Solution Homework 5
1. Given that the temperature of the cosmic microwave background radiation is 2.725 K,
(a) How many photons are there for every cubic centimeter of the universe? Show your derivation!
The number of photons is calculated as such
Z ∞
N=
hn()iρ()d
0
where hn()i is the average occupation number at energy and temperature T and ρ() is the
density of state. In 3−dimension we have
ρ(k)dk = 2
k 2 dk
4πk 2 dk
=
V
3
π2
( 2π
L)
where the 2 appears to account of the 2 possible modes of the photons and V = L3 . For = h̄ω =
h̄ck or k = /h̄c we have
2 d
ρ()d = V 2
π (h̄c)3
therefore we have
2 d
eβ − 1
N
1
= 2
V
π (h̄c)3
Z ∞
N
(kb T )3
= 2
V
π (h̄c)3
Z ∞ 2
x dx
0
with x = β or = x/β we have
0
ex − 1
R
2 dx
We know 0∞ exx −1
' 2.4 and
N
= 4 × 108 m−3 ' 400cm−3
V
1m3 = 106 cm3
(b) What is the entropy density of the cosmic microwave background?
We know the energy density is
u=
U
π2
(kb T )4 .
=
V
15h̄3 c3
At constant volume
ds =
du
4π 2 kb4 2
=
T dT
T
15h̄3 c3
1
or
s=
4π 2 kb4 3
T ' 2.38 × 10−14 J.K −1 m−3
45h̄3 c3
(c) If all of the matter in the universe can be approximated as one hydrogen atom per cubic meter,
what is the entropy density of the matter in the universe?
We assume ideal gas of hydrogen and we have
s=
S
nQ 5
= nkb (ln
+ )
V
n
2
where n = 1m−3 and nQ = (mkb T /2πh̄2 )3/2 . We have
5
s = kb (ln nQ + ) ' 8.92 × 10−22 J.K −1 m−3
2
(d) Compare the answers of (b) and (c) and comment on this.
In this assumption most of the entropy of the universe comes from the cosmic microwave background
radiation.
(e) What is the energy density of the cosmic microwave background today?
The energy density is given by
u=
U
π2
4
−14
=
J.m−3
3 3 (kb T ) = 4.9 × 10
V
15h̄ c
(f) What is the energy density of the matter today? [Use the approximation from part (c)]
For ideal gas
3
u = kb T + mc2 ' 1.5 × 10−10 J.m−3
2
(g) Compare the answers of (e) and (f) and comment on this, including a comparison with what
you said in (d).
In this assumption most of the energy density of the universe comes from the matter.
(h) What is the free energy density of the cosmic microwave background radiation today?
We have f =
F
V
= u − T s ' −1.6 × 10−14 J.m−3
(i) What was the free energy in the universe back in the epoch when the photons and matter
decoupled, assuming the temperature was 3000 K.
f = u − Ts = −
(kb T )4 π 2
' −0.02J.m−3
45h̄3 c3
(j) Assuming that the expansion of the universe has been isentropic, how much smaller was the
radius of the observable universe then compared with now?
If isentropic then S1 = S2 or V1 T13 = V2 T23 or R1 T1 = R2 T2 where Ri is the radius of the universe.
We have
T2
2.725
R1
=
=
' 9 × 10−4
R2
T1
3000
(k) How much work per unit volume did the cosmic microwave background photons do since the
epoch when the temperature was 3000 K?
For isentropic process dU = δW or W =
W
V
= ∆u =
kb4 π 2
(T24
15h̄3 c3
− T14 ) ' −0.06J.m−3
2. Consider a solid at temperature T as composed of N spin=1 nuclei that do not interact with
one another. There are three quantum states accessible to each nucleus, with quantum numbers
m = −1, 0, or +1. Due to electric interactions with the fields in the solid, the m = −1 and the
m = +1 states have the same energy, , and the m = 0 state has zero energy.
(a) What is the partition function?
We have for one spin
z = 1 + 2e−β
and for N spins
Z = z N = (1 + 2e−β )N
(b) What is the free energy?
We have
F = −kb T ln Z = −kb T N ln(1 + 2e−β )
(c) What is the entropy?
We have
S=−
2N
1
∂F
= kb N ln(1 + 2e−β ) +
β
∂T V
T e +2
(d) What is the total energy for << kT ?
We have
U =−
∂ ln Z
2N
= β
∂β
e +2
or
U = F + TS =
2N
+2
eβ
For << kT we have eβ ' 1 and U ' 23 N
(e) What is the heat capacity for E << kT ?
We have
CV =
For << kT we have CV '
22 N
.
