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From: B. Lachele Foley <lfoley.uga.edu>

Date: Fri, 10 Mar 2006 08:47:42 -0700

This is my understanding. Please correct me if I'm wrong.

It isn't analytically zero at lambda=0 for icfe=1 (icfe=2

wasn't available in Amber 8).

The mixing function goes to zero at lambda=1 for higher values

of klambda, but the potential being mixed isn't nicely behaved

-- it diverges at lambda=1. What matters is the combination

of the two functions.

If you look at David Case's example

(http://amber.scripps.edu/tutorial/shirts/index.html), a

favorable combination of cut values and klambda can, indeed,

converge to zero at lambda=1, but they don't necessarily.

I don't think I -need- the DV/DL to go to zero at lambda=1.

Merely being finite is probably ok (if someone who's done more

math on this disagrees, I'd gladly hear their arguments). I

think I just need the value to be finite and to know what the

value is.

:-) lachele

*>I am not sure how this works with icfe=1, but when I set
*

icfe=2, for

*>lambda=0 and lambda=1, the dv/dl values are not printed out.
*

They are

*>analytically zero. When I run with lambda=0.05 or
*

lambda=0.95, I am

*>getting dv/dl values, but they are close to zero. I am doing
*

a dummy to

*>hydrogen and hydrogen to dummy transformations in the TI
*

calculation. So,

*>I am guessing that the same thing happens with icfe=1 when
*

lambda=1,

*>because analytically it is zero.
*

*>
*

*>On Fri, 10 Mar 2006, B. Lachele Foley wrote:
*

*>
*

*>> I've done van der Waals perturbation in Amber 8 and Amber 9.
*

*>> My issue (below) happens in both. I want to know if it can
*

*>> change for Amber 9.
*

*>>
*

*>> The context, here, is disappearing atoms in an explicit box of
*

*>> water by first turning off charges and then by turning off the
*

*>> van der Waals.
*

*>>
*

*>> When I run the charge perturbation, I get DV/DL values printed
*

*>> in the output file for all values of clambda.
*

*>>
*

*>> When I run a vdW perturbation, I get DV/DL values printed in
*

*>> the output file for all values of clambda except 1. There
*

*>> should be a value at 1, even if it's zero. When I say not
*

*>> printed, I mean that "grep DV\/DL blah_1.000.o" yields nada.
*

*>> Do you know why the values aren't printed? Can they be
*

*>> printed? I'm using klambda=6, if that's important.
*

*>>
*

*>> You'd never notice this with quadrature. For my systems,
*

*>> though, having endpoints seems to improve accuracy.
*

*>>
*

*>> :-) Lachele
*

*>> B. Lachele Foley, PhD, '02
*

*>> Assistant Research Scientist
*

*>> Complex Carbohydrate Research Center, UGA
*

*>> 706-542-0263
*

*>> lfoley.uga.edu
*

*>>
*

*>>
*

*>
*

*>--
*

*> Ilyas Yildirim
*

*> ---------------------------------------------------------------
*

*> - Department of Chemisty - -
*

*> - University of Rochester - -
*

*> - Hutchison Hall, # B10 - -
*

*> - Rochester, NY 14627-0216 - Ph.:(585) 275 67 66 (Office) -
*

*> - http://www.pas.rochester.edu/~yildirim/ -
*

*> ---------------------------------------------------------------
*

*>
*

B. Lachele Foley, PhD, '02

Assistant Research Scientist

Complex Carbohydrate Research Center, UGA

706-542-0263

lfoley.uga.edu

Received on Wed Apr 05 2006 - 23:49:40 PDT

Date: Fri, 10 Mar 2006 08:47:42 -0700

This is my understanding. Please correct me if I'm wrong.

It isn't analytically zero at lambda=0 for icfe=1 (icfe=2

wasn't available in Amber 8).

The mixing function goes to zero at lambda=1 for higher values

of klambda, but the potential being mixed isn't nicely behaved

-- it diverges at lambda=1. What matters is the combination

of the two functions.

If you look at David Case's example

(http://amber.scripps.edu/tutorial/shirts/index.html), a

favorable combination of cut values and klambda can, indeed,

converge to zero at lambda=1, but they don't necessarily.

I don't think I -need- the DV/DL to go to zero at lambda=1.

Merely being finite is probably ok (if someone who's done more

math on this disagrees, I'd gladly hear their arguments). I

think I just need the value to be finite and to know what the

value is.

:-) lachele

icfe=2, for

They are

lambda=0.95, I am

a dummy to

calculation. So,

lambda=1,

B. Lachele Foley, PhD, '02

Assistant Research Scientist

Complex Carbohydrate Research Center, UGA

706-542-0263

lfoley.uga.edu

Received on Wed Apr 05 2006 - 23:49:40 PDT

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