ryan d. reece august 11, 2006 - university of pennsylvaniarreece/docs/notes/stat_mech_notes.pdf ·...
TRANSCRIPT
Statistical Mechanics Notes
Ryan D. Reece
August 11, 2006
Contents
1 Thermodynamics 31.1 State Variables . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Inexact Differentials . . . . . . . . . . . . . . . . . . . . . . . 51.3 Work and Heat . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Thermodynamic Potentials . . . . . . . . . . . . . . . . . . . . 71.6 Maxwell Relations . . . . . . . . . . . . . . . . . . . . . . . . 91.7 Heat Capacities . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Ideas of Statistical Mechanics 102.1 The Goal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Boltzmann’s Hypothesis . . . . . . . . . . . . . . . . . . . . . 102.3 Entropy and Probability . . . . . . . . . . . . . . . . . . . . . 112.4 The Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Canonical Ensemble 133.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 The Bridge to Thermodynamics . . . . . . . . . . . . . . . . . 163.3 Example: A Particle in a Box . . . . . . . . . . . . . . . . . . 17
4 Maxwell-Boltzmann Statistics 214.1 Density of States . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 The Maxwell-Boltzmann Distribution . . . . . . . . . . . . . . 234.3 Example: Particles in a Box . . . . . . . . . . . . . . . . . . . 24
5 Black Body Radiation 265.1 Molecular Beams . . . . . . . . . . . . . . . . . . . . . . . . . 265.2 Cavity Radiation . . . . . . . . . . . . . . . . . . . . . . . . . 26
1
5.3 The Planck Distribution . . . . . . . . . . . . . . . . . . . . . 26
6 Thermodynamics of Multiple Particles 276.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.2 Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . 276.3 The Gibbs-Duhem Relation . . . . . . . . . . . . . . . . . . . 28
7 Identical Particles 297.1 Exchange Symmetry of Quantum States . . . . . . . . . . . . 297.2 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.3 Over Counting . . . . . . . . . . . . . . . . . . . . . . . . . . 31
8 Grand Canonical Ensemble 328.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328.2 The Bridge to Thermodynamics . . . . . . . . . . . . . . . . . 338.3 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348.4 Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358.5 Non-relativistic Cold Electron Gas . . . . . . . . . . . . . . . 368.6 Example: White Dwarf Stars . . . . . . . . . . . . . . . . . . 388.7 Relativistic Cold Electron Gas . . . . . . . . . . . . . . . . . . 418.8 Example: Collapse to a Neutron Star . . . . . . . . . . . . . . 42
9 Questions 45
2
Chapter 1
Thermodynamics
1.1 State Variables
Given a function f(x, y), its differential is
df =∂f
∂xdx+
∂f
∂ydy (1.1)
The function is called analytic if its mixed partials are equal.
∂2f
∂x∂y=
∂2f
∂y∂x(1.2)
In this case, the differential of f , equation 1.1, is called an exact differential.Because the mixed partials are equal, Green’s Theorem says that any closedline integral of f in x-y-space is zero.∫∫
:0(∂2f
∂x∂y− ∂2f
∂y∂x
)dx dy =
∮ (∂f
∂xdx+
∂f
∂ydy
)(1.3)
0 =
∮df (1.4)
Suppose we break some closed path into two paths α and β which beginat point Λ and end at point Γ (see Figure 1.1). If f is analytic, the pathintegrals of f along the two paths α and β must be equal because the integralalong the closed path, equal to the difference of the the two path integrals,is zero. ∮
df =
∫α
df −∫
β
df = 0 (1.5)
3
Figure 1.1: Path Independence of State Variables
This means the value of f , is a function of the endpoints alone and indepen-dent of the path.
f =
∫df = f(Λ,Γ) → f(∆x,∆y) (1.6)
Concerning thermodynamics, if x and y are macroscopic variables that wecan measure (temperature, pressure, volume, etc.), and we can find somevariable f with an exact differential in terms of x and y, then f is calleda state variable. State variables have defined values when a system hascome to equilibrium, regardless of what processes (paths) brought the systemthere. The macroscopic variables x and y, are themselves state variables. Anequation of state can be derived relating state variables independent of path,for example the equation of state for an ideal gas1 is
p V = N k T (1.7)
where pressure (p), volume (V ), number of particles (N), and temperature(T ) are all state variables and k is Boltzmann’s constant.
1We prove that this is the equation of state for an ideal gas, using statistical mechanicsin Section 3.3
4
1.2 Inexact Differentials
Suppose we had derived the following differential equation for f
δf = A(x, y)dx+B(x, y)dy (1.8)
and f is not analytic.∂A
∂y6= ∂B
∂x(1.9)
Then δf is an inexact differential, denoted by δ. It may be possible tomultiply δf by some expression called an integrating factor to make an exactdifferential. Consider the following example.
δf = (2x+ yx2)dx+ x3dy (1.10)
We see that δf is indeed inexact.
∂2f
∂y∂x= x2,
∂2f
∂x∂y= 3x2 (1.11)
Choosing an integrating factor of exy gives
dS ≡ exy δf = exy (2x+ yx2)dx+ exy x3dy (1.12)
dS is exact.∂2S
∂x∂y=
∂2S
∂y∂x= (x3y + 3x2)exy (1.13)
We will use this method when we define entropy in Section 1.4.
1.3 Work and Heat
Naıve application conservation of energy requires that the change in theenergy of a system, ∆U , is the work done on it, W . But this ignores thermalprocesses as a means to take and give energy. Introducing a quantity for theheat added to a system, Q, accounts for these thermal processes. This is theFirst Law of Thermodynamics.
