All 16 Chapters Covered
SOLUTIONS
,Chapter 1
1.1. (a) We expand the quantity ln Ω(0)(E1) as a Taylor series in the variable (E1 −
Ē 1 ) and get
ln Ω(0)(E1) ≡ lnΩ1(E1) + ln Ω 2(E2) (E2 = E(0) − E1)
= {ln Ω1 (Ē 1 ) + ln Ω2 (Ē 2 )}+
∂ ln Ω1(E1) ∂ ln Ω2(E2) ∂E2 (E1 − Ē 1 )+
+
( ∂E1 ∂E2 ∂E1 E1=Ē 1 )
2
1 ∂2 ln Ω1(E1) ∂2 ln Ω 2(E2) ∂E2
+ (E1 − Ē 1 )2 + ··· .
2 ∂E12 ∂E 22 ∂E1
E1=Ē 1
The first term of this expansion is a constant, the second term van- ishes
as a result of equilibrium (β1 = β2), while the third term may be written as
1
1 ∂β1 ∂B2 E − Ē 2 1 1 + (E −Ē )2 ,
+
=− 1 1
2 ∂E1 ∂E2 1 1 2 kT 21(Cv)1 kT 2(Cv)2
eq. 2
with T1 = T2. Ignoring the subsequent terms (which is justified if the
systems involved are large) and taking the exponentials, we readily see
that the function Ω0(E1) is a Gaussian in the variable
(E1 −Ē 1 ), with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that
if (Cv)2 >> (Cv)1 — corresponding to system 1 being in thermal con-
tact with a very large reservoir — then the variance becomes simply
kT 2(C v)1, regardless of the nature of the reservoir; cf. eqn. (3.6.3).
(b) If the systems involved are ideal classical gases, then (Cv)1 = 3 N1k 2
and (Cv)2 = 3N2 k; the variance then becomes 3k 2T 2 ·N1N2/(N1 +
2 2
N3 2). 2 Again,
2 if N2 >> N1, we obtain the simplified expression
; cf. Problem 3.18.
2 N 1k T
1.2. Since S is additive and Ω multiplicative, the function f(Ω) must satisfy the
condition
f(Ω1Ω 2) = f(Ω1) + f(Ω 2). (1)
5
,
,6 CHAPTER 1.
Differentiating (1) with respect to Ω1 (and with respect to Ω2), we get
′ ′ ′ ′
Ω2f (Ω1Ω 2) = f (Ω1) and Ω1f (Ω1Ω 2) = f (Ω2),
so that ′ ′
Ω1f (Ω1) = Ω2f (Ω2). (2)
Since the left-hand side of (2) is independent of Ω2 and the right-hand side is
independent of Ω1, each side must be equal to a constant, k, independent of both
Ω1 and Ω2. It follows that f′(Ω) = k/Ω and hence
f (Ω) = k ln Ω + const. (3)
Substituting (3) into (1), we find that the constant of integration is zero.
1.4. Instead of eqn. (1.4.1), we now have
Ω ∝ V (V − v0)(V − 2v0) . . . (V − N − 1v0),
so that
ln Ω = C + ln V + ln (V − v0) + ln (V − 2v0) + . . . + ln (V − N − 1v0),
where C is independent of V . The expression on the right may be written as
X N−1 X N −1 2
jv0 jv0 N
C+N ln V + ln 1 − ≃ C+N ln V + − ≃ C+N ln V − v0
V V 2V .
j=1 j=1
Equation (1.4.2) is then replaced by
P N N 2v 0 N v0
= + = N 1+ , i.e.
kT V 2V 2 V 2V
−1
Nv
PV 1+ 0 = NkT .
2V
Since N v0 << V, (1 + N v0/2V )−1 ≃ 1 − N v0/2V . Our last result then takes the
form: P (V − b) = NkT , where b = 1 N v0.
2
A little reflection shows that v0 = (4π/3)σ3, with the result that
3
1 4π 4π 1
b= N· σ3 = 4N · o .
