Introduction to Thermodynamics and Statistical Physics phần 8 ppt

Write a relation between the temperature of the surface of a planet and its distance from the Sun, on the assumption that as a black body in thermal equilibrium, it reradiates as much th

ε

2

µ

β

α2

¶3/2

dε = n2dn Thus, by introducing the density of states

D (ε) =

(

V 2π 2

¡2m

~ 2

¢3/2

ε1/2ε ≥ 0

one has

log Zgc= 1

2

X

l

∞

Z

−∞

dε D (ε) log (1 + λ exp (−β (ε + El))) (3.98)

3.3.3 Energy and Number of Particles

Using Eqs (1.80) and (1.94) for the energy U and the number of particles

N , namely using

U = −

µ

∂ log Zgc

∂β

¶

η

N = λ∂ log Zgc

one finds that

U = 1

2

X

l

∞

Z

−∞

dε D (ε) (ε + El) fFD(ε + El) , (3.101)

N = 1

2

X

l

∞

Z

−∞

where fFD is the Fermi-Dirac distribution function [see Eq (2.35)]

fFD( ) = 1

3.3.4 Example: Electrons in Metal

Electrons are Fermions having spin 1/2 The spin degree of freedom gives rise

to two orthogonal eigenstates having energies E+and E−respectively In the absent of any external magnetic field these states are degenerate, namely

E+ = E− For simplicity we take E+ = E− = 0 Thus, Eqs (3.101) and (3.102) become

3.3 Fermi Gas

U =

∞

Z

−∞

N =

∞

Z

−∞

Typically for metals at room temperature or below the following holds τ ¿ µ Thus, it is convenient to employ the following theorem (Sommerfeld expan-sion) to evaluate these integrals

Theorem 3.3.1 Let g (ε) be a function that vanishes in the limit ε → −∞, and that diverges no more rapidly than some power of ε as ε → ∞ Then, the following holds

∞

Z

−∞

dε g (ε) fFD(ε)

=

µ

Z

−∞

dε g (ε) +π

2g0(µ) 6β2 + O

µ 1 βµ

¶4

Proof See problem 7 of set 3

With the help of this theorem the number of particles N to second order

in τ is given by

N =

µ

Z

−∞

dε D (ε) +π

2τ2D0(µ)

Moreover, at low temperatures, the chemical potential is expected to be close the the Fermi energy εF, which is defined by

εF= lim

Thus, to lowest order in µ − εFone has

µ

Z

−∞

dε D (ε) =

ε F

Z

−∞

dε D (ε) + (µ − εF) D (εF) + O (µ − εF)2 , (3.108)

and therefore

N = N0+ (µ − εF) D (εF) +π

2τ2D0(εF)

where

ε F

Z

−∞

is the number of electrons at zero temperature The number of electrons N

in metals is expected to be temperature independent, namely N = N0 and consequently

µ = εF− π

2D0(εF)

Similarly, the energy U at low temperatures is given approximately by

U =

∞

Z

−∞

dε D (ε) εfFD(ε)

=

ε F

Z

−∞

dε D (ε) ε

U 0

+ (µ − εF) D (εF) εF+π

2τ2

6 (D

0(εF) εF+ D (εF))

= U0−π

2τ2D0(εF) 6D (εF) D (εF) εF+

π2τ2

6 (D

0(εF) εF+ D (εF))

= U0+π

2τ2

6 D (εF) ,

(3.112) where

U0=

ε F

Z

−∞

From this result one finds that the electronic heat capacity is given by

CV=∂U

∂τ =

π2τ

Comparing this result with Eq (3.81) for the phonons heat capacity, which

is proportional to τ3 at low temperatures, suggests that typically, while the electronic contribution is the dominant one at very low temperatures, at higher temperatures the phonons’ contribution becomes dominant

3.4 Semiconductor Statistics

To be written

3.5 Problems Set 3

3.5 Problems Set 3

1 Calculate the average number of photons N in equilibrium at tempera-ture τ in a cavity of volume V Use this result to estimate the number

of photons in the universe assuming it to be a spherical cavity of radius

1026m and at temperature τ = kB× 3 K

2 Write a relation between the temperature of the surface of a planet and its distance from the Sun, on the assumption that as a black body in thermal equilibrium, it reradiates as much thermal radiation, as it receives from the Sun Assume also, that the surface of the planet is at constant temper-ature over the day-night cycle Use TSun= 5800 K; RSun= 6.96 × 108m; and the Mars-Sun distance of DM−S = 2.28 × 1011m and calculate the temperature of Mars surface