∆U = W +Q (1.14)
5
Considering the traditional example of a cylinder of gas with a piston suchthat we may vary the volume, V , and pressure, p, of the gas, we note thata differential change in volume, dV , does the following differential work onthe system
δW = − p dV (1.15)
For some less ideal system, this can be generalized. If some system has Xj
variables that when varied do work on th system, then there are pj “gener-alized pressures” that satisfy
δW = −∑
j
pj dXj (1.16)
But, equation 1.15 is an inexact differential meaning that the work is depen-dent on the path through P -V -space and not just on the endpoints. That is,it is also dependent on how much heat comes in or out of the system. Thisimplies that the differential heat is also an inexact differential.
dU = δW + δQ (1.17)
1.4 Entropy
Our goal is to make equation 1.17 an exact differential in terms of statevariables. We have seen that in our traditional system of a cylinder of gas,we can write δW in terms of state variables p and V . Now we need to finda way of writing δQ in terms of state variables.
δQ = dU + p dV (1.18)
From the equation of state for an ideal gas (equation 1.7) we have
P =NkT
V(1.19)
P dV =NkT
VdV (1.20)
Then knowing that the energy of an ideal gas2 is
U =3
2NkT (1.21)
2We prove the equation of state and equation 1.21 for an ideal gas in Section 3.3
6
dU =3
2Nk dT (1.22)
Plugging eqautions 1.20 and 1.22 into 1.18 gives
δQ =3
2Nk dT +
NkT
VdV (1.23)
Indeed, δQ is inexact.
∂2Q
∂V ∂T=
∂
∂V
(3
2Nk
)= 0 (1.24)
∂2Q
∂T∂V=
∂
∂T
(NkT
V
)=Nk
V(1.25)
If equation 1.25 had not had that factor of T before taking the derivative,then both mixed partials would equal zero, and the differential would beexact. This motivates us to pick an integrating factor of T−1. We define theentropy, S, to have the differential
dS ≡ δQ
T(1.26)
=3
2
Nk
TdT +
Nk
VdV (1.27)
It is an exact differential
∂2S
∂V ∂T=
∂2S
∂T∂V= 0 (1.28)
meaning S is a state variable. With entropy defined, we can now write dUin terms of state variables.
dU = T dS − p dV (1.29)
1.5 Thermodynamic Potentials
Sometimes it is convenient to create other state variables from our existingones, depending on which variables are held constant in a process. Thethermodynamic potentials, other than the total energy (U), are enthalpy (H),Helmholtz Free Energy (F ), and Gibbs Free Energy (G). They are createdby adding pV and/or subtracting TS from U (see Figure 1.2).
7
Figure 1.2: Mnemonic for the thermodynamic potentials
We can derive the exact differentials of these potentials, simply by differ-entiating their definitions and plugging in our exact differential for dU .
dH = T dS −p dV +p dV + V dp = T dS + V dp (1.30)
dF = T dS − p dV −T dS − S dT = −S dT − p dV (1.31)
dG = T dS −p dV +p dV + V dp−T dS − S dT (1.32)
= V dp− S dT (1.33)
8
1.6 Maxwell Relations
By recognizing that each of the thermodynamic potentials has an exact dif-ferential, we can derive the Maxwell Relations by equating the mixed partials.
∂2U
∂V ∂S=
∂2U
∂S∂V⇒
(∂T
∂V
)S
= −(∂p
∂S
)T
(1.34)
∂2H
∂S∂p=
∂2H
∂p∂S⇒
(∂V
∂S
)p
=
(∂T
∂p
)S
(1.35)
∂2F
∂T∂V=
∂2F
∂V ∂T⇒
(∂p
∂T
)V
=
(∂S
∂V
)T
(1.36)
∂2G
∂p∂T=
∂2G
∂T∂p⇒ −
(∂S
∂p
)T
=
(∂V
∂T
)p
(1.37)
1.7 Heat Capacities
Heat capacity is defined as the amount of differential heat that must besupplied to a system to change the system’s temperature, per the differentialchange in temperature.
C ≡ δQ
dT(1.38)
Because gases expand significantly upon heating, their heat capacity dependson whether pressure or volume is held fixed. For most solids and liquids thedifferences between CV and Cp is less pronounced. From
δQ = dU + p dV (1.39)
we see that when V is held fixed
CV =
(∂U
∂T
)V
= T
(∂S
∂T
)V
(1.40)
To express Cp, it is convenient to note the differential enthalpy.
dH = δQ+ V dp (1.41)
Therefore, when pressure is held fixed, dH = δQ, and we have that
Cp =
(∂H
∂T
)p
= T
(∂S
∂T
)p
(1.42)
9
Chapter 2
Ideas of Statistical Mechanics
2.1 The Goal
The goal of statistical mechanics is to form a theoretical foundation for theempirical laws of thermodynamics based on more fundamental physics (i.e.quantum mechanics).
2.2 Boltzmann’s Hypothesis
Boltzmann’s insightful hypothesis to define entropy in terms of the numberof microstates.
S = kB ln Ω (2.1)
One way to see why this is a convenient definition is to recognize that thismakes entropy extensive, that is, it is proportional to the number of particlesor subsystems in our system, N . Other extensive state variables are U , V ,and trivially N . If subsystem 1 has Ω1 microstates and entropy S1, andsubsystem 2 has Ω2 microstates and entropy S2, then the combined systemhas ΩT = Ω1 × Ω2 different microstates. The entropy of the combined systemis
S = kB ln ΩT (2.2)
= kB ln(Ω1 × Ω2) (2.3)
= kB ln Ω1 + kB ln Ω2 (2.4)
= S1 + S2 (2.5)
10
All extensive quantities are proportional to each other, with the constantof proportionality equal to the corresponding partial derivative. Therefore,instead of writing U in differential form as in equation 1.29, we may write
U =
(∂U
∂V
)S
V +
(∂U
∂S
)V
S (2.6)
= −p V + T S (2.7)
2.3 Entropy and Probability
Suppose a huge number of systems, N , are in thermal contact. N1 of whichare in mircrostate 1, N2 in microstate 2, etc. The number of ways to makethis arrangement is
Ω =N !
