[gmx-users] PME constant shifts. (Sergio Perez)
ggroenh at gwdg.de
Fri Mar 29 13:57:40 CET 2019
If you're using PME, you need to also include the reciprocal contribution.
Dear gmx comunity,
I am trying to calculate the electrostatic interaction of my system for
force field development. My system consists of 7 positively charged
particles interacting (system A) with another positively charged particle
(system B). I calculate the electrostatic interaction by doing reruns of
the trajectory in which some of them have been removed.
And to my surprise I get negative energies. I know the PME method has its
"With the Verlet cut-off scheme, the PME direct space potential is shifted
by a constant such that
the potential is zero at the cut-off. This shift is small and since the net
system charge is close to
zero, the total shift is very small, unlike in the case of the
Lennard-Jones potential where all shifts
add up. We apply the shift anyhow, such that the potential is the exact
integral of the force."
Since for the system with just one charge I get a possitive E_1q(coul-LR),
I imagine the LR-coul term is shifted (eventhough I have to use the group
cut-off scheme and not the verlet). The problem is why don't all these
shifts cancel? Is there a way to not add the shifts?
I leave bellow the .mdp parameters. Note that for the VDW part I need to
use user-tables and therefore I need to use the group scheme.
Thank you very much!
; neighbour search
cutoff-scheme = group
nst-list = 10
verlet-buffer-tolerance = 0.005
ns-type = grid
pbc = xyz ; xyz or xy
rvdw = 1.4
rlist = 1.4
vdwtype = user ; PME or User to look for table /user
DispCorr = No ; corrections to energy and pressure or No
coulombtype = PME ; User if look for table
rcoulomb = 1.4
pme-order = 4
fourierspacing = 0.2
ewald-geometry = 3d ; 3d or 3dc
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