3 Calculate the Helmholtz free energy F of photon gas having total energy

U and volume V and use your result to show that the pressure is given by

p = U

4 Consider a photon gas initially at temperature τ1and volume V1 The gas

is adiabatically compressed from volume V1to volume V2in an isentropic process Calculate the final temperature τ2 and final pressure p2

5 Consider a one-dimensional lattice of N identical point particles of mass

m, interacting via nearest-neighbor spring-like forces with spring constant

mω2(see Fig 3.2) Denote the lattice spacing by a Show that the normal mode eigen-frequencies are given by

ωn= ωp

where kn= 2πn/aN , and n is integer ranging from −N/2 to N/2 (assume

N À 1)

6 Consider an orbital with energy ε in an ideal gas The system is in thermal equilibrium at temperature τ and chemical potential µ

a) Show that the probability that the orbital is occupied by n particles

is given by

pF(n) = exp [n (µ − ε) β]

for the case of Fermions, where n ∈ {0, 1}, and by

pB(n) = {1 − exp [(µ − ε) β]} exp [n (µ − ε) β] , (3.118) where n ∈ {0, 1, 2, }, for the case of Bosons

b) Show that the variance (∆n)2=D

(n − hni)2E

is given by

for the case of Fermions, and by

for the case of Bosons

7 Let g (ε) be a function that vanishes in the limit ε → −∞, and that diverges no more rapidly than some power of ε as ε → ∞ Show that the following holds (Sommerfeld expansion)

I =

∞

Z

−∞

dε g (ε) fFD(ε)

=

µ

Z

−∞

dε g (ε) +π

2g0(µ) 6β2 + O

µ 1 βµ

¶4

8 Consider a metal at zero temperature having Fermi energy εF, number

of electrons N and volume V

a) Calculate the mean energy of electrons

b) Calculate the ratio α of the mean-square-speed of electrons to the square of the mean speed

α =

­

v2®

c) Calculate the pressure exerted by an electron gas at zero tempera-ture

9 For electrons with energy ε À mc2 (relativistic fermi gas), the energy

is given by ε = pc Find the fermi energy of this gas and show that the ground state energy is

E (T = 0) = 3

10 A gas of two dimensional electrons is free to move in a plane The mass

of each electron is me, the density (number of electrons per unit area) is

n, and the temperature is τ Show that the chemical potential µ is given by

µ = τ log

· exp

µ nπ~2

meτ

¶

− 1

¸

3.6 Solutions Set 3

3.6 Solutions Set 3

1 The density of states of the photon gas is given by

dg = V ε

2

Thus

π2}3c3

∞

Z

0

ε2

eε/τ − 1dε

= V³ τ

}c

´3

where

α = 1

π2

∞

Z

0

x2

The number α is calculated numerically

For the universe

N = 4π

3

¡

1026m¢3µ

1.3806568 × 10−23J K−13 K 1.05457266 × 10−34J s2.99792458 × 108m s−1

¶3

× 0.24359

2 The energy emitted by the Sun is

and the energy emitted by a planet is

Eplanet= 4πR2

planetσBT4

The fraction of Sun energy that planet receives is

πR2

planet

4πD2

M −S

and this equals to the energy it reradiates Therefore

πR2

planet

4πD2 ESun= Eplanet,

Tplanet=

r

RSun

2D TSun, and for Mars

TMars=

r 6.96 × 108m

2 × 2.28 × 1011m5800 K = 226 K

3 The partition function is given by

Z =Y

n

∞

X

s=0

exp (sβ}ωn) =Y

n

1

1 − exp (−β}ωn) , thus the free energy is given by

F = −τ log Z = τX

n

log [1 − exp (−β}ωn)] Transforming the sum over modes into integral yields

F = τ π

Z ∞

0

dnn2log [1 − exp (−β}ωn)] (3.132)

= τ π

Z ∞

0

dnn2log

·

1 − exp

µ

−β}πcnL

¶¸

,

or, by integrating by parts

F = −1

3

}π2c L

Z ∞

0

3

exp³

β}πcn L

´

− 1 = −

1

3U , where

U =π

2τ4V

Thus

p = −

µ

∂F

∂V

¶

τ

= U

4 Using the expression for Helmholtz free energy, which was derived in the previous problem,

F = −U

3 = −π

2τ4V 45}3c3 , one finds that the entropy is given by

3.6 Solutions Set 3

σ = −

µ

∂F

∂τ

¶

V

=4π

2τ3V 45}3c3 Thus, for an isentropic process, for which σ is a constant, one has

τ2= τ1

µ

V1

V2

¶1/3

Using again the previous problem, the pressure p is given by

p = U

3V ,

thus

p = π

2τ4

45}3c3 ,

and

p2= π

2τ4 45}3c3

| {z }

p 1

µ

V1

V2

¶4/3

5 Let u (na) be the displacement of point particle number n The equations

of motion are given by

m¨u (na) = −mω2{2u (na) − u [(n − 1) a] − u [(n + 1) a]} (3.135) Consider a solution of the form

