finds
P
∂S
kN
=
=
,
T
∂V
V
or, in the more familiar ideal gas law form, P V = N kT .
2. (25) Consider a spin-1 paramagnetic atom that can have three possible values for its z-component of angular momentum:
m = +1, 0, −1. The atom is in a uniform magnetic field which means its energy depends on its z-component of angular
momentum, i.e. E = −m , where is some energy scale that depends on the magnetic field and the magnetic moment
of the atom. The atom is in thermal contact with a reservoir at temperature, T .
a. Calculate the single atom partition function.
Solution: Z =
P
m
e−βEm = e−β(−1) + e−β(0) + e−β(+1) = 1 + 2 cosh(β)
b. Find the average energy of one atom as a function of temperature.
Solution: The average energy is obtained from the partition function as follows:
Ē = −
sinh(β)
1 ∂Z
= −2
.
Z ∂β
1 + 2 cosh(β)
1
c. For this atom in a 1 T magnetic field, ' 6.0 × 10−5 eV. What is the probability that the atom is in its lowest
energy state (m = +1) at a temperature of 4 K?
Solution: First calculate β =
6×10−5 eV
(8.617×10−5 eV/K)(4K)
= 0.1741. The probability is given by
Pm=+1 =
e−β
= 0.3927.
1 + 2 cosh(β)
3. (20) When a common star burns up its nuclear fuel, it cools and contracts until it becomes a white dwarf which is
prevented from collapsing further by the degenerate (T = 0) electron pressure.
(a) Assume that the nuclei making up the star have an equal number of protons and neutrons. Let mN the average
mass a nucleon and let me be the mass of an electron (mN >> me ). Derive an expression for the Fermi momentum,
pF = h̄kF , of the electrons for a white dwarf star with radius, R? , and mass, M? . (Since the star is electrically neutral,
the number of electrons, Ne , equals the number of protons.) Give your answer in terms of R? and M? and physical
contants.
Solution: The number of electrons is
M?
=
Ne =
2mN
kF
Z
2V
0
4πk 2
V kF3
dk
=
(2π)3
3π 2
The Fermi momentum is given by
h̄
pF = h̄kF =
R?
9πM?
8mN
1/3
.
3GM 2
(b) The gravitational potential energy of a white dwarf star of uniform density is UG = − 5R?? . Treating only the
kinetic energy of the degenerate electrons, the total energy of the star is Etotal = 35 Ne EF + UG , where EF is the
nonrelativistic Fermi energy (p2F /(2me )). From the total energy, find an expression for the equilibrium radius of the
white dwarf star (i.e. the radius which minimizes the total energy).
∂ETotal
∂R?
= 0, or
"
#
2/3 5/3 ∂
3GM?2
M?
GM?2
3
3 h̄2
9π
2
N e EF −
.
=
− 3 +
∂R? 5
5R?
5 4me 8
mN
R?
R?2
Solution: The equilibrium condition is
Solving for R? gives:
R? =
h̄2
2me GM?2
M?
mN
5/3 9π
8
2/3
.
(c) Evaluate the equilibrium radius for a white dwarf of one solar mass, M? = 2.0 × 1030 kg.
Solution: Substituting values for all the constants gives: R? = 7.22 × 106 m.
4. (30) Consider a thin sheet of material of area A with N atoms which can be approximated as a two-dimensional
membrane with an area density of nA = N/A. Let cs be the speed of sound for the membrane. Treat only one phonon
d2 k
polarization; so the density of states is given by A (2π)
2 , where A is the area of the membrane.
(a) Find an expression for the 2-d cut-off wave number, kmax , that gives the correct area density, nA .
Solution: Integrating the density of states in 2-dimensions gives:
Z
Z 2π Z km ax
2
d2 k
A
Akmax
N =A
θ(k
−
k)
=
dθ
dk k =
.
max
2
2
(2π)
(2π) 0
4π
0
2
where θ(x) is the Heaviside step function (θ(+) = 1, θ(−) = 0). Solving gives kmax = (4πnA )1/2 , where nA = N/A.
(b) Find an expression for the 2-d Debye temperature (i.e. the temperature such that kB TD = Ephonon = h̄cs kmax ).
Solution: The Debye temperature is defined by kTD = h̄cs kmax which gives TD =
h̄cs (4πnA )1/2
.
k
(c) Write down the integral expression for the 2-d phonon energy at temperature, T (do not evaluate it).
Solution: The average 2-d phonon energy at temperature, T, is
U=
A
(2π)
(2π)2
Z
kmax
dk k(h̄cs k)
0
1
eβh̄cs k
−1
.
(d) Calculate the phonon contribution to the internal energy in the high temperature limit.
Solution: In the high temperature limit one can expand the denominator term to find: eβh̄cs k −1 ' 1+βh̄cs k +. . .−1 =
βh̄cs k. Substituting this into the integral gives:
Z
A
h̄cs
U'
2π
kmax
dkk 2
0
2
1
A kmax
=
.
βh̄cs k
2πβ 2
(e) Calculate the phonon contribution to the internal energy in the low temperature limit.
Solution: In the low-temperature limit, β is large; so eβh̄cs k >> 1. Thus,
U'
A
h̄cs
2π
Z
kmax
dkk 2 e−βh̄cs k =
0
A k3
T 3.
π (h̄cs )3
(f) (Extra credit) Now lets do the numbers. Suppose the membrane is graphine. The area density of graphine is
nA = 42.2 atoms/nm2 and its speed of sound is cs = 12, 450 m/s. Evaluate your expressions to find the following values
in the indicated units:
Solution: Substituting the values from the previous page, find:
p
i) kmax (in nm−1 ): kmax = 4π(42.2/m3 ) = 23.03 nm−1 .
ii) Debye temperature (in K): TD = 2.189 × 103 K.
iii) Phonon contribution to the heat capacity (in eV/K) per atom in the high temperature limit evaluated at T=5000
cv
K: cv = ∂U
∂T . Using the high temperature limit of U calculated calculated in part (d), we have N = k, independent of
T.
iii) Phonon contribution to the heat capacity (in eV/K) in the low temperature limit evaluated at T= 5 K. Instead, if
we use the low-energy limit in part (e) we find: cNv = 6.3 × 10−5 k.
3