CN as constraint for free energy calculation

[Marcella Iannuzzi]

Dear Mengzi, 

The variable as you defined it takes the average of the coordination of the 
ATOMS_FROM.
If you want to use the sum of the coordinations of each of the specified 
atoms in ATOMS_FROM, you need to define a COMBINED collective variable, 
where you set the sum of coordinations of single atoms.
Kind regards
Marcella


On Thursday, June 7, 2018 at 10:53:32 AM UTC+2, Mengzi Zhou wrote:
>
> Dear all，
>           Recently I use CN as the constraint for free energy calculation 
> for Mg2+ ---  sulfate anion ----Mg2+ in water. CN is defined as the 
> coordination number of two Mg atoms to four O atoms in SO4(2-). The 
> below is related part in input file
> &COORDINATION
>           ATOMS_FROM 7 25
>           ATOMS_TO 15 16 17 18
>           ND 18
>           NN 10
>           R0 [angstrom] 2.7
> &END COORDINATION
> …………
> &CONSTRAINT
>          &COLLECTIVE
>                     TARGET 1.0
>                     INTERMOLECULAR
>                     COLVAR 1
>          &END COLLECTIVE
>          &LAGRANGE_MUTIPLIERS
>                      FILENAME=force1.0
>                      &EACH
>                              MD 1
>                      &END EACG
>          &END LAGRANGE_MUTIPLIERS
> &END CONSTRAINT
>
> At first, I set TARGET 1.9, because I thought this constraint is for the 
> sum of both Mg, and by a free AIMD run two Mg are both collected to SO4. 
> But I checked the result of TARGET 1.9 and found SO4 structure got 
> distorted largely. It seems that each Mg is under the constraint of 1.9 not 
> the sum of 2 Mg. So I retry with TARGET 1.0. This time, the trajectory is 
> realistic. Both Mg are collected to SO4 just like what happens in free AIMD 
> run. 
> I want to ask when several atoms are in ATOMS_FORM or KINDS_FROM section, 
> whether the TARGET constant is for each one of them or the sum of them? 
> From my tests, it seems the former one. If so, what should I do to make the 
> sum of them be the constraint? or it cannot be done by methodology due to 
> more than one degree of freedom? 
>
> Sincerely,
> Mengzi.
>
>
>
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