Periodic boundary condition requires that

thus

kn= 2πn

Substituting in Eq 3.135 yields

−mω2nu (na) = −mω2£

2u (na) − u (na) e−ika− u (na) eika¤

, (3.139) or

ωn= ωp

2 (1 − cos kna) = 2ω¯¯¯

¯sinkn2a¯¯¯

6 In general using Gibbs factor

p (n) = exp [n (µ − ε) β]

X

n 0

where β = 1/τ , one finds for Fermions

pF(n) = exp [n (µ − ε) β]

where n ∈ {0, 1}, and for Bosons

pB(n) = exp [n (µ − ε) β]

∞

X

n 0 =0

exp [n0(µ − ε) β]

= {1 − exp [(µ − ε) β]} exp [n (µ − ε) β] ,

(3.143) where n ∈ {0, 1, 2, } The expectation value of hni in general is given by

hni =X

n 0

n0p (n0) =

X

n 0

n0exp [n (µ − ε) β]

X

n 0

exp [n0(µ − ε) β] , (3.144) thus for Fermions

and for Bosons

hniB= {1 − exp [(µ − ε) β]}

∞

X

n 0 =0

n0exp [n0(µ − ε) β]

= {1 − exp [(µ − ε) β]} exp [(µ − ε) β]

(1 − exp [(µ − ε) β])2

exp [(ε − µ) β] − 1.

(3.146)

In general, the following holds

τ

µ

∂ hni

∂µ

¶

τ

=

X

n 00

(n00)2exp [n (µ − ε) β]

X

n 0

exp [n0(µ − ε) β] −







X

n 0

n0exp [n0(µ − ε) β] X

n 0

exp [n0(µ − ε) β]







2

=­

n2®

− hni2=D

(n − hni)2E

(3.147)

3.6 Solutions Set 3 Thus

(∆n)2F = exp [(ε − µ) β]

(exp [(ε − µ) β] + 1)2 = hniF(1 − hniF) , (3.148) (∆n)2B= exp [(ε − µ) β]

(exp [(ε − µ) β] − 1)2 = hniB(1 + hniB) (3.149)

7 Let

G (ε) =

ε

Z

−∞

Integration by parts yields

I =

∞

Z

−∞

dε g (ε) fFD(ε)

= [G (ε) fFD(ε)|∞−∞

=0

+

∞

Z

−∞

dε G (ε)

µ

−∂f∂εFD

¶ ,

(3.151) where the following holds

µ

−∂f∂εFD

¶

β(ε −µ)

¡

eβ(ε −µ)+ 1¢2 = β

4 cosh2³

β

2(ε − µ)´ (3.152) Using the Taylor expansion of G (ε) about ε − µ, which has the form

G (ε) =

∞

X

n=0

G(n)(µ)

yields

I =

∞

X

n=0

G(n)(µ) n!

∞

Z

−∞

β (ε − µ)ndε

4 cosh2³

β

Employing the variable transformation

and exploiting the fact that (−∂fFD/∂ε) is an even function of ε−µ leads to

I =

∞

X

n=0

G(2n)(µ) (2n)!β2n

∞

Z

−∞

x2ndx

With the help of the identities

∞

Z

−∞

dx

∞

Z

−∞

x2dx

4 cosh2 x2 =

π2

one finds

I = G (µ) +π

2G(2)(µ) 6β2 + O

µ 1 βµ

¶4

=

µ

Z

−∞

dε g (ε) +π

2g0(µ) 6β2 + O

µ 1 βµ

¶4

(3.159)

8 In general, at zero temperature the average of the energy ε to the power

n is given by

hεni =

ε F

R

0

dε D (ε) εn

ε F

R

0

dε D (ε)

where D (ε) is the density of states

D (ε) = V

2π2

µ 2m

~2

¶3/2

thus

hεni = ε

n F 2n

a) Using Eq (3.162) one finds that

hεi = 3εF

b) The speed v is related to the energy by

v =

r 2ε

3.6 Solutions Set 3 thus

α = hεi

­

ε1/2®2 =

ε F

2 +1

¿

ε−1/2F

2 +1

À2 = 16

c) The number of electrons N is given by

N =

ε F

Z

0

dε D (ε) = D (εF)

