38
Thermal and Statistical Physics Tom Charnock
Thermal and Statistical Physics
Tom Charnock
Contents
1 Thermodynamics 21.1 Zeroth Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2.1 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3.1 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.2 Isothermal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.3 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3.4 Adiabatic Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.4 Cyclic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.1 Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4.2 Reversible Systems and Processes . . . . . . . . . . . . . . . . . . . . . . . . 51.4.3 Heat Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.4 Carnot’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.5 Kelvin-Clausius Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5.1 Clausius Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.6 Thermodynamic Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6.1 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6.2 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6.3 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6.4 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6.5 Maxwell Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6.6 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.7 Joule Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7.1 Flow Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7.2 Joule-Kelvin Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.8 Phase Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.8.1 Availability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.8.2 Gibbs Free Energy for Phase Equilibrium . . . . . . . . . . . . . . . . . . . 10
1.9 Third Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.9.1 Nernst’s Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.10 Cooling by Demagnetisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.10.1 Curie’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2 Statistics 122.1 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.1.1 A Priori Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1.2 Prosteriori Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 Macrostates vs. Microstates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.1 Macrostates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2.2 Microstates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 Principle of Equal A Priori Probability . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.1 One Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2
2.3.2 Two Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.3 Many Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.4 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.4 Statistics of Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4.1 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.5 Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5.1 Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5.2 Micro-Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.6 Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6.1 1D Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6.2 3D Random Walk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.7 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.8 Equipartition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.8.1 Failure of Equipartition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.8.2 Factorisation of Z . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3 Quantum Thermodynamics 183.1 Quantum Mechanical Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.1.1 Schrödinger Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.1.2 Dirac Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.2 Particle in a 1D Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.3 k Space and Density of States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.4 Distribution of Speeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.4.1 Particles in a Classical Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.4.2 Speed of a Classical Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.5 Mean Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5.1 Particles from an Oven . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5.2 Mean Free Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.6 Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.6.1 Doppler Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.6.2 Collision Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.6.3 Lifetime Broadening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.7 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.7.1 Theory of Rayleigh and Jeans (1907) . . . . . . . . . . . . . . . . . . . . . . 233.7.2 Wein’s Displacement Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.7.3 Planck’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.8 Photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.8.1 Phonons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4 Extensive and Intensive Variables 274.1 Extensive Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.1.1 Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.1.2 Chemical Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.1.3 Reactions in the Sun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Grand Canonical Ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
5 Bosons and Fermions 315.1 Fermi Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5.1.1 Fermi-Dirac Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.2 Bose Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.2.1 Composite Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.2.2 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2.3 Bose Condensed Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2.4 Bose-Einstein Condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.2.5 Laser Cooling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
1
Thermodynamics
1.1 Zeroth Law of ThermodynamicsThe zeroth law states that if A and C are in equilibrium with B, then A must also be in thermalequilibrium with C.
1.2 First Law of ThermodynamicsThe first law of thermodynamics states that energy is conserved when heat is taken into account.The total energy is a function of the work done and the thermal energy so that:
δU = −δW +−δQ
Where −δW = −pδV is the inexact equation for the work done. For reversible processes δU =−δQrev +−δW which can be defined in terms of the entropy as:
δU = TδS − pδV
This form is exact and is made of conjugate pairs{p, V
}and
{S, T
}, which both have the dimen-
sions of energy.
1.2.1 Heat CapacityThe heat required is the work done to a gas on the energy in a system. This means the thermalenergy can be calculated from −δQ = δU − −δW . As the change in energy can be found fromδU =
(∂E∂T
)VδT +
(∂E∂V
)TδV the inexact equation for the thermal energy of the system is:
−δQ =
(∂E
∂T
)V
δT +
[(∂E
∂V
)T
+ p
]δV
The heat capacity is the change of temperature of the thermal energy C =−δQδT at a constant
volume δV = 0 and so the heat capacity can be defined as CV =(−δQ
δT
)V
=(∂E∂T
)V
. At constantpressure the heat capacity is given by the whole heat capacity equation.
1.3 Ideal GasIn a monatomic gas there is 1
2kBT per degree of freedom which leads to the energy of the gas beingE = 3RT
2 per mole. A low pressure gas can be modelled well as an ideal gas.
2
1.3.1 Ideal Gas Law
pV = nRT
The ideal gas law states that when the temperature is constant then the pressure and volume varyas an inverse of the other. This leads to the idea of isotherms.
1.3.2 Isothermal Expansion
VpδV
T1
T2
p
Figure 1.1: Two Isotherms at Different Temperatures
The area under the curve is the work done given by W = Fδr = pAδr = pδV . This means theisothermal expansion comes purely from the work done and there is no temperature exchange.
1.3.3 Heat Capacity
If the ideal gas law is taken at one mole pV = RT and so ∂V∂T p
= Rp . Cp − CV = R which leads
to the definition of the ratio of heat capacities. γ =Cp
CV= 1 + R
CV, which for a monatomic gas
CV = 3R2 and Cp = 5R
2 and γ = 53 . For a diatomic gas, where there an extra 2 degrees of freedom
CV = 5R2 and Cp = 7R
2 and γ = 75 . The ratio of heat capacities is useful to explain adiabatic
expansion.
1.3.4 Adiabatic Expansion
There is no change in thermal energy in adiabatic expansion so δU = −δW = −pδV . As δU = CV δTthen −pδV = CV δT can be equated by substituting p into the ideal gas equation and integratingto obtain lnT = − R
CVlnV + constant, which can be rearranged to obtain:
pV γ = constant
3
V
Adiabats
Isotherms
p
Figure 1.2: Isothermic and Adiabatic Expansion
When a gas moves around the shaded area then the process which occurs is that of the CarnotCycle.
1.4 Cyclic Processes1.4.1 Carnot Cycle
V
A
B
CD
p
Figure 1.3: Carnot Cycle
The work done in a Carnot Cycle is given by the area under the shaded top curve (A → B → C)and the work done to the gas is the area under the shaded bottom curve (A → D → C). Thismeans the the total work done by the gas is WABC − WADC . The isothermal expansion fromA → B is given by QH = RTH ln
(VB
VA
)which is followed by adiabatic expansion from B → C
where the temperature change is TH
TC=(
VC
VB
)γ−1
. Then there is the return isothermal expansion
from C → D which is QC = RTC ln(
VC
VD
)and finally the adiabatic expansion from D → A which
is TC
TH=(
VA
VD
)γ−1
. By combining the adiabatic expression and the isothermal expressions:
(VCVB
)=
(VDVA
)(QH
QC
)=
(THTC
)
4
As the work is W = QH −QC then the efficiency of the cycle can be defined as:
η =W
QH= 1− QC
QH= 1− TC
TH
1.4.2 Reversible Systems and ProcessesA reversible process is so slow that the system goes through an infinite sequence of equilibriumstates (which are all infinitely close). The system must be absent of any friction, turbulence andacceleration. In a reversible process the mass is split into an infinite number of infinitely smallmass elements which are added individually.
1.4.3 Heat EnginesA heat engine is a cyclic process where heat is converted to work.
W
QCQH
TCTH
Figure 1.4: Heat Engine
Carnot Engine
The Carnot engine is the most efficient engine possible where η = 1− TC
TH. All other engines cannot
be as efficient as the Carnot engine unless they themselves are Carnot engines.
Carnot Refrigerator (Heat Pump)
The efficiency of a Carnot refrigerator is ηref = QC
W = TC
TH−TCand also cannot be surpassed by any
other refrigerator. The Carnot refrigerator takes in the least amount of work to move heat from alow temperature to a high temperature.
1.4.4 Carnot’s TheoremThe sole effect of an engine is to pump heat from cold to hot. This violates the Clausius statementand so it cannot be true, and so another engine is therefore not allowed to be more efficient thanthe Carnot engine.
