1549380323-Statistical Mechanics Theory and Molecular Simulation

Pressure tensor estimator 235

H=

∑

i

∑

α,β

1

mi

πi,αh−αβ^1 pi,β−L=

∑

i

∑

α,β,γ

1

mi

πi,αh−αβ^1 πi,γh−γβ^1 −L. (5.7.14)

Since the kinetic energy term inLis just 1/2 of the first term in eqn. (5.7.14), the
Hamiltonian becomes

H=

∑

i

∑

α,β,γ

πi,απi,γh−αβ^1 h−γβ^1

2 mi

+U(hs 1 ,...,hsN). (5.7.15)

The pressure tensor in the canonical ensemble is given by

Pαβ(int)=

1

Q(N,h,T)

kT

det(h)

∑

γ

hβγ

∂Q(N,h,T)

∂hαγ

=

kT

det(h)

1

Q(N,h,T)

∫

dNπdNs

∑

γ

hβγ

(

−β

∂H

∂hαγ

)

e−βH

=−

〈

1

det(h)

∑

γ

hβγ

∂H

∂hαγ

〉

. (5.7.16)

Eqn. (5.7.16) requires the derivative of the Hamiltonian with respectto an arbitrary
element ofh. This derivative must be obtained from eqn. (5.7.15), which requires
more index bookkeeping. Let us first rewrite the Hamiltonian using a different set of
summation indices:

H=

∑

i

∑

μ,ν,λ

πi,μπi,νh−μλ^1 h−νλ^1

2 mi

+U(hs 1 ,...,hsN). (5.7.17)

Computing the derivative with respect tohαγ, we obtain

∂H

∂hαγ

=

∑

i

∑

μ,ν,λ

πi,μπi,ν

2 mi

(

∂h−μλ^1

∂hαγ

h−νλ^1 +h−μλ^1

∂h−νλ^1

∂hαγ

)

+

∂

∂hαγ

U(hs 1 ,...,hsN). (5.7.18)

In order to proceed, we will derive an identity for the derivative of the inverse of a
matrix M(λ) with respect to an arbitrary parameterλ. We start by differentiating the
matrix identity
M(λ)M−^1 (λ) = I (5.7.19)

with respect toλto obtain

dM

dλ

M−^1 + M

dM−^1

dλ

= 0. (5.7.20)

Solving eqn. (5.7.20) for dM−^1 /dλyields

dM−^1

dλ

=−M−^1

dM

dλ

M−^1. (5.7.21)

Applying eqn. (5.7.21) to eqn. (5.7.18), we obtain