Problems in Statistical Physics and Thermodynamics
1. Calculation of equation of state
Helmholtz free energy of a gas is given by
F(V,T) = −
1
3
C ·V ·T4
,
where C is a constant. Calculate equation of state of a given gas.
Solution: We shall start from the definition of Helmholtz free energy
F(V,T) = E −TS,
from which
dF = −pdV −SdT.
Because this is an exact differential,
∂F
∂V T
= −p
should hold, and, consequently
1
3
C ·T4
= p,
which is the equation of state we were looking for.
2. Gamma function
Gamma function is defined by
Γ(n) :=
∞
0
dt exp(−t)tn−1
.
(a) Prove that
Γ(n+1) = nΓ(n),
(b) evaluate Γ(n), n ∈ N,
(c) evaluate
Γ n+
1
2
, n ∈ N.
Solution: We will start with 2a, because we will use the formula later. From the definition, we shall
express Γ(n+1) and manipulate the expression using integration by parts
Γ(n+1) =
∞
0
dt exp(−t)tn
= −tn
exp(−t)|∞
0
0
+n
∞
0
dt tn−1
exp(−t) = nΓ(n).
To calculate 2b, we shall start with Γ(1):
Γ(1) =
∞
0
dt exp(−t) = −exp(−t)|∞
0 = 1.
Using expression from 2a, we calculate values of gamma function for other natural numbers
Γ(2) = Γ(1+1) = 1·Γ(1) = 1,
Γ(3) = 2·Γ(2) = 1·2,
Γ(4) = 3·Γ(3) = 1·2·3 = 6,
Γ(5) = 4·Γ(4) = 1·2·3·4 = 24;
1
we can obtain formula for general n:
Γ(n+1) = n·(n−1)·····3·2·1 = n!.
We will work out the formula 2c in a same way: we will start with Γ(1/2):
Γ
1
2
=
∞
0
dt exp(−t)t− 1
2 = 2
∞
0
dsexp −s2
=
√
π,
where we have introduced a substitution t = s2 in the last integral. We continue with evaluation of values
of gamma function for other n:
Γ
3
2
= Γ
1
2
+1 =
1
2
Γ
1
2
=
1
2
√
π,
Γ
5
2
=
3
2
Γ
3
2
=
1
2
·
3
2
√
π =
3
4
√
π,
Γ
7
2
=
5
2
Γ
5
2
=
1
2
·
3
2
·
5
2
√
π =
15
8
√
π,
Γ
9
2
=
7
2
Γ
7
2
=
1
2
·
3
2
·
5
2
·
7
2
√
π =
105
16
√
π;
for a general n
Γ n+
1
2
=
2n−1
2
·
2n−3
2
·····
7
2
·
5
2
·
3
2
·
1
2
√
π =
(2n−1)!!
2n
√
π.
3. Stirling’s formula
Using the gamma function, approximate ln(n!) for large n.
Solution:
We will use formula Γ(n+1) = n! from the previous problem. We will rearrange the Gamma function
Γ(n+1) =
∞
0
dt exp(−t)tn
=
∞
0
dt exp(−t)·exp[nln(t)] =
∞
0
dt exp[nln(t)−t].
After a substitution t = n+x
∞
0
dt exp[nln(t)−t] =
∞
−n
dx exp[nln(n+x)−n−x]
≈
∞
−n
dt exp nln(n)−n−
x2
2n
= exp[nln(n)−n]
∞
−n
dt exp −
x2
2n
.
Because the Gaussian function significantly differs from zero just in an interval with width much smaller
than n, we can approximate the lower bound of integral by −∞. Then we can easily integrate the Gaussian
function
exp[nln(n)−n]
∞
−n
dt exp −
x2
2n
=
+∞
−∞
dt exp −
x2
2n
=
√
2πnexp[nln(n)−n].
4. Multidimensional calculations
Determine volume and surface area of n-dimensional sphere.
Solution:
There are two ways how to solve this. One can either calculate the Jacobian matrix of the transformation
2
to n-dimensional spherical coordinates or to try some other way. In 3D case, the sphere is described as a
set of points that fulfill
B3
(R) : x2
1 +x2
2 +x2
3 ≤ R2
, (1)
with a boundary described by
S2
(R) : x2
1 +x2
2 +x2
3 = R2
. (2)
The general form in a higher dimension d:
Bd
(R) : x2
1 +x2
2 +···+x2
d ≤ R2
, (3)
and the boundary
Sd−1
(R) : x2
1 +x2
2 +···+x2
d = R2
. (4)
We define a volume in a general from; in 1D the volume corresponds to the length of a line, in 2D it
correspond to area of circle, etc.,
Vol S1
(R) = 2πR, (5)
Vol S2
(R) = 4πR2
. (6)
Since the unit of volume corresponds to unit of length raised to a power of d, we can write the volume in
terms of unit sphere volume,
Vol Sd−1
(R) = Rd−1
Vol Sd−1
. (7)
I a special case of d = 1 a d = 2 we have:
Vol S1
= 2π, (8)
Vol S2
= 4π. (9)
So, we have to find the volume of unit sphere. Let us assume space Rd with coordinates x1, x2, ...,xd let
r be the radial coordinate,
r2
= x2
1 +x2
2 +···+x2
n. (10)
Let us calculate an integral
Id =
Rd
dx1 dx2 ...dxd exp −r2
. (11)
First method: we divide an integral into d parts each of which is integrated separately, namely
Id =
d
∏
i=1
+∞
−∞
dxie−x2
i =
√
π
d
= π
d
2 . (12)
Second method: let us divide Rd into thin spherical shells. The volume given by r is Sd−1(r) sphere and
the volume of shell between r and r +dr is given by Sd−1(r) multiplied by dr. As a result,
Id =
∞
0
drVol Sd−1
(r) exp −r2
= Vol Sd−1
∞
0
drrd−1
exp −r2
(13)
=
1
2
Vol Sd−1
∞
0
dt exp(−t)t
d
2 −1
, (14)
where we used (7) and performed a substitution t = r2. The last integral can be written in terms of gamma
function,
Id =
1
2
Vol Sd−1
Γ
d
2
. (15)
3
This integral should be also equal to Id = π
d
2 , as we have already calculated. As a result, the unit sphere
volume is equal to
Vol Sd−1
=
2π
d
2
Γ d
2
. (16)
We can easily determine Vol Bd :
Vol Bd
=
1
0
drVol Sd−1
(r) = Vol Sd−1
1
0
drrd−1
= Vol Sd−1 rd
d
1
0
=
Vol Sd−1
d
, (17)
therefore
Vol Bd
=
2π
d
2
dΓ d
2
=
π
d
2
d
2 Γ d
2
=
π
d
2
Γ 1+ d
2
. (18)
5. Occupation numbers of hydrogen energy levels
Let us assume that the hydrogen atom exists in a level with principal quantum number n = 3. Assuming
that the level occupation numbers are given by microcanonical distribution, calculate a probability that
the atom exists in states with the same orbital quantum number l.
Solution: We shall first determine the number of states for each state described by a given l. For n = 3 the
possible values of orbital quantum number are l = {0,1,2}. Each state with a given l is split according
to a given magnetic quantum number m = {−l,−l +1,...,l −1,l}. Furthermore, each of these states is
split according to spin quantum number s = ±1/2. Now the number of possible states with given l is :
• l = 0: m = {0}, s = {±1/2}, this gives two states,
• l = 1: m = {−1,0,+1}, s = {±1/2}, this gives six states,
• l = 2: m = {−2,−1,0,+1,+2}, s = {±1/2}, what gives ten states.
The probability is given by a formula
wi =
# states with l = i
# all possible states
,
from which we can easily find
w0 =
1
9
, w1 =
1
3
, w2 =
5
9
.
6. Expression for entropy
Entropy of an isolated system is given by
S = kB lnΩ, (19)
where Ω is the number of microstates. For a closed system
S = −kB ∑
n
wn lnwn. (20)
Show that these formulae give the same result for an isolated system.
Assume that the system can be divided into a physical object A and thermostat A′, which form an isolated
system. Calculate the entropy of a) the whole isolated system A+A′ and b) sum of entropies of A and A′
and show that they yield the same result.
Solution: For an isolated system, according to the postulate of a priori equal probabilities wn = 1/Ω.
Consequently
S = −kB ∑
n
wn lnwn = −kB
Ω
∑
n=1
1
Ω
ln
1
Ω
= kB lnΩ
Ω
∑
n=1
1
Ω
= kB lnΩ.
4
Let assume the system composed of two subsystems A and A′. Energy of the isolated system is given by
the sum of energies of these subsystems, E0 = E +E′ = const. Let us now calculate the entropy in both
cases. I the case a), Ω(E0 − En) state of A′ system correspond to each n state of A system. Taking the
logarithm of the number of states and expanding to the first order in En
kB ln[Ω(E0 −En)] ≈ kB lnΩ(E0)−
∂
∂E
kB lnΩ(E)En = kB lnΩ(E0)−
∂S
∂E
En = kB lnΩ(E0)−
En
T
, (21)
after exponentiation we arrive at
Ω(E0 −En) ≈ Ω(E0)exp −
En
kBT
. (22)
The total entropy of the isolated system is then
S0 = kB ln ∑
n
Ω(E0)exp −
En
kBT
= kB lnΩ(E0)+kB ln∑
n
exp −
En
kBT
= kB lnΩ(E0)+kB lnZ.
(23)
In the case b) we have to sum up the entropies of both subsystems
S0 = S+S′
= −kB ∑
n
wn lnwn +kB lnΩ(E0 −E) =
−kB ∑
n
1
Z
exp −
En
kBT
−lnZ −
En
kBT
+kB lnΩ(E0)−
∂kB lnΩ(E0)
∂E0
E =
kB lnZ +
1
T ∑
n
1
Z
exp −
En
kBT
En +kB lnΩ(E0)−
E
T
= kB lnZ +
E
T
+kB lnΩ(E0)−
E
T
=
kB lnΩ(E0)+kB lnZ. (24)
This gives the same formula at the end.
7. Heat capacity
Show that cV is given by the fluctuation of energy,
cV =
1
kBT2
∆E2
.
Solution: Mean energy is given by
E = ∑
n
wnEn,
En is constant for V = const., therefore we can rewrite the heat capacity as
cV =
∂E
∂T V
= ∑
n
En
∂wn
∂T V
.
The probability corresponding to each state is
wn = exp
F −En
kBT
,
from which we obtain using differentiation
∂wn
∂T V
=
∂
∂T
F −En
kBT V
exp
F −En
kBT
=
T ∂F
∂T V
−F +En
kBT2
exp
F −En
kBT
.
Now we can apply thermodynamics relation for Helmholtz free energy
F = E −TS, dF = −pdV −SdT,
5
giving us
∂F
∂T V
= −S,
and
E = F +TS;
inserting this to the formula for derivative of wn with respect to temperature
∂wn
∂T V
=
En −E
kBT2
exp
F −En
kBT
=
En −E
kBT2
wn.
The heat capacity takes the form of
cV = ∑
n
Enwn
En −E
kBT2
=
1
kBT2 ∑
n
E2
n wn −E ∑
n
Enwn =
1
kBT2
E2
−E2
,
from which we derive the final relationship
cV =
1
kBT2
∆E2
.
8. Fraction of nitrogen molecules in individual states
