3.20, the dynamics of a mode amplitude q is the same as the dynamics of an harmonic oscillator having angular frequency ωκ= cκ... 3.1.2 Partition FunctionWhat is the partition function o
Trang 1Using pV = Nτ yields
η = 1 +τd− τc
τb− τa
= 1 + p2(Vd− Vc)
Along the isentropic process pVγ is constant, where γ = Cp/Cv, thus
η = 1 +p2
p1
³
p 1
p 2
´1 γ (Va− Vb)
Vb− Va = 1 −
µ
p2
p1
¶γ−1 γ
28 No heat is exchanged in the isentropic processes, thus the efficiency is given by
η = 1 + Ql
Qh
= 1 +Qc→d
Qa →b
= 1 +cV (τd− τc)
cV (τb− τa) .
(2.297) Since τ Vγ −1 remains unchanged in an isentropic process, where
γ = cp
one finds that
or
τc
τb
= τd
τa
=
µ
V2
V1
¶1 −γ
thus
η = 1 −
µ
V2
V1
¶1 −γ
29 Let VA1= NτA/p (VB1= NτB/p) be the initial volume of vessel A (B) and let VA2 (VB2) be the final volume of vessel A (B) In terms of the final temperature of both vessels, which is denoted as τf, one has
VA2= VB2= Nτf
Trang 2The entropy of an ideal gas of density n = N/V , which contains N particles, is given by
σ = N
µ
lognQ
n +
5 2
¶
where
nQ=
µMτ
2π~2
¶3/2
or as a function of τ and p
σ = N
Ã
log
¡ M
2π~ 2
¢3/2
τ5/2
5 2
!
Thus the change in entropy is given by
∆σ = σfinal− σinitial
= 5N
2 log
τ2f
τAτB
(2.307)
In general, for an isobaric process the following holds
Q = W + ∆U = p (V2− V1) + cV (τ2− τ1) , (2.308) where Q is the heat that was added to the gas, W the work done by the gas and ∆U the change in internal energy of the gas Using the equation
of state pV = N τ this can be written as
Since no heat is exchanged with the environment during this process the following holds
QA+ QB = 0 ,
where
thus
τf =τA+ τB
and therefore
∆σ = 5N
2 log
(τA+ τB)2 4τAτB
Trang 3In the first part of this chapter we study two Bosonic systems, namely photons and phonons A photon is the quanta of electromagnetic waves whereas a phonon is the quanta of acoustic waves In the second part we study two Fermionic systems, namely electrons in metals and electrons and holes in semiconductors
3.1 Electromagnetic Radiation
In this section we study an electromagnetic cavity in thermal equilibrium 3.1.1 Electromagnetic Cavity
Consider an empty volume surrounded by conductive walls having infinite conductivity The Maxwell’s equations in SI units are given by
∇ × H = 0
∂E
∇ × E = −µ0
∂H
and
where 0 = 8.85 × 10−12F m−1 and µ0 = 1.26 × 10−6N A−2 are the permit-tivity and permeability respectively of free space, and the following holds
0µ0= 1
where c = 2.99 × 108m s−1 is the speed of light in vacuum
In the Coulomb gauge, where the vector potential A is chosen such that
Trang 4the scalar potential φ vanishes in the absence of sources (charge and current), and consequently both fields E and H can be expressed in terms of A only as
and
The gauge condition (3.6) and Eqs (3.7) and (3.8) guarantee that Maxwell’s equations (3.2), (3.3), and (3.4) are satisfied
∇ × E = −∂ (∇ × A)∂t = −µ0
∂H
∇ · H = 1
where in the last equation the general vector identity ∇ · (∇ × A) = 0 has been employed Substituting Eqs (3.7) and (3.8) into the only remaining nontrivial equation, namely into Eq (3.1), leads to
∇ × (∇ × A) = −c12∂
2A
Using the vector identity
and the gauge condition (3.6) one finds that
∇2A= 1
c2
∂2A
Consider a solution in the form
where q (t) is independent on position r and u (r) is independent on time t The gauge condition (3.6) leads to
From Eq (3.14) one finds that
q∇2u= 1
c2ud
2q
Trang 5Multiplying by an arbitrary unit vector ˆnleads to
¡
∇2u¢
· ˆn
u· ˆn =
1
c2q
d2q
The left hand side of Eq (3.18) is a function of r only while the right hand side is a function of t only Therefore, both should equal a constant, which is denoted as −κ2, thus
and
d2q
where
Equation (3.19) should be solved with the boundary conditions of a per-fectly conductive surface Namely, on the surface S enclosing the cavity we have H · ˆs = 0 and E × ˆs = 0, where ˆs is a unit vector normal to the surface
To satisfy the boundary condition for E we require that u be normal to the surface, namely, u = ˆs(u · ˆs) on S This condition guarantees also that the boundary condition for H is satisfied To see this we calculate the integral of the normal component of H over some arbitrary portion S0 of S Using Eq (3.8) and Stoke’s’ theorem one finds that
Z
S 0(H · ˆs) dS = µq
0
Z
S 0[(∇ × u) · ˆs] dS
= q
µ0
I
C
u·dl ,
(3.22) where the close curve C encloses the surface S0 Thus, since u is normal to the surface one finds that the integral along the close curve C vanishes, and therefore
Z
Since S0 is arbitrary we conclude that H · ˆs =0 on S
Each solution of Eq (3.19) that satisfies the boundary conditions is called
an eigen mode As can be seen from Eq (3.20), the dynamics of a mode amplitude q is the same as the dynamics of an harmonic oscillator having angular frequency ωκ= cκ
Trang 63.1.2 Partition Function
What is the partition function of a mode having eigen angular frequency ωκ?
