About a meaning of epsilon and sigma in units lj for DPD

Dear LAMMPS users,

I have one question about a meaning of epsilon and sigma in units lj for DPD calculation. In the ordinary DPD method, all units are non-dimentionalized by kBT and rc=1, but in the units lj in LAMMPS, all units are non-dimentionalized by epsilon and sigma. Could you tell me the meaning of epsilon and sigma in units lj for DPD calculation? When I use units lj for DPD calculation. Are those equal to kBT and rc?

In general, time of DPD is expressed by tDPD = rc/(kBT/m)^1/2, but in the units lj, it’s different. Time is reduced by epsilon, sigma and m. So, how can I compare between tDPD and time which expressed by units lj in LAMMPS?

Best,

Hironori Sakai

Dear LAMMPS users,

I have one question about a meaning of epsilon and sigma in units lj for
DPD calculation. In the ordinary DPD method, all units are
non-dimentionalized by kBT and rc=1, but in the units lj in LAMMPS, all
units are non-dimentionalized by epsilon and sigma. Could you tell me the
meaning of epsilon and sigma in units lj for DPD calculation? When I use
units lj for DPD calculation. Are those equal to kBT and rc?

In general, time of DPD is expressed by tDPD = rc/(kBT/m)^1/2, but in the
units lj, it's different. Time is reduced by epsilon, sigma and m. So, how
can I compare between tDPD and time which expressed by units lj in LAMMPS?

as for the relation between sigma and rc. that is up to you. if you set rc
to be 1.0 then they are the same. if you set it to something like 2.5, then
they are not, but the relation is obvious.

for epsilon you can ​just look at the documentation of the units command​.
at temperature T*=1.0 you have epsilon = kBT

axel.