Nitrogen atom nucleus 7N14 has a nuclear spin I = 1. Let us assume that that diatomic molecule N2 excite
just rotational states at normal temperatures, but not the vibrational ones. Let us neglect the dynamics of
electrons. Determine fraction of „ortho“ and „para“ molecules in a gas composed of nitrogen molecules.
(„Ortho“ – symmetric spin state, „Para“ – antisymmetric spin state). How the fraction of molecules in
individual states behaves when the temperature of gas approaches zero?
Solution: 7N14 nucleus has a bosonic spin of I = 1; the total wave function of system of two nuclei
should be symmetric. The rotational quantum number J should be odd number for ortho-state to ensure
symmetry of the total wave function. In a para state, J should be even number. Rotational energy of
nitrogen molecule is
EJ =
¯h2
2H
J(J +1), J ∈ N0,
where H is angular momentum. The ratio of population of individual states is
#population of para-nitrogen
#population of ortho-nitrogen
=
∑odd J(2J +1)exp − ¯h2
2HkBT J(J +1)
∑even J(2J +1)exp − ¯h2
2HkBT J(J +1)
·
I +1
I
,
where I is nitrogen nuclear spin. For
¯h2
HkBT
≪ 1
we can replace sums by integrals obtaining
∑
sudé J
(2J +1)exp −
¯h2
2HkBT
J(J +1) =
∞
∑
m=0
(4m+1)exp −
¯h2
2HkBT
2m(2m+1) ≈
∞
0
dm(4m+1)exp −
¯h2
2HkBT
2m(2m+1) =
=
∞
0
dx exp −
¯h2
HkBT
x =
HkBT
¯h2
. (25)
6
The second integral can be evaluated in the same way:
∑
odd J
(2J +1)exp −
¯h2
2HkBT
J(J +1) =
∞
∑
m=0
(4m+3)exp −
¯h2
2HkBT
(2m+1)(2m+2) =
∞
0
dm(4m+3)exp −
¯h2
2HkBT
(2m+1)(2m+2) =
= exp
−¯h2
HkBT
∞
0
dm(4m+3)exp −
¯h2
HkBT
(2m2
+3m) =
= exp
−¯h2
HkBT
∞
0
dyexp −
¯h2
HkBT
y =
HkBT
¯h2
exp
−¯h2
HkBT
. (26)
This after the substitution gives
#para-nitrogen population
#ortho-nitrogen population
=
I +1
I
exp
¯h2
HkBT
≈
I +1
I
;
because I = 1 the ratio is
#para-nitrogen population
#ortho-nitrogen population
=
2
1
. (27)
To calculate the ratio for the case T → 0, we will use
¯h2
HkBT
≫ 1;
in this case the exponential function approaches zero, consequently, we will use just the first term of each
sum,
∑
even J
(2J +1)exp −
¯h2
2HkBT
J(J +1) =
∞
∑
m=0
(4m+1)exp −
¯h2
2HkBT
2m(2m+1) ≈ 1,
∑
odd J
(2J+1)exp −
¯h2
2HkBT
J(J +1) =
∞
∑
m=0
(4m+3)exp −
¯h2
2HkBT
(2m+1)(2m+2) ≈ 3exp −
¯h2
HkBT
.
The population ratio of individual molecules is then
#para-nitrogen population
#ortho-nitrogen population
=
I +1
3I
exp
¯h2
HkBT
→ ∞;
this implies that all nitrogen molecules appear in para-state for temperatures tending to zero.
9. Wien’s displacement law
Derive Wien’s displacement law from the Planck’s law.
Solution: Planck’s law reads
B(ν,T) =
8πν2
c3
hν
exp hν
kBT −1
. (28)
Wien’s displacement law is written in terms of wavelengths, consequently, we will transform (28) into
wavelength space using
B(ν,T)dν = B(λ(ν),T)
∂ν(λ)
∂λ
B(λ,T)
dλ;
the transformation yields
B(λ,T) =
8πch
λ5
1
exp hc
kBTλ −1
. (29)
7
Now we can determine the maximum of the function. Differentiating with respect to wavelength gives
−5+
hc
kBTλ
1
1−exp − hc
kBTλ
= 0.
We denote
xmax =
hc
kBTλmax
,
where subscript „max“ denotes a point, where the function has its maximum. This is given by the solution
of equation
xmax
1−exp(−xmax)
= 5.
From this the wavelength corresponding to the maximum is
λmax =
b
T
, (30)
where
b =
hc
kBxmax
.
10. Stefan–Boltzmann law
Derive the Stefan–Boltzmann law for the amount of energy radiated by a black body per unit area and
per unit of time.
Solution: The specific intensity is defined as
δE = I(ν,n)cosδ ·dΩ·dS·dt ·dν. (31)
Therefore, the specific intensity is given by the amount of radiative energy that passes through a surface
area dS in solid angle dΩ in time dt in frequency range dν. Here δ is an angle between n and dS. During
the time dt the surface intercepts radiation from a volume
dV = dS·cosδ ·c·dt.
The radiation has energy density ε(ν) and comes from all directions. Therefore from dΩ comes
cε(ν)cos(δ)·
dΩ
4π
dν ·dS·dt,
which is equalt to δE above. Moreover, in equilibrium I(ν,n) = B(ν) and therefore
B(ν) =
c
4π
ε(ν).
The total energy emited by the unit surface element per unit of time and frequency is
F(ν) =
δE
dSdt dν
=
Ω+
dΩ
c
4π
ε(ν)cosδ =
c
4π
ε(ν)
2π
0
dϕ
π
2
0
dδ sin(δ)cos(δ)
2π·1
2
=
c
4
ε(ν).
Inserting the energy density
F(ν) =
c
4
8πν2
c3
hν
exp hν
kBT −1
=
2πν2
c2
hν
exp hν
kBT −1
,
after integrating over all frequencies
F = σT4
. (32)
8
11. Rayleigh–Jeans law
Derive Rayleigh–Jeans law using the equipartition theorem.
Solution: Rayleigh–Jeans law describes the equilibrium electromagnetic radiation in a closed cavity.
According to the equipartition theorem, this corresponds to the system of oscillators with energy
Ev = 2·
1
2
kBT = kBT.
Energy of a system of oscillators is given by the sum of energies of individual oscillators. We can approximate
the summation by integration. From the solution for a particle in a box we have
ki =
2πni
Li
,
then the number of states in intervals ∆ni is given by
∆nx∆ny∆nz = Lx ·Ly ·Lz
V
∆kx∆ky∆kz
(2π)3
.
The total energy of all particles can be written as
E = ∑
k
Ek ≈ 2
R3
d3
k
(2π)3
VEk = V
2π
0
dϕ
π
0
dθ
∞
0
dk′ k′2
(2π)3
2kBT = V
∞
0
dk2kBT
k2
(2π)3
4π,
where we perform a substitution k = 2πν/c, yielding
E ≈ V
∞
0
dν 2kBT ·4π
ν2
c3
, (33)
from which
εν ≈
8πν2
c3
kBT. (34)
12. Influence of Sun on Earth
Let us assume that both Sun and Earth radiate as black bodies in empty space. Temperature of Sun is
TS = 6000K. Let us assume that the temperature on Earth is the same everywhere. The radius of Earth is
RE = 6·108 cm and the Earth-Sun distance is d = 1.5·1011 m.
(a) Estimate temperature of Earth.
(b) Determine the radiative force of Sun on Earth.
(c) Compare the results with interplanetary chondrites of spherical shape. Chondrites can efficiently
transfer heat and can be regarded as black bodies. Their radius is d = 0.1cm and let us assume that
they move at the same distance as our Earth, that is d.
Solution:
(a) The total radiative flux coming from our Sun is
F = 4πR2
s σT4
⊙;
the radiative flux per unit of surface at the distance d (Earth) is
F
S
=
4πR2
s σT4
⊙
4πd2
,
therefore the Earth intercepts
F⊕ =
R2
s σT4
⊙
d2
πR2
⊕.
9
Because we assume that the Earth is in a state of thermodynamic equilibrium, it emits as a black
body,
F⊕ = 4πR4
⊕σT4
⊕,
therefore
R2
s σT4
⊙
d2
πR2
⊕ = 4πR4
⊕σT4
⊕,
T⊕ =
R⊙
2d
T⊙ = 290K. (35)
(b) Because the radius of Earth is much smaller than its distance from Sun, we can estimate the force
directly from
F =
F⊕
c
=
R2
s σT4
⊙
d2 πR2
⊕
c
= 6×108
N.
(c) In this case we cane use the same equation as derived above,
T = 290K, F = 1.7×10−11
N.
13. Harmonic oscillator
Determine eigenvectors of linear harmonic oscillator in coordinate representation.
Solution: We will use creation and annihilation operators. Let us remind that ˆa|0 = 0|0 . We will
determine the coordinate representation of this vector
q| ˆa|0 = q|0|0 .
Because the annihilation operator can be written in terms of ˆx and ˆpx as
ˆa =
1
√
2
( ˆq+i ˆp),
where the dimensionless operators are
ˆq =
mω
¯h
ˆx, ˆq =
1
m¯hω
ˆpx.
We shall substitute in a previous equation obtaining
q| ˆa|0 = q
1
√
2
( ˆq+i ˆp) 0 =
1
√
2
( q| ˆq|0 +i q| ˆp|0 ).
The operator can act on the left state,
1
√
2
( q| ˆq|0 +i q| ˆp|0 ) =
1
√
2
q q 0 +
1
√
2
d
dq
q 0 =
1
√
2
qh0(q)+
1
√
2
d
dq
h0(q) = 0.
We obtain equation for the eigenfunction of a given state |0 as
qh0(q)+
d
dq
h0(q) = 0. (36)
Solution is
h0(x) = Cexp −
mω
2¯h
x2
.
C constant can be determined from the norm of eigenvector
dx[h0(x)]2
= 1.
Then
C2
=
mω
π¯h
. (37)
10
The final eigenfunction is
h0(x) = 4 mω
π¯h
exp −
mωx2
¯h
. (38)
To determine the coordinate representation of eigenstate with nonzero n, we shall use the identity ˆa† |n =√
n+1|n+1 . Fro n = 1 we have
q| ˆa†
|0 = q 0 ,
then
q
1
√
2
( ˆq−i ˆp) 1 = q 0 .
We shall rewrite the left hand side
1
√
2
q q 0 +
1
√
2
d
dq
q 0 = q 1 ,
from which
h1(q) =
1
√
2
qh0(q)+
d
dq
h0(q) . (39)
The eigenfunction hn(q) can be written in terms of recursion formula
hn(q) =
1
√
2
qhn−1(q)+
d
dq
hn−1(q) . (40)
14. System of harmonic oscillators
Determine thermodynamic properties of the system of N distinguishable classical harmonic oscillators
with frequency ω.
Solution: The energy of a system of harmonic oscillators is
E =
N
∑
n=1
p2
n
2m
+
1
2
mω2
q2
n . (41)
We will calculate the statistical sum first,
Z =
1
hN
R2N
dN
q·dN
p·exp −
N
∑
n=1
p2
n
2mkBT
+
mω2q2
n
2kBT
=
1
hN
R2N
dN
p·dN
q·
N
∏
n=1
exp
p2
n
2mkBT
+
mω2q2
n
2kBT
=
1
hN
N
∏
n=1 R
dpn ·dqn ·exp
p2
n
2mkBT
+
mω2q2
n
2kBT
=
1
hN
∞
0
dp·exp
p2
2mkBT
N


