Statistical Physics: November 28, 2012
Solutions for the Homework 10
Problem 7.66: Consider a collection of 10, 000 atoms of rubidium-87, confined inside a box of volume
(10−5 m)3 .
(a) Calculate ǫ0 , the energy of the ground state. (Express your answer in both joules and
electron-volts.)
Solution: From the equation 7.118,
ǫ0 =
3h2
= 1.14085 × 10−32 J = 7.12063 × 10−12 eV.
8mL2
(1)
(b) Calculate the condensation temperature, and compare kTc to ǫ0 .
Solution: From the equation 7.126,
Tc
h2
= (0.527) ×
2πmk
N
V
2/3
= 8.57849 × 10−8 K.
(2)
Then, the ratio of kTc to ǫ0 is
kTc
ǫ0
= 103.816.
(3)
(c) Suppose that T = 0.9Tc . How many atoms are in the ground state? How close is the chemical
potential to the ground-state energy? How many atoms are in each of the (threefold-degenerate)
first excited states?
Solution: From the equation 7.129,
N0 =
"
1−
T
Tc
3/2 #
N = 1461.85.
(4)
Approximate that the number of atoms in the ground state is sufficiently greater than 1, so we
can use the equation 7.119:
ǫ0 − µ =
kT
= 7.29179 × 10−34 J = 4.55118 × 10−15 eV.
N0
(5)
Now, to calculate the number of atoms in the first exited state, I need to calculate the energy of
first exited state and use it.
ǫ1 =
N1 =
6h2
= 2ǫ0 ,
8mL2
1
= 87.3226,
(ǫ
−µ)/kT
1
e
−1
1
(6)
⇒
3N1 = 261.968.
(7)
Statistical Physics: November 28, 2012
(d) Repeat parts (b) and (c) for the case of 106 atoms, confined to the same volume. Discuss
the conditions under which the number of atoms in the ground state will be much greater than
the number in the first excited state.
Solution: Same step for (b) and (c) with only difference is the number of atoms. Then,
2/3
h2
N
Tc = (0.527) ×
= 1.84818 × 10−6 K
2πmk V
kTc
= 2236.66
ǫ0
"
3/2 #
T
N = 146185
N0 = 1 −
Tc
ǫ0 − µ =
N1 =
kT
= 1.57097 × 10−34 J = 9.80522 × 10−16 eV
N0
1
= 1985.15,
⇒
3N1 = 5955.45.
(ǫ
−µ)/kT
1
e
−1
(8)
(9)
(10)
(11)
(12)
Critical temperature is higher than before and the number of atoms in the ground state increases
simply 100 times higher than before. But the energy difference between the ground state energy
and chemical potential decreases and it leads that the number of atoms in the first excited state
is not 100 times, just small factor times higher than before. This means that the number ratio
N1 /N0 will increase if the total number of atom increases.
Problem 7.68: Calculate the condensation temperature for liquid helium-4, pretending that the
liquid is a gas of noninteracting atoms. Compare to the observed temperature of the superfluid
transition, 2.17K. (The density of liquid helium-4 is 0.145 g/cm3 .)
Solution: Use the equation 7.126:
Tc
h2
= (0.527) ×
2πmk
N
V
2/3
= 3.13318 K.
(13)
Helium-4 has some real effect and it disturb the condensation at the predicted temperature, so
the experimental critical temperature is smaller than the theoretical critical temperature.
Problem 7.70: Figure 7.37 shows the heat capacity of a Bose gas as a function of temperature. In
this problem you will calculate the shape of this unusual graph.
(a) Write down an expression for the total energy of a gas of N bosons confined to a volume V ,
in terms of an integral (analogous to equation 7.122).
Solution: Similar to the procedure on the textbook from the equation 7.121 to 7.123,
Z ∞
X
ǫ
ǫs
=
dǫ
g(ǫ) (ǫ−µ)/kT
U =
(ǫs −µ)/kT − 1
e
e
−
1
0
alls
Z ∞
2πm 3/2
2
ǫ3/2
√
=
dǫ.
V
h2
π
e(ǫ−µ)/kT − 1
0
(14)
(15)
(b) For T < Tc you can set µ = 0. Evaluate the integral numerically in this case, then differentiate
the result with respect to T to obtain the heat capacity. Compare to Figure 7.37.