N1! N2! N3! . . .(2.8)
The entropy of the entire ensemble of systems is
SN = k ln Ω = k lnN !− k∑
i
lnNi! (2.9)
Using Stirling’s approximation that lnN ≈ N lnN −N for large N gives
SN = k(N lnN −N −∑
i
Ni lnNi +∑
i
Ni) (2.10)
Noting that N =∑
iNi simplifies this to
SN = k
(∑i
Ni lnN −∑
i
Ni lnNi
)(2.11)
= k∑
i
Ni(lnN − lnNi) (2.12)
= − k∑
i
Ni lnNi
N(2.13)
= − k N∑
i
Ni
NlnNi
N(2.14)
The probability of finding any given system in the i-th microstate is Pi =Ni/N . Therefore,
SN = − k N∑
i
Pi lnPi (2.15)
11
The entropy per system, S = SN/N , is
S = − k∑
i Pi lnPi (2.16)
2.4 The Procedure
• Determine the spectrum of energy levels and their degeneracies usingquantum mechanics or some other means.
• Calculate the partition function.
• Use “the bridge to thermodynamics” (F = −kB T lnZ in CanonicalEnsemble or Φ = −kB T ln Ξ in the Grand Canonical Ensemble) tocalculate any desired thermodynamic variable.
12
Chapter 3
Canonical Ensemble
3.1 Definition
Suppose system A is in contact with a reservoir (heat bath) R and has cometo equilibrium with a temperature T . The total number of microstates inthe combined system is
Ω = ΩA(UA)× ΩR(UR) (3.1)
where UA and UR denote the energy in system A and the reservoir respec-tively. We choose to write quantities in terms of the energy in system A.Ei is the energy of the i-th energy level of system A. If the energy levels ofsystem A have degeneracy gi, the total number of microstates with UA = Ei
isΩ(Ei) = gi × ΩR(UT − Ei) (3.2)
where UT = UA + UR is the total energy of the combined system.From equation 1.29, we have that
dU = TdS − pdV (3.3)
⇒(∂S
∂U
)V
=1
T(3.4)
Combining this with the statistical definition of entropy, S = k ln Ω, gives
1
kT=
(∂ ln ΩR
∂UR
)V
(3.5)
13
When energy is added to the reservoir, the temperature does not change.This is the property that defines a heat bath. Therefore, we can integrateequation 3.5 to get
CUR
kT= ln ΩR (3.6)
where C is some integration constant.
ΩR = eC eUR/kT (3.7)
Ω(Ei) = gi eC e(UT−Ei)/kT (3.8)
The total number of states is
Ωtotal =∑
j
Ω(Ej) (3.9)
= eC eUT /kT∑
j
gj e−Ej/kT (3.10)
where j indexes the energy levels of A. The probability that A has an energyEi is
P (Ei) =Ω(Ei)
Ωtotal
(3.11)
=gi eC e(UT−Ei)/kT
eC eUT /kT∑
j gj e−Ej/kT(3.12)
=gi e
−Ei/kT∑j gj e−Ej/kT
(3.13)
=gi e
−Ei/kT
Z(3.14)
where we defined the partition function, Z, as
Z =∑
j
gj e−Ej/kT (3.15)
We could have chosen to index the microstates instead of the energy levels, inwhich case the probability of the system being in the i-th microstate wouldsimply be
Pi = eEi/kT
Z(3.16)
14
Figure 3.1: Boltzmann Distribution for nondegenerate energy levels
and the partition function would be
Z =∑
j e−Ej/kT (3.17)
This probability distribution is called the Boltzmann Distribution. The par-tition function serves as a normalization factor for the distribution. Fornondegenerate states and finite T , lower energy levels are always more likely.As T approaches infinity, all states are equally likely. A large number ofdistinguishable systems in thermal contact can be treated as a heat bath forany one of the individual systems in the group. This is called the Canon-ical Ensemble, in which case the the distribution of energies should fit theBoltzmann Distribution.
15
3.2 The Bridge to Thermodynamics
Plugging in the Boltzmann Distribution to equation 2.16 gives
S = − k∑
i
Pi lnPi (3.18)
= − k∑
i
e−Ei/kT
Zln
(e−Ei/kT
Z
)(3.19)
= k∑
i
e−Ei/kT
Z
(Ei
kT+ lnZ
)(3.20)
=1
T
∑i
e−Ei/kT
ZEi + k lnZ
*
1∑i
e−Ei/kT
Z(3.21)
=1
T
∑i
e−Ei/kT
ZEi + k lnZ (3.22)
Recognizing that the first term in equation 3.22 contains an express for theaverage energy of any one of the systems
U = 〈E〉 =∑
i
Pi Ei =∑
i
e−Ei/kT
ZEi (3.23)
equation 3.22 can be rewritten as
S =U
T+ k lnZ (3.24)
We can now calculate the Helmholtz Free Energy, F , from equation 3.24
U − TS = − k T lnZ (3.25)
F = − k T lnZ (3.26)
It is now apparent that the partition function is more than just a normaliza-tion factor for the probability distribution of energies. If one can calculatethe partition function of some system, then F can be calculated using equa-tion 3.26. From F one can determine relations between the thermodynamicstate variables.
16
Another way to see how Z is directly related to state variables, specificallyU , is the following. For convenience, define β ≡ 1/kT .
U = 〈E〉 (3.27)
=1
Z
∑i
EiPi (3.28)
=1
Z
∑i
Ei e−βEi (3.29)
Noting that
Z =∑
i
e−βEi (3.30)
∂Z
∂β= −
∑i
Ei e−βEi (3.31)
means that equation 3.29 can be written as
U = − 1
Z
∂Z
∂β(3.32)
= − ∂ lnZ
∂β(3.33)
3.3 Example: A Particle in a Box
Imagine our system is a single particle confined to a one-dimensional box.This system is replicated a huge number of times and the systems are all inthermal contact making a Canonical Ensemble.