∞
0
dq exp
mω2q2
2kBT


N
=
1
hN
2mkBTπ ·
2πkBT
mω2
N
=
2πkBT
hω
N
=
kBT
¯hω
N
. (42)
Helmholtz free energy can be determined from
F = −kBT ln(Z) = −kBT ln
kBT
¯hω
N
= −NkBT ln
kBT
¯hω
. (43)
The pressure and entropy are
p = −
∂F
∂V T
= 0, (44)
11
S = −
∂F
∂T V
= NkB ln
kBT
¯hω
+1 . (45)
Energy of the system is
E = F +TS = NkBT.
15. Density distribution in an atmosphere
Determine the density distribution in a gas column with a cross-section A in homogeneous gravitational
field. Let us assume that the gas is composed from indistinguishable particles with mass m.
Solution: We will start with calculation of the canonical partition function, which can be written as
Z =
1
N!(2π¯h)3
R3N ×Ω
d3N
p·d3N
q·exp −∑
n
p2
n
2mkBT
+
mgq
kBT
,
where Ω = {[x,y,z]; x ∈ [−L,L], y ∈ [−L,L], z ∈ [0,∞),L ∈ R+}. One-particle partition function is
Z =
1
(2π¯h)3


R
dp·exp −
p2
2mkBT
Ω
dq·exp −
mgq
kBT

;
the first integral leads to Gaussian function,
R
dp·exp −
p2
2mkBT
= (2πmkBT)3.
The second integral gives
Ω
dq·exp −
mgq
kBT
=
L
−L
dx
L
−L
dy
A
∞
0
dz exp −
mgz
kBT
kBT
mg
= A
kBT
mg
.
Consequently,
Z =
1
(2π¯h)3
(2πmkBT)
3
2 A
kBT
mg
. (46)
The probability of finding the particle in a phase space d3
p·d3
q is given by
dwn =
1
Z
d3
p·d3
q
(2π¯h)3
exp −
p2
2mkBT
exp −
mgq
kBT
=
mg
(2πmkBT)
3
2 AkBT
exp −
p2
2mkBT
exp −
mgq
kBT
d3
p·d3
q.
Our aim is to obtain the probability density in real space q, therefore we shall integrate wn over the
momentum space,
p
dwn =
mg
AkBT
exp −
mgq3
kBT
P
·d3
q,
where P denotes the probability density of finding the particle at the position q. The particle number
density is
n =
Nmg
AkBT
exp −
mgq3
kBT
. (47)
The mass density is
ρ(z) = ρ(0)·exp −
mgq3
kBT
. (48)
12
16. Heat capacity of a gas I
Let us study a gas composed of diatomic molecules. We shall calculate heat capacity per molecule. We
shall account just vibrational motion of molecules with energy given by
En = ¯hω n+
1
2
. (49)
Calculate the partition function, determine the Helmholtz free energy, and the heat capacity. Determine
the approximate behaviour at low and high temperatures.
Solution: The partition function is
Z =
∞
∑
n=0
exp −
−En
kBT
=
∞
∑
n=0
exp −
¯hω n+ 1
2
kBT
= exp −
¯hω
2kBT
∞
∑
n=0
exp −
¯hω
kBT
n . (50)
Using the sum of a geometric sequence we derive
Z =
exp − ¯hω
2kBT
1−exp − ¯hω
kBT
=
2
2 exp ¯hω
2kBT −exp − ¯hω
2kBT
=
1
2sinh ¯hω
2kBT
. (51)
From the partition function we determine the Helmholtz free energy
F = −kBT ln(Z) = kBT ln 2sinh
¯hω
2kBT
.
We will estimate the energy first,
E = F +TS = F −T
∂F
∂T V
.
After substitution and differentiation
E = kBT ln 2sinh
¯hω
2kBT
−kBT ln 2sinh
¯hω
2kBT
+
¯hω
2
cotanh
¯hω
2kBT
=
¯hω
2
cotanh
¯hω
2kBT
.
From this for the heat capacity follows
cV =
∂E
∂T V
=
¯hω
2T
2
1
kB sinh2 ¯hω
2kBT
. (52)
The limiting behaviour at low and high temperatures is
• low-temperature approximation: we will approximate the sinh function using exponentials
cV =
∂E
∂T V
=
¯hω
T
2
1
kB exp ¯hω
2kBT −exp − ¯hω
2kBT
2
.
The first term dominates, therefore
cV ≈
1
kB
¯hω
T
2
exp −
¯hω
2kBT
. (53)
• high-temperature approximation: We perform first-order Taylor expansion of sinh, from which
cV ≈
¯hω
T
2
1
kB
¯hω
kBT
2
= kB. (54)
13
17. Heat capacity of a gas II
Let us study a gas composed from diatomic molecules. Calculate heat capacity per mole of a given gas.
You may account just rotational movement of the molecules with energy given by
Ej,m =
¯h2
j( j +1)
2I
, (55)
where I is molecular moment of inertia. The partition function cannot be calculated analytically, therefore
express this quantity in the limit of high and low temperatures.
Solution: The partition function can be written as
Z = ∑
j
gj exp −
Ej
kBT
, (56)
where gj is the degeneracy factor of a given energy level. After substitution (55) we derive
Z =
∞
∑
j=0
(2j +1)exp −
¯h2
j( j +1)
2IkBT
. (57)
The sum cannot be evaluated analytically, therefore we express the quantity in the limit of high and low
temperatures.the degeneracy factor,
• high-temperature limit: in this case
¯h2
2IkBT
≪ 1,
therefore (57) is in fact left Riemann sum. Therefore, we can approximate
∞
∑
j=0
(2j+1)exp −
¯h2
j( j +1)
2IkBT
≈
∞
0
d j(2j+1)exp −
¯h2
j( j +1)
2IkBT
=
∞
0
dz exp −
¯h2
z
2IkBT
=
2IkBT
¯h2
.
From this the Helmholtz free energy is
F = −kBT ln
2IkBT
¯h2
,
energy
E = kBT,
and heat capacity
cV = kB. (58)
• low-temperature limit: for low temperatures
¯h2
2IkBT
≫ 1;
the exponential tends quickly to zero, and we can account just limited number of summands. Let us
take just tow; in this case the partition function can be approximated as
∞
∑
j=0
(2j +1)exp −
¯h2
j( j +1)
2IkBT
≈ 1+3exp −
¯h2
IkBT
.
We shall proceed as previously estimating the Helmholtz free energy, energy, and heat capacity. The
energy is
E =
3¯h2
I
1
3+exp ¯h2
IkBT
,
and heat capacity is
cV =
3¯h4
kBT2I2
1
3exp − ¯h2
2kBTI +exp ¯h2
2kBTI
2
. (59)
14
18. Unit testing
Show that pressure and energy density share the same unit.
Solution: we shall proceed from definition
[p] =
[F]
[S]
=
kg·m·s−2
m2
= kg·m−1
·s−2
,
[e] =
[E]
[V]
=
kg·m2 ·s−2·
m3
= kg·m−1
·s−2
.
19. Relativistic particles
Calculate density of states of relativistic particles and find the limiting formulae for classical and ultrarelativistic
particles.
Solution: The density of states is given by
ρ(E) =
gV
πd
1
2d
Vol Sd−1 [k(E)]d−1
dE
dk
, (60)
where g is the degeneracy factor, d is the dimension of space, and Vol Sd−1 surface area of d − 1
dimensional sphere. In our case d = 3. Dispersion relation E(k) is given by
E = m2c4 + ¯h2
k2c2,
from which
k =
√
E2 −m2c4
¯hc
.
We substitute in 60) obtaining
ρ(E) =
4πgV
(2π¯h)3
1
c3
E E2 −m2c4. (61)
• Classical limit: Energy is given by
Ecl ≈ E −mc2
= m2c4 + ¯h2
k2c2 −mc2
= mc2
1+
¯h2
k2
m2c2
−mc2
≈
mc2
1+
1
2
¯h2
k2
m2c2
−mc2
=
¯h2
k2
2m
. (62)
• Ultra-relativistic limit – in this case E ≫ mc2 and we can neglect the second term in square root
ρ(E) =
4πgV
(2π¯h)3
1
c3
E2
.
20. Maxwell–Boltzmann distribution
Show that it is possible to derive Maxwell–Boltzmann distribution of velocities from the grand-canonical
distribution.
Solution: The number of bosons in energy interval (E,E +dE) is
dN =
ρ(E)dE
exp E−µ
kBT +1
.
In a classical case
exp −
E − µ
kBT
≪ 1,
therefore we can neglect one in the denominator. After the substitution of ρ(E) we arrive at
dN =
4πgV
(2π¯h)3
√
2m3E exp
E − µ
kBT
dE.
15
The expression can be transformed into velocity space
dE =
∂E
∂v
dv = mvdv,
using classical expression for kinetic energy E = 0.5·mv2
dw =
4πgV
(2π¯h)3
√
2m3
m
2
exp
0.5mv2 − µ
kBT
mv2
dv =
4πgV
(2π¯h)3
m3
v2
exp −
v2
2mkBT
exp −
µ
kBT
dv.
The chemical potential can be derived from the equation for one particle
1 = N =
gV
(2π¯h)3
(2πmkBT)
3
2 F3
2
µ
kBT
=
gV
(2π¯h)3
(2πmkBT)
3
2
1
Γ 3
2
∞
0
dx
x
1
2
exp x− µ
kBT +1
.
The exponential is significantly higher than one, therefore we can approximate
∞
0
dxx
1
2 exp −x+
µ
kBT
≈ exp
µ
kBT
∞
0
dxx
1
2 exp(−x) = exp
µ
kBT
Γ
3
2
.
We substitute in the previous formula
gV
(2π¯h)3
exp
µ
kBT
= (2πmkBT)− 3
2 .
This gives
dw = 4π
m
2πkBT
3
2
v2
exp
v2
2kBT
dv. (63)
21. Derivation of the Planck’s law
Show that the Planck’s law can be derived from grandcanonical distribution of particles with µ = 0.
Solution: The number of bosons in energy interval is
dN =
ρ(E)dE
exp E−µ
kBT −1
.
For photons µ = 0 holds and their energy is equal to E = hν. Transforming to frequencies, we derive the
energy density
EdN =
8πv2
c3
hν
exp hν
kBT −1
. (64)
22. Bosonic a Fermionic integral
Prove that
If(m) =
∞
0
dx·
xm−1
exp(x)+1
= 1−21−m
ζ(m)·Γ(m), (65)
and
Ib(m) =
∞
0
dx·
xm−1
exp(x)−1
= ζ(m)·Γ(m), (66)
where ζ(m) is the Riemann function
ζ(m) =
∞
∑
k=1
1
km
.
16
Solution: We can manipulate If(m) to
∞
0
dx
xm−1
exp(x)+1
=
∞
0
dxxm−1
exp(−x)
1
1+exp(−x)
;
the fraction can be expanded as
1
1±x
= 1∓x+x2
∓..., (67)
which gives
∞
0
dxxm−1
exp(−x)
1
1+exp(−x)
=
∞
0
dxxm−1
exp(−x)
∞
∑
k=0
(−1)k
exp(−kx) =
∞
∑
k=0
∞
0
dx(−1)k
exp[−(1+k)]xm−1
.
In sum we substitute k′ = k+1, take out terms that do not depend on variable x, and perform a substitution
y = k′x
∞
∑
k′=0
∞
0
dx(−1)k′
exp[−(1+k′
)]xm−1
=
∞
∑
k′=1
∞
0
dx(−1)k′
−1
exp(−xk′
)xm−1
=
∞
∑
k′=1
(−1)k−1
k′m
∞
0
dy exp(−y)ym−1
.
Because the integral does not depend on k′, the sum can be evaluated separately. We can separate positive
and negative contributions to the sum, which gives
∞
∑
k′=1
(−1)k−1
(k′)m
=
∞
∑
l=1
1
(2l −1)m
−
∞
∑
l=1
1
(2l)m
.
This can be manipulated in the following way
∞
∑
l=1
1
(2l −1)m
−
∞
∑
l=1
1
(2l)m
=
1
1m
+
1
3m
+
1
5m
+···+
1
(2l −1)m
+...−
1
2m
−
1
4m
−
1
6m
−...−
1
(2l)m
−··· =
1
1m
+
1
2m
+
1
3m
+
1
4m
+···+
1
km
−2
1
2m
+
1
4m
+···+
1
(2l)m
+... =
∞
∑
k=1
1
km
−2
∞
∑
l=1
1
(2l)m
=
∞
∑
k=1
1
km
−21−m
∞
∑
l=1
1
lm
= 1−21−m
∑
k=1
1
km
. (68)
The function is then
∞
∑
k′=1
(−1)k−1
k′m
∞
0
dy exp(−y)ym−1
= 1−21−m
∑
k=1
1
km
∞
0
dy exp(−y)ym−1
= (1−21−m
)ζ(m)Γ(m). (69)
The equality can be proven analogously – we expand according to (67)
Ib(m) =
∞
0
dx·
xm−1
exp(x)−1
=
∞
∑
k=0
∞
0
dxxm−1
exp[−x(1+k)] =
∞
∑
k=1
∞
0
dxxm−1
exp(−xk) =
∞
∑
k=1
1
km
∞
0
dyym−1
exp(−y) = ζ(m)Γ(m). (70)
17
23. Properties of functions
We shall define
Bn(y) =
1
Γ(n)
∞
0
dx
xn−1
exp(x−y)−1
, (71)
and
Fn(y) =
1
Γ(n)
∞
0
dx
xn−1
exp(x−y)+1
. (72)
Prove that
dBn+1(y)
dy
= Bn(y),
and
dFn+1(y)
dy
= Fn(y).
Solution: We will start with
dBn+1(y)
dy
=
d
dy

 1
Γ(n+1)
∞
0
dx
xn
exp(x−y)−1

 =
1
Γ(n+1)
∞
0
dxxn d
dy
1
exp(x−y)−1
.
Because the following identity holds
d
dy
1
exp(x−y)−1
= −
d
dx
1
exp(x−y)−1
,
we can substitute into the integral and integrate by parts
−
1
Γ(n+1)
∞
0
dxxn d
dx
1
exp(x−y)−1
=
1
Γ(n+1)



−
xn
exp(x−y)−1
∞
0
0
+
∞
0
dx
n·xn−1
exp(x−y)−1



.
Using Γ(n+1) = nΓ(n) we derive
dBn+1(y)
dy
d
dy
=
1
Γ(n)
∞
0
dx
xn−1
exp(x−y)−1
= Bn(y). (73)
The fermionic relations can be proven analogously.
24. Bose–Einstein condensate in 2D?
Prove that in there is no Bose–Einstein condensate in d = 2.
Solution: We shall estimate the density of states in 2d:
ρ(E) =
2πgSm
(2π¯h)2
.
Landau potential is
Ω = −
∞
0
dE


∞
0
dE′
ρ(E′
)

