Hi Berk,
the difference remains when utilizing double precision:
free energy code on with lambda=0.00: Coulomb (SR) = -3.23064e+05
free energy code off: Coulomb (SR) = -3.22653e+05
Here are the full double precision values:
; Free energy control stuff
free_energy = no
Energies (kJ/mol)
Angle Proper Dih. Ryckaert-Bell. LJ-14 Coulomb-14
2.91166e+03 5.90834e+02 1.07820e+03 7.70226e+02 6.22283e+03
LJ (SR) LJ (LR) Disper. corr. Coulomb (SR) Coul. recip.
4.94952e+04 -1.84944e+03 -6.38472e+02 -3.22653e+05 -7.12639e+04
Potential Kinetic En. Total Energy Temperature Pressure (bar)
-3.35336e+05 5.67661e+04 -2.78570e+05 3.00586e+02 6.48848e-01
Cons. rmsd () Cons.2 rmsd ()
0.00000e+00 0.00000e+00
; Free energy control stuff
free_energy = yes
init_lambda = 0.00
delta_lambda = 0
sc_alpha =0.0
sc-power =1.0
sc-sigma = 0.3
Angle Proper Dih. Ryckaert-Bell. LJ-14 Coulomb-14
2.91166e+03 5.90834e+02 1.07820e+03 7.70226e+02 6.22283e+03
LJ (SR) LJ (LR) Disper. corr. Coulomb (SR) Coul. recip.
4.94952e+04 -1.84944e+03 -6.38472e+02 -3.23064e+05 -7.12639e+04
Potential Kinetic En. Total Energy Temperature Pressure (bar)
-3.35747e+05 5.67544e+04 -2.78993e+05 3.00524e+02 -4.14918e+01
dVpot/dlambda dEkin/dlambda dG/dl constr. Cons. rmsd () Cons.2 rmsd ()
1.29815e+03 0.00000e+00 0.00000e+00 0.00000e+00 0.00000e+00
I'll try to develop a simplet system to see if this is reproducible
between systems or is system specific.
I am willing to post my topologies, but what would be most useful at
this point is some assistance confirming if anybody else encounters
this difference.
Thanks,
Chris.
--- original message ---
Hi,
I would also expect the same results.
I have no clue what kind of system you have.
But check that it is not an accuracy issue.
Could you recalculate the energies in double precision.
Berk
Date: Tue, 27 Jan 2009 20:00:23 -0500
From: chris.neale at utoronto.ca
To: gmx-users at gromacs.org
Subject: [gmx-users] free_energy = yes & init_lambda = 0.00 does not
yield identical energies to free_energy = no.
Hello,
when applying the free energy code, I expected that the
instantaneous energy of a given conformation when free_energy = yes
& init_lambda = 0.00 would be identical to that when free_energy =
no. However, I have found this not to be the case in zero-step
mdrun. This is the same result that I get from gmx v 3.3.1, 3.3.3,
and 4.0.3. Is this expected?
With free_energy = yes & init_lambda = 0.00:
Coulomb (SR) -3.23064e+05
With free_energy = no:
Coulomb (SR) -3.22654e+05
and all other energy components are identical. The Coulomb (SR)
difference seems significant to me. Note that the free_energy = no
values are identical to those that I get when I use a topology that
has not been modified to add a B-state so I don't think that my
A-state values are incorrect.
## free_energy related .mdp options:
free_energy = yes
init_lambda = 0.00
delta_lambda = 0
sc_alpha =0.0
sc-power =1.0
sc-sigma = 0.3
## Other .mdp options:
nsteps = 0
tinit = 0
dt = 0.004
integrator = sd
comm_mode = linear
nstcomm = 1
comm_grps = System
nstlog = 2500
nstlist = 5
ns_type = grid
pbc = xyz
coulombtype = PME
rcoulomb = 0.9
fourierspacing = 0.12
pme_order = 4
vdwtype = cut-off
rvdw_switch = 0
rvdw = 1.4
rlist = 0.9
DispCorr = EnerPres
Pcoupl = Berendsen
pcoupltype = isotropic
compressibility = 4.5e-5
ref_p = 1.
tau_p = 4.0
tcoupl = Berendsen
tc_grps = System
tau_t = 0.1 ref_t = 300.
annealing = no
gen_vel = no
unconstrained-start = yes
gen_temp = 300.
gen_seed = 9896
constraints = all-bonds
constraint_algorithm= lincs
lincs-iter = 1
lincs-order = 6
Thank you,
Chris.
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