Introduction to Thermodynamics and Statistical Physics phần 3 pdf

Consider a system which has two single particle states both of the same energy.. We suppose that the energy of the system is much higher infinitely higher when both states are occupied..

a) Show that α is given by

α = a

2

·

1 + tanh

µ

F a 2τ

¶¸

b) Show that in the limit of high temperature the spring constant is given approximately by

19 A long elastic molecule can be modelled as a linear chain of N links The state of each link is characterized by two quantum numbers l and n The length of a link is either l = a or l = b The vibrational state of a link

is modelled as a harmonic oscillator whose angular frequency is ωafor a link of length a and ωb for a link of length b Thus, the energy of a link is

En,l=

½

}ωa

¡

n +1 2

¢ for l = a }ωb

¡

n +1 2

¢

n = 0, 1, 2,

The chain is held under a tension F Show that the mean length hLi of the chain in the limit of high temperature T is given by

hLi = Naωωb+ bωa

b+ ωa + N

F ωbωa(a − b)2 (ωb+ ωa)2 β + O

¡

β2¢

where β = 1/τ

20 The elasticity of a rubber band can be described in terms of a one-dimensional model of N polymer molecules linked together end-to-end The angle between successive links is equally likely to be 0◦or 180◦ The length of each polymer is d and the total length is L The system is in thermal equilibrium at temperature τ Show that the force f required to maintain a length L is given by

f = τ

dtanh

−1 L

21 Consider a system which has two single particle states both of the same energy When both states are unoccupied, the energy of the system is

zero; when one state or the other is occupied by one particle, the energy

is ε We suppose that the energy of the system is much higher (infinitely higher) when both states are occupied Show that in thermal equilibrium

at temperature τ the average number of particles in the level is

where µ is the chemical potential and β = 1/τ

22 Consider an array of N two-level particles Each one can be in one of two states, having energy E1 and E2 respectively The numbers of particles

in states 1 and 2 are n1and n2respectively, where N = n1+ n2(assume

n1 À 1 and n2 À 1) Consider an energy exchange with a reservoir at temperature τ leading to population changes n2→ n2−1 and n1→ n1+1 a) Calculate the entropy change of the two-level system, (∆σ)2LS b) Calculate the entropy change of the reservoir, (∆σ)R

c) What can be said about the relation between (∆σ)2LSand (∆σ)Rin thermal equilibrium? Use your answer to express the ration n2/n1as

a function of E1, E2 and τ

23 Consider a lattice containing N sites of one type, which is denoted as A, and the same number of sites of another type, which is denoted as B The lattice is occupied by N atoms The number of atoms occupying sites of type A is denoted as NA, whereas the number of atoms occupying atoms

of type B is denoted as NB, where NA+ NB = N Let ε be the energy necessary to remove an atom from a lattice site of type A to a lattice site of type B The system is in thermal equilibrium at temperature τ Assume that N, NA, NBÀ 1

a) Calculate the entropy σ

b) Calculate the average number hNBi of atoms occupying sites of type B

24 Consider a microcanonical ensemble of N quantum harmonic oscillators

in thermal equilibrium at temperature τ The resonance frequency of all oscillators is ω The quantum energy levels of each quantum oscillator is given by

εn= }ω

µ

n +1 2

¶

where n = 0, 1, 2, is integer The total energy E of the system is given by

E = }ω

µ

m +N 2

¶

where

m =

N

X

l=1

and nl is state number of oscillator l

a) Calculate the number of states g (N, m) of the system with total energy }ω (m + N/2)

b) Use this result to calculate the entropy σ of the system with total energy }ω (m + N/2) Approximate the result by assuming that N À

1 and m À 1

c) Use this result to calculate (in the same limit of N À 1 and m À 1) the average energy of the system U as a function of the temperature

τ

25 The energy of a donor level in a semiconductor is −ε when occupied

by an electron (and the energy is zero otherwise) A donor level can be either occupied by a spin up electron or a spin down electron, however, it cannot be simultaneously occupied by two electrons Express the average occupation of a donor state hNdi as a function of ε and the chemical potential µ

1.8 Solutions Set 1

1 Final answers:

a) N !

(N

2)!(N

2)!

