1549380323-Statistical Mechanics Theory and Molecular Simulation

Spatial distribution functions 165

that the volume differentiation cannot be easily performed. The task would be made
considerably simpler if the volume dependence could be moved into theintegrand by
some means. In fact, we can achieve this by a simple change of variables in the integral.
The change of variables we seek should render the limits independentof the box size.
In a cubic box of lengthL, for example, the range of all of the integrals is [0,L], which
suggests that if we introduce new Cartesian coordinates

si=

1

L

ri i= 1,...,N, (4.6.50)

all of the integrals range from 0 to 1. The coordinatess 1 ,...,sN are calledscaled
coordinates. For orthorhombic boxes, the transformation can be generalizedto

si=

1

V^1 /^3

ri, (4.6.51)

whereV is the volume of the box. Performing this change of variables inZ(N,V,T)
yields

Z(N,V,T) =VN

∫

ds 1 ···dsN exp

[

−βU

(

V^1 /^3 s 1 ,...,V^1 /^3 sN

)]

. (4.6.52)

The volume derivative ofZ(N,V,T) may now be easily computed as

∂Z

∂V

=

N

V

Z(N,V,T)

−βVN

∫

ds 1 ···dsN

1

3

V−^2 /^3

[N

∑

i=1

si·

∂U

∂(V^1 /^3 si)

]

e−βU(V

1 / (^3) s 1 ,...,V 1 / (^3) sN)

=

N

V

Z(N,V,T) +

β

3 V

∫

dr 1 ···drN

[N

∑

i=1

ri·Fi

]

e−βU(r^1 ,...,rN), (4.6.53)

where, in the last line, we have transformed back tor 1 ,...,rN. Thus,

1

Z

∂Z

∂V

=

N

V

+

β

3 V

〈N

∑

i=1

ri·Fi

〉

, (4.6.54)

so that the pressure becomes

P=

NkT

V

+

1

3 V

〈N

∑

i=1

ri·Fi

〉

. (4.6.55)

Again, using the fact thatNkT/V = (1/ 3 V)〈

∑N

i=1p

2

i/mi〉, eqn. (4.6.55) becomes

P=

1

3 V

〈N

∑

i=1

[

p^2 i

mi

+ri·Fi

]〉

. (4.6.56)