Charged solute, thermodynamic integration, couple-intramol = yes

GROMACS version: Any version
GROMACS modification: No

Dear everyone,

Given that I am annihilating charges and van der waals of a charged solute in a solvent using thermodynamic integration (TI) and couple-intramol = yes, I will have to compute the free energy contribute of the solute in vacuum so to subtract it to the free energy difference of desolvation computed that way. The TI is carried out with a certain value for rlist, rcoulomb, rvdw , a neutralized box (addition of counterions) and PME.

Now, the run in vacuum of the solute is the problem. Knowing that gromacs puts a background charge in case of net charge of the simulation box ( DOI: 10.1021/ct400626b ), I have to avoid this so to not bias this contribute. As a consequence, I varied the “three rs” abovementioned to a value that is larger than the maximum distance between two atoms of the solute (I will call this “conditionA”), and turned off PME with hard cutoff, as this way the discontinuity in the force at that point will never be experienced. I couldn’t do differently, since without conditionA I would either have to use PME and get the bias referenced by the DOI or, without PME, completely miss the long range interactions.

A very subtle detail is in the fact that there is a different potential shift for both charges and VdWs in using a different cutoff range. Anyway, in TI, the derivative of the energy with respect to lambda should not suffer the addition of a constant to the potential to be derived. Is everything said so far correct? I would then proceed more confidently.

I have a second question. The majority of publications are oriented to DeltaG, but in condensed phase that is similar to DeltaA. What about TI on a way more simple whole-NVT equilibration + production: would you consider the DeltaA coming from it comparable to the corresponding DeltaG? It’s not just a matter of A and G being different in the theory, but also that the statistical sampling is different since we’re not in the thermodynamic limit - hence NVT and NPT are non-equivalent ensembles - given the inevitably limited sampling.

Thank you for your time and your interest, wish you a great day.

Sincerely,
Jacopo