¡1

2

¢N

b) 0

2 Final answers:

a) ¡5

6

¢N

b) ¡5

6

¢N −1 1

6

c) In general

∞

X

N =0

N xN−1 = d

dx

∞

X

N =0

xN= d dx

1

1 − x =

1 (1 − x)2 , thus

¯

N = 1

6

∞

X

N =0

N

µ 5 6

¶N −1

=1 6

1

¡

1 −56

¢2 = 6

3 Let W (m) be the probability for for taking n1 steps to the right and

n2 = N − n1 steps to the left, where m = n1− n2, and N = n1+ n2 Using

n1=N + m

n2=N − m

one finds

W (m) = ¡N +mN !

2

¢

!¡N−m

2

¢

!p

N +m

2 qN−m2

It is convenient to employ the moment generating function, defined as

φ (t) =­

etm®

In general, the following holds

φ (t) =

∞

X

k=0

tk

k!

­

mk® ,

thus from the kth derivative of φ (t) one can calculate the kth moment

of m

­

mk®

= φ(k)(0)

Using W (m) one finds

φ (t) =

N

X

m= −N

W (m) etm

=

N

X

m= −N

N !

¡N +m

2

¢

!¡N−m

2

¢

!p

N+m

2 qN−m2 etm ,

or using the summation variable

n1=N + m

one has

φ (t) =

N

X

n 1 =0

N !

n1! (N − n1)!p

n 1qN−n1et(2n1 −N)

= e−tN

N

X

n 1 =0

N !

n1! (N − n1)!

¡

pe2t¢n 1

qN−n1

= e−tN¡

pe2t+ q¢N

Using p = q = 1/2

φ (t) =

Ế

et+ e−t 2

ảN

= (cosh t)N Thus using the expansion

(cosh t)N= 1 + 1

2!N t

2+ 1 4![N + 3N (N − 1)] t4+ Oâ

t5đ , one finds

hmi = 0 ,

Ẻ

m2Ẽ

= N ,

Ẻ

m3Ẽ

= 0 ,

Ẻ

m4Ẽ

= N (3N − 2)

4 Using the binomial distribution

W (n) = N!

n! (N − n)!p

n

(1 − p)N−n

= N (N − 1) (N − 1) Ứ (N − n + 1)

n

(1 − p)N−n

∼

= (Np)

n

n! exp (−Np) 5

a)

∞

X

n=0

W (n) = e−λ

∞

X

n=0

λ n

n! = 1 b) As in Ex 1-6, it is convenient to use the moment generating function

φ (t) =Ẻ

etnẼ

=

∞

X

n=0

etnW (n) = e−λ

∞

X

n=0

λnetn

n! = e

−λX∞

n=0

(λet)n n! = exp

ê

λâ

et− 1đô

Using the expansion

expê

λâ

et− 1đô

= 1 + λt +1

2λ (1 + λ) t

2+ Oâ

t3đ , one finds

hni = λ

c) Using the same expansion one finds

Ẻ

n2Ẽ

= λ (1 + λ) , thus

D

(∆n)2E

=Ẻ

n2Ẽ

− hni2= λ (1 + λ) − λ2= λ

­

R2®

=

*Ã N X

n=1

rn

!2+

=

N

X

n=1

­

r2n®

|{z}

=l 2

n 6=m

hrn· rmi

| {z }

=0

= N l2

7

20

X

n=11

20!

n! (20 − n)!0.2

n

× 0.820−n= 5.6 × 10−4 (1.144)

8 Using

N+− N−= M

one has

N+= N

2

µ

1 + M mN

¶

N−= N

2

µ

1 −mNM

¶

or

N+= N

N−= N

where

x = M

mN .

The number of states having total magnetization M is given by

Ω (M ) = N !

N+!N−! =

N !

£N

2 (1 + x)¤

!£N

2 (1 − x)¤

Since all states have equal probability one has

f (M ) = Ω (M )

Taking the natural logarithm of Stirling’s formula one finds

log N! = N log N − N + O

µ 1 N

¶

thus in the limit N À 1 one has

log f = − log 2N+ N log N − N

−

· N

2 (1 + x)

¸ log

· N

2 (1 + x)

¸ +

· N

2 (1 + x)

¸

−

· N

2 (1 − x)

¸ log

· N

2 (1 − x)

¸ +

· N

2 (1 − x)

¸

= −N log 2 + N log N

−

· N

2 (1 + x)

