Hello
I've previously posted questions about my efforts to calculate
absolute dG of benzene to T4-lysozyme using TI in Amber 10. I will
follow the work by Roux, as cited in the Amber manual (JCTC 2 p1255).
This means that I want to add a restraint to my benzene molecule, but
I would also like to do an energy decomposition. The restraint energy
should not be included in the dv/dl energies so I set dvdl_norest to
1. Now I have run a few lambdas on the van der Waals-part, i.e.
perturbing an uncharged benzene to nothing but I'm confused about the
output of sander. Here is a part of the output for one of the lambdas
DV/DL, AVERAGES OVER 100000 STEPS
NSTEP = 100000 TIME(PS) = 320.000 TEMP(K) = -0.17 PRESS = 0.0
Etot = -8.9561 EKtot = -8.9561 EPtot = -22.2126
BOND = 0.0000 ANGLE = 0.0000 DIHED = 0.0000
1-4 NB = 0.0000 1-4 EEL = 0.0000 VDWAALS = 8.8226
EELEC = 0.0000 EHBOND = 0.0000 RESTRAINT = -31.0353
EAMBER (non-restraint) = 8.8226
DV/DL = -53.2479
Ewald error estimate: 0.3153E-14
------------------------------------------------------------------------------
CHECK DECOMP - TOTAL ENERGIES (w/ REST)
INTERNAL= 0.0000
VDWAALS = 8.8135 EEL = 0.000
In a usual TI run, without restraint the EPTot is equal to DVDL, as
EPTot = V1-V0. But EPtot is also, in this case, VDWAALS+RESTRAINT as
it should be. But then I see that DV/DL, in this case is
EPtot+RESTRAINT=VDWAALS+2*RESTRAINT, although I do NOT want to include
the restraint energy in the dv/dl energy. In my book, if the only
thing I'm perturbing is the van der Waals-energy, the dv/dl energy
should be equal to the van der Waals-energy, i.e. ~8 kcal in this
case. Am I wrong?
And then about the decomposition, if I understand it correctly, the
total VDWAALS part in the decomposition should add up to the VDWAALS
energy in the DV/DL summary. But here I see a small difference on the
second decimal. Is this just an numerical issue, or I'm wrong in my
interpretation?
Best regards,
Samuel
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Received on Fri Feb 19 2010 - 01:30:03 PST
