[gmx-users] LJ-14 energy
jalemkul at vt.edu
Thu Jul 16 13:50:20 CEST 2015
On 7/16/15 7:08 AM, Ming Tang wrote:
> Dear Justin,
> Thanks. 1. The total Lennar-Jones 12-6 energy of the whole system is the
> addition of LJ-14, LJ-SR and LJ-LR. Analogously, the coulomb energy is the
> addition of coulomb-14, coulomb-SR and coulomb-LR. Is my understanding of
> your explanation right? Besides, in Gromacs, is the potential energy the
> addition of G96Bond, G96Angle, Proper-Dih, Improper-Dih, LJ-14, Coulomb-14,
> LJ-SR, LJ-LR, Coulomb-SR and Coulomb-LR? You are right. That's why I don't
Well, that's the exact definition of an additive functional form, so yes.
> get the LJ-LR and Coulomb-LR options when using g_energy. Suddenly, I
> realised that my understanding of the twin-range approach was totally wrong.
> For cutoff-scheme = verlet, rlist cannot be smaller than rvdw = rcoulomb, and
> When using twin-range method, either coulombtype = cut-off with rcoulomb ≥
> rlist or vdwtype = cut-off with rvdw ≥ rlist. Does this mean that for
> twin-range approach, rlist = rcoulomb = rvdw when cutoff-scheme = verlet?
> However, even if I set vdw-modifier = potential- switch, the rlistlong larger
> than rlist is reset to be equal to rlist by gromacs, and there is not LJ-SR
> option when using g_energy. I am totally lost. Can you help to tell me how
> does twin-range work when cutoff-scheme = verlet? Another question is I used
> the following code to make my cut-off parameters be consistent with those
> used for the triple-range cut-off scheme in the paper defining Gromos 54a7
> force field:
> coulombtype = reaction-field coulomb-modifier = potential-shift
> rcoulomb-switch = 0.8 rcoulomb = 1.4 epsilon_rf = 61
> Currently, I think it is wrong. It seems that rlist = 0.8 is reasonable. But
> rlist should not be smaller than rcoulomb when cutoff-scheme = verlet. How to
> achieve this triple-range cut-off scheme?
I honestly have no idea. I don't use GROMOS force fields any more. With the
group scheme, it is easy, but with Verlet I have no idea what the equivalent
> 2. The pull rate I use is 4e-5/ps. I thought it is small so that the energy
> of water can be treated as constant during the pull process. I forgot to
> consider the water-protein interaction and that my pull rate is much larger
> than the pull rates used in the experiments which is in order of micrometre
> per second and is much more relevant to physiological pulling rate. You are
> right, I cannot treat the trends of the system's energy as those of the
> protein's energy. But how can I get the energy of the protein? Is it feasible
> to set all the atoms of the protein as one energy group, and do rerun to
> achieve this?
The energy of the protein is not a useful quantity. It is a force
field-specific quantity. Energy differences and free energies are valuable.
Justin A. Lemkul, Ph.D.
Ruth L. Kirschstein NRSA Postdoctoral Fellow
Department of Pharmaceutical Sciences
School of Pharmacy
Health Sciences Facility II, Room 629
University of Maryland, Baltimore
20 Penn St.
Baltimore, MD 21201
jalemkul at outerbanks.umaryland.edu | (410) 706-7441
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