TC
TH
W
Q′H
Q′C
QH
QC
CarnotOther
Figure 1.5: Carnot Pump
5
WQ′
H> W
QHand so QH −Q′
H > 0. From this it can be seen that all reversible engines must have thesame efficiency as that of a Carnot engine. For a Carnot engine and a different refrigerator, thenas the Carnot engine is more efficient, heat would be pumped from cold to hot, which violates thesecond law of thermodynamics.
Kelvin Statement
No process is possible whose sole result is the complete conversion of heat to work.
Clausius Statement
No process is possible whose sole result is the complete transfer of heat from cold to hot.
1.4.5 Kelvin-Clausius StatementIf all the heat were transformed into work to drive a Carnot refridgerator then heat would bepumped from cold to hot which violates the Kelvin statement and leads to the violation of theClausius statement. If heat was pumped from cold to hot with work outputted then the Clausiusstatement would be violated and this leads to the violation of the Kelvin Statement. Given thereare two contradictions then it can be said that the Clausius and Kelvin statements are the same.
1.5 EntropyFor a reversible process ∆Q
T = QH
TH− QC
TC , which for the Carnot Cycle ∆Q = 0. Any reversiblecycle, as an approximation, can be idealised as a series of Carnot cycles. This gives the entropy.
S =
∮ −δQrev
T
For an ideal monatomic gas E = 32RT and so −δQ = 3
2RδT + RTδV . As this equation is inexact then
the solution is not very useful. By converting to entropy the equation becomes exact.
δS =−δQT
=3
2RδT
T+R
δV
V= R
(3
2δ(lnT ) + δ(lnV )
)
1.5.1 Clausius InequalityFor irreversible engines QH
QC≤ TH
TCand so δS ≥ δQ
T . For an isolated system δQ = 0 and so δS ≥ 0,which is the law of increase in entropy.
1.6 Thermodynamic Potentials1.6.1 EnergyThe first law of thermodynamics δU = TδS − pδV gives rise to the thermodynamic definitions oftemperature and pressure.
δU =
(∂E
∂S
)V
δS −(∂E
∂V
)S
δV
1.6.2 EnthalpyFor isolated, fixed boundary conditions δU = 0. This leads to the enthalpy where H = E + pVand so δH = TδS + V δp.
δH =
(∂H
∂S
)p
δS −(∂H
∂p
)S
δp
6
1.6.3 Helmholtz Free EnergyThe Helmholtz Free Energy is F = E − TS which leads to δF = −SδT − pδV . The HelmholtzFree Energy is particularly useful for linking between thermodynamics and statistical physics.
δF = −(∂F
∂T
)V
δT −(∂F
∂V
)T
δV
1.6.4 Gibbs Free EnergyThe Gibbs Free Energy is the most complete and so describes real measurable systems most usefully.It is G = E − TS + pV , which can be written δG = −SδT + V δp.
δG = −(∂G
∂T
)p
δT +
(∂G
∂p
)T
1.6.5 Maxwell RelationsUsing the relations from the thermodynamic potentials, Maxwells relations can be used to combinethe quantities:
(∂T∂V
)S=−
(∂p∂S
)V=⇒
(∂2E∂S∂V
)(
∂T∂p
)S=
(∂V∂S
)p=⇒
(∂2H∂S∂p
)(∂S∂V
)T=
(∂V∂T
)V=⇒
(∂2F∂T∂V
)−(
∂S∂p
)T=
(∂V∂T
)p=⇒
(∂2G∂T∂p
)
1.6.6 Heat CapacityThe heat capacity at constant volume is:
CV = T
(∂S
∂T
)V
And the heat capacity at constant pressure is:
Cp = T
(∂S
∂T
)p
The heat capacity can be obtained from the Helmholtz Free Energy by dividing by δT and multi-plying by T . This means that:
Cp − CV = T
(∂p
∂T
)V
(∂V
∂T
)p
Isobaric Expansivity and Isothermal Compressibility
The isobaric expansivity is the expansion at constant pressure and is given by βp = 1V
(∂V∂T
)p.
The isothermal compressibility is the compression at constant temperature and is given by κT =
− 1V
(∂V∂p
)T
. Using the reciprocity theorem it can be seen(
∂p∂T
)V=
βp
κTand so Cp − CV =
V Tβ2p
κT.
This is valid always, for both ideal and non-ideal gases.
7
Adiabatic Compressibility
The compressibility at constant entropy is κS = − 1V
(∂V∂p
)S
. This allows γ to be found as:
κTκS
=
(∂V∂p
)T(
∂V∂p
)S
=
(∂p∂S
)S
(∂S∂V
)p(
∂p∂T
)V
(∂T∂V
)p
=
(∂S∂T
)p(
∂S∂T
)V
=Cp
CV= γ
1.7 Joule Expansion
Figure 1.6: Box with Particles Separate by a Semipermeable Plug
Joule expansion is an irreversible process in which a volume of gas is contained in one half ofcontainer at constant temperature. When the other half of the container is exposed then the gasexpands and the increase in entropy can be calculated. The entropy of the surrounding remainsconstant but the entropy in the box increases by δS = R ln
(V2
V1
)and so the total entropy of the
universe increases by this amount. As this is none zero, it shows that the process is irreversible. Ifthe Joule Coefficient is given by CJ =
(∂T∂V
)E
then using the reciprocity theorem it can be extendedto: (
∂T
∂V
)E
= − 1
CV
(∂E
∂V
)T
At a constant temperature this can be substituted into the first law of thermodynamics to get:
CJ = − 1
CV
[T
(∂p
∂T
)V
− p
]For ideal gases pV = nRT and so CJ = 0, but for gases interacting via Van der Waals p = nRT
V−b −aV2
and so CJ = − aCV V 2 . This shows that when the gas expands in the box it will cool down due to
the work done against the attractive Van der Waals forces.
1.7.1 Flow Process
Qpr Wpr
p2p1 E1V1 E2V2
Figure 1.7: Flow Process
The law of energy conservation states E1+p1V1+Qpr+Wpr = E2+p2V2 which leads to H2−H1 =Qpr +Wpr so δH = −δQpr + −δWpr. In the case where Qpr = 0 and Wpr = 0 then the processconserves enthalpy.
8
1.7.2 Joule-Kelvin Expansion
p1
p2
Figure 1.8: Two Cylinders With Pressures p1 and p2 Exerted
The Joule-Kelvin expansion describes the temperature change when a gas is throttled through avalve, whilst the temperature of the system is kept constant. This method is used to cool gases tolow temperature. The Joule-Kelvin Coefficient is
(∂T∂p
)H
and is obtained from the enthalpy δH =
TδS + V δp. As(∂H∂T
)p= T
(∂S∂T
)p= Cp and
(∂T∂p
)H
= − 1Cp
(∂H∂T
)T
and(
∂H∂p
)T= −T
(∂V∂T
)p+ V .
(∂T
∂p
)H
= − 1
Cp
[−T
(∂V
∂T
)p
+ V
]
For ideal gases where pV = nRT the Joule-Kelvin coefficient is zero. For real gases, when p2 < p1there is heating and the Joule-Kelvin coefficient is negative. At low pressures the Joule-Kelvincoefficient is positive and cooling occurs.
1.8 Phase Equilibrium
1.8.1 Availability
The availability is δA = δU − TδS + pδV ≤ 0 where the equilibrium of the system occurs whenδA = 0. The minimum position of equilibrium is normally found, because the maximum is generallyunstable. The availability can be used to calculate the thermodynamics of the system.
First Law of Thermodyanics
For a thermally isolated system with a fixed volume then δA = δU .
Enthalpy
For a thermally isolated system with a constant pressure then δA = δU + pδV = δH
Helmholtz Free Energy
For a system with a fixed volume and a constant temperature then δA = δU − TδS = δF .
Gibbs Free Energy
For a system with constant pressure and constant temperature then δA = δU − TδS + pδV = δG.
