222 Isobaric ensembles
Noting that the thermodynamic relations in eqn. (5.2.5) can also be written in
terms of the entropyS=S(N,P,H) as
1
T=
(
∂S
∂H
)
N,P,
〈V〉
T
=
(
∂S
∂P
)
N,H,
μ
T=
(
∂S
∂N
)
V,H, (5.3.6)
the thermodynamics can be related to the number of microscopic states by the anal-
ogous Boltzmann relation
S(N,P,H) =kln Γ(N,P,H), (5.3.7)so that eqns. (5.3.6) can be expressed in terms of the partition function as
1
kT=
(
∂ln Γ
∂H)
N,P,
〈V〉
kT=
(
∂ln Γ
∂P)
N,H,
μ
kT=
(
∂ln Γ
∂N)
V,H. (5.3.8)
The partition function for the isothermal-isobaric ensemble can be derived in much
the same way as the canonical ensemble is derived from the microcanonical ensemble.
The proof is similar to that in Section 4.3 and is left as an exercise (see Problem 5.1).
As an alternative, we present a derivation of the partition functionthat parallels
the development of the thermodynamics: We will make explicit use of the canonical
ensemble.
Consider two systems coupled to a common thermal reservoir so that each system
is described by a canonical distribution at temperatureT. Systems 1 and 2 haveN 1
andN 2 particles respectively withN 2 ≫N 1 and volumesV 1 andV 2 withV 2 ≫V 1.
System 2 is coupled to system 1 as a “barostat,” allowing the volume tofluctuate such
that the internal pressureP of system 2 functions as an external applied pressure to
system 1 while keeping its internal pressure equal toP(see Fig. 5.1). The total particle
number and volume areN=N 1 +N 2 andV=V 1 +V 2 , respectively. LetH 1 (x 1 ) be
the Hamiltonian of system 1 andH 2 (x 2 ) be the Hamiltonians of system 2. The total
Hamiltonian isH(x) =H 1 (x 1 ) +H 2 (x 2 ).
If the volume of each system were fixed, the total canonical partition function
Q(N,V,T) would be
Q(N,V,T) =CN
∫
dx 1 dx 2 e−βH^1 (x^1 )+H^2 (x^2 )=g(N,N 1 ,N 2 )CN 1∫
dx 1 e−βH^1 (x^1 )CN 2∫
dx 2 e−βH^2 (x^2 )∝Q 1 (N 1 ,V 1 ,T)Q 2 (N 2 ,V 2 ,T), (5.3.9)
whereg(N,N 1 ,N 2 ) is an overall normalization constant. Eqn. (5.3.9) does not produce
a proper counting of all possible microstates, as it involves only one specific choice ofV 1
andV 2 , and these volumes need to be varied overallpossible values. A proper counting,
therefore, requires that we integrate over allV 1 andV 2 , subject to the condition that
V 1 +V 2 =V. SinceV 2 =V−V 1 , we only need to integrate explicitly over one of the