2 3 3 2
1.5. This problem is essentially solved in Appendix A; all that remains to be done is
to substitute from eqn. (B.12) into (B.11), to get
X (πε ∗1/2 /L) 3 (πε ∗1/2/L)2
(ε∗) = V∓ S.
1 6π2 16π
, 7
Substituting V = L3 and S = 6L2, we obtain eqns. (1.4.15 and 16). The
expression for T now follows straightforwardly; we get
1 ∂ ln Ω k ∂ ln Ω k R+ N k Nhν
=k = = ln = ln 1+ ,
T ∂E N hν ∂R N
hν R hν E
so that
hν Nhν
T= ln . 1+
k E
For E >> Nhν, we recover the classical result: T = E/Nk.
1.9. Since the function S(N,V,E) of a given thermodynamic system is an ex- tensive
quantity, we may write
V E
S(N, V, E) = Nf V , E = Nf (v, ε) v= ,ε = .
N N N N
It follows that
∂f −V −E
N ∂S =N f+N · +N ∂f ·
∂N V,E ∂v ε N2 ∂ε v N 2 ,
∂S ∂f ∂f 1
V = VN ∂S. = EN · .
∂V N,E ∂v · ∂E
ε N,V ∂ε v N
Adding these expressions, we obtain the desired result.
1.11. Clearly, the initial temperatures and the initial particle densities of the two gases
(and hence of the mixture) are the same. The entropy of mixing may, therefore, be
obtained from eqn. (1.5.4), with N1 = 4NA and N2 = NA. We get
(∆S)∗ = k[4N A ln(5/4) + NA ln 5]
= R[4 ln(5/4) + ln 5] = 2.502 R,
which is equivalent to about 0.5 R per mole of the mixture.
1.12. (a) The expression in question is given by eqn. (1.5.3a). Without loss of
generality, we may keep N1, N2 and V1 fixed and vary only V2 . The first and
second derivatives of this expression are then given by
N 1 + N2 N 1 + N2 N2
k and k − + (1a,b)
V1 + V2 N2
− V2 (V1 + V2)2 V 22
respectively. Equating (1a) to zero gives the desired condition, viz. N1V2 =
N2V 1, i.e. N1/V1 = N2/V2 = n, say. Expression (1b) then reduces to
n n knV1
k − + = > 0.
V1 + V2 V2 V 2 (V 1 + V 2)
Clearly, (∆S)1≡2 is at its minimum when N1/V1 = N2/V2, and it is
straightforward to check that the value at the minimum is zero.
,8 CHAPTER 1.
(b) The expression now in question is given by eqn. (1.5.4). With N1 = αN and
N2 = (1 − α)N , where N = N1 + N2 (which is fixed), the expression for
(∆S)∗/k takes the form
−αN ln α − (1 − α)N ln (1 − α).
The first and second derivatives of this expression with respect to α are
N N
[−N ln α + N ln(1 − α)] and − − (2a,b)
α 1 −α
respectively. Equating (2a) to zero gives the condition α = 1/2, which reduces
(2b) to −4N. Clearly, (∆S)∗/k is at its maximum when N1 = N2 = (1/2)N , and it
is straightforward to check that the value at the maximum is N ln 2.
1.13. Proceeding with eqn. (1.5.1), with T replaced by Ti, it is straightforward to see
that the extra contribution to ∆S, owing to the fact that T1 6= T2, is given by the
expression
3 3
N1k ln (Tf /T1) + N2k ln(Tf /T2),
2 2
where Tf = (N1T1 + N2T2)/(N1 + N2). It is worth checking that this expression
is always greater than or equal to zero, the equality holding if and only if T1 = T2.
Furthermore, the result quoted here does not depend on whether the two gases
were different or identical.
1.14. By eqn. (1.5.1a), given on page 24 of the text, we get
3
(∆S)v = Nk ln(Tf /Ti).
2
Now, since PV = NkT , the same equation may also be written as
3 5 2πmkT
S = Nk ln kT Nk + ln . (1b)
+2 3 h2
P
It follows that
5 5
(∆S)P = Nk ln(Tf / Ti) = (∆S)V.
