AMBER Archive (2007)Subject: AMBER: Calculating dipole moments
From: Austin B. Yongye (ayongye_at_chem.uga.edu)
Date: Wed Nov 28 2007 - 13:33:41 CST
Dear amber users,
I would like to calculate the dipole moments of a trisaccharide in
explicit water during the first equilibration stage of a thermodynamic
integration calculation.
When I use this input:
Ligand_cap_coul trisaccharide
&cntrl
ntr=0,
nstlim=20000, nscm=2000,
ntx=5, irest=1, ntb=2, ntpr=100, tempi=300.0, ig=974651,
ntp=1, taup=1.0,
dt=0.0001, nrespa=1,
ntt=1, temp0=300.0, tautp=2.0,
ntc=2, ntf=2, tol=0.000001,
ntwr=10000, ntwx=100,
scnb=1.0, scee=1.0, dielc=1, cut=10.0,
icfe=1,
/
&dipoles
RES 1 4
END
END
The computed dipole moment is 122 +/- 32D.
However, when I use this other input for a single gas-phase structure,
Minimization everything
&cntrl
imin=0, dielc=1,ntb=0,
maxcyc=1,
scee=1.0, scnb=1.0, cut=12.0,
drms=0.01,nmropt=0,
&end
&dipoles
RES 1 4
END
END
The dipole moment is 5.5D, which makes sense when compared with the
HF/6-31G* and B3LYP/6-31++G(2d,2p) values of 4.8D and 4.1D, respectively.
Please does anyone have any suggestions as to why the dipole moment from
the TI calculation is inconsistent with the QM values?
Thanks,
Austin-
Austin B. Yongye
Complex Carbohydrate Research Center,
University of Georgia,
315 Riverbend Road,
Athens, GA 30602.
Phone: 706 542 0263
"...then, we learn better in a free spirit of curiosity, than under fear
and compulsion..." -St. Augustine of Hippo
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