[gmx-users] PME

Michael Brunsteiner mbx0009 at yahoo.com
Wed Apr 6 21:01:37 CEST 2011


You CAN, in fact calculate the contribution of the reciprocal part
of the PME energy to the binding energy between two components in 
a heterogeneous system, its just quite tedious...
say, your system is molecules A and B for which you want to know
the interaction energy, and the rest of the system, typically
the solvent, we call C.
Now your total Reciprocal Coulomb energy will have six parts: 
ER_tot = ER_AA + ER_BB + ER_CC + ER_AB + ER_AC + ER_BC
but these parts are NOT given in the gromacs output as they
cannot be calculated DIRECTLY, you have to calculate
them by setting the charges on A, B, or C (or combinations thereof)
to zero (there is a tool for setting the charges in a tpr file
to zero) and then do more runs with: "mdrun -rerun" based on the 
original trajectory to get the required contributions.

then E_AB = ER_C0 - ER_A0C0 - ER_B0C0

(or something like it, do double check that formula, i can't be bothered
thinking it through now ... here ER_A0C0, for example,  is the reciprocal
part of the coulomb energy with charges in groups A and C set to zero, etc)

this being said ... it's tedious, time-consuming, and error-prone
(you need to use double precision and save a lot of frames to
get reasonably accurate numbers)
You might be better off using reaction field, or PME and simply
ignore the reciprocal part altogether (if your molecules A, B
are NOT charged and have no permanent and large dipole moment
you might get away with the latter)

What Justin said is correct, PME (or any other Ewald-like
method, PPPM, FMA, etc) is standard these days, and for a good reason.
However, different properties are affected to a different
extent by neglecting the long range interactions, and for 
what you want to calculate it might be OK for getting at least
a qualitative answer, as long as you use PME for the actual MD.
(I'd be VERY surprised if everybody who did LIE in the last 10
years went through the trouble outlined above)

have fun!


Elisabeth wrote:
> Hello Justin,
> Several days ago you answered my question about calculating nonbonded 
> terms:
> Question: If I want to look at nonboded interactions only, do I have to 
> add  Coul. recip.  to [ LJ (SR)  + Coulomb (SR) ] ?
> Answer: The PME-related terms contain both solute-solvent, 
> solvent-solvent, and potentially solute-solute terms (depending on the 
> size and nature of the solute), so trying to interpret this term in some 
> pairwise fashion is an exercise in futility.
> my question is if I want to add up nonbonded related terms to get inter 
> molecular energies, do I have to add Coul. recip. or it is already 
> included in Coulomb (SR)? 

They are separate energy terms.  The PME mesh terms is "Coul. recip." and the 
short-range interactions (contained within rcoulomb, calculated by a modified 
switch potential) are "Coulomb (SR)."

> and also, for a A-B system, I have been using energy groups to extract  
> solute-solvent, solvent-solvent, solute-solute terms. Did you mean that 
> applying doing so with PME as electrostatics treatment is not correct?

PME has been consistently shown to be one of the most accurate long-range 
electrostatics methods and is widely used, but in your case is preventing you 
from extracting the quantity you're after (if it can even be reasonably defined 
at all).  Using energygrps will not resolve the problem I described above.  The 
"Coul. recip." term contains long-range energies between (potentially) A-B, A-A, 

and B-B, depending on the nature of what A and B are.  The only terms that are 
decomposed via energygrps are the short-range terms, which are calculated 
pairwise.  Thus, with PME, there is no straightforward way to simply define an 
"intermolecular energy" for a heterogeneous system.  You might be able to define 

such a term for a completely homogeneous system (which also assumes that the 
sampling has converged such that the charge densities etc are uniform...but I'm 
sort of thinking out loud on that), but not one that is a mixture.


> Thanks for your help!
> Best,


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