9
1.8.2 Gibbs Free Energy for Phase Equilibrium
m1
m2
Figure 1.9: Container with Gas Separated by a Porous Plug
The Gibbs free energy of each gas separated by a porous plug is G = m1g1 +m2g2 where g is thespecific Gibbs free energy g = G
M . If an amount δm gets transferred from m1 to m2 then the Gibbsfree energy of the transfer is G = −δmg1 + δmg2. In equilibrium g1 = g2 so G = 0. On a pressurevs. temperature graph the line where g1 = g2 is the line of coexistence where two states can existsimultaneously.
TT T + δT
p
p
p+ δp
Phase 3
Phase 2
Phase 1
Figure 1.10: Phase Transition
For phase 2 the change in Gibbs free energy is δG2 =(∂G2
∂T
)pδT +
(∂G2
∂p
)Tδp and for phase 1
δG1 =(∂G1
∂T
)pδT +
(∂G1
∂p
)Tδp. At the line of coexistence δG1 = δG2. Rearranging the equations
such that:
δp
δT= −
(∂G1
∂T
)p−(∂G2
∂T
)p(
∂G1
∂p
)T−(
∂G2
∂p
)T
Clausius-Clapeyron Relation
By multiplying the top and bottom by T this gives:
δp
δT=
(S1 − S2)
(V1 − V2)=
L
V1 − V2
Where L is the latent heat capacity. For vapour transitions then V2 − V1 = V where V is thevolume of the larger phase.
10
1.9 Third Law of Thermodynamics1.9.1 Nernst’s PostulateAs a system approaches absolute zero all processes cease and the entropy of the system approachesa minimum value.
S(T2)− S(T1) =
∫ T2
T1
CVδT
T
In the limit that T → 0 the heat capacities go to zero as(
∂S∂p
)T= 0 and
(∂V∂T
)p= 0
1.10 Cooling by DemagnetisationIf a system is placed in a magnetic field B then δU = TδS − pδV +Bδm. Assuming the system isisovolumic then δU = TδS +Bδm.
1.10.1 Curie’s LawCurie’s Law sates that B = µo(H +M) where M ≪ H for most paramagnets. As M = χmH thenit can be written as M = χm
BµoM
. The magnetic susceptibility is χm = aT and so M = aB
µoT. The
demagnetisation of the system can be calculated using the reciprocity theorem. CB = T(∂S∂T
)B
which gives the adiabatic demagnetisation(∂T∂B
)S
= − TCB
(∂S∂B
)T
. The coefficient of adiabaticdemagnetisation can now be defined: (
∂T
∂B
)S
=1
CB
aB
µoT
This shows that by reducing the magnetic field the temperature will drop. Magnetising the systemat a constant temperature will cause the entropy to drop and then removing the field at constantentropy will reduce the temperature to achieve a lower temperature.
11
2
Statistics
2.1 Probability2.1.1 A Priori ProbabilityA priori probability is the probability that can be calculated with previous knowledge, and it hasno influence on the outcome.
Lottery
Six number are picked from one to forty-nine. To get the top prize all six must match. Theprobability of getting the top prize is given by 1
49C6≈ 1
14,000,000 . The probability of getting thebottom prize is found by the chance of having three correct and three incorrect numbers multipliedby the amount of ways the number could be arranged. 1
6C3×43C3= 1
57
Poker
Five cards are picked out of fifty-two cards. This means the number of arrangements of cards thatcould be picked up are 52C5. A full hand is two of the same number card and three of a different,same number card. The number of ways of getting this is given by getting two different numbersof a certain suit which is given by 13C2 × 2 and then of pulling three of the same number of anysuit and two of the other number for any suit. 1
13C2×2×4C3×4C2≈ 1
700 . A four of a kind is given bythe number of ways to pick any number in any suit, by the chance of getting the same number 3more times. 1
13C2×2×4 = 14165 . This shows it is about six times more likely to get a full house than
a four of a kind.
2.1.2 Prosteriori ProbabilityIf a population has 0.01% chance of catching a rare disease and the chance that a test to detectthe disease is accurate is 99.99% then:
Disease TestPositive 0.01% Positive 99.99%Positive 0.01% Negative 0.01%Negative 99.99% Positive 0.01%Negative 99.99% Negative 99.99%
A priori probability states that the probability of having the disease is 0.01%. If the test sayspositive then the posteriori probability of having the disease is 50%. This is because the chances ofthe test being incorrect and not having the disease is the same as the chance of having the diseaseand the test being correct.
12
2.2 Macrostates vs. Microstates2.2.1 MacrostatesA thermodynamic state of a system characterised by macroscopic functions of state, such as tem-perature, pressure and volume is called a macrostate.
2.2.2 MicrostatesA specific quantum state of a system which specifies all the relevant information of all the particles,such as all of the particles positions and velocities is a microstate.
2.3 Principle of Equal A Priori ProbabilityWhen an isolated system reaches equilibrium, all of the microstates accessible are equally probable.
2.3.1 One Particle
Figure 2.1: Single Particle Microstates
The chance of the particle being on the right or the left is given by 12C1
= 12 .
2.3.2 Two Particles
Figure 2.2: Two Particle Microstates
The probability of each of these microstates is P = 14C1
= 14 and so Pll =
14 = Prr and because
the particles are identical Plr = Prl =14 this microstate has a probability of 1
2 .
2.3.3 Many ParticlesFor any system there are 2N microstates available with a probability of 1
2Neach, where N = 2n.
There is only one microstate where all the particles are on one side, and so the probability of thismicrostate is 1
2N. The probability of having equal amounts of particles on the left and right is
Pnn =NCn
2N. This shows that the probability of the microstate occuring is proportional to the
number of microstates.
2.3.4 EntropyFor two systems, A and B, which are in contact with each other so that energy can be exchangedδUA = −δUB . The microsystems of A are WA and the microsystems of B are WB. There is no
13
change in the microstates due to energy, ∂(WA·WB)∂E = 0. This means that ∂WA
∂EAEB − ∂WB
∂EBWA = 0
which can be can be rearranged to get 1WA
δWA
δUA
δ lnWA
δUA= 1
kBTA. This is the zeroth law:
1
T= kB
(δ lnW
δU
)V
As 1T =
(∂S∂E
)V
then δU = TδS − pδV and so:
S = kBT lnW
This states that if system A has a lower temperature than system B then the energy must benegative. This means that the hot must flow to cold as indicated by the second law of ther-modynamics. For many states then the probability is given by Pin = Win
Wtotso the entropy is
S = kB (lnPin + lnWtot).
Joule Expansion
Where there are two states, such as the Joule Expansion then, δS = kB ln(
P1
P2
)and as Pin
Ptot=(12
)Nthe entropy for this system is δS = kB ln 2.
Lattice Model
In the lattice model of a gas the volume of the gas is V , which is localised on M sites. If each atomis treated as a point particle, such that all the particles N can occupy the same site. The entropyof the gas is S = NkB lnM and as m = V
δV then:
S = NkB (lnV − ln δV )
For a constant internal energy a variation in volume V at constant δV gives a change in pressureby:
p =TδS
δV=NkBT
VWhich is the ideal gas equation.
2.4 Statistics of FluctuationsThe number of microstates in a system is given by:
W (M) =N CN1 =N !
(N −M)!N1!
The multiplication can be converted into an addition by using natural logarithms. lnN ! =∑N lnNδN ≈
∫ N
1ln δN
Stirlings Approximation
Stirlings approximation states that
lnN ! = N(lnN − 1)
2.4.1 ProbabilityThe probability of choosing one microstate in the system using Stirling’s Approximation is P(M) =−N ln 2 +N lnN − (N −M) ln(N −M) −M lnM . If the probability of choosing the microstateis given by M = N
2 (1 + α) then ln(P(α)) = −N2 [(1 + α) ln(1 + α) + (1− α) ln(1− α)] + constant.