ε1/2F

ε F

Z

0

dε ε1/2=D (εF)

ε1/2F

2

3ε

3/2

F , thus

εF= ~2

2m

µ 3π2N V

¶2/3

and therefore

U = 3N

5

~2

2m

µ 3π2N V

¶2/3

Moreover, at zero temperature the Helmholtz free energy F = U −

τ σ = U , thus the pressure is given by

p = −

µ

∂F

∂V

¶

τ ,N

= −

µ

∂U

∂V

¶

τ ,N

=3N

5

~2

2m

µ 3π2N V

¶2/3

2 3V

= 2N εF

5V .

(3.168)

9 The energy of the particles are

where p = ~k, and k = πn

L, n =q

n2

x+ n2+ n2 and ni= 1, 2,

Therefore

N = 2 ×18×43πn3F= π

3n

3

nF=

µ3N

π

¶1/3

εF= ~cπ

L

µ 3N π

¶1/3

= ~cπ

µ 3N πV

¶1/3

The energy is given by

E (T = 0) =

ε F

Z

0

εg (ε) dε = L

3

π2~3c3

Z ε F

0

3

π2~3c3 ×ε

4 F

4 =

1 4

L3

π2~3c3ε3F× εF=

= 1 4

L3

π2~3c3

3π2~3c3N

L3 εF= 3

4N εF

10 The energy of an electron having a wave function proportional to exp (ikxx) exp (ikyy) is

~2

2me

¡

k2x+ ky2¢

For periodic boundary conditions one has

kx= 2πnx

ky= 2πny

Ly

where the sample is of area LxLy, and nx and ny are both integers The

number of states having energy smaller than E0 is given by (including

both spin directions)

2π2meE0

~2

LxLy

thus, the density of state per unit area is given by

D (E) =

½ me

π~ 2 E > 0

Using Fermi-Dirac function

where β = 1/τ , the density n is given by

n =

Z ∞

−∞

D (E) f (E) dE

= me

π~2

Z ∞

0

dE

1 + exp [β (E − µ)]

= meτ

π~2 log¡

1 + eβµ¢

,

(3.178)

3.6 Solutions Set 3 thus

µ = τ log

· exp

µ nπ~2

meτ

¶

− 1

¸

4 Classical Limit of Statistical Mechanics

In this chapter we discuss the classical limit of statistical mechanics We dis-cuss Hamilton’s formalism, define the Hamiltonian and present the Hamilton-Jacobi equations of motion The density function in thermal equilibrium is used to prove the equipartition theorem This theorem is then employed to analyze an electrical circuit in thermal equilibrium, and to calculate voltage noise across a resistor (Nyquist noise formula)

4.1 Classical Hamiltonian

In this section we briefly review Hamilton’s formalism, which is analogous to Newton’s laws of classical mechanics Consider a classical system having d degrees of freedom The system is described using the vector of coordinates

¯

Let E be the total energy of the system For simplicity we restrict the dis-cussion to a special case where E is a sum of two terms

E = T + V ,

where T depends only on velocities, namely T = T³·

¯

q´ , and where V depends only on coordinates, namely V = V (¯q) The notation overdot is used to express time derivative, namely

·

¯

q =

µ

dq1

dt ,

dq2

dt , ,

dqd dt

¶

We refer to the first term T as kinetic energy and to the second one V as potential energy

The canonical conjugate momentum pi of the coordinate qi is defined as

pi= ∂T

The classical Hamiltonian H of the system is expressed as a function of the vector of coordinates ¯q and as a function of the vector of canonical conjugate momentum variables

namely

and it is defined by

H =

d

X

i=1

4.1.1 Hamilton-Jacobi Equations

The equations of motion of the system are given by

˙qi= ∂H

∂pi

(4.7)

˙

pi= −∂H∂q

i

where i = 1, 2, d

4.1.2 Example

q

V(q)

m

q

V(q)

m

Consider a particle having mass m in a one dimensional potential V (q) The kinetic energy is given by T = m ˙q2/2, thus the canonical conjugate mo-mentum is given by [see Eq (4.3)] p = m ˙q Thus for this example the canon-ical conjugate momentum equals the mechancanon-ical momentum Note, however, that this is not necessarily always the case Using the definition (4.6) one finds that the Hamiltonian is given by