We have seen that the mode amplitude has the dynamics of an harmonic oscillator having angular frequency ωκ Thus, the quantum eigenenergies of the mode are
where s = 0, 1, 2, is an integer1 When the mode is in the eigenstate having energy εs the mode is said to occupy s photons The canonical partition function of the mode is found using Eq (1.69)
Zκ =
∞
X
s=0
exp (−sβ~ωκ)
1 − exp (−β~ωκ) .
(3.25) Note the similarity between this result and the orbital partition function ζ
of Bosons given by Eq (2.36) The average energy is found using
hεκi = −∂ log Z∂β κ
= ~ωκ
eβ~ω κ − 1 .
(3.26) The partition function of the entire system is given by
Z =Y
κ
and the average total energy by
U = −∂ log Z∂β =X
κ
3.1.3 Cube Cavity
For simplicity, consider the case of a cavity shaped as a cube of volume
V = L3 We seek solutions of Eq (3.19) satisfying the boundary condition
1 In Eq (3.24) above the ground state energy was taken to be zero Note that
by taking instead ε s = (s + 1/2) ~ω κ , one obtains Z κ = 1/2 sinh (β}ω κ /2) and
hε n i = (~ω κ /2) coth (β}ω κ /2) In some cases the offset energy term ~ω κ /2 is very important (e.g., the Casimir force), however, in what follows we disregard it.
Trang 7that the tangential component of u vanishes on the walls Consider a solution having the form
ux=
r
8
Vaxcos (kxx) sin (kyy) sin (kzz) , (3.29)
uy =
r
8
Vaysin (kxx) cos (kyy) sin (kzz) , (3.30)
uz=
r
8
Vazsin (kxx) sin (kyy) cos (kzz) (3.31) While the boundary condition on the walls x = 0, y = 0, and z = 0 is guaranteed to be satisfied, the boundary condition on the walls x = L, y = L, and z = L yields
kx= nxπ
ky = nyπ
kz= nzπ
where nx, ny and nz are integers This solution clearly satisfies Eq (3.19) where the eigen value κ is given by
κ =q
k2
Alternatively, using the notation
one has
κ = π
where
n =q
n2
Using Eq (3.21) one finds that the angular frequency of a mode characterized
by the vector of integers n is given by
ωn=πc
In addition to Eq (3.19) and the boundary condition, each solution has to satisfy also the transversality condition ∇ · u = 0 (3.16), which in the present case reads
Trang 8Thus, for each set of integers {nx, ny, nz} there are two orthogonal modes (polarizations), unless nx= 0 or ny= 0 or nz= 0 In the latter case, only a single solution exists
3.1.4 Average Energy
The average energy U of the system is found using Eqs (3.26), (3.28) and (3.39)
n
~ωn
eβ~ω n− 1
= 2τ
∞
X
n x =0
∞
X
n y =0
∞
X
n z =0
αn
eαn− 1 ,
(3.42) where the dimensionless parameter α is given by
α = β~πc
The following relation can be employed to estimate the dimensionless param-eter α
α = 2.4 × 10−3
L
cm
τ
300 K
In the limit where
the sum can be approximated by the integral
U ' 2τ4π8
∞
Z
0
dn n2 αn
Employing the integration variable transformation [see Eq (3.39)]
n = L
allows expressing the energy per unit volume U/V as
U
V =
∞
Z
0
Trang 9uω= ~
c3π2
ω3
This result is know as Plank’s radiation law The factor uω represents the spectral distribution of the radiation The peak in uωis obtained at β~ω0= 2.82 In terms of the wavelength λ0= 2πc/ω0 one has
λ0
µm= 5.1
µ
T
1000 K
¶−1
0.2 0.4 0.6 0.8 1 1.2 1.4
The function x3/ (ex− 1)
The total energy is found by integrating Eq (3.48) and by employing the variable transformation x = β~ω
U
V =
τ4
c3π2~3
∞
Z
0
x3dx
ex− 1
| {z } π4 15
= π
2τ4
15~3c3
(3.51) 3.1.5 Stefan-Boltzmann Radiation Law
Consider a small hole having area dA drilled into the conductive wall of an electromagnetic (EM) cavity What is the rate of energy radiation emitted from the hole? We employ below a kinetic approach to answer this question Consider radiation emitted in a time interval dt in the direction of the unit vector ˆu Let θ be the angle between ˆuand the normal to the surface of the hole Photons emitted during that time interval dt in the direction ˆucame from the region in the cavity that is indicated in Fig 3.1, which has volume
Trang 10dAcosθ
cdt
uˆ θ
dAcosθ
cdt
uˆ
Fig 3.1 Radiation emitted through a small hole in the cavity wall.