The time independent Schodinger wave equation in one dimension is
− h2
2m
∂2
∂2xψ(x) + V (x)ψ(x) = Eψ(x) (3.34)
For a particle in a square potential of width L and with infinite barriers, theenergy eigenvalues are
En =h2π2
2mL2n2, n ∈ 1, 2, 3, 4 . . . (3.35)
17
For convenience, we define the constant ε to be
ε ≡ h2π2
2mL2, En = ε n2 (3.36)
Now that we know the spectrum of energies of the system, we can calcu-late1 the partition funciton.
Z =∞∑
n=1
e−En/kT (3.37)
=∞∑
n=1
e−εn2/kT (3.38)
≈∫ ∞
0
e−εn2/kTdn (3.39)
=
√ε
kT
√π
2(3.40)
=
√L2mkT
2πh2 (3.41)
If we define the de Broglie wavelength2 as
λD ≡
√2πh2
mkT(3.42)
then we can write the partition function as
Z =L
λD
(3.43)
We can now calculate the Helmholtz Free Energy, and other macroscopicthermodynamic quantities.
F = − kT ln(Z) = − kT ln
(√L2mkT
2πh2
)(3.44)
= − k
2T ln
(L2mkT
2πh2
)(3.45)
1In equation 3.40, we use the integral∫∞0
e−x2=√
π2
2According to de Broglie, a particle in an eigenstate of momentum with momentum phas a corresponding wavelength λ = h/p. A thermal distribution of particles therefore hasa distribution of wavelengths. λD represents the average wavelength.
18
The entropy is
S = −(∂F
∂T
)V
= −(∂F
∂T
)L
(3.46)
=k
2ln
(L2mkT
2πh2
)+k
2T
(2πh2
L2mkT
)
(L2mk
2πh2
)(3.47)
=k
2
[ln
(L2mkT
2πh2
)+ 1
](3.48)
Using equation 1.40, the heat capacity is
CV = T
(∂S
∂T
)V
= T
(∂S
∂T
)L
(3.49)
= Tk
2(
2πh2
L2mkT
)
(L2mk
2πh2
)(3.50)
=k
2(3.51)
What if the particle is in a three-dimensional box? Solving this basicquantum mechanical problem gives the spectrum of energies.
En1,n2,n3 = ε (n21 + n2
2 + n23) (3.52)
where ε is the same as we defined in equation 3.36. The partition function is
Z =∞∑
n1=1
∞∑n2=1
∞∑n3=1
e−ε(n21+n2
2+n23)/kT (3.53)
=
(∞∑
n1=1
e−εn21/kT
)(∞∑
n2=1
e−εn22/kT
)(∞∑
n3=1
e−εn23/kT
)(3.54)
=
(L2mkT
2πh2
)3/2
, V = L3 (3.55)
= V
(mkT
2πh2
)3/2
=V
λ3D
(3.56)
The Helmholtz Free Energy is
F = − kT lnZ (3.57)
= − kT lnV +3
2ln
(mkT
2πh2
)(3.58)
19
The pressure is
p = −(∂F
∂V
)T
=kT
V(3.59)
We have derived an equation of state for this system!
p V = k T (3.60)
Imagine our box contains N particles instead of one. If the particles do notinteract, this defines an ideal gas, then the pressure of the system is justN times the pressure due to a single particle, giving equation 3.59 an extrafactor of N . This gives the equation of state for an ideal gas.
p V = N k T (3.61)
Back to the case of only one particle, the entropy of the system is
S = −(∂F
∂T
)V
= k
[lnV +
3
2ln
(mkT
2πh2
)+
3
2
](3.62)
The heat capacity is
CV = T
(∂S
∂T
)V
(3.63)
= T3k
2 (
2πh2
L2mkT
)
(L2mk
2πh2
)(3.64)
=3k
2(3.65)
We see that the heat capacity of a particle in a three-dimensional box isthree times that for a a particle in a one-dimensional box. This makes sensebecause each dimension of translational motion independently adds to thecapacity of the system to store energy in thermal motion. Each translationdegree of freedom gives a k
2term to the heat capacity for a system with
constant volume.Because the heat capacity is not dependent on temperature, we can easily
integrate it from zero to the some temperature to find out how much thermalenergy is stored in the system.∫ T
0
(∂U
∂T ′
)V
dT ′ =
∫ T
0
CV dT ′ (3.66)
U = CV T (3.67)
U =3
2kT (3.68)
20
Chapter 4
Maxwell-Boltzmann Statistics
4.1 Density of States
We know that a free particle in a three-dimensional box has energy eigen-states:
ψn1,n2,n3(x, y, z) = C sin
(n1πx
Lx
)sin
(n2πy
Ly
)sin
(n3πz
Lz
)(4.1)
= C sin (kxx) sin (kyy) sin (kzz) (4.2)
with energies
En1,n2,n3 =h2π2
2m
(n2
1
L2x
+n2
2
L2y
+n2
3
L2z
)(4.3)
=h2
2m
(k2
x + k2y + k2
z
)(4.4)
These states correspond to lattice points in a space, k-space, whose axisesare kx, ky, and kz. A vector in this space is called a wave vector
~k = kx x+ ky y + kz z (4.5)
with a magnitude, or wavenumber 1
k =√k2
x + k2y + k2
z (4.6)
1We will now use kB to denote Boltzmann’s constant in order not to confuse it withthe magnitude of a wave vector k.
21
Figure 4.1: k-space
In polar coordinates,
θ = arccos
(kz
k
), φ = arctan
(ky
kx
)(4.7)
volume element in k-space = k2 sin θ dk dθ dφ (4.8)
density of points in k-space =LxLyLz
π3=V
π3(4.9)
Therefore, the number of states in a differential volume is
dN =V
π3k2 sin θ dk dθ dφ (4.10)
Integrating dθ and dφ gives the number of states in a differential shell ofthickness dk.
dN =V
π3k2 dk
∫ π2
0
dφ
∫ π2
0
sin θ dθ (4.11)
=V
π3k2 dk · π
2· 1 =
V
2π2k2 dk (4.12)
22
Figure 4.2: Density of points in k-space
We define that density of states, D(k), such that
dN =V
2π2k2 dk = D(k) dk (4.13)
In this discussion, we have ignored the possibility of there being multiplespin states at a single point in k-space. For future reference, we define D0(k)to be the density of states ignoring any spin degeneracy.