Expanding ln(1 + α) ≈ α+O(α2) gives ln(P(α)) = −Nα2 + constant and so the probability is:
P(α) ∝ e−Nα2
Which is similar to the Gaussian distribution where f = ex2
2σ2 and so σ = 1√N
.
14
2.5 EnsembleAn ensemble is a collection of identical systems, whose statistical fluctuations are cancelled outupon averaging. They are a necessary concept to connect statistics and thermodynamics.
2.5.1 Canonical EnsembleCanonical ensembles have a fixed volume, number of particles and temperature.
Gibbs Entropy
The Gibbs entropy is a more powerful version of the Boltzmann entropy. It states that:
S = −kB∑i
Pi lnPi
When the probability is Pi =1W then the Boltzmann Expression is recovered.
Probability
Using the mean energy where U =∑
i PiEi and the Gibbs entropy in the Helmholtz free energyequation allows the probability to be found. As the system is constrained to
∑i Pi = 1 then a
Lagrange multiplier can be included so that minimising the free energy determines the value of thevariables.
∂F
∂Pi+∂F
∂λ= 0
F =∑i
PiEi + kBTPi lnPi + λ∑i
Pi
This gives Ei+λ+kBT (lnPi + 1) = 0 and so the probability is Pi = e− Ei
kBT e−1+ λ
kBT . Calculatingthe value of λ gives Z−1 =
∑i e
EikBT which is defined as the partition function. The probability is:
Pi =e− Ei
kBT
Z
2.5.2 Micro-Canonical EnsembleMicro-canonical ensembles have a fixed energy, volume and number of particles.
Figure 2.3: Micro-Canonical System
A micro-canonical ensemble with N atoms where n are excited has W = N !(N−n)!n! microstates.
Using Stirling’s Approximation the entropy is S = kB [N lnN − (N − n) ln(N − n)− n lnn] and1T = ∂S
∂E where E = εn gives the probability of:
P(ε) =n
N=
1
eε
kBT + 1
This probability is the same as for the canonical ensemble.
15
2.6 Random Walk2.6.1 1D Random WalkIf a particle can take N
2 − s steps to the left and N2 + s steps to the right such that L = 2sa then
the number of microstates is:W =
N !(N2 − s
)!(N2 + s
)!
And so the entropy is S ≈ −kB L2
2Na2 .
2.6.2 3D Random WalkWhere there are three dimensions then the number of microstates increases to:
W =3N !(
N2 − s
)!(N2 + s
)!
And so the entropy is S ≈ −3kBL2
2Na2 . The Helmholtz free energy of the system is F = 3kBTL2
2Na2
which is the energy of a spring with a spring constant of k = 3kBTNa2 .
2.7 Helmholtz Free Energy
Placing the probability Pi =e− Ei
kBT
Z into the Gibbs entropy gives:
S = −kB∑i
P⟩
(− Ei
kBT− lnZ
)Expanding this out gives:
F = U − TS = −kBT lnZ
Which links microscopic statistics to thermodynamic quantities.
−kBT lnZ = pδV − SδT
Pressure
p = −(∂F
∂V
)T
p = kB
(∂(T lnZ)
∂V
)T
Entropy
S = −(∂F
∂T
)V
S = kB
(∂(T lnZ)
∂T
)V
= kB
(lnZ + T
∂ lnZ
∂T
)
Heat Capacity
CV = T
(∂S
∂T
)CV = kBT
(∂2(T lnZ)
∂T 2
)V
16
2.8 EquipartitionEquipartition shows that kBT is always included in energy. If the energy is E = αq2 then thepartition function shows that:
Z =
∫ ∞
−∞e− αq2
kBT δq =
√πkBT
α
The expectation for the energy is:
⟨U⟩ =∫αq2e
− αq2
kBT
Z
⟨U⟩ = kBT
2
This shows that if the energy can be expressed in a quadratic form then the average internal energyis kBT
2 per degree of freedom.
2.8.1 Failure of EquipartitionThe failure of equipartition arrises because a summation cannot truly be approximated as anintegral.
2.8.2 Factorisation of ZThe total energy of a particle is given by Etot = Etrans + Erot + Evib. Each of these energies areindependent of the other, therefore the partition function is:
Z =∑all
= e− E
kBT =∑i
e−Etrans
kBT
∑j
e−Erot
kBT
∑k
e−Evibe
kBT = Ztrans · Zrot · Zvib
The Helmholtz free energy can be calculated from Ftot = Ftrans + Frot + Fvib.
17
3
Quantum Thermodynamics
3.1 Quantum Mechanical Formulation3.1.1 Schrödinger FormulationIn the Schrödinger formulation the state of the system is represented by a wavefunction ψ and itscomplex conjugate ψ∗. The wavefunction must satisfy the equation of motion:
ih∂ψ
∂t= Hψ
The observable values of an operator are the eigenvalues of Aϕ = aϕ, where the operator must behermitian with real eigenvalues. The eigenfunctions satisfy the orthonormality relations:∫
ϕ∗iϕjδ3r = δij
Any wavefunction can be written in terms of a linear combination of the eigenstates of an operator,which corresponds to an observable:
ψ =∑i
ci(t)ϕi(r˜)The probability of a measurement giving a value of ai is |ci(t)|2 where:
ci(t) =
∫ϕ∗(r˜)ψ(r˜, t)δr˜
3.1.2 Dirac FormulationDirac introduced an abstract complex vector space which generalises the vectors and tensors ofordinary space. There is a complex vector called a ket |·⟩ and its complex conjugate vector calleda bra ⟨·|.
Addition
|A⟩+ |B⟩ = |C⟩
Scalar Product
⟨A|B⟩ = C
Where C is complex.
Operator Multiplication
G|A⟩ = |B⟩
18
Systems
The state of the system is described by abstract vectors |ψ(t)⟩ and the dynamical variables arerepresented by linear operators A. The equation of motion can be written:
ih∂
∂t|ψ⟩ = H|ψ⟩
The observables are the eigenvalues of the operators A|ϕo⟩ = ai|ϕ⟩ where the eigenvectors satisfythe orthogonality condition ⟨ϕi|ϕj⟩ = δij . Any state can be written in terms of the eigenstates ofϕi:
|ψ⟩ =∑i
ci(t)|ϕi⟩
The probability is given by |ci(t)|2 where:
ci(t) = ⟨ϕi|ψ(t)⟩
The Dirac formulation is more concise and useful because it allows a state with no particles in tobe described.
3.2 Particle in a 1D BoxIn quantum mechanics, a particle in a 1D box can be modelled by the Schrödinger Equation:
− h2
2m
∂2ψ
∂x2+ V ψ = Eψ
The solution to this equation can be found by inspection to be ψn ∝ sin(nπxL
)and so the energy
of the system is En = n2h2π2
2mL2 . The partition function is Z =∑
n e− En
kBT and when En
kBT is verysmall then the summation can be approximated as an integral so:
Z =
∫ ∞
0
e− En
kBT =
√πkBT
4En=
L√2πh2
mkBT
=L
λD
λD is the thermal de Broglie wavelength. If a particle has an energy 12mv
2 ≈ kBT then m2v2 ≈2mkBT which means the momentum is p ≈
√2mkBT so the de Broglie equation is:
λ =h
mv=
√h
2mkBT
If T is small, so that λD is the same size as the box, then the particle will interact with the entirebox at once. The Helmholtz free energy is F = −kBT ln L
λD, the entropy is S = kB
(ln L
λD+ 1
2
)and so the heat capacity is CV = kB
2 . The energy in one dimension is kBT2 which scales up for
more dimensions. A closed system with N particles obeys the rules of quantum mechanics.
3.3 k Space and Density of StatesIf the quantum wavenumber is defined as k = nπ
L and δk = πLδn then the partition function
becomes:Z =
∫ ∞
0
D(k)e− ε
kBT δk
Where D(k)δk is the number of quantum states in the range of k → k + δk. In one dimensionD(k) = L
π , in two dimensions D(k) = A4π2 and in three dimensions D(k) = V
8π3 The density ofenergy states is the number of states for a value of the dispersion relation, ε, which is the quantisedenergy.
