[gmx-users] Setting attractive part (C6 parameter) zero in non-bonded parameters
Suman Chakrabarty
suman at sscu.iisc.ernet.in
Sat Jul 20 14:48:19 CEST 2013
Hello,
Following the Section 5.3.2 of the manual, I have tried to set the C6
term of the non-bonded (Lennard-Jones) interaction to zero as follows:
"When sigma and epsilon need to be supplied (rules 2 and 3), it
would seem it is impossible to have a non-zero C12 combined with a
zero C6 parameter. However, providing a negative sigma will do
exactly that, such that C6 is set to zero and C12 is calculated
normally. This situation represents a special case in reading the
value of sigma, and nothing more."
Unfortunately when I try this the potential energy of the system becomes "nan":
Step Time Lambda
0 0.00000 0.00000
Energies (kJ/mol)
LJ (SR) Coulomb (SR) Coul. recip. Potential Kinetic En.
nan -1.76446e+05 -1.26632e+03 nan 2.30786e+04
Total Energy Conserved En. Temperature Pressure (bar)
nan nan 2.25008e+02 nan
With the positive sign for sigma in the topology file, the same
structure gives the following energetics:
Energies (kJ/mol)
LJ (SR) Coulomb (SR) Coul. recip. Potential Kinetic En.
4.56607e+04 -1.76446e+05 -1.26632e+03 -1.32051e+05 2.30777e+04
Total Energy Conserved En. Temperature Pressure (bar)
-1.08974e+05 -1.08974e+05 2.24999e+02 3.52545e+04
=========================================================
Here I am trying to look at a single solute atom (I change the sign of
the sigma value for this atom only) in a box of TIP4P water.
Strangely, I encounter this "nan" problem only for comb-rule=3.
Whereas, for comb-rule=2, energetics are reasonable, but it appears
that sigma_ij between the solute and solvent becomes very small (as
evident for rdf of water oxygen atoms around this solute atom). This
would imply that the negative value of sigma_i for the solute is being
retained as it is, so in 0.5*(sigma_i + sigma_j) they cancel each
other to large extent.
It seems there is a bug in the treatment of the negative sigma values
as described in the manual. Please let me know if you think I am
making a mistake somewhere. Thanks.
Best regards,
Suman.
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