D0(k) ≡ V2π2 k
2 (4.14)
In general, if there are gs spin states with the same energy at the same pointin k-space, the density of states is
D(k) = gs D0(k) (4.15)
We can now write down the partition function, using k to integrate overall states.
Z =
∫ ∞
0
D(k) e−Ek/kBT dk (4.16)
4.2 The Maxwell-Boltzmann Distribution
If we allow multiple particles to occupy the same state, the probability thei-th state is occupied if we “toss” N particles among the states at random
23
is2
Pocp = 1− P (Ni = 0) (4.17)
= 1− (1− Pi)N (4.18)
= 1− 1 +NPi −N(N − 1)
2P 2
i + . . . (4.19)
where Ni denotes the number of particles occupying the i-th state and Pi isthe probability that a particle gets tossed into the i-th states. If the densityof states is sufficiently large, and the temperature high, then Pi 1, andwe can ignore higher terms in Pi. In this regime, it does not even matter ifwe allow multiple particles to occupy the same state, be them fermions orbosons, because the probability of multiple occupancies is so low.
Pocp ' N Pi (4.20)
Using this approximation, the number of states occupied in a differentialk-shell is
dNocp = Pocp dN (4.21)
=N e−Ek/kBT
ZD(k) dk (4.22)
= fMB(k) dk (4.23)
where we have defined the Maxwell-Boltzmann Distribution, fMB(k), to be
fMB(k) = N D(k) e−Ek/kBT
Z(4.24)
4.3 Example: Particles in a Box
Back to our example of particles in a box, we can plug in the partitionfunction from equation 3.56 and the density of states from equation 4.13 intoequation 4.22.
fMB(k) dk = N e−Ek/kBT
(λ3
D
V
)(V
2π2k2
)dk (4.25)
=Nλ3
D
2π2k2 e−Ek/kBT dk (4.26)
2We have used that the probability of r successful tosses out of N tosses, witha probability p for the success of each toss, is given by the binomial distribution.P (x = r) =
(Nr
)pr (1− p)(N−r)
24
Let us say we want to calculate the average velocity squared of the particles.We first have to calculate⟨
k2⟩
=
∫∞0k2 fMB(k) dk∫∞
0fMB(k) dk
(4.27)
=
∫∞0k4 e−Ek/kBT dk∫∞
0e−Ek/kBT dk
(4.28)
=
∫∞0k4 e−h2k2/2mkBT dk∫∞
0e−h2k2/2mkBT dk
(4.29)
Substitute3 x2 ≡ h2k2
2mkBT
⟨k2⟩
=2mkBT
h2
∫∞0x4 e−x2
dx∫∞0e−x2 dx
(4.30)
=2mkBT
h2
(3
√π
8
)
(
4√π
)(4.31)
=3mkBT
h2 (4.32)
In the non-relativistic limit,
⟨v2⟩
=〈p2〉m2
=h2
m2
⟨k2⟩
(4.33)
=3kBT
m(4.34)
For each particle, the average energy is
U =1
2m⟨v2⟩
=3
2kBT (4.35)
which is the same result we got for the ideal gas using a different method inequation 3.68.
3We used the integrals∫∞0
x4 e−x2dx = 3
√π
8 and∫∞0
e−x2dx =
√π
4 .
25
Chapter 5
Black Body Radiation
5.1 Molecular Beams
5.2 Cavity Radiation
5.3 The Planck Distribution
26
Chapter 6
Thermodynamics of MultipleParticles
6.1 Definition
Now we will consider systems consisting of distinguishable types of particles,where all the particles in a given type are identical. In addition to theexchange of work and heat between subsystems, now the number of particlesof one type can be transfered to another type. One example of this is acylinder of gas with a semipermeable membrane dividing the cylinder. Thereare two types of gas molecules that are distinguishable by which side of themembrane they are on. Another example is a single container of gas holdingA type molecules, B type molecules, and a third AB type that is the productof some kind of combination of A and B type molecules.
6.2 Chemical Potential
Introducing the ability for particles to change type within a system gives anew way to exchange energy. We define the chemical potential, µ, to be theinternal energy gained by a system when the average number of particles, N ,of some type increases by one. This means we must add a term to equation2.7.
U = −p V + T S + µ N (6.1)
27
Therefore,
µ =
(∂U
∂N
)V,S
(6.2)
and we can add this term to equation 1.29, giving
dU = T dS − p dV + µ dN (6.3)
This also leads to a noteworthy equation for the Gibbs Free Energy:
G ≡ U − T S + p V = µ N (6.4)
µ =
(∂G
∂N
)T,P
=G
N(6.5)
Flipping around equation 6.3 gives a relation for µ that is often convenient.
dS =1
TdU +
p
TdV − µ
TdN (6.6)
µ = −T(∂S
∂N
)U,V
(6.7)
6.3 The Gibbs-Duhem Relation
Differentiating equation 6.1 directly gives
dU = T dS + S dT − p dV − V dP + µ dN +N dµ (6.8)
but then using equation 6.3, we cancel some terms, leaving
0 = S dT − V dp+N dµ (6.9)
This is called the Gibbs-Duhem Relation. We can generalize this for a systemwith various work variables Xi and particle types indexed by j.
0 = S dT −∑
i
Xi dpi +∑
j
Nj dµj (6.10)
28
Chapter 7
Identical Particles
7.1 Exchange Symmetry of Quantum States
Suppose we have a system of two identical particles. Because they are iden-tical, the square of the wavefunction that describes the state should be un-changed if two particles are exchanged. X denotes the exchange operator.
|ψ(x1, x2)|2 = |Xψ(x1, x2)|2 (7.1)
= |ψ(x2, x1)|2 (7.2)
This implies that the exchanged wavefunctions differ only by a complex factorwhose modulus squared is one.