D(ε) = D(k)δk
δε
19
The partition function for the density of energy states is:
Z =
∫ ∞
0
D(ε)e− ε
kBT δε
For a highly relativistic particle the dispersion relation is ε(k) = hck so the density of energy statesin three dimensions is D(ε) = V
2π2ε2
h3c3.
Internal Energy
U =
∫Ef(ε)D(ε)δε
Density of Particles
n =
∫f(ε)D(ε)δε
3.4 Distribution of Speeds3.4.1 Particles in a Classical GasThe Boltzmann probability distribution shows that the probability of a particle being in a singleparticle state with an energy ε(k) is:
Pk =e
ε(k)kBT
Z
If there are N particles then the probability that any of the particles are in a particular state is:
nk = NPk =Ne
ε(k)kBT
Z
This is valid as long as nk ≪ 1. nk is the average number of particles in the state. The averagenumber of particles f(k)δk in the range between k → k + δk is:
f(k)δk =Ne
ε(k)kBT
ZD(k)δk
The particles are distributed into states with different wavevectors. This distribution depends onthe number of spacial dimensions of the density of states , and the Boltzmann probability factor.For three dimensions the average number of particles is:
f(k)δk =
(Nλ3D2π2
)k2e
ε(k)kBT δk
3.4.2 Speed of a Classical GasFor an ideal gas with particles moving in one dimension the distribution of speeds can be foundfrom the k-states. The momentum is mv = p = hk and so the kinetic energy is εk = h2k2
2m = mv2
2 .In the range of speeds v → v + δv then:
n(v)δv = f(k)δk = f(k)
(δk
δv
)δv = 2N
√m
2mkBTe− mv2
2kBT δv
This shows that the most likely speed is zero and the higher speeds are less likely due to theBoltzmann factor reducing the probability of a particle occupying the higher energy states.
20
Three Dimensions
In terms of k the number of particles with a value of k between k → δk is:
f(k)δk) =
(Nλ3D2π2
)k2e
ε(k)kBT δk
The distribution of speeds can be calculated by substituting hk = mv.
n(v)δv = 4πN
(m
2πkBT
) 32
v2e− mv2
2kBT δv
The speed can only be positive in this case. n(v) is the product of the probability of a given speedand the density of states at a given speed.
3.5 Mean ValuesThe Maxwell-Boltzmann distribution can be used to find the mean value.
A =
∫∞0f(k)A(k)δk∫∞0f(k)δk
Finding k is useful for finding the mean speed v and the mean squared speed v2. From this themean kinetic energy of a system can be calculated. If the function is:
f(k) =
(2πh2
mkBT
) 32(N
2π2
)k2e
− h2k2
2mkBT
The mean wavevector is:
k =
∫∞0k3e
− h2k2
2mkBT δk∫∞0k2e
− h2k2
2mkBT δk
By substituting in x2 = h2k2
2mkBT the mean wavevector is k =√
8mkBTπh2 . The means speed is
v = hkm =
√8kBTπm . In the same way the mean square wavevector can be calculated k2 = 3mkBT
h2 so
the root mean squared speed is√v2 =
√3kBTm and the kinetic energy is 1
2mv2 = h2k2
2m = 32kBT .
3.5.1 Particles from an OvenParticles leaving an oven are more likely to be moving faster because the faster particles can leavethe oven more quickly. The flux of the particles leaving the oven can be calculated from the volumeswept out by the particles:
f(v)δv =vn(v)δv
4V
The flux of the particles leaving the oven with a speed v is n(v) which is proportional to v and sothe total flux per unit area is:
v =
∫∞0vn(v)δv∫∞
0n(v)δv
The function found by the distribution of speeds of emerging particles is:
f(v) =Nλ3Dm
3
8π2V h3v3e
− mv2
2kBT
21
Miller and Kusch
The Miller and Kusch experiment was used to measure the distribution of speeds. This was doneby firing particles from an oven into a rotating drum. The faster travelling atoms were foundtowards the centre of the drum, whilst the slower ones were found near to the outside of the drum.
Slower Atoms
Faster Atoms
Figure 3.1: Miller and Kusch Experiment
The rate at which the atoms leave the oven is(
n(v)δvV
)vA4 , where the factor of a quarter is to take
into account the atoms which are blocked or travelling in the wrong direction. The rate can bewritten:
g(v) =Av
4Vn(v)δv
3.5.2 Mean Free Path
R
Figure 3.2: Collsion Cross-Section
Real gas molecules experience collisions. The number of collisions depend on the size, speed andnumber density of the particle. The average distance travelled by each molecule is the mean freepath ℓ. If the molecules are modelled as spheres with a radius of R then a collision will occur whenmolecules come within a distance of 2R. The average distance needed for molecule to collide isℓ = 1
4πR2n = 1σn where σ is the collision cross-section. n = N
V is the number density which can becalculated from the ideal gas equation.
3.6 Broadening3.6.1 Doppler BroadeningThe classical theory of equipartition of energy states that there is a thermal energy of kBT
2 foreach degree of freedom. If the translational motion of the particle is taken into consideration then12mv
2 = 12kBT , and so the mean thermal speed is v =
√kBTm . If the particle is also emitting
light from the transition of energy hνo then the thermal motion of the particles causes a Dopplershift ν − νo = ±vo v
c . The ± is due to the component of motion being towards or away from theobserver, which is determined by the Maxwell-Boltzmann distribution. By combining the Doppler
22
shift equation with the Maxwell-Boltzmann distribution it can be seen that the Doppler shift hasa Gaussian lineshape.
n(v)δv = 2N
√m
2kBTe− m
2kBT (c(ν−νo)
νo)2 c
νoδν = g(ν)δν
The standard deviation is frequency is ∆ν = νo
√kBTmc2 and the full width at half maximum is
∆νFWHM = 2νo
√2kBT ln 2
mc2 . The Doppler linewidth at standard pressure and temperature isgenerally several orders of magnitude larger than the natural linewidth, but can be reduced bycooling the gas.
3.6.2 Collision BroadeningThe excited state of a particle can be interrupted by collisions. This means that the process oflight emission is also interrupted. This leads to broadening of the linewidth. The rate at whichthe collisions occur is 1
τ = vℓ = pσ
√8
πmkBT , which means that reducing the pressure decreases thelinewidth.
3.6.3 Lifetime BroadeningLight is emitted when an excited state drops to the ground state by spontaneous emission. Therate at which this occurs is the radiative lifetime. The energy-time uncertainty principle showsthat ∆E∆t ≥ h
2 . If the spontaneous radiative lifetime is τ = ∆t then there is an uncertainty in theenergy of the energy of the excited state by a minimum of h
2τ . This means there is an uncertaintyin the frequency of the emission of ∆νlife = ∆E
2h = 14πτ . This lifetime broadening is intrinsic to
the transition and therefore cannot be reduced.
3.7 Blackbody RadiationThe walls of an oven can emit electromagnetic radiation. The rate of radiation of energy is givenby Stefan’s Law for a blackbody.
δQ
δt= σAT 4
3.7.1 Theory of Rayleigh and Jeans (1907)Rayleigh imagined that the oven walls have harmonic oscillators with moving charges which radiateelectromagnetic waves. The oscillators have a temperature T and an average energy of kBT . Theseare in thermal equilibrium with electromagnetic waves of the same frequency. The energy of theelectromagnetic wavevectors from k → k+δk is U(k)δk = 2kBTD(k)δK where the two comes fromthe two transverse polarisations for electromagnetic waves. The total energy of the electromagneticwaves in three dimensions is:
U =V kBT
π2
∫ ∞
0
k2δk = ∞
This states that the total energy diverges and therefore is nonsense. This is called the UltravioletCatastrophe. The internal energy in terms of wavelengths is:
U(λ)δλ =V kBT
πk2δk
δλδλ =
V kBT
π
(2π
λ
)3
(−δλ)
3.7.2 Wein’s Displacement LawWein’s Displacement Law states that the maximum U(λ) occurs at λm where λmT = 2.9×10−3mK.The peaks all occur on a λ−4 line for large values of λ, but fails at low values of λ. At T = 290Kthe wavelength emitted is λ = 10× 10−6m. At T ≈ 6000K (the temperature at the surface of thesun) the wavelength is λm = 500nm which is approximately yellow light.