The average energy in that region can be found using Eq (3.51) Integrating over all possible directions yields the total rate of energy radiation emitted from the hole per unit area
J = 1
dAdt
1 4π
π/2
Z
0
dθ sin θ
2π
Z
0
dϕ U
VVθ
= π
2τ4
15~3c2
1 4π
π/2
Z
0
dθ sin θ cos θ
2π
Z
0
dϕ
1/4
= π
2τ4
60~3c2
(3.53)
In terms of the historical definition of temperature T = τ /kB[see Eq (1.92)] one has
where σB, which is given by
σB = π
2k4
B
60~3c2 = 5.67 × 10−8 W
is the Stefan-Boltzmann constant
Trang 11m mω2 m mω2 m mω2 m mω2 m
Fig 3.2 1D lattice.
3.2 Phonons in Solids
In this section we study elastic waves in solids We start with a one-dimensional example, and then generalize some of the results for the case
of a 3D lattice
3.2.1 One Dimensional Example
Consider the 1D lattice shown in Fig 3.2 below, which contains N ’atoms’ having mass m each that are attached to each other by springs having spring constant mω2 The lattice spacing is a The atoms are allowed to move in one dimension along the array axis In problem 5 of set 3 one finds that the normal mode angular eigen-frequencies are given by
ωn= ωp
2 (1 − cos kna) = 2ω
¯
¯
¯sinkn2a
¯
¯
where a is the lattice spacing,
kn= 2πn
and n is integer ranging from −N/2 to N/2
0 0.2 0.4 0.6 0.8
-1 -0.8 -0.6 -0.4 -0.2 0.2 0.4x0.6 0.8 1
The function |sin (πx/2)|
Trang 12What is the partition function of an eigen-mode having eigen angular fre-quency ωn? The mode amplitude has the dynamics of an harmonic oscillator having angular frequency ωn Thus, as we had in the previous section, where
we have discussed EM modes, the quantum eigenenergies of the mode are
where s = 0, 1, 2, is an integer When the mode is in an eigenstate having energy εs the mode is said to occupy s phonons The canonical partition function of the mode is found using Eq (1.69)
Zκ =
∞
X
s=0
exp (−sβ~ωκ)
1 − exp (−β~ωκ) .
(3.59) Similarly to the EM case, the average total energy is given by
U =
N/2
X
n= −N/2
}ωn
where β = 1/τ , and the total heat capacity is given by
CV=∂U
∂τ =
N/2
X
n= −N/2
(β}ωn)2exp (β}ωn)
High Temperature Limit In the high temperature limit β}ω ¿ 1 (β}ωn)2exp (β}ωn)
therefore
Low Temperature Limit In the low temperature limit β}ω À 1 the main contribution to the sum in Eq (3.61) comes from terms for which
|n| N/β}ω Thus, to a good approximation the dispersion relation can be approximated by
ωn= 2ω
¯
¯
¯sinkn2a
¯
¯
¯ ' 2ω
¯
¯kna 2
¯
¯
Moreover, in the limit N À 1 the sum in Eq (3.61) can be approximated
by an integral, and to a good approximation the upper limit N/2 can be substituted by infinity, thus
Trang 13N/2
X
n= −N/2
(β}ωn)2exp (β}ωn) [exp (β}ωn) − 1]2 ' 2
∞
X
n=0
¡ β}ω2πNn¢2
exp¡ β}ω2πNn¢
£ exp¡ β}ω2πNn¢
− 1¤2
' 2
Z ∞
0
dn
¡ β}ω2πNn¢2
exp¡ β}ω2πNn¢
£ exp¡ β}ω2πNn¢
− 1¤2
= N
π
τ
}ω
Z ∞
0
dx x
2exp (x) (exp (x) − 1)2
π 2 /3
= Nπ
3
τ }ω .
