[gmx-users] microcanonical

karamyog singh karamyog.singh at gmail.com
Thu May 18 11:14:08 CEST 2006

I know where the minimum is and I said exactly what you did. However I
interpreted the potential becoming 0 as the point before r-min. The
potential once becomes 0 before minimum potential value. I thought you were
referring to that point and that is why i got confused and asked you further

anyway, thanks a lot for your help. Hope I get my desired result. :) Thanks
a lot!

On 5/18/06, Mark Abraham <Mark.Abraham at anu.edu.au> wrote:
> karamyog singh wrote:
> > I placed the atoms randomly means i used random atomic positions, not
> > velocities. The velocities were 0 initially and then I generated
> > velocities at 0.01 K.(1st MD run to get a stable configuration) After
> > that I subjected this structure that I got to NVE.(2nd MD run) Now over
> > here I will try with no velocity generation. Thank you for pointing out
> > this mistake. I should have realised this. One more thing, is generating
> > velocities in the MD run correct?
> It isn't meaningful to have velocities before starting an MD run, so you
> want to generate them at some point before any MD, then equilibrate as
> appropriate.
> > I will try what you are saying. I was doing it till the r-min beacuse at
> > that point the forces become 0.Minimum potential should imply 0 force.
> > So if the forces are zero and I increase my simulation time for 1st MD
> > run, I get a stabler structure i.e. the forces between atoms should be
> > nearly equal to zero. However my forces wil never be zero. Wil I ever
> > get a frozen structure for NVE?
> The minimum of the potential is where -F=dU/dx=0, not where U=0 - after
> all the latter depends where you take your convention of zero energy. A
> normal (e.g. 6,12) LJ potential has a minimum below zero and then
> converges up to it (e.g. Fig 4.1 in gromacs manual). You want your
> cutoff to be large enough that your potential is getting close to zero
> under the standard convention that zero is at infinite separation -
> because the cutoff means that the potential effectively is zero. My
> point is that a cut-off at the minimum gives you a sharp discontinuity
> for your effective potential, which means you are doing something other
> than you want to do.
> Mark
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