23
3.7.3 Planck’s LawPlanck imagined that an oven was made with harmonic oscillators in the walls, which emit andabsorb electromagnetic waves, as Rayleigh and Jeans did, but he stated that the oscillators couldonly have energy in quanta, 0, ε, 2ε, 3ε, .... The total energy is Mε amongst N oscillators. Mε =UN = NU . The total number of positions the oscillators can have is:
W =(N +M − 1)!
(N − 1)!M !≈ (N +M)!
N !M !
The approximation can only be made when N is huge. The entropy can then be found fromS = kB lnW and Stirling’s Approximation.
kB lnW = kB ((N +M) ln(N +M)− (N +M)−N ln(N) +N −M ln(M) +M)
S = kBN
[(1 +
M
N
)(ln(N)) + ln
(1 +
M
N
))− ln(N)− M
Nln(M)
]As U
ε = MN the entropy can be written as:
S = kBN
[(1 +
U
ε
)ln
(1 +
U
ε
)− U
εln
(U
ε
)]The temperature can be calculated using
(∂S∂U
)V= 1
T
1
T=kBε
ln( εU
+ 1)
And so the internal energy is:U =
ε
eε
kBT − 1
Including the density of state this can be extended to Planck’s Law
U(λ) =8πhc
λ5(e
hcλkBT − 1
)
λλm
U(λ)
Figure 3.3: Planck’s Law
This equation agrees with Wien’s Displacement Law when U(λ) is at a maximum. λmT = hckB4.965
3.8 PhotonsEinstein proposed that the electromagnetic waves themselves were quantised as particles. Thesephotons have an energy E = hν such that EN = Nhν. This gives rise to the photoelectric effectwhere a single quantum of light removes one electron from a metal. If the energy of the state isE = hck with wavevector k = 2π
λ then a solution would be of the form:
ϕk = A sin(nπxL
)sin(nπyL
)sin(nπzL
)24
The partition function of this type of wave can be found from:
Zk =∞∑n
e−nhck
kBT = 1 + e− hck
kBT + e− 3hck
kBT + e− 3hck
kBT + ...
The partition function must converge and so, by substituting in e− hck
kBT = x then:
Zk = 1 + x+ x2 + x3 + ...
Zkx = x+ x2 + x3 + x4 + ...
So as Zk(1− x) = 1 the partition function is:
Zk =1
1− e− hck
kBT
For many modes of wavevectors the total partition function is the sum of all the partition functions.The Helmholtz free energy can be calculated, where there is an extra factor of two due to thepolarisations of electromagnetic waves.
F = −2kBT∑k
lnZk
Including the density of states gives:
F =V kBT
π2
∫ ∞
0
k2 ln(1− e
− hckkBT
)δk
By substituting in x = hckkBT and using the standard integral
∫∞0x2 ln(1 − e−x)δx = −π4
45 theHelmholtz free energy is:
F =V (kBT )
4π2
45(hc)3
This can then be used to calculate thermodynamic quantities.
3.8.1 PhononsEinstein proposed that, for a solid with N atoms with N modes of oscillation, the internal energyis given by:
U =3Nhω
3hω
kBT − 1
And the heat capacity is:
CV =δ
δT
(3Nhω
ehω
kBT − 1
)This did not fit experimental results. Debye extended this to a range of 3N modes. The wavevectorsrun from 0 → kD, which is a cutoff with a speed s, which is the speed of sound for the material.
3N =
∫ kD
0
D(k)δk =3V
2π2
∫ kD
0
k2δk =V
2π2k3D
kD =3
√6π2N
V
Above kD the law of Dulong and Petit takes over at higher temperatures. The law of Dulong andPetit states that if there are N atoms in a solid with 3N modes of oscillation then each degree offreedom has an energy of U = kBT and a heat capacity CV = 3NkB . The energy per mode forthe Debye model is:
U =3V
2π2
∫ kD
0
hsk
ehskkBT − 1
δk
25
The internal energy due to phonons can be calculated by substituting x = hskkBT and converting
hskkBT = θ
T . If θT ≫ 1 then the upper limit of the integral can be replaced with ∞ and therefore
solved by standard integral∫∞0
x3δxex−1 = π4
15 . The internal energy is:
U =V π2
10
(kBT )4
(hs)3
And the heat capacity is:
CV =4V π2
10
k4BT3
(hs)3
If θT ≪ 1 then the exponential can be expanded and O(x2) ignored to give:
U = 3NkBT
And a heat capacity:CV = 3NkB
Average Number of Phonons
If the phonon wave is ϕk = A sin(n1πxL
)sin(n2πyL
)sin(n3πyL
)then the probability of n particles is:
Pn =e−nhsk
kBT
Zk
The average number of phonons is then found by:
n(k) =− δZk
δ hskkBT
Zk=
1
ehskkBT − 1
This means the average energy can be calculated:
U = hskn(k) =hsk
ehskkBT − 1
Which is of the same form as the Planck distribution. The total number of phonons can becalculated:
N = 3
∫ kD
0
k2δk
ehskkBT − 1
The energy of the phonons is:
U =3V
2π2
∫ kD
0
hsk3δk
ehslkBT − 1
The total number of phonons, and photons, can fluctuate even if the system is closed becausephonons and photons are non-conservative.
26
4
Extensive and Intensive Variables
4.1 Extensive QuantitiesExtensive quantities are proportional to then number of particles in the system N . The internalenergy, volume, number, entropy and heat capacity are extensive and so U
N , VN ,NN , SN and CV
N areconstant. All chemical processes can be described by:
δS =
(∂S
∂U
)V,N
δU +
(∂S
∂V
)U,N
δV +
(∂S
∂N
)U,V
δN
4.1.1 Chemical PotentialAs pressure and temperature can be defined from the entropy, so can the chemical potential:
µ = −T(∂S
∂N
)U,V
And so the first law of thermodynamics can be rewritten:
δU = −δQ+−δW = µδN − pδV + TδS
NA
SystemA SystemB
NB
Figure 4.1: System with Two Chambers Separated by a Semi-Permeable Membrane
The total number of particle in the whole system is given by NA + NB = N which is constant.The change of number of particles from A to B is δNA = −δNB and so the change in entropy is:
δS =
(∂SA
∂NA
)U,V
δNA +
(∂SB
∂NB
)U,V
δNB =
((∂SA
∂NA
)U,V
−(∂SB
∂NB
)U,V
)δNA
The chemical potential for system A is µA = −T(
∂SA
∂NA
)U,V
and for system B µB = −T(
∂SB
∂NB
)U,V
so the total entropy can be given by:
δS = −µA − µB
TδNA
27
The entropy is at a maximum when the system is in equilibrium and µA = µB so δS = 0. IfµA > µB then δNA must be negative so that δS > 0. The particles in the system will move tolower the value of µ. Photons and phonons are not conserved and so µ = 0.
Partition Function
The partition function for distinguishable particle is ZN = Z1 ×Z1 ×Z1 × ... = ZN1 but as Sackur
and Tetrode stated, for indistinguishable particles the partition function is ZN =ZN
1
N ! .