2 3
A numerical verification of this result is straightforward. It
should be noted that, quite generally,
(∆S)P T (∂S / ∂T )P CP
= = =γ
(∆S)V T (∂S / ∂T )V CV
which, in the present case, happens to be 5/3.
, 9
1.15. For an ideal gas, CP − CV = nR, where n is the number of moles of the gas.
With CP /CV = γ, one gets
CP = γnR / (γ − 1) and CV = nR / (γ − 1).
For a mixture of two ideal gases,
n 1R n 2R + n )R.
f1 f2 (n
CV =
γ1 + = + 1 2
−1 γ2 − 1 γ1 − 1 γ2 − 1
Equating this to the conventional expression (n1 + n2)R/(γ − 1), we get the
desired result.
1.16. In view of eqn. (1.3.15), E − TS + PV = µN. It follows that
dE − TdS − SdT + PdV + VdP = µdN + Ndµ.
Combining this with eqn. (1.3.4), we get
−SdT + VdP = Nd µ, i.e. dP = (N / V )dµ + (S / V )dT .
Clearly, then,
(∂P / ∂µ)T = N / V and (∂P / ∂T )µ = S / V.
Now, for the ideal gas
( 3/2
)
NkT N h2
P= and µ = kT ln ;
V V 2πmkT
see eqn. (1.5.7). Eliminating (N/V ), we get
3/2
2πmkT
P = kT eµ/kT ,
h2
which is the desired expression. It follows quite readily now that for this system
1
∂P = P.
∂µ T kT
which is indeed equal to N/V , whereas
" (
5 5 )#
∂P = P−
µ
P= — ln N h2 3/2
Nk
∂T µ 2T kT 2
2 V 2πmkT V
which, by eqn. (1.5.1a), is precisely equal to S/V .
, 10 CHAPTER 1.
Chapter 2
2.3. The rotator in this problem may be regarded as confined to the (z = 0)- plane and its
position at time t may be denoted by the azimuthal angle ϕ. The conjugate variable
pϕ is then mρ2 ϕ̇ , where the various symbols have their usual meanings. The energy
of rotation is given by
1
E = m(ρϕ̇)2= p2 ϕ / 2mρ2.
2
Lines of constant energy in the (ϕ, pϕ)-plane are “straight lines, running parallel to the
ϕ-axis from ϕ = 0 to ϕ = 2π”. The basic cell of area h in this plane is a “rectangle
with sides ∆ϕ = 2π and ∆pϕ = h/2π”. Clearly, the eigenvalues of pϕ, starting with pϕ
= 0, are n~ and those of E are
n2~ 2/2I, where I = mρ2 and n = 0, ±1, ±2, . . .
The eigenvalues of E obtained here are precisely the ones given by quan- tum
mechanics for the energy “associated with the z-component of the rotational motion”.
2.4. The rigid rotator is a model for a diatomic molecule whose internuclear distance r may
be regarded as fixed. The orientation of the molecule in space may be denoted by the
angles θ and ϕ, the conjugate variables being pθ = mr 2 θ̇ and pϕ = mr 2 sin2 θϕ̇ . The
energy of rotation is given by
1 p 2θ p2ϕ M2
E = m(rθ̇)2 + 1 + =
m(r sin θϕ̇)2 = 2 2mr2 2mr sin θ
2 2 2I
,
2
where I = mr 2 and M 2 = p2 + p2 / sin2 θ .
θ ϕ
The “volume” of the relevant region of the phase space is given by the
R ′
integral dpθdpϕdθ dϕ, where the region of integration is constrained
by the value of M . A little reflection shows that in the subspace of pθ
and pϕ we are restricted by an elliptical boundary with semi-axes M and M sin θ, the
enclosed area being πM 2 sin θ. The “volume” of the relevant region, therefore, is
π 2π
Z Z
(πM 2 sin θ)dθ dϕ = 4π2M 2.