 1
exp E−µ
kBT −1
,
where ∞
0
dE′
ρ(E′
) =
2πgSmE
(2π¯h)2
.
18
The potential Ω is equal to
Ω =
2πgSm
(2π¯h)2
∞
0
dE
E
exp E−µ
kBT −1
.
Number of particles is in the 2D case equal to (taking into account µ → 0)
N =
∞
0
dE
ρ(E)
exp E−µ
kBT −1
=
2πgmS
(2π¯h)2
kBT
∞
0
dx
1
exp(x)−1
= .
2πgmS
(2π¯h)2
kBTζ(1)Γ(1).
Because ζ(1) → ∞, any number of particles can appear in the state with E = 0, and there is no BoseEinstein
condensate.
25. Electron-positron gas
At temperatures kBT ≈ mec2 electron-positron pairs are created. Determine equilibrium number of e−
and e+. Solution: We shall start with.
∑
i
γiµi = 0. (74)
From the equation that describes the reaction follows that the number of electrons and positrons is the
same, therefore
µ− + µ+ = 0 = 2µ,
that is, the chemical potential is equal zero. For ultra-relativistic fermions
dNE =
4πgV
(2π¯h)3
1
c3
E2dE
exp E−µ
kBT +1
,
therefore the number of particles is
NE =
4πgV
(2π¯h)3
1
c3
dE
E2
exp E−µ
kBT +1
=
4πgV
(2π¯h)3
(kBT)3
c3
Γ(3)F3
µ
kBT
.
For the case µ = 0 we derive
F3
µ
kBT
= 1−2−2
ζ(3) =
3
4
·ζ(3).
Number of particles is
N+ = N− =
3πgV
4(π¯h)3
(kBT)3
c3
ζ(3).
For g = 2 in particular
N+ = N− =
3V
2π2
kBT
¯hc
3
ζ(3). (75)
26. Number of particles of bosonic gas
From the Landau potential of nonrelativistic bosonic gas
Ω = −kBT
gV
(2π¯h)3
(2πmkBT)
3
2 B5
2
µ
kBT
, (76)
determine number of particles N.
Solution: The number of particles is given by
N = −
∂Ω
∂µ T,V
,
19
substituting from (76) we obtain
N = −kBT
gV
(2π¯h)3
(2πmkBT)
3
2
∂
∂µ
B5
2
µ
kBT
.
From the previous problem can use the identity
dBn+1(y)
dy
= Bn(y),
from which
dBn+1
µ
kBT
dµ
=
1
kBT
Bn
µ
kBT
.
The number of particles is then
N =
gV
(2π¯h)3
(2πmkBT)
3
2 B3
2
µ
kBT
. (77)
27. Adiabatic process equation for photon gas
Determine the adiabatic process equation for photon gas in terms of p and V.
Solution: Helmholtz free energy of photon gas is
F = −
4
3c
σT4
V,
the following relations hold
∂F
∂V T
= −p,
∂F
∂T V
= −S.
From these equations the entropy is
S =
16
3
σT3
V,
and the temperature is
T =
4 3cp
4σ
,
after the substitution
S =
4
5
4
3
1
4
c
3
4 p
3
4 σ
1
4 V = const..
After using more suitable constant, we derive p
3
4 V = const., what can be manipulated into the form
pV
4
3 = const. (78)
28. Fermi gas degeneration
Rewrite the degeneracy condition of Fermi gas kBT ≪ εF in terms of wavelength of the de Broglie wave
corresponding to the thermal motion and Fermi wavelength.
Solution: We substitute into the formula for λT
λT =
2π¯h2
mkBT
≫
2π¯h2
mεF
.
The Fermi wavelength is
λF =
2π¯h
pF
=
2π¯h
√
2mε
=
4π2 ¯h
2mε
,
The inequality can be rewritten as
λT =
2π¯h2
mkBT
≫
2π¯h2
mεF
≈ λF.
From this follows
λT ≫ λF. (79)
20
29. Heat capacity III
Determine the heat capacity cV of fermionic gas and prove the validity of classical limit for cV /N.
Solution: The energy of classical fermionic gas is
E =
3
2
gV
λ3
T
kBTF5
2
µ
kBT
, (80)
and number of particles is
N =
gV
λ3
T
F3
2
µ
kBT
, (81)
where
λT =
2π¯h2
mkBT
. (82)
From this the energy could be written as
E =
3
2
NkBT
F5
2
µ
kBT
F3
2
µ
kBT
.
We denote
Fn
µ
kBT
= Fn.
Heat capacity is
cV =
∂E
∂T N,V
=
3
2
NkB
F5
2
F3
2
+
3
2
NkBT
∂F5
2
∂T F3
2
−F5
2
∂F3
2
∂T
F3
2
2
,
where the differentiation of Fn with respect to T is
∂Fn
∂T V,N
=
∂
∂T
µ
kBT
F′
n =
∂µ
∂T V,N
1
kBT
−
µ
kBT2
Fn−1.
Substituting into the formula for cV
cV =
3
2
NkB
F5
2
F3
2
+
3
2
N
∂µ
∂T V,N
−
µ
T


1−
F5
2
F1
2
F3
2
2


.
The number of particles does not depend on temperature,
∂N
∂T V,N
=
3
2
F3
2
+
1
kB
∂µ
∂T V,N
−
µ
kBT
F1
2
= 0.
From this the temperature derivative of chemical potential is
∂µ
∂T V,N
=
µ
T
−
3
2
kB
F3
2
F1
2
. (83)
After substitution
cV =
3
2
N


1−
F5
2
F1
2
F3
2
2


 −
3
2
kB
F3
2
F1
2
+
3
2
NkB
F5
2
F3
2
=
3
2
NkB



F5
2
F3
2
−
3
2
N
F3
2
F1
2


1−
F5
2
F1
2
F3
2
2






.
21
In a classical case we can approximate Fn by
Fn ≈ exp
µ
kBT
, (84)
then all ratios of F functions are equal to one and we derive the classical limit
cV ≈
3
2
NkB. (85)
Another approach: We shall calculate cV,N first. In
cV,µ = T
∂S
∂T V,µ
,
we express entropy as S = S(N(µ,V,T),V,T), therefore
T
∂S
∂T V,µ
= T
∂S(N(µ,V,T),V,T)
∂T V,µ
= T
∂S
∂N V,µ
∂N
∂T V,µ
+T
∂S
∂T V,N
.
We shall use Maxwell’s relation that follows from the Helmholtz free energy
dF = −SdT − pdV + µdN,
from which
∂S
∂N V,T
= −
∂µ
∂T N,V
.
Because the number of particles is constant
dN =
∂N
∂µ T,V
+
∂N
∂T µ,V
= 0,
we derive
∂µ
∂T N,V
= −
∂µ
∂N T,V
∂N
∂T µ,V
.
We substitute partial derivatives into cV,µ
cV,µ = T
∂N
∂µ
−1
T,V
∂N
∂T
2
µ,V
+T
∂S
∂T V,N
cV,N
.
This gives for the heat capacity
cV,N = cV,µ −T
∂N
∂T
2
µ,V
∂N
∂ µ T,V
. (86)
Now we can substitute results from the statistical physics for fermionic gas. Landau potential is
Ω = −
gV
λ3
T
kBTF5
2
µ
kBT
. (87)
The entropy is from (87)
S = −
∂Ω
∂T V,µ
=
−gVkB
λ3
T
5
2
F5
2
−
µ
kBT
F3
2
.
From this the heat capacity cV,µ is
cV,µ = T
∂S
∂T V,µ
= −
gVkB
λ3
T
15
4
F5
2
−3
µ
kBT
F3
2
+
µ2
k2
BT2
F1
2
.
22
We calculate derivative from (81)
∂N
∂T V,µ
=
gV
λ3
T
3
2T
F3
2
−
µ
kBT2
F1
2
.
Finally,
∂N
∂µ V,T
=
gV
λ3
T
1
kBT
F1
2
.
We substitute this into (86)
cV,N =
gVkB
λ3
T

 15
4
F5
2
−3
µ
kBT
F3
2
+
µ2
k2
BT2
F1
2
−T2

 9
4T2
F2
3
2
F1
2
−
3µ
kBT3
F3
2
+
µ2
k2
BT4
F1
2



.
We divide by kBN and substitute (81)
cV,N
kBN
=
15
4
F5
2
F3
2
−3
µ
kBT
+
µ2
k2
BT2
F1
2
F3
2
−
9
4
F3
2
F1
2
+
3µ
kbT
−
µ2
k2
BT2
F1
2
F3
2
.
In a classical case one can approximate Fn by (84), therefore
cV,N
kBN
=
15
4
−3
µ
kBT
+
µ2
k2
BT2
−
9
4
+3
µ
kBT
−
µ2
k2
BT2
.
All terms except 3/2 cancel out and we finally arrive at
cV,N
kBN
=
3
2
, (88)
what is the same result as in previous case.
30. Classical limit
Prove the validity of the classical limit
E =
3
2
NkBT,
Solution: We use (80) and (81) once again. We approximate Fn as (84). Then the energy is
E =
3
2
gV
λ3
T
kBTF5
2
µ
kBT
≈
3
2
gV
λ3
T
kBTF5
2
exp
µ
kBT
,
and analogously for the number of particles
N ≈
gV
λ3
T
exp
µ
kBT
.
This gives for E ·N/N,
E =
3
2
kBTN. (89)
31. Relativistic fully degenerate fermionic gas
Calculate:
(a) density of states,
(b) Fermi energy, Fermi momentum,
(c) number of particles,
(d) energy,
(e) Landau potential,
(f) equation of state
23
for relativistic (in case 31f ultra-relativistic) fully degenerated fermionic gas. Evaluate number of particles,
energy, and the Landau potential in terms of Fermi energy.
Solution: The assumption of fully degenerated fermionic gas means for temperature T → 0. In this case
lim
T→0+
1
exp E−µ
kBT +1
=



