[gmx-users] Re: Bonds - force constant for given Beads

Thomas Schlesier schlesi at uni-mainz.de
Mon Nov 26 20:23:58 CET 2012

A good start might be:
Phys. Chem. Chem. Phys., 2011, 13, 10437–10448
This paper is about hybrid-models (mixing CG and AA). But they discuss 
'boltzmann inversion' and 'force matching', which are both methods to 
obtain CG-potentials.
Since they use small molecules it focusses on nonbonded interactions, 
but one can probably transfer the methods to bonded interactions.

Am 26.11.2012 18:45, schrieb gmx-users-request at gromacs.org:
> the distance distribution should be given by the Boltzmann factor of the
> potential energy function between the beads, assigning V(r)=0 for the most
> probable distance in your histogram. that's what you get when you take a
> molecule in vacuum and for instance you compare the dihedral distribution
> with the dihedral potential, the distribution is just exp[-U(theta)/RT],
> except maybe for an additive constant.
> you should be aware that the distribution may change appreciably depending
> on the environment, so this approach may be tricky: you may include
> implicit solvent effects on your bonded parameters and then you would end
> up with a forcefield that counts twice the solvent effects on the internal
> structure if you add explicit solvent in your model system.
> I hope it helps.
> Andre
> On Mon, Nov 26, 2012 at 2:06 PM, Steven Neumann<s.neumann08 at gmail.com>wrote:
>> >Dear Gmx Users,
>> >
>> >
>> >I am planning to build coarse grained model based on the all atom
>> >simulation. I created (using VMD) beads representing 2-4 atoms of my
>> >protein chain. I want to extract bonded parameters. The equilibrium
>> >lenght for bonds (between specified beads) would be the average over
>> >the equilibrium from all atom simulation using g_dist between Centre
>> >of Mass of group of atoms belonging to given bead.
>> >
>> >My question: How can I extract the force constant for the bonds from
>> >all atom simulation between those beads?
>> >
>> >Thank you,
>> >
>> >Steven

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