Re: [AMBER] why is TAUTP scaled by "e" ?

From: Jose Borreguero <borreguero.gmail.com>
Date: Fri, 16 Oct 2009 15:13:42 -0400

I was fitting the time evolution of the temperature to tempi+(temp0-tempi)*(
1-exp(-t/tautp) ). I'm using SHAKE, so this fitting is definitely wrong.
So, it was by pure coincidende that the ratio of effective and input tautp's
was close to "e".
Thanks for the Langeving tip

On Fri, Oct 16, 2009 at 9:30 AM, case <case.biomaps.rutgers.edu> wrote:

> On Thu, Oct 15, 2009, Jose Borreguero wrote:
> >
> > >From my simulations, it seems that the time evolution of the temperature
> > upon heating the system from 300K to 550K with TAUTP=5000psec (very slow
> > heating) in the input file results in the system being heated with an
> > effective TAUTP=13570=e*5000
>
> I suspect it has nothing to do with "e". The Berendsen procedure changes
> the kinetic energy, which is then partitioned into potential energy and
> kinetic energy parts. For a harmonic system, changes in <V> are equal
> to those in <K.E>; so, other things being equal, the rate of temperature
> change will be about two times slower than the value of tautp (since half
> of
> the added kinetic energy will get siphoned off to changes in V). But a
> water system (with all water bonds held rigid by SHAKE) will be rather
> anharmonic, with a different partitioning between kinetic and potential
> energies.
>
> I vaguely recall doing tests 20 years ago on a very harmonic system, which
> did
> have a factor of about 2, but I don't recall what "about" meant in this
> context in quantitative terms.
>
> And, as Ross points out, use of the Berendsen thermostat is not generally
> recommended.
>
> ...dac
>
>
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Received on Fri Oct 16 2009 - 12:30:03 PDT
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