Computational Physics

242 Molecular dynamics simulations

or subtracting a constant to the potential does not alter the forces and hence the
dynamics of the system) leading to the following expression for the total configur-
ational energy for a collection of particles with charge (or mass)qilocated atqi,
i=1,...,N:

U=

∑

R

∑

i<j

qiqj

|ri−rj+R|

−

∑

i<j

qiqj

∑′

R

1

R

. (8.130)

Here

∑

i<jdenotes a sum overiandjrunning from 1 toNwith the restrictioni<j;
furthermore,

∑

Rdenotes a sum over the locationsRof the system replicas, the
prime with the second sum denoting exclusion ofR= 00 0. From now on we shall
restrict ourselves to charge-neutral systems with

∑

iqi=0, for which the second
term in(8.130)vanishes. The system then has a dipole moment and the leading term
in computing the total energy in PBC is the result of the dipole–dipole interactions
between the replicas. Evaluating the sum over the replicas is a difficult problem
even for charge-neutral systems and it will be addressed in the next subsection. In
Section 8.7.2we shall then see how the forces can be evaluated more efficiently
than in the conventional MD approach where we must sum over all pairs.

8.7.1 The periodic Coulomb interaction

The total configurational energy of the charge-neutral system is given by

U=

∑

R

∑

i<j

qiqj

|ri−rj+R|

;

∑

i

qi=0. (8.131)

It is assumed here that the particles are point particles, that is, their charge distri-
bution is given by a delta-functionρi(r) =qiδ(r−ri). In most realistic cases
there will be additional short range interactions preventing the particles from
approaching each other too closely. We now apply a Fourier transform as defined
inEqs. (4.104)–(4.105)to(8.131). We have

1

r

=

∫

d^3 k

( 2 π)^3

4 π

k^2

eik·r. (8.132)

Substituting this into (8.131) and using

∑

R

eik·R=

( 2 π)^3

V

∑

K

δ^3 (k−K) (8.133)

whereVis the volume of the system andKare reciprocal lattice vectors, we obtain

U=

1

V

∑

K
= 000

∑

i<j

eiK·rij

K^2

qiqj. (8.134)