van der Waals equation 171We can equate the two expressions forA(1)by further expanding the natural log
in eqn. (4.7.11) using
ln(1 +x) =∑∞
k=1(−1)k−^1xk
k. (4.7.13)
Substituting eqn. (4.7.13) into eqn. (4.7.11) gives
A(1)(N,V,T) =−
1
βln〈e−βU^1 〉 0=−
1
βln(
1 +
∑∞
l=1(−β)l
l!〈U 1 l〉 0)
=−
1
β∑∞
k=1(−1)k−^11
k(∞
∑
l=1(−β)l
l!〈U 1 l〉 0)k. (4.7.14)
Equating eqn. (4.7.14) to eqn. (4.7.12) and canceling an overall factor of 1/βgives
∑∞k=1(−1)k−^11
k(∞
∑
l=1(−β)l
l!〈U 1 l〉 0)k
=∑∞
k=1(−β)kωk
k!. (4.7.15)
In order to determine the unknown coefficientsωk, we equate like powers ofβon
both sides. Note that this will yield an expansion in powers such as〈U 1 〉kand〈U 1 k〉,
consistent with the perturbative approach we have been following.To see how the
expansion develops, consider working to first order only and equating theβ^1 terms
on both sides. On the right side, theβ^1 term is simply−βω 1 /1!. On the left side, the
term withl= 1,k= 1 is of orderβ^1 and is−β〈U 1 〉 0 /1!. Thus, equating these two
expressions allows us to determineω 1 :
ω 1 =〈U 1 〉 0. (4.7.16)The coefficientω 2 can be determined by equating terms on both sides proportional
toβ^2. On the right side, this term isβ^2 ω 2 /2!. On the left side, thel= 1,k= 2 and
l= 2,k= 1 terms both contribute, giving
β^2
2(
〈U 12 〉 0 −〈U 1 〉^20
)
.
By equating the two expressions, we find that
ω 2 =〈U 1 〉^20 −〈U 12 〉 0 =〈
(U 1 −〈U 1 〉 0 )^2
〉
0. (4.7.17)
Interestingly,ω 2 is related to the fluctuation inU 1 in the unperturbed ensemble. This
procedure can be repeated to generate as many orders in the expansion as desired. At
third order, for example, the reader should verify thatω 3 is given by
ω 3 =〈U 13 〉 0 − 3 〈U 1 〉 0 〈U 12 〉 0 + 2〈U 1 〉^30. (4.7.18)