# [lammps-users] Coulomb Potential

LAMMPS community,

pair_style coul/cut specifies that the Coulomb potential is:
E = (C/dielectric) * (q_i*q_j)/r_ij
where C is an energy-conversion constant, etc.

My simulation uses units lj, therefore, from the manual I believe that the default dielectric equals 1. However, I am unclear what “C” equals. My question is what is the exact form of the Coulomb potential that LAMMPS uses? Also, for future use, can you please inform me how to find this on my own - I’m sure it’s somewhere in the source files.

The reason I ask is that my simulation results indicate that I may be off by a factor of 4pi or eps_0 or some combination thereof. If the energy-conversion constant C is not equal to unity, which I’ve been assuming, then I should be able to correct my problem.

Brian

brian,

LAMMPS community,
pair_style coul/cut specifies that the Coulomb potential is:
E = (C/dielectric) * (q_i*q_j)/r_ij
where C is an energy-conversion constant, etc.
My simulation uses units lj, therefore, from the manual I believe that the
default dielectric equals 1. However, I am unclear what "C" equals. My
question is what is the exact form of the Coulomb potential that LAMMPS
uses? Also, for future use, can you please inform me how to find this on my
own - I'm sure it's somewhere in the source files.
The reason I ask is that my simulation results indicate that I may be off by
a factor of 4pi or eps_0 or some combination thereof. If the
energy-conversion constant C is not equal to unity, which I've been
assuming, then I should be able to correct my problem.

you need a unit dependend conversion factor.
there are actually quite a few of them.
those are all set in Update::set_units() starting
at line 99 of update.cpp.

cheers,
axel.

The C factor is 1 for LJ units (see update.cpp as Axel said, for qqr2e = 1).
If this is not consistent with what the units doc page says for the
units of charge and energy for LJ units then let me know. I'm fuzzy
on remembering what reduced charge means in LJ units.

Steve

Forgot to respond a while back - Thanks for the help this is exactly what I needed.

-B