1 E < µ
1
2 E = µ
0 E > µ.
(a) Density of states can be derived from
n(p)dp =
V
(2π¯h)3
gdp,
in this case
n(p)dp =
V
π2 ¯h3
p2
dp.
(b) Because fully degenerated fermionic gas populates levels up to state with Fermi energy or momentum
pF, we obtain after integration of density of states
N =
V
π2¯h3
p3
F
3
.
From this the Fermi momentum follows as
pF = (3π2
)
1
3
N
V
1
3
¯h.
The Fermi energy is
εF = p2
Fc2 +m2
0c4.
(c) We will express the density of states in terms of energy,
n(ε)dε =
V
π2¯h3
√
ε2 −m2c4
c3
εdε.
After substitution ε → mc2t we obtain
n(t)dt =
V mc2 3
π2¯h3
c3
t2 −1tdt.
The number of particles can be obtained by integration with respect to t from 1 to εF/mc2 (which
corresponds to from mc2 to εF before substitution) and by multiplying by the above term,
N =
V mc2 3
π2¯h3
c3
εF
mc2
1
dtt t2 −1,
what gives
N =
V
3π2¯h3
c3
ε2
F −m2
c4
3
2
.
(d) The energy is given by
U =
∞
mc2
dN E =
V
π2¯h3
εF
mc2
dε
√
ε2 −m2c4
c3
=
V mc2 4
π2¯h3
εF
mc2
1
dt t2
−1
1
2
t2
.
24
The integral can be evaluated using substitution t = cosh(x), which, after some manipulation gives
V mc2 4
π2¯h3
εF
mc2
1
dt t2
−1
1
2
t2
=
V mc2 4
8π2¯h3
t 2t2
−1 t2 −1−ln t + t2 −1
t=
εF
mc2
t=1
.
The result can be cast in the form of
U =
V mc2 4
8π2 ¯h3

 εF
mc2
2
ε2
F
(mc2)2
−1
ε2
F
(mc2)2
−1−ln

 εF
mc2
+
ε2
F
(mc2)2
−1



.
(e) We can proceed analogously as in previous case to determine the Landau potential Ω. We will
evaluate
Ω = −
V mc2 4
3π2 ¯h3
εF
mc2
1
dt t2
−1
3
2
.
We integrate as in previous problem obtaining
Ω = −
V mc2 4
8π2 ¯h3

 εF
mc2
2
3
ε2
F
(mc2)2
−1
ε2
F
(mc2)2
−1+ln

 εF
mc2
+
ε2
F
(mc2)2
−1



.
(f) We shall rewrite the formulae obtained above in the case of ultra-relativistic gas. Number of particles
is
N =
V mc2 3
3π2¯h3
εF
mc2
3
,
Landau potential is
Ω = −
V mc2 4
12π2 ¯h3
εF
mc2
4
,
and energy
U =
V mc2 4
4π2¯h3
εF
mc2
4
.
From the identity Ω = −PV follows
P =
mc2 4
12π2 ¯h3
εF
mc2
4
,
or, in terms of number of particles,
P =
3
1
3
4
N
V
4
3
π
2
3 ¯h.
This gives P ∝ ρ
4
3 .
32. Fluctuations
Determine fluctuation of number of particles in the case of grandcanonical distribution ∆N2 = N2 − N 2
.
Evaluate in the case of nonrelativistic bosonic and fermionic gas.
Solution: We shall start manipulating the formula for N
N = ∑
n,N
Nwn,N = ∑
n,N
N
exp −
En,N−µN
kBT
Ξ
25
Now we differentiate the exponential with respect to the chemical potential
∂ exp −
En,N−µN
kBT
∂µ
=
N
kBT
exp −
En,N − µN
kBT
.
Then
∑
n,N
N
exp −
En,N−µN
kBT
Ξ
=
kBT
Ξ ∑
n,N
∂
∂µ
exp −
En,N − µN
kBT
,
the differentiation can be taken out of the sum,
kBT
Ξ
∂
∂µ ∑
n,N
exp −
En,N − µN
kBT
=
kBT
Ξ
∂Ξ
∂µ
.
We proceed analogously to evaluate N2
N2
= ∑
n,N
N2
wn,N = ∑
n,N
N2
exp −
En,N−µN
kBT
Ξ
.
Then
∑
n,N
N2
exp −
En,N−µN
kBT
Ξ
=
kBT
Ξ ∑
n,N
N
∂
∂µ
exp −
En,N − µN
kBT
,
the differentiation can be taken out of the sum and we can expand Ξ
kBT
Ξ
∂
∂µ ∑
n,N
N exp −
En,N − µN
kBT
Ξ
Ξ
=
kBT
Ξ
∂
∂µ
( N Ξ) =
kBT
Ξ
∂ N
∂µ
Ξ+
kBT
Ξ
N
∂Ξ
∂µ
.
Because we have already determined N , we can rewrite
kBT
∂ N
∂µ
+
kBT
Ξ
N
∂Ξ
∂µ
= kBT
∂ N
∂µ
+ N 2
.
From this follows
N2
− N 2
= kBT
∂ N
∂µ
.
We apply this to the case of bosonic gas; the number of particles is
N =
gV
λ3
T
B3
2
µ
kBT
,
then
(∆N)2
=
gV
λ3
T
B1
2
µ
kBT
.
This can be written as
(∆N)2
N2
=
1
N
B1
2
µ
kBT
B3
2
µ
kBT
.
A similar equation can be written also for fermionic gas except the fact that we shall replace B by F.
Therefore, the resulting formula is
(∆N)2
N2
=
1
N
F1
2
µ
kBT
F3
2
µ
kBT
.
26
33. Virial theorem
Derive the virial theorem from the classical mechanics.
Solution: The virial theorem can be derived for the movement of particles in central force field. Let us
study
G = ∑
i
pi ·ri,
where we sum over all particles of the system. Total derivative is
dG
dt
= ∑
i
˙ri ·pi +∑
i
˙pi ·ri. (90)
We can manipulate the first term on the right hand side in the following way
∑
i
˙ri ·pi = ∑
i
mi˙ri · ˙ri = ∑
i
miv2
i = 2T.
The second term can be rewritten as
∑
i
˙pi ·ri = ∑
i
Fi ·ri.
Equation (90) then takes the form of
d
dt ∑
i
p·ri = 2T +∑
i
Fi ·ri. (91)
Taking the mean value by integrating from to τ and divide by τ
1
τ
τ
0
dt
dG
dT
≡
d G
dt
= 2 T + ∑
i
Fi ·ri .
Left hand side can be rewritten as
1
τ
[G(τ)−G(0)] = 2 T + ∑
i
Fi ·ri . (92)
Left hand side tends to zero for periodic motion or for spatially confined system (in this case also G is
bounded from above), therefore for large τ we obtain
T = −
1
2 ∑
i
Fi ·ri . (93)
We can rewrite the right hand term. The force acting on a given particle is given by the mutual interaction
of particles and by pressure,
−
1
2 ∑
i
Fi ·ri = −
1
2
− ∑
i
ri ·∇Π −P
S
dA n·r .
The diversion: Euler’s homogeneous function theorem
A function f of N variable is homogeneous of degree k if
f (tx1,tx2,··· ,txN) = tk
f (x1,x2,··· ,xN).
The Euler’s theorem states that the sum of product of the partial derivatives of homogeneous function
and the variables is equal to the given function multiplied by the degree,
N
∑
n=1
xn
∂ f (x1,x2,··· ,xN)
∂xn
= k f (x1,x2,··· ,xN). (94)
27
End of the diversion.
Let us assume that the potential energy Π is a homogeneous function of coordinates of degree n. In this
case
−
1
2
− ∑
i
ri ·∇Π −P
S
dA n·r =
1
2
(n Π −3PV).
We substitute this to the right hand side of (93), what gives the virial theorem in the form of
2 T −n Π −3PV = 0. (95)
34. Application of the virial theorem
Determine the equation of state using the virial theorem.
Solution: The mean kinetic energy is from the equipartition theorem equal to
T =
3
2
kBT.
We shall evaluate the right hand side term of (93). From the definition of the perfect gas follows that the
mutual interaction of particle is rare in comparison with interaction with walls of the vessel. This is why
we can replace summation with integral over the gas surface. The differential of the force is
dFi = −PndA,
or
1
2 ∑
i
Fi ·ri = −
P
2
dAn·r,
where P is a pressure due to the particle flux, n is vector of the normal and dA is surface element. This
can be rewritten using the divergence theorem as
dAn·r = dV ∇·r = 3V.
We substitute the expression for kinetic energy and potential into (93)
3
2
NkBT =
3
2
PV, (96)
from which follows
NkBT = PV, (97)
This is the state equation. The same result can be derived from (95).
35. Application of the virial theorem II
Studied system contains N weakly interacting particle and its temperature is high enough to use classical
mechanics. Each particle has mass of m and oscillates around the equilibrium position. Determine the
heat capacity for the following cases:
(a) The return force is proportional to deviation x from the equilibrium position.
(b) The return force is proportional to x3.
Calculate without evaluating the corresponding integrals.
Solution: We will use (95) without surface forces, that is with P = 0.
(a) Because the force is proportional to r, the potential is proportional to Π ∝ r2. Consequently, the
potential is a homogeneous function of second order. Therefore, the virial theorem is
2 T −2 Π = 0,
and
T = Π .
28
We insert this result into the law of conservation of energy U = T + Π
U = 2 T = 3NkBT,
from which
U = 3NkBT.
(b) In this case Π ∝ r4, therefore
2 T −4 Π = 0,
from which
U =
9
4
NkBT.
36. Lattice vibrations
Determine the heat capacity due to the lattice vibrations for
(a) Debye model,
(b) Einstein model
of the lattice.
Solution: The energy of vibrations is
u =
1
V
∑i Ei exp(−βEi)
∑i exp(−βEi)
. (98)
Let us assume harmonic lattice composed of N ions, which can be regarded as a system of 3N oscillators.
The contribution of ordinary mode with frequency ωs(k) to the energy is quantized,
En = nks +
1
2
¯hωs(k), (99)
where nks ∈ 0,1,2,···. The total energy is given by a sum of energies of individual modes,
E = ∑
ks
nks +
1
2
¯hωs(k).
Let us define f
f =
1
V
ln ∑
i
exp(−βEi) .
We can easily prove that
u = −
∂ f
∂β
.
Substituting for En, we obtain
f =
1
V
ln ∑
nks
exp −β ∑
ks
¯hωs(k) nks +
1
2
=
1
V
ln ∏
ks
∑
nks
exp −β ¯hωs(k) nks +
1
2
=
1
V
ln