ψ(x1, x2) = eiθ ψ(x2, x1) (7.3)
If we exchange the particles twice, we have done nothing. Therefore thewavefunctions should be identical.
ψ(x1, x2) = X 2ψ(x1, x2) (7.4)
⇒ ei2θ = 1 (7.5)
which only has solutions θ = . . .− 2π,−π, 0, π, 2π . . .⇒ eiθ = ±1.Particles with an exchange factor of +1, i.e. symmetric under exchange,
are called bosons. Those with an exchange factor of −1, i.e. antisymmetricunder exchange, are called fermions.
29
Larger numbers of identical bosons or fermions must still obey the appro-priate exchange symmetry. The ways to construct a wavefunction for severalidentical particles with the correct symmetry are the following:
ψBose(x1, x2, x3 . . .) = ψi(x1) ψj(x2) ψk(x3) . . .+ permutations of i, j, k . . .(7.6)
ψFermi(x1, x2, x3 . . .) =
∣∣∣∣∣∣∣∣∣ψi(x1) ψi(x2) ψi(x3)ψj(x1) ψj(x2) ψj(x3) . . .ψk(x1) ψk(x2) ψk(x3)
.... . .
∣∣∣∣∣∣∣∣∣ (7.7)
The determinate in equation 7.7 is called a Slater Determinate and gives thecorrect signs for the total wavefunction to be asymmetric under exchange.
Consider a system of two fermions, for example two electrons bound toan atom. The Slater Determinate gives
ψFermi(x1, x2) = ψi(x1) ψj(x2)− ψi(x2) ψj(x1) (7.8)
Immediately one can see that if the two particles are in the same state (i = j),ψFermi ≡ 0. This explains the Pauli Exclusion Principle, that no two fermionscan exist in the same quantum state.
7.2 Spin
There is a very important result from relativistic quantum mechanics calledthe Spin-Statistics Theorem, which is beyond the scope of this work. It holdsthat fermions have half-integer spin and bosons have integer spin.
fermions: s = 1/2, 3/2, 5/2 . . . (7.9)
bosons: s = 0, 1, 2, 3 . . . (7.10)
The total wavefunction must obey the exchange symmetry.
ψtotal = φ(space) χ(spin) (7.11)
This allows fermions to exist in a symmetric spacial state, as long as theircombined spin state is antisymmetric. For two spin 1/2 fermions, the com-bined spin states with spin, s, and z-projection of the spin, m, are χ = |s m〉.
30
s = 1 symmetric triplet:
|1 1〉 = ↑↑ (7.12)
|1 0〉 = ↑↓ + ↓↑ (7.13)
|1 − 1〉 = ↓↓ (7.14)
s = 0 antisymmetric singlet:
|0 0〉 = ↑↓ − ↓↑ (7.15)
The spin singlet allows two electrons (which are fermions) to exist in thesame spacial state (orbital) without violating Pauli’s Exclusion Principle.
ψee(x1, x2) = φi(x1) φi(x2) |0 0〉 (7.16)
7.3 Over Counting
31
Chapter 8
Grand Canonical Ensemble
8.1 Definition
Suppose system A is in contact with a with a huge reservoir with whichthe system can exchange both heat and particles. We use ΩR to denotethe number of microstates in the some macrostate specified by a number ofparticles and a temperature. Plugging S = kB ln ΩR into equation 6.7 gives
−µkB T
=
(∂ ln ΩR
∂NR
)UR,VR
(8.1)
Similar to the Canonical Ensemble in equation 3.5, we also have that
1
kBT=
(∂ ln ΩR
∂UR
)VR,NR
(8.2)
The reservoir is so large that its temperature and chemical potential do notsignificantly change when heat or particles are added to it. Therefore, wemay trivially integrate both equations. They have the simultaneous solution(where C is some combined constant of integration):
ln ΩR = C +UR − µNR
kB T(8.3)
ΩR = eC e(UR−µNR)/kBT (8.4)
The total number of microstates of the combined system is
Ω = ΩA(UA, NA)× ΩR(UR, NR) (8.5)
32
The number of microstates with system A in its i-th microstate is
Ωi = 1× ΩR(UT − UA, NT −NA) = ΩR(UT − Ei, NT −Ni) (8.6)
where UT = UA + UR and NT = NA +NR.
Ωi = eC e(UT−Ei−µNT +µNi)/kBT (8.7)
Since UT and NT are conserved, they can be absorbed into the leading con-stant.
Ωi = C ′ e−(Ei−µNi)/kBT (8.8)
The total number of states is
Ωtotal =∑
j
Ωj (8.9)
The probability that A is in the i-th state is
Pi =Ωi
Ωtotal
(8.10)
=C ′ e−(Ei−µNi)/kBT
C ′∑
j e−(Ej−µNj)/kBT
(8.11)
=e−(Ei−µNi)/kBT
Ξ(8.12)
where we defined the Grand Partition Function as
Ξ =∑
j e−(Ej−µNj)/kBT (8.13)
8.2 The Bridge to Thermodynamics
Plugging in the distribution function in equation 8.12 into equation 2.16 gives
S = − kB
∑i
Pi lnPi (8.14)
= − kB
∑i
Pi ln
(e−(Ei−µNi)/kBT
Ξ
)(8.15)
= − kB
∑i
Pi
[−(Ei − µNi)
kBT− ln Ξ
](8.16)
33
Recognizing that
U =∑
i
Pi Ei and N =∑
i
Pi Ni (8.17)
we have that
S =U
T− µN
T+ kB ln Ξ (8.18)
Rearranging this, we define the Grand Potential to be
Φ ≡ U − T S − µ N (8.19)
Φ = −kB T ln Ξ (8.20)
Differentiating Φ and plugging in the differential of U from equation 6.3gives
dΦ = dU − T dS − S dT − µ dN −N dµ (8.21)
= T dS − p dV +µ dN −T dS − S dT −µ dN −N dµ (8.22)
= − p dV − S dT −N dµ (8.23)
By taking various partial derivatives, we can calculate p, S, and N from Φ.We can then calculate any thermodynamic state variable we want using thevarious Maxwell Relations.