2
Statistical Physics: November 28, 2012
Solution: Set µ = 0 and then,
U
CV
=
2
√
π
=
2
√
π
=
2
√
π
=
5
√
π
3/2
ǫ3/2
dǫ
eǫ/kT − 1
0
Z ∞ 3/2
x
2πm 3/2
5/2
V (kT )
dx
2
h
ex − 1
0
2πm 3/2
V (kT )5/2 × (1.78329),
h2
2πmkT 3/2
V k × (1.78329).
h2
2πm
h2
V
Z
∞
To plot the graph, make a dimensionless variable:
"
# "
!
#
CV
5
2πmkTc 3/2
2πmkT 3/2
√
=
V k × (1.78329) /
2.612
V ×k
Nk
h2
h2
π
5 × (1.78329) T 3/2
√
.
=
Tc
2.612 π
(16)
(17)
(18)
(19)
(20)
(21)
Then, the plot of heat capacity is below blue lines in figure.
CV  Nk
2.5
2.0
1.5
1.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
t =T Tc
(c) Explain why the heat capacity must approach 23 N k in the high-T limit.
Solution: By the equipartition theorem at high temperature, monatomic gas behave like ideal
gas and it should have the heat capacity 3N k/2.
(d) For T > Tc you can evaluate the integral using the values of µ calculated in Problem 7.69.
Do this to obtain the energy as a function of temperature, then numerically differentiate the
result to obtain the heat capacity. Plot the heat capacity, and check that your graph agrees with
Figure 7.37.
Solution: First, find µ from the result of the problem 7.69 (a). After that, we can explain
the total energy using by the total number of atoms N with the variable changes: T /Tc = t,
3
Statistical Physics: November 28, 2012
x = ǫ/kTc and m = µ/kTc . Then,
Z ∞
2
ǫ3/2
2πm 3/2
U = √
V
dǫ
h2
π
e(ǫ−µ)/kT − 1
0
Z ∞
2πmkTc 3/2
x3/2
2
V
kT
dx
= √
c
h2
π
e(x−m)/t − 1
0
Z ∞
2
x3/2
√ N kTc
=
dx
2.612 π
e(x−m)/t − 1
0
(22)
(23)
(24)
From this, we can calculate the heat capacity but it’s hard to see analytically. So, use some
computation program. Then, the plot of heat capacity is above red lines in figure. (See details
in code section.)
Problem 7.72: For a gas of particles confined inside a two-dimensional box, the density of states is
constant, independent of ǫ (see Problem 7.28). Investigate the behavior of a gas of noninteracting bosons in a two-dimensional box. You should find that the chemical potential remains
significantly less than zero as long as T is significantly greater than zero, and hence that there
is no abrupt condensation of particles into the ground state. Explain how you know that this
is the case, and describe what does happen to this system as the temperature decreases. What
property must g(ǫ) have in order for there to be an abrupt Bose-Einstein condensation?
Solution: Since the density of states is constant, the total number of atoms N is
Z ∞
1
dǫ.
N = g
e(ǫ−µ)/kT − 1
0
(25)
When µ = 0, integrand of above equation is infinite at the ground state. But the number of
atoms is finite, so that is not possible. To modify this, the chemical potential should be negative
value. Divergence of the integrand is independent to the temperature, so the chemical potential
will be negative value whatever the temperature is.
2.0
1.5
T high
1.0
0.5
T low
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
Suppose that the temperature goes to zero. Then excited atoms continuously move to the ground
state. In order to get the abrupt Bose-Einstein condensation, set µ = 0 and to make the number
of atoms finite, set the density of state g(ǫ) goes to zero when ǫ goes to zero, similarly to the
three-dimensional case.