θ=0 ϕ=0
1
,2
The number of microstates available to the rotator is then given by 4π2M 2/h2, which is precisely
(M/~)2. At the same time, the number of microstates associated with the quantized value M
2
= j(j + 1)~2 may be estimated as j
1 h 2 —M − j−
2
i 1 + 1
3
M
j− 2 = j +
1 j j+ 1 = 2j + 1.
~2 j+ 1
2
2 2 2 2
This is precisely the degeneracy arising from the eigenvalues that the az- imuthal quantum
number m has, viz. j, j − 1, . . . , −j + 1, −j.
2.6. In terms of the variables θ and L(= mℓ2θ), the state of the simple pendu- lum is given by,
see eqns. (2.4.9),
θ = (A/ℓ) cos(ωt + ϕ), L = −mℓωA sin(ωt + ϕ),
with E = 1
mω
2
2A 2
and τ = 2π/ω. The trajectory in the (θ, L)-plane is given by the
equation
θ2 L2
+ = 1,
(A/ℓ) (mℓ ωA)2
2
which is an ellipse — just like in Fig. 2.2. The enclosed area turns out to be πmωA2,
which is precisely equal to the product Eτ .
2.7. Following the argument developed on page 70 of the text, the number of microstates for
a given energy E turns out to be
1
Ω(E) = (R + N − 1)!/ R!(N − 1)!, R = E − N ~ω /~ω. (1)
2
For R >> N, we obtain the asymptotic result
Ω(E) ≈ RN−1 / (N − 1)!, where R ≈ E / ~ω. (3.8.25a)
The corresponding expression for Γ(E; ∆) would be
(E / ~ω) N−1 ∆ EN−1∆
Γ(E; ∆) ≈ ·
= . (1)
(N − 1)! ~ω (N − 1)!(~ω)N
The “volume” of the relevant region of the phase space may be derived from the integral
Z ′ YN X 1N 2 1
(dq dp ), with kq + p2 ≤ E.
i=1
i i
2
i
2m i
i=1
This is equal to, see eqn. (7a) of Appendix C,
1
2 2N 1 πN 2π N
(2m) 2 N · EN = EN ,
k N! ω N!
3
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SOLUTIONS
,Chapter 1
1.1. (a) We expand the quantity ln Ω(0)(E1) as a Taylor series in the variable (E1 −
Ē 1 ) and get
ln Ω(0)(E1) ≡ lnΩ1(E1) + ln Ω 2(E2) (E2 = E(0) − E1)
= {ln Ω1 (Ē 1 ) + ln Ω2 (Ē 2 )}+
∂ ln Ω1(E1) ∂ ln Ω2(E2) ∂E2 (E1 − Ē 1 )+
+
( ∂E1 ∂E2 ∂E1 E1=Ē 1 )
2
1 ∂2 ln Ω1(E1) ∂2 ln Ω 2(E2) ∂E2
+ (E1 − Ē 1 )2 + ··· .
2 ∂E12 ∂E 22 ∂E1
E1=Ē 1
The first term of this expansion is a constant, the second term van- ishes
as a result of equilibrium (β1 = β2), while the third term may be written as
1
1 ∂β1 ∂B2 E − Ē 2 1 1 + (E −Ē )2 ,
+
=− 1 1
2 ∂E1 ∂E2 1 1 2 kT 21(Cv)1 kT 2(Cv)2
eq. 2
with T1 = T2. Ignoring the subsequent terms (which is justified if the
systems involved are large) and taking the exponentials, we readily see
that the function Ω0(E1) is a Gaussian in the variable
(E1 −Ē 1 ), with variance kT 2 (Cv )1 (Cv )2 /{(Cv )1 + (Cv )2 }. Note that
if (Cv)2 >> (Cv)1 — corresponding to system 1 being in thermal con-
tact with a very large reservoir — then the variance becomes simply
kT 2(C v)1, regardless of the nature of the reservoir; cf. eqn. (3.6.3).
(b) If the systems involved are ideal classical gases, then (Cv)1 = 3 N1k 2
and (Cv)2 = 3N2 k; the variance then becomes 3k 2T 2 ·N1N2/(N1 +
2 2
N3 2). 2 Again,
2 if N2 >> N1, we obtain the simplified expression
; cf. Problem 3.18.