Most notably, we can calculate the average number of particles in thesystem.
N = −(∂Φ
∂µ
)T,V
(8.24)
8.3 Fermions
For our system we choose the i-th single quantum state of a fermion, a statewith a specific wave vector and spin state. Pauli’s exclusion principle forbidsmore than one particle to exist in this state. Therefore, the spectrum ofenergies for this system is only 0, Ei, 0 for the state being unoccupied,and Ei, being the energy of that state, for when it is occupied. Then thegrand partition function for this system is
Ξ = 1 + e−(Ei−µ)/kBT (8.25)
34
The grand potential is
Φ = − kB T ln(1 + e−(Ei−µ)/kBT
)(8.26)
The average number of particles in this state is
N = −(∂Φ
∂µ
)T,V
(8.27)
=1
e(Ei−µ)/kBT + 1(8.28)
Now considering all of the quantum states, if we use the magnitude ofthe wave vector, k, to index the energy levels, and let the density of states,D(k), handle the spin and other degeneracies, we can write the distributionof occupied states for fermions, called the Fermi-Dirac Distribution.
fFD(k) = D(k)
e(Ek−µ)/kBT +1(8.29)
8.4 Bosons
We choose our system to be the i-th single quantum state of a boson, a statewith a specific wave vector and spin state. Any nonnegative integer numberof bosons can occupy a single state. Therefore, the spectrum of energies forour system is 0, Ei, 2Ei, 3Ei . . .. The grand partition function is1
Ξ = 1 + e−(Ei−µ)/kBT + e−2(Ei−µ)/kBT + e−3(Ei−µ)/kBT + . . . (8.30)
=∞∑
j=0
(e−(Ei−µ)/kBT
)j(8.31)
=1
1− e−(Ei−µ)/kBT(8.32)
where (Ei − µ) must be positive for the series to converge. The grand po-tential is
Φ = kB T ln(1− e−(Ei−µ)/kBT
)(8.33)
1Recognize that the sum in equation 8.31 is a geometric series.
35
The average number of particles in this state is
N = −(∂Φ
∂µ
)T,V
(8.34)
=1
e−(Ei−µ)/kBT − 1(8.35)
Just like we did for fermions, if we use k to index the energy levels, and letD(k) handle the spin and other degeneracies, we can write the distributionof occupied states for bosons, called the Bose-Einstein Distribution.
fBE(k) = D(k)
e(Ek−µ)/kBT−1(8.36)
8.5 Non-relativistic Cold Electron Gas
Electrons have a spin degeneracy of 2. Therefore, the density of states of anelectron gas is
D(k) = 2 D0(k) =V
π2k2 (8.37)
As temperature approaches zero, a the particles in a fermion gas stack in theallowed states, starting at the ground states. This means that the particlespile up in the states in the corner of k-space closest to the origin, making aneighth of a sphere. The surface of that sphere is called the Fermi surface.The states on that surface all have the same radius or wavenumber, kF ,called the Fermi wavenumber. In order to calculate the Fermi wavenumber,we equate the number of electrons,Ne, with number of states, stacking themin the corner.
Ne =
∫ kF
0
D(k) dk =V
π2
∫ kF
0
k2 dk =V kF
3π2(8.38)
kF =
(3π2 Ne
V
)1/3
= (3π2n)1/3 (8.39)
where n is the number density of electrons. In the non-relativistic limit, theenergy of a state with a wavenumber k is
Ek =p2
2m=h2k2
2me
(8.40)
36
Therefore, the energy of the states on the Fermi surface, called the Fermienergy, is
EF =h2
2me
(3π2n)2/3 (8.41)
The total internal energy of the electron gas at T = 0 is
U =
∫ kF
0
Ek D(k) dk (8.42)
=
∫ kF
0
(h2k2
2me
)(V k2
π2
)dk (8.43)
=h2V
2π2me
∫ kF
0
k4 dk (8.44)
=h2V
10π2me
k5F (8.45)
From equations 8.39 and 8.41
k3F = 3π2n = 3π2 Ne
V(8.46)
k2F =
2meEF
h2 (8.47)
Therefore,
U =h2
V
10π2me
(3π
2 Ne
V
)(2meEF
h2
)(8.48)
=3
5EFNe (8.49)
Even at T=0, an electron gas has positive internal energy and therefore apositive pressure. To calculate this degeneracy pressure, we write the explicitdependence of U on V . Plugging equation 8.39 into 8.45 gives
U =h2V
10π2me
(3π2 Ne
V
)5/3
(8.50)
=h2
10π2me
(3π2Ne
)5/3V −2/3 (8.51)
37
p = −(∂U
∂V
)T,Ne
(8.52)
=2
3
h2
10π2me
(3π2 Ne
V
)5/3
(8.53)
=2
3
h2k2F
10π2me
(3π
2n)
(8.54)
=2
5EF n (8.55)
Notice that the degeneracy pressure increases with the number density, andthat the Fermi energy also increases with the number density.
8.6 Example: White Dwarf Stars
When a star can no longer perform fusion of its matter, creating heat andpressure to suppress its collapse, the star collapses until the electron degen-eracy pressure increases enough to halt the collapse. Such a star is called awhite dwarf.
In order to calculate the stable radius, R, of a white dwarf, we write Uas a function of R.
