[gmx-users] simulation in vacuum

Xavier Periole X.Periole at rug.nl
Wed Sep 17 18:43:52 CEST 2008

Actually in vaccum the RF is meaningless ... use cutoff.

On Wed, 17 Sep 2008 18:49:38 +0200
  Thomas Schlesier <schlesi at uni-mainz.de> wrote:
>             >>/ Thank you.           > />/ But i have one last question: > 
>/>/For epsilon_rf i use the relative permittivity of the medium. I simulate 
>>in > />/vacuum so epsilon_rf would be 1? > /GROMOS ff is not parameterized 
>for vaccum simulations of the b-something > version ... have look at the 
>paper. > > > So i should use ffG43b1 instead of the other GROMOS96 force 
>fields?     That would be the right choice for the GROMOS43 ff series.     > 
>In the GROMACS manual i found that the cut-off-distance must at least be 1.4 
>>nm for the GROMOS96 force field.     Never believe what is written in a 
>manual :)) It is always better to keep the values that are used for 
>parameterization (the one I gave you) when using a force field. The idea that 
>bigger = better does not work ... you change the balance of forces and thus 
>the properties of the force field.     > So can i use: type = cut-off; rlist 
>= 1.4 and r_x = 1.7? > (the problem is that the only reference i find for 
>ffG43b1 is the GROMOS >manual, and that's not free avaible)     Any paper 
>from van Gunsteren group or Alan Mark groujp (GROMOS developers) would 
>describe the simulation setup: no need of the manual for this.     > Thanks 
>for an answer > Thomas > > _______________________________________________ > 
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>http://www.gromacs.org/mailing_lists/users.php  I looked in some older papers 
>but found no values for epsilon_r and _rf. From older posts of this mailing 
>list I got the impression that epsilon_r = 1 and epsilon_rf the value of the 
>relative permitivity of the medium. In the case of a simulation in vacuum 
>epsilon_rf would be also 1 and then I have no correction from the reaction 
>field, also grompp tells me that epsilon_r = epsilon_rf would be meaningless 
>with a reaction field. So I'm puzzeld.
> If both epsilon values have the same value, is it then equal to normal 
>cut-offs (type = cut-off)?
> Sorry for the many questions.
> Thomas

XAvier Periole - PhD

Molecular Dynamics Group / NMR and Computation
University of Groningen
The Netherlands

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