# Re: AMBER: Question about TI integral method

From: David A. Case <case.scripps.edu>
Date: Sun, 23 Jan 2005 22:41:26 -0800

On Sun, Jan 23, 2005, tanc.uci.edu wrote:

> I performed an almost the same perturbation
> simulation with yours used 80 windows (both in water and vacuum),

Thanks for your input and questions. I'll try to answer what I can here.

> why did you
> integrate the dV/d_lam from 0.05 to 0.98 instead of 0 to 1?

I *did* integrate from zero to one, just using the lambda values indicated
in the tutorial. For example, the integral from 0 to 0.1 was approximated
as 0.1 * [<dV/dl> at 0.05]. And so on. I should probably modify the term
"trapezoidal rule", as that may be a bit misleading.

> this question because I performed an almost the same perturbation
> simulation with yours used 80 windows (both in water and vacuum),
> i.e. clambda = 0, 0.0125, ... 0.9875, 1. I totally got 80 values of
> dV/d_lam ( because when clambda = 1, Sander did not print out the
> value of dV/d_lam). Then I integrated them with trapezoidal rule and
> got the result of -5.37kal/mol in water and -3.80kal/mol in vacuum.
> Clearly, my final result (-3.80-(-5.37) = 1.57kal/mol )of vdw
> contribution of toluene's solvation free enery is much smaller then
> yours(2.70kal/mol) and Shirts et al (2.45kal/mol).

The fact that your vacuum result is almost the same as mine (-3.80 vs. -3.74
kcal/mol) is hopeful. But when you look at the values of <dV/dl> in water at
various values of lambda (shown below compared to mine) it is clear that,
although the overall shape of the curves is quite close, your values are
generally more positive than mine, by roughly 1 kcal/mol. And hence your
final result is also more positive than mine, if we use the sign convention of
the tutorial, which is (water-gas), not (gas-water).

This has the "flavor" of a difference in how the long-range vdW correction was
handled, i.e. that maybe your results don't include the long range vdW
correction(?). What version of Amber did you use for these calculations? [Of
course this is quite speculative; there may be other reasons for the
discrepancies.]

> when I picked out the dV/d_lams of the following clambda (0.05,
> 0.15,...0.85,0.9, 0.9125... 0.9625, 0.975) and integrated them with
> trapezoidal rule, I got the result of (-3.80-(-7.02) ) 3.22kal/mol .

When I take your data, and use just the subset of lambda values you give
above, I get an integrated free energy of -5.45 for the water part, which
is quite close to you "full" value of -5.37. So I think maybe you are doing
the integration differently from me (perhaps not going fully from 0 to 1)?
My conclusion is (using your data) that using 14 windows vs. using 80 has
less than 0.1 kcal/mol effect on the final answer.

> My result of charging free energy of toluene is :
> -1.62(vacuum)-(+0.62)(water) = -2.24kal/mol.

This suggests that we may have different charges. I get +1.75(water) and
-0.46 vacuum, for a total (vacuum-water) of -2.18; the total is close to
yours, but the individual components are very different. What ever slight
differences we have, though, seem to cancel in water vs. gas-phase, so that
the overall change is nearly the same; this is not unexpected.

>
> dV/d_lam of 80 clambdas : (for the second leg, removing the vdW terms)
lamda chunhu dac
-------------------------------
> 0.0 47.0559
> 0.05 33.0724 32.33
> 0.10 22.4820
> 0.15 13.5812 13.13
> 0.20 7.4047
> 0.25 1.6973 0.88
> 0.30 -1.4059
> 0.35 -5.3718 -6.75
> 0.40 -9.6055
> 0.45 -13.1140 -11.32
> 0.50 -12.2951
> 0.55 -14.7040 -15.96
> 0.60 -16.9654
> 0.65 -19.0062 -19.25
> 0.70 -19.1727
> 0.75 -20.1404 -21.56
> 0.80 -23.5859
> 0.85 -19.2507 -23.40
> 0.90 -17.2204 -20.55
> 0.925 -15.1130 (-15.08) (dac result is interpolated)
> 0.95 -16.4275 (-13.74) "
> 0.975 -9.6216 (-10.29) "
1.00 0.0 0.0

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David A. Case                     |  e-mail:      case.scripps.edu
Dept. of Molecular Biology, TPC15 |  fax:          +1-858-784-8896
The Scripps Research Institute    |  phone:        +1-858-784-9768