1549380323-Statistical Mechanics Theory and Molecular Simulation

Thermodynamics 461

E=−

∂

∂β

lnQ(L,T) =

1

Q(L,T)

∂Q(L,T)

∂β

. (12.3.18)

SinceQ(L,T) is expressible using only cyclic paths, these are all we need to calculate
E. Taking the derivative of eqn. (12.2.26) with respect toβ, we obtain the following
expression for the energy:

E=

1

Q(L,T)

lim

P→∞

(

mP

2 πβ ̄h^2

)P/ 2 ∫

dx 1 ···dxPεP(x 1 ,...,xP)

×exp

{

−

1

̄h

∑P

k=1

[

mP

2 β ̄h

(xk+1−xk)^2 +

β ̄h

P

U(xk)

]}∣∣

∣

∣

∣

xP+1=x 1

= lim

P→∞

〈εP(x 1 ,...,xP)〉f, (12.3.19)

where

εP(x 1 ,...,xP) =

P

2 β

−

∑P

k=1

mP

2 β^2 ̄h^2

(xk+1−xk)^2 +

1

P

∑P

k=1

U(xk). (12.3.20)

Therefore,εP(x 1 ,...,xP) is an estimator for the energy, and the average〈Hˆ〉P =
〈ε(x 1 ,...,xP)〉fconverges to the true thermodynamic energyEin the limitP→∞.
Similarly, we can obtain an estimator for the one-dimensional “pressure,” which
we will denote Π, from the thermodynamic relation

Π =kT

∂lnQ

∂L

=

kT

Q

∂Q

∂L

. (12.3.21)

As was done in Section 4.6.3, the one-dimensional “volume”Lis made explicit by
introducing scaled variablessk=xk/Linto the path integral for the partition function,
which yields

Q(L,T) = lim

P→∞

(

mP

2 πβ ̄h^2

)P/ 2

LP

∫

ds 1 ···dsP

×exp

[

−

1

̄h

∑P

k=1

(

mP

2 β ̄h

L^2 (si+1−si)^2 +

β ̄h

P

U(Lsi)

)]∣∣

∣

∣

∣

sP+1=s 1

. (12.3.22)

Eqn. (12.3.22) can now be differentiated with respect toLand transformed back to
the original path variablesx 1 ,...,xPto yield

Π =

1

Q(L,T)

lim

P→∞

(

mP

2 πβ ̄h^2

)P/ 2 ∫

dx 1 ···dxPPP(x 1 ,...,xP)

×exp

{

−

1

̄h

∑P

k=1

[

mP

2 β ̄h

(xk+1−xk)^2 +

β ̄h

P

U(xk)

]}∣∣

∣

∣

∣

xP+1=x 1