karamyog.singh at gmail.com
Fri May 19 17:11:45 CEST 2006
I want a constant energy simulation at 0.01 K . Is it possible. I cannot
keep increasing my cut off because for that i have to increase my box size.
I have tried so many times but the temperature increases and neither is the
energy constant. :(
On 5/19/06, karamyog singh <karamyog.singh at gmail.com> wrote:
> I guess since after the first run, the forces between atoms is not
> zero(only negligible), when i do the 2nd run, the atoms start moving under
> the action of these forces and since there is no temperature scaling or
> pressure scaling, we can or might see an increase in KE as in my case. So an
> increase might be correct.
> Am I right? Please answer this because I am an undergraduate student and
> this will clear my misconception, if any. :)
> Thanks in advance.
> On 5/18/06, Mark Abraham <Mark.Abraham at anu.edu.au> wrote:
> > karamyog singh wrote:
> > > Well, I am back. Remember my first and second mdruns. In my second,
> > > after stablizing my initial configuration in the 1st md run, I used
> > cut
> > > off upto 6.3 whereas the potential minima is at 0.309. 6.3 is quite
> > > large a distance compared to 0.309. In the second run, the kinetic
> > > energy increases which is correct but the total energy doesnt seem to
> > > remain constant even when there are 3 million steps involved.
> > Why is increasing KE correct?
> > > How is this constant energy mode? KE increases(accepted) but the total
> > > never seems to be constant. I cant keep increasing my no. of steps.
> > > moreover how can so many vibrations be introduced at a temp. of 0.001K.
> > It's well known (as I said a few emails ago) that cut-offs in NVE will
> > cause system heating - that's why people tend to use NVT if they are
> > using cut-offs. Expected NVE behaviour for this sort of system as a
> > function of cut-off is outside my experience, but you can use g_energy
> > to judge how fast KE and thus total energy is increasing as a function
> > of simulation time. If that rate varies with the cutoff, then I'd say
> > there's the problem.
> > Mark
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