¸ log

· N

2 (1 + x)

¸

−

· N

2 (1 − x)

¸ log

· N

2 (1 − x)

¸

=

µ

−N2

¶ ½

−2 logN2 + (1 + x) log

· N

2 (1 + x)

¸ + (1 − x) log

· N

2 (1 − x)

¸¾

=

µ

−N2

¶ ·

−2 logN2 + (1 + x)

µ logN

2 + log (1 + x)

¶ + (1 − x)

µ logN

2 + log (1 − x)

¶¸

=

µ

−N2

¶ µ log¡

1 − x2¢

+ x log1 + x

1 − x

¶

(1.151) The function log f (x) has a sharp peak near x = 0, thus we can

approx-imate it by assuming x ¿ 1 To lowest order

log¡

1 − x2¢

+ x log1 + x

1 − x = x

2+ O¡

x3¢

thus

f (M ) = A exp

µ

2

2m2N

¶

where A is a normalization constant, which is determined by requiring

that

1 =

∞

Z

−∞

Using the identity

∞

Z

−∞

exp¡

−ay2¢

dy =

r π

one finds

1

A =

∞

Z

−∞

exp

µ

2

2m2N

¶

dM = m√

f (M ) = 1

m√ 2πN exp

µ

2

2m2N

¶

The expectation value is giving by

hMi =

∞

Z

−∞

and the variance is given by

D

(M − hMi)2E

=­

M2®

=

∞

Z

−∞

M2f (M ) dM = m2N (1.159)

9 The probability to have n steps to the right is given by

W (n) = N!

n! (N − n)!p

a)

hni =

N

X

n=0

N!n n! (N − n)!p

= p ∂

∂p

N

X

n=0

N!

n! (N − n)!p

nqN−n

= p ∂

∂p(p + q)

N

= pN (p + q)N−1= pN Since

X = an − a (N − n) = a (2n − N) (1.162)

we find

hXi = aN (2p − 1) = aN (p − q) (1.163) b)

­

n2®

=

N

X

n=0

N !n2

n! (N − n)!p

nqN−n

=

N

X

n=0

N !n (n − 1) n! (N − n)!p

nqN−n+

N

X

n=0

N !n n! (N − n)!p

nqN−n

= p2 ∂

2

∂p2

N

X

n=0

N ! n! (N − n)!p

n

qN−n+ hni

= p2 ∂

2

∂p2(p + q)N+ hni = p2N (N − 1) + pN

D

(n − hni)2E

= p2N (N − 1) + pN − p2N2= N pq , (1.164) and

D

(X − hXi)2E

10 The total energy is given by

E =kx

2

2 +

m ˙x2

ka2

where a is the amplitude of oscillations The time period T is given by

T = 2

Z a

−a

dx

˙x = 2

r m k

Z a

−a

dx

√

a2− x2 = 2π

r m

thus

f (x) = 2

T | ˙x| =

1

π√

11 The six experiments are independent, thus

σ = 6 ×

µ

−23ln2

3−13ln1

3

¶

= 3.8191

12 The random variable X obtains the value n with probability pn = qn, where n = 1, 2, 3, , and q = 1/2

a) The entropy is given by

σ = −

∞

X

n=1

pnlog2pn= −

∞

X

n=1

qnlog2qn =

∞

X

n=1

nqn This can be rewritten as

σ = q ∂

∂q

∞

X

n=1

qn= q ∂

∂q

µ 1

1 − q − 1

¶

(1 − q)2 = 2 b) A series of questions is of the form: ”Was a head obtained in the 1st time ?”, ”Was a head obtained in the 2nd time ?”, etc The expected number of questioned required to find X is

1

2× 1 +1

4× 2 +1

8× 3 + = 2 , which is exactly the entropy σ

13

a) Consider an infinitesimal change in the variable z = z (x, y)

δz =

µ

∂z

∂x

¶

y

δx +

µ

∂z

∂y

¶

x

For a process for which z is a constant δz = 0, thus

0 =

µ

∂z

∂x

¶

y

(δx)z+

µ

∂z

∂y

¶

x

Dividing by (δx)z yields

µ

∂z

∂x

¶

y

= −

µ

∂z

∂y

¶

x

(δy)z (δx)z

= −

µ∂z

∂y

¶

x

µ∂y

∂x

¶

z

= −

³

∂y

∂x

´

z

³

∂y

∂z

´

x

(1.171) b) Consider a process for which the variable w is kept constant An infinitesimal change in the variable z = z (x, y) is expressed as (δz)w=

