Spatial distribution functions 167β
3 V〈N
∑
i=1ri·Fi〉
=
β
6 V〈
∑
i,j,i 6 =jrij·fij〉
=
β
6 V Z∫
dr 1 ···drN
∑
i,j,i 6 =jrij·fij
e−βUpair(r^1 ,...,rN). (4.6.64)As we saw in the derivation of the internal energy, all of the integrals can be made
identical by changing the particle labels. Hence,
β
3 V〈N
∑
i=1ri·Fi〉
=
βN(N−1)
6 V Z∫
dr 1 ···drNr 12 ·f 12 e−βUpair(r^1 ,...,rN)=
β
6 V∫
dr 1 dr 2 r 12 ·f 12[
N(N−1)
Z
∫
dr 3 ···drNe−βUpair(r^1 ,...,rN)]
=
β
6 V∫
dr 1 dr 2 r 12 ·f 12 ρ(2)(r 1 ,r 2 )=
βN^2
6 V^3∫
dr 1 dr 2 r 12 ·f 12 g(2)(r 1 ,r 2 ). (4.6.65)Moreover,
f 12 =−∂Upair
∂r 12=−u′(|r 1 −r 2 |)(r 1 −r 2 )
|r 1 −r 2 |=−u′(r 12 )r 12
r 12, (4.6.66)
whereu′(r) = du/dr, andr 12 =|r 12 |. Substituting eqn. (4.6.66) into the ensemble
average gives
β
3 V〈N
∑
i=1ri·Fi〉
=−
βN^2
6 V^3∫
dr 1 dr 2 u′(r 12 )r 12 g(2)(r 1 ,r 2 ). (4.6.67)As was done for the average energy, we change variables using eqn. (4.6.14), which
yields
β
3 V〈N
∑
i=1ri·Fi〉
=−
βN^2
6 V^3∫
drdRu′(r)rg ̃(2)(r,R)=−
βN^2
6 V^2∫
dru′(r)r ̃g(r)=−
βN^2
6 V^2∫∞
0dr 4 πr^3 u′(r)g(r). (4.6.68)Therefore, the pressure becomes