∏
ks
exp −β ¯hωs(k)
2
1−exp[−β ¯hωs(k)]



=
1
V ∑
ks
−
β ¯hωs(k)
2
−ln[1−exp(−β ¯hωs(k))] .
The energy is given by the expression above
u = −
1
V ∑
ks
−
¯hωs(k)
2
−
¯hωs(k)exp(−β ¯hωs(k))
1−exp(−β ¯hωs(k))
=
1
V ∑
ks
¯hωs(k)
1
2
+
1
exp(β ¯hωs(k))−1
.
This can be rewritten as
u =
1
V ∑
ks
¯hωs(k) nks +
1
2
, (100)
29
where
nks =
1
exp(β ¯hωs(k))−1
.
From the comparison with (99) follows that nks excitational number of ordinary mode ks at temperature
T. The classical energy of lattice is
u = ueq
+
1
V ∑
ks
1
2
¯hωs(k)+
1
V ∑
ks
¯hωs(k)
exp(β ¯hωs(k))−1
.
The specific heat capacity is then
cV =
1
V ∑
ks
∂
∂T
¯hωs(k)
exp(β ¯hωs(k))−1
. (101)
(a) Within Debye model we assume
ωs(k) = cs(k0
)k.
Sum in (101) can be replaced by integral (across the first Brillouin zone)
cV =
∂
∂T ∑
s
dk
(2π)3
¯hωs(k)
exp(β ¯hωs(k))−1
.
This integral can be replaced by an integral across the sphere with radius kD chosen in such a way
that it contains exactly N allowed vectors, where N is the number of ions in the lattice. Volume of
k-space corresponding to one wavevector is (2π)3/V, what requires volume of k-space to fill the
volume (2π)3N/V to fill 4πk3
D/3, so kD is
n =
k3
D
6π2
.
We substitute
cV = ∑
s
∂
∂T
dk
(2π)3
¯hcsk
exp(β ¯hcsk)−1
= ∑
s
∂
∂T
2π
0
dΩ
kD
0
dk
(2π)3
¯hcsk3
exp(β ¯hcsk)−1
and perform a substitution
x =
¯hcsk
kBT
,
obtaining
∑
s
∂
∂T
2π
0
dΩ
kD
0
dk
(2π)3
¯hcsk3
exp(β ¯hcsk)−1
=
1
8π3
kBT
¯h
3
∑
s
dΩ
1
c3
s (Ω)
¯hcskD
kBT
0
dx
x4 exp(x)
(exp(x)−1)2
.
We denote
1
c3
=
1
4π ∑
s
dΩ
1
c3
s
(102)
and for kD holds
ωD = kDc,
for Debye temperature ΘD
kBΘD = ¯hωD = ¯hckD.
After substitution,
cv = 9nkB
T
ΘD
3
ΘD
T
0
dx
x4 exp(x)
(exp(x)−1)2
. (103)
30
The derived equation depends just on ΘD. Debye temperature can be derived using low-temperature
limit. At low temperatures, the heat capacity is
cV =
2π2
5
kB
kBT
¯hc
3
,
where we substitute equation for Debye temperature,
kBΘD = ¯hckD = ¯hc
3
√
6π2n,
after the substitution of ¯hc/kB
cv =
12π4
5
nkB
T
ΘD
3
.
(b) Einstein model is more suitable for optical branch and assumes that
ωs(k) = ωE.
The heat capacity can be derived from (101), where we substitute the dispersion relation
cV =
∂
∂β
1
V
¯hωEN
1
exp ¯hωE
kBT −1
= nkB
¯hωE
kBT
2 exp ¯hωE
kBT
exp ¯hωE
kBT −1
2
.
37. Free particle I
Determine the mean coordinate in the case of 1D movement of free particle confined in the region x ∈
[0,L], the density matrix operator is
ρ(x,x′
,β) =
1
L
exp −
π(x−x′)2
λ2
T
. (104)
Solution: The mean value can be determined from
ˆx = Tr( ˆρ ˆx).
The trace is
Tr( ˆρ ˆx) =
L
0
dx′
x′
| ˆρ ˆx|x′
=
L
0
dx′
∞
−∞
dx x′
| ˆρ|x x|ˆx|x′
,
The first matrix element correspond to the density matrix, and the second element is
x|ˆx|x′
= x x x′
= xδ(x−x′
).
We obtain
ˆx =
L
0
dx′
∞
−∞
dx
1
L
exp −
π(x−x′)2
λ2
T
xδ(x−x′
) =
1
L
L
0
dx′
x′
=
L
2
.
38. Free particle II
Determine matrix elements of the density operator of free particle in the momentum representation.
Solution: Density operator is
ˆρ =
exp −β ˆH
Tr −β ˆH
. (105)
We shall start with the partition function. Free particle Hamiltonian is ˆH = ˆp2/(2m):
Z(T,V,1) = Trexp −β ˆH = ∑
p
φp exp −β ˆH φp = ∑
p
exp −
βp2
2m
.
31
We can approximate the sum by an integral
Z(T,V,1) =
V
(2π)3
dpexp −
βp2
2m
=
V
(2π)3
2mπ
β ¯h2
3
2
=
V
λ3
.
From this the matrix elements of the density operator are
φp′ | ˆρ|φp =
λ3
V
exp
−βp2
2m
δpp′ .
39. Free particle III
Determine the mean value of the Hamiltonian of a free particle using a direct calculation in the momentum
representation from
ˆH = Tr ˆρ ˆH . (106)
Solution: We will write the trace as
Tr ˆρ ˆH = ∑
p
p ˆρ ˆH p = ∑
p,p′
p| ˆρ|p′
p ˆH p = ∑
p,p′
λ3
V
exp
−βp2
2m
δpp′
p2
2m
δpp′ .
We approximate the summation by an integral and rewrite the integral in the spherical coordinates
1
¯h2
¯h2
2m
λ3
V
V
(2π)3
dpp2
exp
−βp2
2m
=
1
2m
λ2 2
(2π)2
∞
0
dkk4
exp
−β p2
2m
.
After the substitution we have
ˆH =
¯h2
λ3
m
1
(2π)2
1
2
2m
β ¯h2
5
2
∞
0
dxx
3
2 exp(−x).
From this follows
ˆH =
3
2
kBT.
40. Free particle IV
Calculate the mean value of the Hamiltonian of a free particle from the partition function (in a framework
of quantum physics)
Z =
V
λ3
T
. (107)
Solution: The mean value of the Hamiltonian is
H =
Tr exp −β ˆH ˆH
Tr exp −β ˆH
= −
∂
∂β
ln Tr exp −β ˆH = −
∂
∂β
ln(Z),
now we can substitute
−
∂
∂β
ln(Z) = 3
∂
∂β
λT = 3
1
2
∂
∂β
(lnβ) =
3
2
1
β
=
3
2
kBT.
41. Two fermions
Calculate the matrix elements of density operator and the partition function for two noninteracting fermi-
ons.
Solution: The wave function should be antisymmetric with respect to interchange of particles, therefore
|p1, p2 =
1
√
2
(|p1, p2 −|p2, p1 ).
32
The matrix elements of density operator are according to (105)
p′
1, p′
2 ˆρ |p1, p2 =
1
2Z
p′
1, p′
2 − p′
2, p′
1 exp −
ˆp2
1 + ˆp2
2
2mkBT
(|p1, p2 −|p2, p1 )
=
1
2Z
exp −
p2
1 +p2
2
2mkBT
p′
1, p′
2 p1, p2 − p′
1, p′
2 p2, p1 − p′
2, p′
1 p1, p2 + p′
2, p′
1 p2, p1
1
2Z
exp −
p2
1 +p2
2
2mkBT
δ p′
1 −p1 δ p′
2 −p2 −δ p′
1 −p2 δ p′
2 −p1 −
δ p′
2 −p1 δ p′
1 −p2 +δ p′
2 −p2 δ p′
1 −p1
=
1
Z
exp −
p2
1 +p2
2
2mkBT
δ p′
1 −p1 δ p′
2 −p2 −δ p′
1 −p2 δ p′
2 −p1 .
The partition function is
Z = Tr exp −β ˆH =
1
2 ∑
p1,p2
( p1, p2|− p2, p1|)exp −
ˆp2
1 + ˆp2
2
2mkBT
(|p1, p2 −|p2, p1 )
=
1
2 ∑
p1,p2
exp −
p2
1 +p2
2
2mkBT
(2−2 p1, p2| p2, p1 ) =
1
2 ∑
p1,p2
exp −
p2
1 +p2
2
2mkBT
[2−2δ (p1 −p2)]
=
1
2 ∑
p1,p2
exp −
p2
1 +p2
2
2mkBT
−
1
2 ∑
p
exp −
p2
2mkBT
Approximating the sum by the integral
1
2
V2
(2π¯h)6
dp1dp2 exp −
p2
1 +p2
2
2mkBT
−
1
2
V
(2π¯h)3
dp exp −
p
2mkBT
=
1
2
V2
(2π¯h)6
(2πmkBT)3
−
1
2
V
(2π¯h)3
(2πmkBT)
3
2 . (108)
This can be rewritten as
λ =
2π¯h2
mkBT
,
and finally
Z =
1
2
V2
λ6
1−
1
2
3
2
λ3
V
.
42. White dwarf
We will study a star made of electron degenerate gas.
(a) Denoting the number of nucleons N estimate the number of electrons (assume 12C a 16O composi-
tion).