µ

∂z

∂x

¶

y

(δx)w+

µ

∂z

∂y

¶

x

Dividing by (δx)wyields

(δz)w

(δx)w =

µ

∂z

∂x

¶

y

+

µ

∂z

∂y

¶

x

(δy)w

or

µ

∂z

∂x

¶

w

=

µ

∂z

∂x

¶

y

+

µ

∂z

∂y

¶

x

µ

∂y

∂x

¶

w

14 We have found in class that

hUi = −

µ

∂ log Zgc

∂β

¶

η

hNi = −

µ

∂ log Zgc

∂η

¶

β

a) Using relation (1.125) one has

hUi = −

µ

∂ log Zgc

∂β

¶

η

= −

µ

∂ log Zgc

∂β

¶

µ

−

µ

∂ log Zgc

∂µ

¶

β

µ

∂µ

∂β

¶

η

= −

µ

∂ log Zgc

∂β

¶

µ

− η

β2

µ

∂ log Zgc

∂µ

¶

β

= −

µ

∂ log Zgc

∂β

¶

µ

+ τ µ

µ

∂ log Zgc

∂µ

¶

β

(1.177) b) Using Eq (1.176) one has

hNi = −

µ

∂ log Zgc

∂η

¶

β

= −

µ

∂µ

∂η

¶

β

µ

∂ log Zgc

∂µ

¶

β

= τ

µ

∂ log Zgc

∂µ

¶

β

,

(1.178)

or in terms of the fugacity λ, which is defined by

one has

hNi = τ

µ

∂ log Zgc

∂µ

¶

β

= τ∂λ

∂µ

∂ log Zgc

∂λ

= λ∂ log Zgc

∂λ .

(1.180)

15 The canonical partition function is given by

where

Z1= exp

µ βε 2

¶ + exp

µ

−βε2

¶

= 2 cosh

µ βε 2

¶

hUi = −∂ log Zc

∂β = −N∂ log Z1

∂β = −N ε

2 tanh

βε

and

2 tanh−1³

−2N εhUi

The negative temperature is originated by our assumption that the en-ergy of a single magnet has an upper bound In reality this is never the case

16 The canonical partition function is given by

where

Z1= exp

µ

−β}ω2

¶X∞ n=0

=

exp³

−β}ω2 ´

1 − exp (−β}ω) =

1

2 sinhβ}ω2 . a)

hUi = −∂ log Z∂β c = −N∂ log Z∂β 1 = N }ω

2 coth

β}ω

b)

D

(∆U )2E

= ∂

2log Zc

∂β2 = N

∂2log Z1

∂β2 =

N¡}ω

2

¢2

sinh2 β}ω 2

(1.188)

17 The canonical partition function is given by

Zc= [exp (βε) + 1 + exp (−βε)]N= [1 + 2 cosh (βε)]N , (1.189) where β = 1/τ

a) Thus the average energy is

hUi = −∂ log Z∂β c = −2N ε sinh (βε)1 + 2 cosh βε , (1.190) b) and the variance is

D

(U − hUi)2E

=∂

2

log Zc

∂β2 = −∂ hUi∂β = 2Nε2 cosh (βε) + 2

[1 + 2 cosh (βε)]2 .

(1.191)

18 Each section can be in one of two possible sates with corresponding

en-ergies 0 and −F a

a) By definition, α is the mean length of each segment, which is given

by

α = a exp (F aβ)

1 + exp (F aβ)=

a 2

·

1 + tanh

Ế

F aβ 2

ảÌ

where β = 1/τ

b) At high temperature F aβ Ò 1 the length of the chain L = Nα is

given by

L = N a

2

·

1 + tanh

Ế

F aβ 2

ảÌ

'N a2

Ế

1 +F aβ 2

ả , (1.193) or

F = k

Ế

L −N a2

ả

where the spring constant k is given by

k = 4τ

19 The average length of a single link is given by

hli =

a exp (βF a)

∞

X

n=0

expê

−β}ωa

â

n +12đô

+ b exp (βF b)

∞

X

n=0

expê

−β}ωb

â

n +12đô

exp (βF a)