2 N 1k T
1.2. Since S is additive and Ω multiplicative, the function f(Ω) must satisfy the
condition
f(Ω1Ω 2) = f(Ω1) + f(Ω 2). (1)
5
,
,6 CHAPTER 1.
Differentiating (1) with respect to Ω1 (and with respect to Ω2), we get
′ ′ ′ ′
Ω2f (Ω1Ω 2) = f (Ω1) and Ω1f (Ω1Ω 2) = f (Ω2),
so that ′ ′
Ω1f (Ω1) = Ω2f (Ω2). (2)
Since the left-hand side of (2) is independent of Ω2 and the right-hand side is
independent of Ω1, each side must be equal to a constant, k, independent of both
Ω1 and Ω2. It follows that f′(Ω) = k/Ω and hence
f (Ω) = k ln Ω + const. (3)
Substituting (3) into (1), we find that the constant of integration is zero.
1.4. Instead of eqn. (1.4.1), we now have
Ω ∝ V (V − v0)(V − 2v0) . . . (V − N − 1v0),
so that
ln Ω = C + ln V + ln (V − v0) + ln (V − 2v0) + . . . + ln (V − N − 1v0),
where C is independent of V . The expression on the right may be written as
X N−1 X N −1 2
jv0 jv0 N
C+N ln V + ln 1 − ≃ C+N ln V + − ≃ C+N ln V − v0
V V 2V .
j=1 j=1
Equation (1.4.2) is then replaced by
P N N 2v 0 N v0
= + = N 1+ , i.e.
kT V 2V 2 V 2V
−1
Nv
PV 1+ 0 = NkT .
2V
Since N v0 << V, (1 + N v0/2V )−1 ≃ 1 − N v0/2V . Our last result then takes the
form: P (V − b) = NkT , where b = 1 N v0.
2
A little reflection shows that v0 = (4π/3)σ3, with the result that
3
1 4π 4π 1
b= N· σ3 = 4N · o .
2 3 3 2
1.5. This problem is essentially solved in Appendix A; all that remains to be done is
to substitute from eqn. (B.12) into (B.11), to get
X (πε ∗1/2 /L) 3 (πε ∗1/2/L)2
(ε∗) = V∓ S.
1 6π2 16π
, 7
Substituting V = L3 and S = 6L2, we obtain eqns. (1.4.15 and 16). The
expression for T now follows straightforwardly; we get
1 ∂ ln Ω k ∂ ln Ω k R+ N k Nhν
=k = = ln = ln 1+ ,
T ∂E N hν ∂R N
hν R hν E
so that
hν Nhν
T= ln . 1+
k E
For E >> Nhν, we recover the classical result: T = E/Nk.
1.9. Since the function S(N,V,E) of a given thermodynamic system is an ex- tensive
quantity, we may write
V E
S(N, V, E) = Nf V , E = Nf (v, ε) v= ,ε = .
N N N N
It follows that
∂f −V −E
N ∂S =N f+N · +N ∂f ·
∂N V,E ∂v ε N2 ∂ε v N 2 ,
∂S ∂f ∂f 1
V = VN ∂S. = EN · .
∂V N,E ∂v · ∂E
ε N,V ∂ε v N
Adding these expressions, we obtain the desired result.
1.11. Clearly, the initial temperatures and the initial particle densities of the two gases
(and hence of the mixture) are the same. The entropy of mixing may, therefore, be
obtained from eqn. (1.5.4), with N1 = 4NA and N2 = NA. We get
(∆S)∗ = k[4N A ln(5/4) + NA ln 5]
= R[4 ln(5/4) + ln 5] = 2.502 R,
which is equivalent to about 0.5 R per mole of the mixture.
1.12. (a) The expression in question is given by eqn. (1.5.3a). Without loss of
generality, we may keep N1, N2 and V1 fixed and vary only V2 . The first and
second derivatives of this expression are then given by
N 1 + N2 N 1 + N2 N2
k and k − + (1a,b)
V1 + V2 N2
− V2 (V1 + V2)2 V 22
respectively. Equating (1a) to zero gives the desired condition, viz. N1V2 =
N2V 1, i.e. N1/V1 = N2/V2 = n, say. Expression (1b) then reduces to
n n knV1
k − + = > 0.