(b) Determine energy per electron assuming
i. relativistic gas,
ii. nonrelativistic gas.
(c) Determine energy per nucleon.
(d) Find the minimum energy as a function of radius and show that in ultra-relativistic case the minimum
appears for R → 0.
(e) Determine the limiting mass for ultra-relativistic case for zero total energy.
Solution:
(a) Carbon contains six electrons a twelve nucleons. Oxygen has eight electrons and sixteen nucleons.
Consequently, the number of electrons is half the number of nucleons in both cases.
33
(b) In the case of nonrelativistic gas we obtain (see problem 31)
E =
p5
F
10meπ2¯h3
, N =
V p3
F
3π2¯h3
.
Dividing these results
E
N
=
3
10me
N
V
3π2
¯h3
,
we substitute volume of a sphere
E
N
=
3
10me
9
4
π¯h3
2
3 N
2
3
e
R2
.
For ultra-relativistic case
E =
V mc2 4
4π2¯h3
εF
mc2
4
.
N =
V mc2 3
3π2¯h3
εF
mc2
3
.
Energy per particle can be derived in a similar way as in previous case
E
N
=
3
4
mec2
=
3
4
mec2 Ne
V
3π2 ¯h3
(mec2)3
1
3
=
3
4
mec2 9
4
π
1
3
¯h
mec2
N
1
3
e
R
.
(c) Gravitational potential energy is
EG = −
Gm2
R
.
Mass is
m = Nu,
then
EG
N
= −
GNu2
R
.
(d) The total energy is given by a sum of kinetic and potential energy.
E
Ne
=
Ek
Ne
+
EG
Ne
=
3
10me
9
4
π¯h3
2
3 N
2
3
e
R2
−
4GNeu2
R
.
The necessary condition of extreme is
dE
dR
= 0 =
3
5me
9
4
π¯h3
2
3 N
2
3
e
R3
−
4GNeu2
R2
,
which gives
R =
3¯h2
20meGu2
9
4
π
2
3
N
− 1
3
e ,
from which follows
R ∼ N− 1
3 .
The energy in extremely relativistic case is
E
Ne
=
Ek
Ne
+
EG
Ne
=
3
4
mec2
=
3
4
mec2 Ne
V
3π2 ¯h3
(mec2)3
1
3
−
4GNeu2
R
.
From this follows that the derivative with respect to radius is proportional to R−2 and minimum
energy appears for R → 0, when E → −∞.
34
(e) For Ne > NCr and E = 0 we obtain from the formula above
Ncr =
3hc
16G
3
2 3
2
√
π
1
u3
,
and the critical mass is
mcr = 2Ncru =
3hc
16G
3
2 3
√
π
u2
.
43. Equation for the chemical potential
Determine the equation for the chemical potential for a matter in spherically symmetric gravitational
field. Assume that
µ +kuϕ = µ′
+mc2
+kuϕ = konst.,
where k is number of nucleons per electrons and u is the nucleon mass. Explain the equation. You can
neglect the pressure of nondegenerate matter and electron mass.
Solution: From the instructions follows that
ϕ =
konst.
ku
−
µ
ku
,
or
ϕ =
konst.
ku
−
µ′
ku
−
mc2
ku
.
For spherically symmetric case
1
r2
∂
∂r
r2 ∂ϕ
∂r
= 4πGρ.
We substitute ϕ from previous equation. Only the chemical potential depends on radius. After substitution
and some manipulation we arrive at
1
r2
∂
∂r
r2 ∂µ
∂r
= −4πGk2
u2
n.
We can obtain the same equation after substitution of equation with µ′.
44. Equation for chemical potential in integral form
Integrate the equation for chemical potential assuming
dµ
dr r=0
= 0,
where R is radius of a star. Express the result using total mass of the star.
Solution: We shall start with the equation from the previous problem multiplied by r2
∂
∂r
r2 ∂µ
∂r
= −4πGk2
u2
nr2
.
After integration
R
0
dr
∂
∂r
r2 ∂µ
∂r
= −
R
0
dr4πGk2
u2
nr2
.
The left hand side contains total differentiation and we obtain
R
0
dr
∂
∂r
r2 ∂µ
∂r
= r2 ∂µ
∂r r=R
−r2 ∂µ
∂r r=0
= r2 ∂µ
∂r r=R
.
Right hand side gives
−
R
0
dr4πGk2
u2
nr2
= −4πGku
R
0
drkunr2
= −4πGkunM.
35
We obtain
r2 ∂µ
∂r r=R
= −4πGkunM.
45. Nondimensional form of equation for chemical potential
Rewrite the equation for chemical potential in nondimensional variables
ξ =
r
R
, µ(r) =
1
√
λR
f(ξ),
where
λ =
4
3
k2
π
6u2
(¯hc)3
,
and determine the boundary conditions.
Solution: Number density is
n =
µ3
(¯hc)33π2
,
after the substitution
1
r2
∂
∂r
r2 ∂µ
∂r
= −
4πGk2u2µ3
(¯hc)33π3
= −λµ3
,
and some manipulation gives
1
ξ2
∂
∂ξ
ξ2 ∂ f
∂ξ
= − f3
.
46. Equation of continuity
Derive equation of continuity from the Boltzmann transport equation. Assume that the force is independent
of momentum.
Solution: Boltzmann transport equation is
∂ f
∂t
+v·∇r f +
F
m
·∇p f =
∂ f
∂t coll.
Integrating over momentum
m
Γ
dp
∂ f
∂t
+m
Γ
dpv·∇r f +
Γ
dpF·∇p f = m
∂
∂t
Γ
dp f +m∇r ·
Γ
dpvf +mF
Γ
dp∇p f.
We use Stokes theorem to manipulate the third term
Γ
dp∇p f =
∂Γ
dSp f = 0,
assuming that f tends to zero faster than p2. From
f(r,t) = m·n(r,t) = m
Γ
dp
f(r,p,t)
(2π¯h)3
.
Mean velocity is
u(r,t) = v = Γ
dpvf
Γ
dp f
= Γ
dpvf
(2π¯h)3mn(r,t)
.
After the substitution we derive
m
∂n
∂t
+m∇r ·(nu) = 0.
Equivalently,
∂ρ
∂t
+∇r ·(ρu) = 0.
36
47. Thermal transpiration
Determine the heat flux from the Boltzmann transport equation due to temperature gradient
α = −
dT
dy
, T = T0 −αy. (109)
Solution: The hydrostatic equilibrium holds due to the subsonic motion
p = n0kBT0 = nkB(T0 −αy),
from which
n = n0
T0
T0 −αy
.
In the Boltzmann transport equation
∂ f
∂t
+v·∇r f +
F
m
·∇p f =
∂ f
∂t coll.
,
we shall assume the stationary state. We further assume F = 0 and
∂ f
∂t coll.
= −
f − f0
τ
,
where f0 is equilibrium distribution function, which is assumed to take the form of
f0 = n
2π¯h2
mkBT
3
2
exp −
p2
2mkBT
= n0
2π¯h2
mkB
3
2
T0
(T0 −αy)
5
2
exp −
p2
2mkB(T0 −αy)
.
This simplifies the Boltzmann transport equation substantially
v·∇r f =
f − f0
τ
,
where we shall approximate the gradient of f by the gradient of f0. This gives form f
f = f0 −τv·∇r f0 = f0 −τvy
∂ f0
∂y
.
We substitute into the distribution function,
f = f0 +ατvy
T0
2(T0 −αy)
7
2
p2
mkB(T0 −αy)
−5 n0
2π¯h2
mkB
3
2
exp −
p2
2mkB(T0 −αy)
.
Finally, we can focus on the heat transport flux, which is
qy =
d3
p
(2π¯h)3
p2
2m
vy f.
Because f0 is an eve function, the integral is equal to zero. Within an approximation T0 − αy ≈ T0 the
heat flux is
qy =
ατ
2T0
1
2m
n0
2π¯h2
mkBT0
3
2
d3
p
(2π¯h)3
p2
v2
y
p2
mkBT0
−5 exp −
p2
2mkBT0
.
We shall introduce the spherical coordinates with north pole unusually in the direction of y axis, that is
vy = vcos(θ)
1
m2
1
(2π¯h)3
2π
0
dϕ
π
0
dθ
∞
0
dp
p2
mkBT0
−5 p6
cos2
θ sinθ exp −
p2
2mkBT0
=
1
m2
1
(2π¯h)3
·2π · −
1
3
cos3
δ
π
0
·