∞

X

n=0

expê

−β}ωaâ

n +12đô

+ exp (βF b)

∞

X

n=0

expê

−β}ωbâ

n +12đô (1.196)

=

a exp(βF a)

2 sinhβ}ωa2 + b exp(βF b)

2 sinhβ}ωb2 exp(βF a)

2 sinhβ}ωa2 + exp(βF b)

2 sinhβ}ωb2

To first order in β

hli = aωωb+ bωa

b+ ωa

+F ωbωa(a − b)2 (ωb+ ωa)2 β + O

â

β2đ

The average total length is hLi = n hli

20 The length L is given by

L = N hli ,

where hli is the average contribution of a single molecule to the total

length, which can be either +d or −d The probability of each possibility

is determined by the Boltzmann factor The energy change due to flipping

of one link from 0◦ to 180◦is 2f d, therefore

hli = de

βf d− e−βfd

eβf d+ e−βfd = d tanh (βf d) ,

where β = 1/τ Thus

L = Nd tanh (βf d) ,

or

f = τ

dtanh

−1 L

N d .

21 The grand partition function Zgcis given by

thus

hNi = β1∂ log Z∂µ gc = 2

2 + exp [β (ε − µ)] . (1.199) 22

a)

(∆σ)2LS= log N !

(n2− 1)! (n1+ 1)! − log N !

n2!n1!

= log n2

n1+ 1 ' lognn2

1

(1.200) b)

(∆σ)R= E2− E1

c) For a small change near thermal equilibrium one expects (∆σ)2LS+ (∆σ)R= 0, thus

n2

n1

= exp

µ

−E2− E1 τ

¶

23 The number of ways to select NB occupied sites of type B out of N sites

is N !/n! (N − n)! Similarly the number of ways to select NB empty sites

of type A out of N sites is N!/n! (N − n)!

a) Thus

σ = log

µ

N !

NB! (N − NB)!

¶2

' 2 [N log N − NBlog NB− (N − NB) log (N − NB)]

(1.203) b) The energy of the system is given by U = NBε Thus, the Helmholtz free energy is given by

F = U −τ σ = U −2τ

·

N log N −Uε logU

ε −

µ

N −Uε

¶ log

µ

N −Uε

¶¸

(1.204)

At thermal equilibrium (∂F/∂U )τ= 0, thus

0 =

µ

∂F

∂U

¶

τ

= 1 +2τ ε

· logU

ε − log

µ

N −Uε

¶¸

or

N − NB

NB

= exp³ ε

2τ

´

therefore

hNBi = N

1 + exp¡ε

2τ

¢

Alternatively, one can calculate the chemical potential from the re-quirement

1 = NA

N +

NB

where

NA

exp (βµ)

NB

exp (βµ − βε)

which is satisfied when

µ = ε

thus

hNBi = N

1 + exp¡ε

2τ

24 In general,

g (N, m) = {# os ways to distribute m identical balls in N boxes}

Moreover

{# os ways to distribute m identical balls in N boxes}

= {# os ways to arrange m identical balls and N − 1 identical partitions in a line}

…

…

…

…

a) Therefore

g (N, m) = (N − 1 + m)!

b) The entropy is given by

σ = log(N − 1 + m)!

(N − 1)!m! ' (N + m) log (N + m) − N log N − m log m ,

(1.212)

or in terms of the total energy E = }ω (m + N/2)

σ =

·

N +

µ E }ω −

N 2

¶¸

log

·

N +

µ E }ω −

N 2

¶¸

− N log N −

µ E }ω −

N 2

¶ log

µ E }ω −

N 2

¶

(1.213) c) The temperature τ is given by

1

τ =

∂σ

∂E

= 1 − ln2E+N }ω2}ω

ln2E2}ω−N}ω− 1 }ω

= 1

}ωln

µ 2E + N }ω 2E − N}ω

¶

(1.214)

In the thermodynamical limit (N À 1, m À 1) the energy E and its average U are indistinguishable, thus

exp

µ }ω τ

¶

=2U + N }ω

or

U = N}ω

2 coth

}ω

25 The grand canonical partition function is given by

where λ = exp (βµ) is the fugacity, thus

hNdi = λ∂ log ζ∂λ = 2λe

βε

1 + 2λeβε = 1

1 +12e−β(ε+µ) (1.218)