H.2 Euler-Maclaur in Sum Formula
18.1 Derivation of the Boltzmann Distribution
We examine aconfiguration{Ni} = N1,N2,. . .,Nκ of the system such thatN1particles are in a quantum state2with energyε1,N2particles are in a quantum state with energyε2, etc. Such a configuration is subject to the constraints
1Degeneracy arises when there are stationary states of the subsystems having the same energy but a different set of quantum numbers.
2For brevity we use a single index to denote a quantum state but in fact many quantum numbers may be necessary. Moreover, there can be degeneracy if different quantum states have the same energy.
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286 THERMAL PHYSICS
i
Ni=N (18.1)
and
i
Niεi=E, (18.2)
where E is the total energy of the system. Since the particles are distinguishable, the number of ways of making a given configuration is3
W{Ni}:= N!
N1!N2! ã ã ãNκ!. (18.3) We proceed to maximize W{Ni}, considered to be a function of the Ni, subject to the constraints expressed by Eqs. (18.1) and (18.2). Since lnxis a monotonically increasing function ofx, we actually maximize lnW subject to these same constraints. To handle the constraints, we introduce Lagrange multipliersβandαand solve the problem
∂
∂Nj[lnW{Ni} −βE−αN] =0. (18.4) By virtue of the Lagrange multipliers, allNjin Eq. (18.4) can be regarded as independent variables, which will turn out to be functions ofβ andα. We can then chooseβ andαto satisfy the constraints.
In order to differentiate lnW{Ni}, we use Stirling’s approximation (see Appendix A) and obtain
∂
∂Nj
lnW{Ni} ∼ ∂
∂Nj
NlnN −
i
NilnNi
= −lnNj
N . (18.5)
Thus Eq. (18.4) becomes
−lnNj
N −βεj−α=0. (18.6)
The solution to Eq. (18.6) is
Nj
N =e−αe−βεj. (18.7)
Applying the constraint Eq. (18.1) we obtain
1=
j
Nj
N =e−α
j
e−βεj, (18.8)
which results in
e−α=1/z, (18.9)
3Here,Wplays the same role asfor the microcanonical ensemble, but we use a different notation because corresponds to constrained values ofEandN. In the present case, these constraints are replaced by Eqs. (18.1) and (18.2). Ultimately we will specify the temperatureT and then determineEfrom the probabilitiespi of occupation of the quantum states.
where
z=
j
e−βεj (18.10)
is known as thepartition function.4Thus Eq. (18.7) becomes pi:=Ni
N =e−βεi
z , (18.11)
where we have also introduced the symbolpi, the probability of occupation of theith state of a particle.
The internal energy can now be determined from Eq. (18.2) to be5 U:= E =N
i
piεi=N
i
εie−βεi z = −N
z
∂z
∂β = −N∂lnz
∂β . (18.12)
To obtain the entropy, we use6
S=kBlnW{Ni} (18.13)
withNigiven by Eq. (18.11). With the aid of Stirling’s approximation, Eq. (18.13) becomes S=kB
NlnN−
i
NilnNi
= −kB
i
Niln(Ni/N)= −NkB
i
pilnpi. (18.14) We now proceed to identify the remaining Lagrange multiplierβ. In principle, we could do this by specifying the total energy and solving Eq. (18.2) forβ, withNi given by Eq. (18.11), but this would necessitate solving a complicated transcendental equation.
Instead, we suppose that our system is in equilibrium at fixed temperatureT and appeal to thermodynamics to identifyβ. We do this by relating the above expressions forU and Sby means of the thermodynamic equation dU = TdS−pdV which holds at constant N for a system that can do reversible workpdV.7From Eq. (18.12), the differential of the internal energy is
dU=N
i
εidpi+N
i
pidεi=N
i
εidpi+N
i
pi∂εi
∂V dV, (18.15)
4In Eq. (18.10),zis the partition function for an individual particle. We reserve the symbolZfor the partition function of the whole system that we shall later relate toz.
5Since thepiare probabilities, Eq. (18.12) actually gives the most probable valueEof energy which we identify with the internal energyUthat we will ultimately compute from a knowledge of the temperature.
6To get the entropy, we should really compute the logarithm of the total number of microstates by summing all values of lnW{Ni} that are compatible with the constraints. Instead, we approximate this sum by its overwhelmingly largest term.
7See Section 19.1.3 for a similar treatment for a more general system.
288 THERMAL PHYSICS
where we have assumed that the energies of the states depend on the volume of the system. From Eq. (18.14), the differential of the entropy is
dS= −NkB
i
(lnpi+1)dpi= −NkB
i
lnpidpi
= −NkB
i
(−βεi−lnz)dpi=NkBβ
i
εidpi, (18.16)
where we have used
idpi = 0 because
ipi = 1. By combining Eq. (18.15) with Eq. (18.16) we obtain
dU= 1
kBβdS+N
i
pi∂εi
∂V dV. (18.17)
Comparison with dU=TdS−pdVshows that β= 1
kBT. (18.18)
We also obtain a useful equation for the pressure, namely p= −N
i
pi∂εi
∂V, (18.19)
which by means of Eq. (18.17) with dS = 0 is seen to be equivalent top = −(∂U/∂V)S,N
withU =N
ipiεi. By using Eq. (18.11) to rewrite lnpi, the entropy given by Eq. (18.14) can be written in the form
S=U
T +NkBlnz. (18.20)
Equation (18.20) can then be combined with the equationF =U−TSto deduce a useful formula for the Helmholtz free energy
F= −NkBTlnz= −N
β lnz. (18.21)
In Section 19.1.3, this derivation will be generalized to anensemble of complicated systems (instead of a collection of weakly interacting particles). Such an ensemble is known as a canonical ensemble and allows for each complicated system to consist of many interacting particles. For such complicated systems, determination of the quantum states and the resulting partition functions can be quite difficult.
18.1.1 Summary of Results
The probabilitypiof occupation of theistate is pi:=Ni
N =e−βεi
z , (18.22)
whereβ=1/(kBT),kBis Boltzmann’s constant,Tis the absolute temperature, and z=
j
e−βεj (18.23)
is the partition function. The entropy is S= −NkB
i=1
pilnpi= −NkBβ2 ∂
∂β lnz
β
. (18.24)
The internal energy is
U=N
i=1
piεi= −N ∂
∂βlnz (18.25)
and the Helmholtz free energy is
F= −N
β lnz. (18.26)
In solving problems, one usually proceeds as follows:
• Determine the subsystem statesihaving energiesεifrom a model or from experimental data.
• Calculate the partition functionzand deduce the Helmholtz free energy F by using Eq. (18.26).
• Obtain the entropy fromS= −(∂F/∂T)V,N or from Eq. (18.24).
• Obtain the internal energy fromU=F+TSor from Eq. (18.25).
• Obtain the chemical potential per particle fromμ=(∂F/∂N)T,V.
• If the dependence of theεion volumeV is known, determine the pressure fromp =
−(∂F/∂V)T,N.
In following this procedure, it should be recognized that Eq. (18.26) yieldsFas a func- tion of its natural variablesT,V, andN, whereN =N/NAis the mole number of particles, NAbeing Avogadro’s number. We therefore recover the usual thermodynamic description of a monocomponent system. The volumeV might enter because each particle occupies a volumeV/N on which the energy levels εi might depend. These particles, although identical, are supposed to be distinguishable by virtue of their position. If all particles were tosharethe same volume and were identical, they would not be distinguishable.
Such would be the case for a monatomic ideal gas, so to treat such a system the above equations would have to be modified.