Helmholtz Free Energy
Using this partition function and Stirling’s approximation the Helmholtz free energy is:
F = −N(kBT ln(Z1)− ln(N) + 1)
If a particle has and energy E = ∆+ h2k2
2m , where ∆ is a shift in the kinetic energy by a field suchas ∆ = mg˜h or ∆ = −eE˜ · r˜, then the partition function is Z1 =
∑k e
∆+ h2k2
2m kBT . When thedensity of states is included then
Z1 = e− ∆
kBT
∫ ∞
0
D(k)e− h2k2
2mkBT δk = e− ∆
kBT V
(mkBT
2πh2
) 32
The Helmholtz free energy is:
FN = −N
[−∆+ kBT ln
(V
N
(mkBT
2πh2
) 32
)+ 1
]
For fixed VN the Helmholtz free energy is proportional to N and so is extensive. The entropy can
be calculated and so the chemical potential can be found.
µ = ∆− kBT ln
(1
n
(mkBT
2πh2
) 32
)
4.1.2 Chemical Reactions
C +D ⇀↽ CD
The entropy for this reaction is S = SC(NC)+SD(ND)+SCD(NCD) and so the change in entropycan be defined as δS = δNC
T (µCD + µC − µD). Which means that there is chemical equilibriumwhen µC + µd = µCD. If µC + µD > µCD then δS > 0 and so δNC < 0.
2H2 +O2 ⇀↽ 2H2O
As there are three particles on the left side of the reaction and only two on the right then they canbe represented by stoichiometric constraints.
νH2H2 + νO2O2 ⇀↽ νH2OH2O
As νO2 = 1 = δNO2 , νH2 = 2 = δNH2 and νH2O = −2 = δNH2O the entropy can be written as:
δS =δNO2
T(2µH2O − µO2 − 2µH2)
There is chemical equilibrium when 2µH2O = 2µH2 + µO2 .
28
4.1.3 Reactions in the Sune− + p+ ⇀↽ H + γ
For equilibrium µe + µp = µH + µγ , but as photons are not conserved then µγ = 0. Protons havespin 1
2 and so the density of states increases by 2, to account for spin up and spin down. Electronsalso exhibit the same behaviour. Hydrogen has 2 spin 1
2 components and so the density of statesincreases by 4.
µp = kBT ln
np2
(2πh2
mpkBT
) 32
µe = kBT ln
ne2
(2πh2
mekBT
) 32
µH = ∆+ kBT ln
nH4
(2πh2
mHkBT
) 32
In this case ∆ is the binding energy, which for the case of hydrogen is −13.6eV . The Saha equationcan now be found.
∆+ kBT ln
nH4
(2πh2
mHkBT
) 32
= kBT ln
ne2
(2πh2
mekBT
) 32
+ kBT ln
np2
(2πh2
mpkBT
) 32
As mp ≈ mH and ne = np this equation can be reduced to:
∆
kBTln(nH) = ln(nenp) +
3
2ln
(2πh2
mekBT
)This means the number density of electrons in the sun is:
ne =
√nHe
∆kBT
(mekBT
2πh2
) 32
4.2 Grand Canonical EnsembleThe Gibbs ensemble can be extended by allowing the number to change.
Reservoir System A
Figure 4.2: Grand Canonical Ensemble
The total number of particle and energy is conserved in the entire system such that δNA = −δNR
and δUA = −δUR. The reservoir is huge so that any energy added does not change the temperatureand any particles added does not change the chemical potential. The entropy is a function ofextensive quantities:
δS =
(∂S
∂U
)V,N
δU +
(∂S
∂V
)U,N
δV +
(∂S
∂N
)U,V
δN
29
This means that the microstates of the reservoir is:
ln(WR) = K +UR − µNR
kBT
WR = eKe(UR−µNR)
kBT
This can be expanded using UR = U − UA and NR = N −NA so that:
WR = eKeU
kBT eµNkBTe− (UA−µNA)
kBT = Γe− (UA−µNA)
kBT
It can be assumed that the energy of the state and the number of particles are such that:
Hψi = Eiψi + Nψi = Niψi
Where N is the number operator. The combined number of states is:
W = Γe− (Ei−µNi)
kBT
So the probability of being in a state ψi is:
Pi =Γe
− (Ei−µNi)
kBT
Γ∑
j e−
(Ej−µNj)
kBT
This leads to the definition of the Grand Partition Function:
Ξ =∑j
e−
(Ej−µNj)
kBT
30
5
Bosons and Fermions
All elementary particles are either Fermi particles or Bose particles. Each particle is identical toall other particles of the same sort.
5.1 Fermi ParticlesFermi particles have spin 1
2 . Quarks and Leptons are Fermi particles.
Leptons Quarkse− µ− τ− u s tνe νµ ντ d c b
Fermi particles can only have one particle in a single particle state (neglecting spin). The statesof energy are Ei = 0 and Ni = 0 or Ei = Ei and Ni = 1. The grand partition function for Fermiparticles is:
Ξ =∑j
e−
(Ej−µNj)
kBT = 1 + e− (Ei−µ)
kBT
The probability of there being no particles is:
P0 =1
1− e−
(Ej−µ)
kBT
The probability of there being one particle is:
P1 =1
e(Ei−µ
kBT + 1
And so the average number of particles is ni = 0×P0+1×P1 = 1
e(Ei−µ)kBT +1
. For high temperatures
Ni = e− (Ei−µ)
kBT which is approximately the Boltzmann distribution.
5.1.1 Fermi-Dirac FunctionFor a low density gas at high temperatures, µ < 0 so that e−
µkBT > 1. This means that the average
number can be approximated to n(k) ≈ e− (ε(k)−µ)
kBT . From this it can be assumed that NZ = e
µkBT
and so:
µ = kBT ln
NV
(2πh2
mkBT
) 32
This is the proof of the chemical potential in the classical limit. For a high density gas at lowtemperatures the approximation of the average number cannot be made. µ is positive and T isfinite.
31
n(E)
E
T = 0K
ε > µ
ε < µ
EF
Figure 5.1: Fermi-Dirac Function
At T = 0 the chemical potential gives the Fermi Energy.
Fermi Wavevector kFThe number of particles N in a volume V is given by:
N = (2)spin
∑k
n(k) =2V
2π2
∫ ∞
0
k21
e(ε(k)−µ)
kBT + 1δk
At T = 0 the number of particles is N = Vπ2
∫ kF
0k2δk which is the step function k3F = 3π2N
V . Fornon-relativistic particles pF < mc and so the energy equation is E2 = p2c2 +m2c4. This can beapproximated by Taylor expansion:
E = mc2(1 +
p2
2m2c2+O
(p4
4m4c4
))= mc2 +
p2
2m
EF =h2k2
2m
For relativistic particle pK ∼ mc so the energy is E = pc(1 + m2c2
p2
) 12 so the energy is:
EF = hkF c = pF c
The Fermi temperature can be described by TF = EF
kB. When the temperature of the system is
less than the Fermi temperature then it is degenerate so that the particles fill up the states fromthe lowest first. Theoretical values such as the internal energy of a system at T = 0 can also befound. In the non-relativistic case U = (2)
∑k
h2k2
2m which can be approximated as:
U =V
π2
∫ kF
0
k2h2k2
2mδk =
V h
1− π2mk5F
And as k3F = 3π2NV and k2 = 2mEF
h2 :
U =3
5NEF
Other thermodynamic quantities can also be calculated using:
δU = TδS − pδV + µδN
But as T = 0, δU = µδN − pδV so the pressure is:
p = −(∂U
∂V
)N
=2
5
N
VEF
32
Finite Temperatures
For finite temperatures the step function is smoothed out. There is a change in distribution at thepoint where the step exists at T = 0.
k
δn(k)
T = 0K
Figure 5.2: Change in Distribution at the Step Function
The change in energy is approximately γkBT where γ = π2
4 and the fraction affected is kBTEF
. Thecorrection in the energy is δU = γN (kBT )2
EFso the total energy is:
U =3
5NEF + γN
(kBT )2
EF
The heat capacity can now be obtained CV =2γNk2
BTEF
and the entropy can be calculated:
S = So +2γNk2BT
EF
This is useful because at T = 0 there is only one quantum state so S0 = kB ln(1) = 0 and so S = 0.The Helmholtz free energy can be found from F = U − TS so:
F =3
5NEF − γN(kBT )
2
EF+O(Fn)
And from this the pressure is:
p =2
3
Ef
V
(3
5N +
γN(kBT )2
E2F
)The chemical potential is:
µ = EF
(2− γ
3
(kBT
EF
)2)
T
µ
EF
DegenerateRegime
Maxwell Boltzmann
Figure 5.3: Chemical Potential vs. Temperature
33
5.2 Bose ParticlesBose particle have spin 1. The elementary particles are:
Electromagnetic Interaction Strong Interaction Weak InteractionPhoton Gluon
γ gi W+ W− Zo
Bose particles can have any number of particles in a single state. The number of particles in astate ψi with energies 0, Ei, 2Ei, 3Ei, ...