(3.65) 3.2.2 The 3D Case
The case of a 3D lattice is similar to the case of EM cavity that we have studied in the previous section However, there are 3 important distinctions:
1 The number of modes of a lattice containing N atoms that can move in 3D is finite, 3N instead of infinity as in the EM case
2 For any given vector k there are 3, instead of only 2, orthogonal modes (polarizations)
3 Dispersion: contrary to the EM case, the dispersion relation (namely, the function ω (k)) is in general nonlinear
Due to distinctions 1 and 2, the sum over all modes is substituted by an integral according to
∞
X
n x =0
∞
X
n y =0
∞
X
n z =0
→ 384π
n D Z
0
where the factor of 3 replaces the factor of 2 we had in the EM case Moreover, the upper limit is nD instead of infinity, where nD is determined from the requirement
3
84π
n D
Z
0
thus
nD=
µ
6N
π
¶1/3
Trang 14Similarly to the EM case, the average total energy is given by
n
}ωn
= 3π
2
n D
Z
0
dn n2 }ωn
exp (β}ωn) − 1.
(3.70)
To proceed with the calculation the dispersion relation ωn(kn) is needed Here we assume for simplicity that dispersion can be disregarded to a good approximation, and consequently the dispersion relation can be assumed to
be linear
where v is the sound velocity The wave vector kn is related to n = q
n2
x+ n2+ n2 by
kn= πn
where L = V1/3 and V is the volume In this approximation one finds using the variable transformation
x = β}vπn
that
U = 3π
2
n D
Z
0
dn n2
}vπn L
exp³
β}vπn L
´
− 1
= 3V τ
4
2}3v3π2
x D Z
0
3
exp x − 1.
(3.74) where
xD=β}vπnD
β}vπ¡6N
π
¢1/3
Alternatively, in terms of the Debye temperature, which is defined as
Θ = }v
µ6π2N
V
¶1/3
one has
Trang 15and
U = 9N τ³ τ
Θ
´3ZxD
0
3
As an example Θ/kB = 88 K for Pb, while Θ/kB = 1860 K for diamond Below we calculate the heat capacity CV= ∂U/∂τ in two limits
High Temperature Limit In the high temperature limit xD= Θ/τ ¿ 1, thus
U = 9Nτ³ τ
Θ
´3ZxD
0
3
exp x − 1
' 9Nτ³ τ
Θ
´3ZxD
0
dx x2
= 9Nτ³ τ
Θ
´3x3 D
3
= 3Nτ ,
(3.78) and therefore
CV=∂U
Note that in this limit the average energy of each mode is τ and consequently
U = 3Nτ This result demonstrates the equal partition theorem of classical statistical mechanics that will be discussed in the next chapter
Low Temperature Limit In the low temperature limit xD = Θ/τ À 1, thus
U = 9Nτ³ τ
Θ
´3ZxD
0
3
exp x − 1
' 9Nτ³ τ
Θ
´3Z∞ 0
3
exp x − 1
π 4 /15
= 3π
4
5 N τ
³ τ Θ
´3
,
(3.80)
Trang 16and therefore
CV=∂U
∂τ =
12π4
5 N
³ τ Θ
´3
Note that Eq (3.80) together with Eq (3.75) yield
U
V =
3
2
π2τ4
Note the similarity between this result and Eq (3.51) for the EM case
3.3 Fermi Gas
In this section we study an ideal gas of Fermions of mass m While only the classical limit was considered in chapter 2, here we consider the more general case
3.3.1 Orbital Partition Function
Consider an orbital having energy εn Disregarding internal degrees of free-dom, its grandcanonical Fermionic partition function is given by [see Eq (2.33)]
where
is the fugacity and β = 1/τ Taking into account internal degrees of freedom the grandcanonical Fermionic partition function becomes
ζn=Y
l
(1 + λ exp (−βεn) exp (−βEl)) , (3.85)
where {El} are the eigenenergies of a particle due to internal degrees of freedom [see Eq (2.71)] As is required by the Pauli exclusion principle, no more than one Fermion can occupy a given internal eigenstate and a given orbital
3.3.2 Partition Function of the Gas
The grandcanonical partition function of the gas is given by
Zgc=Y
n