162
3 Ensembles: Systems of Particles
By applying an equivalent argumentation to the constraint (3.4.2b), we obtain the
result
∂ ln 1
∂V 1
V 1 =V 1m
=
∂ ln 2
∂V 2
V 2 =V 2m
≡ ζ ,
(3.4.4b)
which introduces a new parameter that we have designated by the Greek letter zeta,
ζ .
Along the same lines of reasoning that we applied in our development of the
grand ensemble, let us expand ln 2 (( 2 , V 2 ) for the bath in terms of a double Taylor
series expansion
ln 2 (( 2 , V 2 ) = ln 2 (( 2m , V 2m ) +
∂ ln 2
∂∂ 2
2 = 2m
(( 2 − 2m )
+
∂ ln 2
∂V 2
V 2 =V 2m
(V 2 − V 2m ) + · · · ,
(3.4.5)
with the ellipsis representing second- and higher-order partial derivative terms in the
expansion. Now, if we employ the total energy and volume constraints 1 + 2 =
1m + 2m and V 1 + V 2 = V 1m + V 2m , to replace 2 − 2m and V 2 − V 2m by
−(( 1 − 1m ) and −(V 1 − V 1m ), respectively, then utilize the definitions of β and
ζ from Eqs. (3.4.4) in our expression for the Taylor expansion of ln 2 (( 2 , V 2 ) we
find that
ln 2 (( 2 , V 2 ) ln 2 (( 2m , V 2m ) − β(( 1 − 1m ) − ζ(V 1 − V 1m ) + · · · .
Exponentiation of both sides of this equation gives
2 (( 2 , V 2 ) 2 (( 2m , V 2m )e
ββ 2m +ζ V 2m e
−ββ 1 −ζ V 1 .
(3.4.6)
Upon employing Eq. (3.4.6) for 2 (( 2 , V 2 ) in expression (3.4.1b) for (, V ),
we obtain
((, V )
V
0
dV 1
0
d 1 1 (( 1 , V 1 )) 2 (( 2m , V 2m )e
ββ 2m +ζ V 2m e
−ββ 1 −ζ V 1
= 2 (( 2m , V 2m )e
ββ 2m +ζ V 2m
V
0
dV 1
0
d 1 1 (( 1 , V 1 )e
−ββ 1 −ζ V 1 .
Because of the exponentials in 1 and V 1 , we may extend both upper integration
limits to infinity, to give
((, V ) = 2 (( 2m , V 2m )e
ββ 1m +ζ V 1m
∞
0
∞
0
1 (( 1 , V 1 )e
−ββ 1 −ζ V 1 d 1 dV 1 ,
(3.4.7)
3 Ensembles: Systems of Particles
By applying an equivalent argumentation to the constraint (3.4.2b), we obtain the
result
∂ ln 1
∂V 1
V 1 =V 1m
=
∂ ln 2
∂V 2
V 2 =V 2m
≡ ζ ,
(3.4.4b)
which introduces a new parameter that we have designated by the Greek letter zeta,
ζ .
Along the same lines of reasoning that we applied in our development of the
grand ensemble, let us expand ln 2 (( 2 , V 2 ) for the bath in terms of a double Taylor
series expansion
ln 2 (( 2 , V 2 ) = ln 2 (( 2m , V 2m ) +
∂ ln 2
∂∂ 2
2 = 2m
(( 2 − 2m )
+
∂ ln 2
∂V 2
V 2 =V 2m
(V 2 − V 2m ) + · · · ,
(3.4.5)
with the ellipsis representing second- and higher-order partial derivative terms in the
expansion. Now, if we employ the total energy and volume constraints 1 + 2 =
1m + 2m and V 1 + V 2 = V 1m + V 2m , to replace 2 − 2m and V 2 − V 2m by
−(( 1 − 1m ) and −(V 1 − V 1m ), respectively, then utilize the definitions of β and
ζ from Eqs. (3.4.4) in our expression for the Taylor expansion of ln 2 (( 2 , V 2 ) we
find that
ln 2 (( 2 , V 2 ) ln 2 (( 2m , V 2m ) − β(( 1 − 1m ) − ζ(V 1 − V 1m ) + · · · .
Exponentiation of both sides of this equation gives
2 (( 2 , V 2 ) 2 (( 2m , V 2m )e
ββ 2m +ζ V 2m e
−ββ 1 −ζ V 1 .
(3.4.6)
Upon employing Eq. (3.4.6) for 2 (( 2 , V 2 ) in expression (3.4.1b) for (, V ),
we obtain
((, V )
V
0
dV 1
0
d 1 1 (( 1 , V 1 )) 2 (( 2m , V 2m )e
ββ 2m +ζ V 2m e
−ββ 1 −ζ V 1
= 2 (( 2m , V 2m )e
ββ 2m +ζ V 2m
V
0
dV 1
0
d 1 1 (( 1 , V 1 )e
−ββ 1 −ζ V 1 .
Because of the exponentials in 1 and V 1 , we may extend both upper integration
limits to infinity, to give
((, V ) = 2 (( 2m , V 2m )e
ββ 1m +ζ V 1m
∞
0
∞
0
1 (( 1 , V 1 )e
−ββ 1 −ζ V 1 d 1 dV 1 ,
(3.4.7)