 1
mkBT0
1
2
(2mkBT0)
9
2
∞
0
dt t
7
2 exp(−t)−
5
2
(2mkBT0)
7
2
∞
0
dt t
5
2 exp(−t)


=
5
2
α
n0k2
BT0β
m
.
37
We derive for the hear transfer
q = −κ
dT
dr
= κα,
where
κ =
5
2
n0k2
BT0β
m
.
48. Greenhouse effect
Consider a greenhouse formed by placing a horizontal sheet of glas above the ground. The glass is
transparent to radiation with wavelengths below 4 µm, but absorbs a fraction ε of radiation at longer
wavelengths. Suppose a downward flux I of solar radiation. Assume that both earth and glass radiati according
to the Stefan-Boltzmann law. Determine the change of earth temperature due to presence of glass.
Solution The ground warms up to a temperature Tg and emits long-wavelength radiation with upward
flux given by
U = σT4
g .
Part of this flux is absorbed by the glass and reflected back. Equilibrium is reached when the upward
fluxes balance the downward fluxes, that is,
I = (1−ε)U +B = U −B,
where B is a flux emitted by the glass. Solving for B gives B = εU/2 and
σT4
g = U =
I
1− ε
2
, Tg =
I
σ 1− ε
2
1/4
Therefore, Tg is higher by up to 19% (for ε = 1) higher than it would be in the absence of glass (for
ε = 0).
38