U =h2
10π2me
(3π2Ne
)5/3V −2/3, V =
4
3πR3 (8.56)
=h2
10π2me
(3π2Ne
)5/3(
4
3πR3
)−2/3
(8.57)
=9
20
(3
2
)1/3
π2/3 h2
meR2(8.58)
Now we calculate the gravitational potential energy for the star.
dUG =−Gmr
dm (8.59)
m =4
3πr3ρ, dm = 4πr2ρ dr (8.60)
dUG =−Gr
(4
3πr3ρ
)(4πr2ρ
)dr (8.61)
=−16π2
3Gρ2r4 dr (8.62)
38
UG =−16π2
3Gρ2
∫ R
0
r4 dr (8.63)
=−16π2
15Gρ2R5 (8.64)
The density in terms of the number of nucleons, NN , and the mass of anucleon, mN , is
ρ =NN mN
4πR3/3(8.65)
Therefore,
UG =−16π
2
15G
(NN mN
4πR3/3
)2
R5 (8.66)
=−3GN2
Nm2N
5R(8.67)
The total energy, UT , is
UT = U + UG (8.68)
For convenience, we define the constants
A ≡ 9
20
(3
2
)1/3
π2/3 h2
me
(8.69)
B ≡ 3
5GN2
Nm2N (8.70)
UT =A
R2− B
R(8.71)
UT =A
R2− B
R(8.72)
The stable minimum in energy is found by finding the critical point.
dUT
dR=−2A
R3+B
R2= 0 (8.73)
⇒ R =2A
B=
(9π
4
)2/3h2N
5/3e
GN2Nm
2Nme
(8.74)
39
Figure 8.1: Potential well for white dwarf with non-relativistic electrons
Most elements have approximately as many neutrons as protons, and willhave the same number of electrons as protons. Therefore,
N ≡ Ne ≈1
2NN (8.75)
The mass of the star, M , is almost entirely due to nucleons
M ≈ NN mn, m ≡ me ≈1
1800mN (8.76)
⇒ N =M
2mN
=M
2(1800m)=
M
3600m(8.77)
R =
(9π
4
)2/3h2N5/3
GM2m(8.78)
=
(9π
4
)2/3 h2(
M3600 m
)5/3
GM2m(8.79)
=
(9π
4
)2/3
(3600)−5/3 h2
GM1/3 m8/3(8.80)
For a star with the mass of the sun, M = m ≈ 2.0 × 1030 kg, and R ≈7.4× 106 m.
40
The Fermi energy of such a white dwarf can be calculated by combiningthe following equations
EF =h2
2m
(3π2n
)2/3(8.81)
n =N
4πR3/3(8.82)
N =M
2(1800m)(8.83)
which gives that EF ≈ 0.18 MeV. Note that the rest mass energy of an elec-tron is 0.51 MeV, indicating that the most energetic electrons in our examplewhite dwarf star (those on the Fermi surface) are getting near relativistic.
8.7 Relativistic Cold Electron Gas
As we confine an electron gas into a smaller and smaller volume, the allowedquantum states get higher and higher in energy. In a small enough volume,even the lowest energy electrons will be relativistic. In the relativistic limit,the energy of an electron with a wavenumber k is
Ek = pc = hkc (8.84)
So the Fermi energy is
EF = hkF c = hc(3π2n
)1/3(8.85)
41
The total internal energy of the electron gas at T = 0 is
U =
∫ kF
0
Ek D(k) dk (8.86)
=
∫ kF
0
(hkc)
(V k2
π2
)dk (8.87)
=hcV
π2
∫ kF
0
k3 dk (8.88)
=hcV
4π2k4
F (8.89)
=EFV
4π2k3
F (8.90)
=EFV
4π2
(3π
2n)
(8.91)
=3
4EFNe (8.92)
8.8 Example: Collapse to a Neutron Star
Following the same procedure we used in Section 8.6, we write the the internalenergy as a function of radius of the star.
U =hcV
4π2k4
F (8.93)
=hcV
4π2
(3π2Ne
V
)4/3
(8.94)
=hc
4π2
(3π2Ne
)4/3V −1/3 (8.95)
=hc
4π2
(3π2Ne
)4/3(
4
3πR3
)−1/3
(8.96)
=3
4
(3
2
)2/3
π1/3 h c N4/3e R−1 (8.97)
For convenience, we define the constant
C ≡ 3
4
(3
2
)2/3
π1/3 h c N4/3e (8.98)
42
The gravitational potential energy is the same as we calculated in Section8.6. The total energy is
UT = U + UG (8.99)
=C
R− B
R(8.100)
=C −B
R(8.101)
There is no stable radius for a white dwarf star with relativistic electrons. Thecritical point is when C = B. Using the same assumptions from equations8.75, 8.76, and 8.77, we have that
3
4
(3
2
)2/3
π1/3 h c N4/3e =
3
5GN2
Nm2N (8.102)
5
4
(3
2
)2/3
π1/3
(hc
G
)(M
2mN
)4/3
= M2 (8.103)
M2/3 =5
4
(3
2
)2/3(1
2
)4/3
π1/3
(hc
G
)m−4/3N (8.104)
M = 3
(5
16
)3/2√π
(hc
G
)3/2
m−2N (8.105)
The Planck Mass is defined as
mP =
√hc
G(8.106)
We call the critical mass, M , the Chandrasekhar limit, MCh.
MCh = 3
(5
16
)3/2√π m3
P m−2N (8.107)
≈ m3P m−2
N (8.108)
≈ 3× 1030 kg (8.109)
≈ 1.5 m (8.110)
If mass of the star is less than MCh, then C −B is positive and the star willexpand, lowering its energy levels until it is in the non-relativistic regime. If
43
the mass of the star is greater than MCh, then C−B is negative and the starwill continue to collapse. The star is forced to perform an enormous numberof inverse beta-decays, converting its electrons into electron neutrinos thatradiate away and converting its protons into neutrons. Then the star canreach a new equilibrium supported by neutron degeneracy pressure. Such astar is called a neutron star. It is virtually a huge single nucleus consistingof only neutrons. If the neutron degeneracy pressure is not enough to stopthe collapse, the star becomes a black hole.
44
Chapter 9
Questions
• How are the thermodynamic and statistical mechanical definitions ofentropy equivalent?
• U =∑
i Pi Ei
dU =∑
iEi dPi +∑
i Pi dEi
= δQ+ δW ?
• When do we use ZN = ZN1 /N !. Why don’t we in fMB?
45