4
Problem 7.66
HaL
h = 6.62607 * 10^H-34L;
eV = 1.6021766 * 10^H-19L;
rubidiummass = 1.443161 * 10^H-25L;
L = 10^H-5L;
ep0 = 3 h^2  H8 rubidiummass L^2L
ep0  eV
1.14085 ´ 10-32
7.12063 ´ 10-14
HbL
k = 1.38065 * 10^H-23L;
number = 10 000;
T1c = 0.527 * h^2  H2 Pi rubidiummass kL Hnumber  L^3L^H2  3L
k T1c  ep0
8.57849 ´ 10-8
103.816
HcL
T1 = 0.9 T1c;
N0 = H1 - HT1  T1cL^H3  2LL number
k T1  N0
k T1  N0  eV
mu = Solve@ep0 - a == k T1  N0, aD@@1, 1, 2DD;
ep1 = 2 ep0;
N1 = 1  HE^HHep1 - muL  Hk T1LL - 1L
1461.85
7.29179 ´ 10-34
4.55118 ´ 10-15
87.3226
HdL
2
solution_10.nb
number = 10^6;
T1c = 0.527 * h^2  H2 Pi rubidiummass kL Hnumber  L^3L^H2  3L
k T1c  ep0
T1 = 0.9 T1c;
N0 = H1 - HT1  T1cL^H3  2LL number
k T1  N0
k T1  N0  eV
mu = Solve@ep0 - a == k T1  N0, aD@@1, 1, 2DD;
ep1 = 2 ep0;
N1 = 1  HE^HHep1 - muL  Hk T1LL - 1L
1.84818 ´ 10-6
2236.66
146 185.
1.57097 ´ 10-34
9.80522 ´ 10-16
1985.15
Problem 7.68
mol = 6.022141 * 10^H23L;
density = 0.145 * 10^H-3L * 10^H6L;
heliummass = 6.646478 * 10^H-27L;
V = mol heliummass  density;
T2c = 0.527 * h^2  H2 Pi heliummass kL Hmol  VL^H2  3L
3.13318
Problem 7.70
HdL
solution_10.nb
In[9]:=
f@t_D := Integrate@Sqrt@xD  HExp@Hx - mL  tD - 1L, 8x, 0, Infinity<D;
mu@t_D = InterpolatingPolynomial@Prepend@
Table@8t, With@8t1 = t<, FindRoot@f@t1D Š 2315  1000, 8m, -0.01, -0.2<DD@@1, 2DD<,
8t, 1.1, 3, 0.1<D, 81, 0<D, tD;
U@t_D :=
2  H2.612 * Sqrt@PiDL * NIntegrate@x^H3  2L  HExp@Hx - mu@tDL  tD - 1L, 8x, 0, Infinity<D;
u@t_D = InterpolatingPolynomial@Table@8t, U@tD<, 8t, 1, 3, 0.1<D, tD;
c@t_D = D@u@tD, tD;
p1 = Plot@c@tD, 8t, 1, 3<, PlotStyle ® 8Red<, PlotRange ® 880, 3<, 80, 2.5<<D;
p2 = Plot@5 * 1.78329  H2.612 * Pi^H1  2LL * HtL^H3  2L,
8t, 0, 1<, PlotStyle ® 8Blue<, PlotRange ® 880, 3<, 80, 2.5<<D;
Show@p1, p2, GridLines ® 8881, 8Dashed, Gray<<<, 881.5, 8Dashed, Gray<<<<,
AxesLabel ® 8"t=TTc ", "CV Nk"<D
CV  Nk
2.5
2.0
1.5
Out[16]=
1.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
t =T Tc
Problem 7.72
Plot@81  HE^x - 1L, H1  2L  HE^Hx  2L - 1L, H1  3L  HE^Hx  3L - 1L<, 8x, 0, 3<,
PlotStyle ® 8Black, Red, Blue<, Epilog ® [email protected], 1.2<, 80.5, 0.5<<D,
Text@"T high", 81.3, 1.3<D, Text@"T low", 80.4, 0.4<D<, PlotRange ® 880, 3<, 80, 2<<D
2.0
1.5
T high
1.0
0.5
T low
0.0
0.0
In[206]:=
Out[206]=
0.5
1.0
1.5
2.0
2.5
3.0
Table@8t, [email protected] Š Integrate@Sqrt@xD  HExp@Hx - cL  tD - 1L, 8x, 0, Infinity<D,
8c, -0.1, -0.2<D@@1, 2DD<, 8t, 1, 3, 0.2<D
991., -2.50947 ´ 10-9 - 5.23185 ´ 10-15 ä=, 81.2, -0.043972<, 81.4, -0.160954<,
81.6, -0.335558<, 81.8, -0.557827<, 82., -0.820792<, 82.2, -1.1193<,
82.4, -1.44937<, 82.6, -1.80785<, 82.8, -2.19217<, 83., -2.6002<=
3