Hi, Alex,

Thanks so much for your reply.

I means, for each kind of solvent, it has dielectric constant,e.g. 80.4 for water at 20oC.

I set real as unit in my simulation, and i like to set the dielectric constant of water in my simulation as 80.4.

Does that mean i just set dielectric as 80.4?

Thanks.

Houyang

If you look at the Coulombic energy formula on this doc page:

doc/pair_lj.html

the dielectric constant as used in LAMMPS is just a unitless

scale factor on that term, 1 by default. The dielectric command

lets you change the scale factor to another (unitless) value.

I don't know what that means precisely in terms of physical

permittivities, like 80 for water, esp since those values are

temperature-dependent. But I think if you use a standard

water force-field, like CHARMM, TIP3P, etc, and the default

value of 1, you will get standard MD water behavior.

Steve

Hello, Steve,

Thanks so much!

I got it!

If we write the eq. (E=C*qi*qj/(epsilon*r)), we should tell other what value does C has,

can you tell me what value does C has here for unit of “real” or “LJ”?

Thanks & Best wishes.

Houyang

C is just the conversion factor to units of energy (Kcal/mole)

when multiplying 2 charges and dividing by distance.

in update.cpp for real units:

C = force->qqr2e = 332.06371;

which is calculated as follows:

E = c (q^2 / r)

c = force->qqr2e = converts q^2/r to energy

q^2/r is e^2/Ang

eng is Kcal/mole

q^2/r (cgs) = q^2/r (E8 Ang / cm) (1.60219E-19 Coul / e)^2

(statCoul / 3.3356E-10 Coul)^2

E (cgs) = E (4.184E10 erg / Kcal) (mole / 6.022E23 atoms)

c = 332.0697

Steve

Hello, Steve,

I got it.

Steve, Alex, thanks so much again!

Houyang