Ξ = e− (Ei−µ)
kBT + e− (2Ei−µ)
kBT + e− (3Ei−µ)
kBT + ... =1
1− e− (Ei−µ)
kBT
This is true only as long as the geometric series converges e−(Ei−µ)
kBT < 1. This can be used to getthe Bose-Einstein distribution.
∂Ξ
∂µ=
∞∑n=0
n
kBTe−n(εi−µ)
kBT
Multiplying this by kBTΞ gives the average number density.
kBT
Ξ
∂Ξ
∂µ=∑n
nPn = n = kBT∂(lnΞ)
∂µ=
1
e(εi−µ)
kBT − 1
5.2.1 Composite ParticlesParticles made up of more than one fundamental particles can be either Bose or Fermi dependingon the number of Fermi particles. The wavefunction for two identical particles at r˜1 and r˜2 isψ(r˜1, r˜2). If the particles are exchanged then Bose particles have ψ(r˜1, r˜2) = ψ(r˜2, r˜1) and Fermiparticles have ψ(r˜1, r˜2) = −ψ(r˜2, r˜1). The physical quantity |ψ(r˜1, r˜2)|2 is the same for both typesof particles. Protons are an example of a composite particle. It is made up of three quarks, two upand one down. Each of the quantum chromodynamic colours must be different to make the wholeprotons colourless.
DD
UUUU
Figure 5.4: Proton
On swapping three Fermi particles the wavefunction changes sign three times (−1)(−1)(−1) = −1,which means the composite is a Fermi particle. K+ mesons have an up and an anti-strange particle.
UU
SS
Figure 5.5: K+ Meson
34
On swapping two Fermi particles the wavefunction changes sign twice (−1)(−1) = 1 which showsthe K+ Meson is a Bose particle.
5.2.2 Eigenvaluesp12 is the operator which exchanges particles:
p12ψ(r˜1, r˜2) = ψ(r˜2, r˜1)p12p12ψ(r˜1, r˜2) = p12ψ(r˜2, r˜1) = ψ(r˜1, r˜2)
p212 has an eigenvalue of 1. This means the eigenvalue of p12 are λ = ±1. For Bose particles λ = 1and for Fermi particles λ = −1. Two particle states can be made from single particle states suchas the 1s orbital of the Hydrogen atom.
p12 ψBose
(r˜1, r˜2) = ϕi(r˜1)ϕj(r˜) + ϕj(r˜1)ϕi(r˜2)This means that when i = j two Bose particles are in the state ϕi(r˜).
p12 ψFermi
= ϕi(r˜1)ϕj(r˜2)− ϕj(r˜1)ϕi(r˜2) = − ψFermi
(r˜1, r˜2)If i = j then there is no quantum state ψ(r˜1, r˜2) = 0. This gives arise to the Pauli exclusionprinciple which states that two particles cannot be in the same state (with spin included then twoparticles be in the same state, but not more).
5.2.3 Bose Condensed GasesThe state |106, 0, 0, 0, 0, 0, 0, ....⟩ has 106 particles in state ϕi, could be a system such as a laserwhere µ = 0 or a Bose condensed gas where µ = 0. The average number in a state ϕk(r˜) is:
n(k) =1
eε(k)−µkBT − 1
As long as ε(k) − µ > 0 then eε(k)−µkBT > 1 and the series can be summed. The grand partition
function is:Ξk = 1 + e
− (ε(k)−µ)kBT + e
− (2ε(k)−µ)kBT + ...
By substituting x = e− (ε(k)−µ)
kBT the grand partition function is Ξ = 1+x+x2 + ... and so the seriesconverges when x < 1. This is valid for low density gases at high temperatures.
5.2.4 Bose-Einstein CondensationA gas ofN Bose particles with µ > 0 in a single particle state is ϕk(r˜) = A sin(kxx) sin(kyy) sin(kzz).Provided ε(k)− µ > 0 the number of particles is:
N =V
2π2
∫ ∞
0
k2
eh2k2
2mkBT − µkBT − 1
δk
Substituting in x = h2k2
2mkBT and solving numerically for negative values of µ lets the critical tem-perature T = TC be calculated. The standard integral
∫∞0
x2δxex2−1
= 3√πζ8 .
TC =h2
2m
(N
V
) 23
× 4√πζ
8
For T < TC and µ = 0, if there is No particles in the lowest energy state then the number ofparticles is:
N = No +V
2π2
(2mkBT
h2
) 32 3
√πζ
8
35
The ratio between the numbers at the lowest state and the number at TC is NNo
NTC=(
TTC
) 32 so that
No
N = 1−(
TTC
) 32
5.2.5 Laser CoolingRubidium or Sodium atoms need to be cooled without touching the walls which contains them.To do this the atoms are placed in a magnetic trap and lasers are used to cool them.
Recoil
A single laser beam has photons all in the same state, for example, |1020, 0, 0, 0, ...⟩. The momentumof photons in the x-direction is p = hkx. There is a Doppler shift as seen by the atom:
νL =
(1 + vx
c
1− vxc
) 12
νo ≈(1 +
vxc
)νo
Which means that the Doppler shift of the velocity in the x-direction is vx = c(
νL−νo
νo
). There is
a Maxwellian distribution of speeds, vx in the atoms of the gas.
n(vx) = Ke− mv2
2kBT
In terms of frequencies this is:n˜(νL) = Ke
− mc2
2kBT (νL−νo
νo)2
This includes Doppler broadening of the absorption spectrum. After one absorption the atomsreduce their speed in the x-direction to
(vx − hk
m
)and a photon is spontaneously emitted at a
rate of τ − 1. After two absorptions the speed is reduced to(vx − 2hk
m
)and another photon is
spontaneously emitted, and this carries on. The intensity of the laser is large so the absorptionrate is large. In contrast the emission rate from the atom is small as it involves just one emittedphoton (there is no stimulated emission from surrounding atoms). The force exerted in one cycleis F = − h
λτ . The kinematic speed equation is v = vi − hmλτ and so when v = 0 the time taken for
an atom to come to rest is t = mλτh vi. The distance travelled by the atom is:∫ ts
0
vδt = vsts −ht2s
2mλτ
s =mλτ
2hv2i
For sodium atoms the rate of emission is τ = 16× 10−9s and the mass is m = 23× 1.66× 10−27kg.The stopping time for sodium is 0.18 × 10−3s and the stopping distance is 3 × 10−2m. Thereare six lasers with their beams crossing at the position of the gas in the ±x,±y and ±z axessimultaneously. The limit to the decrease in speed is m∆n < hk, so the lowest possible kineticenergy achieved by laser cooling is E = h2
2mλ . This is the Doppler limit. The lowest temperaturefrom this is:
T =h2
2mkBλ2
For sodium atoms this is approximately 10−6K. This can be cooled further by Sisyphus cooling.When a low enough temperature is achieved then Bose-Einstein condensation can be observed.
36