V1 + V2 V2 V 2 (V 1 + V 2)
Clearly, (∆S)1≡2 is at its minimum when N1/V1 = N2/V2, and it is
straightforward to check that the value at the minimum is zero.
,8 CHAPTER 1.
(b) The expression now in question is given by eqn. (1.5.4). With N1 = αN and
N2 = (1 − α)N , where N = N1 + N2 (which is fixed), the expression for
(∆S)∗/k takes the form
−αN ln α − (1 − α)N ln (1 − α).
The first and second derivatives of this expression with respect to α are
N N
[−N ln α + N ln(1 − α)] and − − (2a,b)
α 1 −α
respectively. Equating (2a) to zero gives the condition α = 1/2, which reduces
(2b) to −4N. Clearly, (∆S)∗/k is at its maximum when N1 = N2 = (1/2)N , and it
is straightforward to check that the value at the maximum is N ln 2.
1.13. Proceeding with eqn. (1.5.1), with T replaced by Ti, it is straightforward to see
that the extra contribution to ∆S, owing to the fact that T1 6= T2, is given by the
expression
3 3
N1k ln (Tf /T1) + N2k ln(Tf /T2),
2 2
where Tf = (N1T1 + N2T2)/(N1 + N2). It is worth checking that this expression
is always greater than or equal to zero, the equality holding if and only if T1 = T2.
Furthermore, the result quoted here does not depend on whether the two gases
were different or identical.
1.14. By eqn. (1.5.1a), given on page 24 of the text, we get
3
(∆S)v = Nk ln(Tf /Ti).
2
Now, since PV = NkT , the same equation may also be written as
3 5 2πmkT
S = Nk ln kT Nk + ln . (1b)
+2 3 h2
P
It follows that
5 5
(∆S)P = Nk ln(Tf / Ti) = (∆S)V.
2 3
A numerical verification of this result is straightforward. It
should be noted that, quite generally,
(∆S)P T (∂S / ∂T )P CP
= = =γ
(∆S)V T (∂S / ∂T )V CV
which, in the present case, happens to be 5/3.
, 9
1.15. For an ideal gas, CP − CV = nR, where n is the number of moles of the gas.
With CP /CV = γ, one gets
CP = γnR / (γ − 1) and CV = nR / (γ − 1).
For a mixture of two ideal gases,
n 1R n 2R + n )R.
f1 f2 (n
CV =
γ1 + = + 1 2
−1 γ2 − 1 γ1 − 1 γ2 − 1
Equating this to the conventional expression (n1 + n2)R/(γ − 1), we get the
desired result.
1.16. In view of eqn. (1.3.15), E − TS + PV = µN. It follows that
dE − TdS − SdT + PdV + VdP = µdN + Ndµ.
Combining this with eqn. (1.3.4), we get
−SdT + VdP = Nd µ, i.e. dP = (N / V )dµ + (S / V )dT .
Clearly, then,
(∂P / ∂µ)T = N / V and (∂P / ∂T )µ = S / V.
Now, for the ideal gas
( 3/2
)
NkT N h2
P= and µ = kT ln ;
V V 2πmkT
see eqn. (1.5.7). Eliminating (N/V ), we get
3/2
2πmkT
P = kT eµ/kT ,
h2
which is the desired expression. It follows quite readily now that for this system
1
∂P = P.
∂µ T kT
which is indeed equal to N/V , whereas
" (
5 5 )#
∂P = P−
µ
P= — ln N h2 3/2
Nk
∂T µ 2T kT 2
2 V 2πmkT V
which, by eqn. (1.5.1a), is precisely equal to S/V .
, 10 CHAPTER 1.
Chapter 2
2.3. The rotator in this problem may be regarded as confined to the (z = 0)- plane and its
position at time t may be denoted by the azimuthal angle ϕ. The conjugate variable
pϕ is then mρ2 ϕ̇ , where the various symbols have their usual meanings. The energy
of rotation is given by
1
E = m(ρϕ̇)2= p2 ϕ / 2mρ2.