9kb T 2
∂U
∂U ∂β
22 N eβ
=
=
∂T V
∂β V ∂T
kb T 2 (eβ + 2)2
3. Each atom in a fixed lattice of hydrogen ions can be in any one of four possible states: (i) the
ground state (H), with 1 electron and an energy of = −13.6eV ; (ii) an excited state (H ∗ ) with
one electron and an energy of ∗ = −3.4eV ; (iii) a positive ion state (H + ) with no electrons and
an energy of 0 = 0.0eV ; and (iv) a negative ion state (H − ) with 2 electrons and an energy of
2 = −0.7eV . What is the chemical potential at which the average number of electrons per atom
is unity?
The grand partition function for one atom is
∗
Zg = 1 + eβ(µ−) + eβ(µ− ) + eβ(2µ−2 )
and the occupation number for one atom is
∗
eβ(µ−) + eβ(µ− ) + 2eβ(2µ−2 )
hni =
Zg
If hni = 1 we have
µ=
2
= −0.35eV
2
4. A certain Johns Hopkins University teaching assistant has opened a business, ”Daring Damien’s
Parking Garage.” He charges c dollars to park a car for an entire day in one of his N parking
spaces. Each parking space holds only one car at a time. The garage is only open to Johns Hopkins
students. The TA wants to estimate his profit in advance. To do so, he assumes that the entire
Johns Hopkins student body acts as a joint money and car reservoir for the garage. The average
daily pocket money of a student then acts as an effective value of ”temperature” (kT), while the
abundance of cars can be represented as a ”chemical potential”, µ. The average pocket money is
known to be kT = p dollars, and the abundance of cars can be estimated as µ = 0.69p dollars.
(a) What does the TA estimate for the average number of cars parked in the garage per day?
One parking space can be either occupied at an energy c or not at zero energy. So the grand
partition function for one parking space is
Zg = 1 + e(0.69p−c)/p
and the occupancy number is
hni =
1
.
e(c−0.69p)/p + 1
For N parking spots we have
hN i =
N
e(c−0.69p)/p
+1
.
(b) As the average pocket money increases, does the estimated number of cars parked in the garage
increase or decrease?
As p increases e(c−0.69p)/p decreases therefore hN i increases. The more the students have money
the more they can pay!
(c) How many dollars per day, d, does the TA expect to bring in?
d=
cN
e(c−0.69p)/p
+1
(d) The daring TA wishes to maximize his profits. The garage is fully automated and has no
operating expenses. What does Damien choose for the price per car, c?
We maximize d with respect of c
cN (c−0.69p)/p
∂d
N
p e
− (c−0.69p)/p
=0
= (c−0.69p)/p
∂c
e
+ 1 (e
+ 1)2
⇒
c
− 1 = 2e−c/p
p
⇒ c ' 1.46p
5. A molecule of hemoglobin contains 4 distinguishable sites capable of binding O2 molecules.
Each site can bind no more than one molecule. The binding energy of an empty site is zero. When
occupied, the energy level of a site is −0 − n1 where n is the number of remaining sites on the
same molecule that are also occupied by O2 molecules. A collection of N0 hemoglobin molecules are
brought into thermal and diffusive contact with a reservoir of O2 molecules with chemical potential
µ and fixed temperature T (i.e., they are immersed in a large volume of blood plasma with a fixed
concentration of dissolved O2 ).
(a) How many O2 molecules are bound to the hemoglobin molecules after equilibrium is attained?
We have 5 cases: 0, 1, 2, 3, or 4 O2 molecules. There are C4n possibilities to fix n molecules of O2
on 4 different sites. Therefore the grand partition function for one molecule of hemoglobin is
Zg =
4
X
C4n eβ(nµ+0 +(4−n)1 ) = 1 + 4eβ(µ+0 +31 ) + 6eβ(2µ+0 +21 ) + 4eβ(3µ+0 +1 ) + eβ(4µ+0 )
n=0
The average number of molecule bound is
hN i =
4eβ(µ+0 +31 ) + 12eβ(2µ+0 +21 ) + 12eβ(3µ+0 +1 ) + 4eβ(4µ+0 )
Zg
(b) What is the probability that a particular hemoglobin molecule has exactly three oxygen
molecules bound to it?
P (n = 3) =
4eβ(3µ+0 +1 )
Zg
(c) What is the probability that a particular site on a given hemoglobin molecule is occupied?
We call A the event ”the site is occupied”. We call Bn the events ”there are n O2 molecules”. We
have
P (A) = P (A/B1 ) + P (A/B2 ) + P (A/B3 ) + P (A/B4 )
P (A/B1 ) =
eβ(µ+0 +31 )
Zg
, P (A/B2 ) =
⇒ P (A) =
3eβ(2µ+0 +21 )
,
Zg
P (A/B3 ) =
3eβ(3µ+0 +1 )
,
Zg
P (A/B4 ) =
eβ(µ+0 +31 ) + 3eβ(2µ+0 +21 ) + 3eβ(3µ+0 +1 ) + eβ(4µ+0 )
Zg
eβ(4µ+0 )
Zg