232 Isobaric ensembles
=
1
∆(N,P,T)
∫
dh[det(h)]−^2 e−βPdet(h)
kT
det(h)∑^3
γ=1hβγ(
∂Q
∂hαγ)
N,T.(5.6.10)
An integration by parts can be performed as was done in Section 5.4,and, recognizing
that the boundary term vanishes, we obtain
〈Pαβ(int)〉=−
kT
∆(N,P,T)
∫
dh∑^3
γ=1∂
∂hαγ{
[det(h)]−^2 e−βPdet(h)
kT
det(h)hβγ}
Q(N,h,T)=−
kT
∆(N,P,T)∫
dh∑^3
γ=1{
−3[det(h)]−^4
∂det(h)
∂hαγhβγ−βP[det(h)]−^3∂det(h)
∂hαγ
hβγ+ [det(h)]−^3∂hβγ
∂hαγ}
e−βPdet(h)Q(N,h,T).(5.6.11)In order to proceed, we need to know how to calculate the derivative of the deter-
minant of a matrix with respect to one of its elements. The determinant of a matrix
M can be written as det(M) = exp[Tr ln(M)]. Taking the derivative of this expression
with respect to an elementMij, we obtain
∂[det(M)]
∂Mij= eTr ln(M)Tr[
M−^1
∂M
∂Mij]
= det(M)∑
k,lMkl−^1
∂Mlk
∂Mij, (5.6.12)
where the trace has been written out explicitly. The derivative∂Mlk/∂Mij=δilδkj.
Thus, performing the sums overkandlleaves
∂[det(M)]
∂Mij= det(M)Mji−^1. (5.6.13)Applying eqn. (5.6.13) to eqn. (5.6.11), and using the fact that∑
∑ γ∂hβγ/∂hαγ=
γδβαδγγ= 3δβα, it can be seen that the first and last terms in the curly brackets
of eqn. (5.6.11) cancel, leaving
〈Pαβ(int)〉=
kT
∆(N,P,T)∫
dhβPδαβe−βPdet(h)Q(N,h,T) =Pδαβ, (5.6.14)which states that, on the average, the pressure tensor should be diagonal with each
diagonal element equal to the external applied pressureP. This is the generalization
of the pressure virial theorem of eqn. (5.4.4). In a similar manner, the generalization
of the work virial in eqn. (5.4.8) can be shown to be
〈P
(int)
αβ det(h)〉+kTδαβ=P〈det(h)〉δαβ, (5.6.15)according to which the average〈P
(int)
αβ det(h)〉is diagonal.