2
Lines of constant energy in the (ϕ, pϕ)-plane are “straight lines, running parallel to the
ϕ-axis from ϕ = 0 to ϕ = 2π”. The basic cell of area h in this plane is a “rectangle
with sides ∆ϕ = 2π and ∆pϕ = h/2π”. Clearly, the eigenvalues of pϕ, starting with pϕ
= 0, are n~ and those of E are
n2~ 2/2I, where I = mρ2 and n = 0, ±1, ±2, . . .
The eigenvalues of E obtained here are precisely the ones given by quan- tum
mechanics for the energy “associated with the z-component of the rotational motion”.
2.4. The rigid rotator is a model for a diatomic molecule whose internuclear distance r may
be regarded as fixed. The orientation of the molecule in space may be denoted by the
angles θ and ϕ, the conjugate variables being pθ = mr 2 θ̇ and pϕ = mr 2 sin2 θϕ̇ . The
energy of rotation is given by
1 p 2θ p2ϕ M2
E = m(rθ̇)2 + 1 + =
m(r sin θϕ̇)2 = 2 2mr2 2mr sin θ
2 2 2I
,
2
where I = mr 2 and M 2 = p2 + p2 / sin2 θ .
θ ϕ
The “volume” of the relevant region of the phase space is given by the
R ′
integral dpθdpϕdθ dϕ, where the region of integration is constrained
by the value of M . A little reflection shows that in the subspace of pθ
and pϕ we are restricted by an elliptical boundary with semi-axes M and M sin θ, the
enclosed area being πM 2 sin θ. The “volume” of the relevant region, therefore, is
π 2π
Z Z
(πM 2 sin θ)dθ dϕ = 4π2M 2.
θ=0 ϕ=0
1
,2
The number of microstates available to the rotator is then given by 4π2M 2/h2, which is precisely
(M/~)2. At the same time, the number of microstates associated with the quantized value M
2
= j(j + 1)~2 may be estimated as j
1 h 2 —M − j−
2
i 1 + 1
3
M
j− 2 = j +
1 j j+ 1 = 2j + 1.
~2 j+ 1
2
2 2 2 2
This is precisely the degeneracy arising from the eigenvalues that the az- imuthal quantum
number m has, viz. j, j − 1, . . . , −j + 1, −j.
2.6. In terms of the variables θ and L(= mℓ2θ), the state of the simple pendu- lum is given by,
see eqns. (2.4.9),
θ = (A/ℓ) cos(ωt + ϕ), L = −mℓωA sin(ωt + ϕ),
with E = 1
mω
2
2A 2
and τ = 2π/ω. The trajectory in the (θ, L)-plane is given by the
equation
θ2 L2
+ = 1,
(A/ℓ) (mℓ ωA)2
2
which is an ellipse — just like in Fig. 2.2. The enclosed area turns out to be πmωA2,
which is precisely equal to the product Eτ .
2.7. Following the argument developed on page 70 of the text, the number of microstates for
a given energy E turns out to be
1
Ω(E) = (R + N − 1)!/ R!(N − 1)!, R = E − N ~ω /~ω. (1)
2
For R >> N, we obtain the asymptotic result
Ω(E) ≈ RN−1 / (N − 1)!, where R ≈ E / ~ω. (3.8.25a)
The corresponding expression for Γ(E; ∆) would be
(E / ~ω) N−1 ∆ EN−1∆
Γ(E; ∆) ≈ ·
= . (1)
(N − 1)! ~ω (N − 1)!(~ω)N
The “volume” of the relevant region of the phase space may be derived from the integral
Z ′ YN X 1N 2 1
(dq dp ), with kq + p2 ≤ E.
i=1
i i
2
i
2m i
i=1
This is equal to, see eqn. (7a) of Appendix C,
1
2 2N 1 πN 2π N
(2m) 2 N · EN = EN ,
k N! ω N!
3
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