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From: Lekpa Duukori <duukori.gmail.com>

Date: Wed, 14 Jul 2010 15:29:18 -0600

Thanks Jason. I have used the quasi-harmonic facility with ptraj but I

wanted to compare that to what I would get from the Schlitter formula.

Lepka

On Wed, Jul 14, 2010 at 1:30 PM, Jason Swails <jason.swails.gmail.com>wrote:

*> Hello,
*

*>
*

*> I believe this corresponds to the quasi-harmonic approximation to the
*

*> entropy, right? I think if you do it correctly, ptraj will calculate the
*

*> entropy for you. This is actually done in MMPBSA.py by setting entropy=1 in
*

*> the input file. You don't have to use that script, but the code will show
*

*> you how to do it.
*

*>
*

*> Good luck!
*

*> Jason
*

*>
*

*> --
*

*> Jason Swails
*

*> Quantum Theory Project,
*

*> University of Florida
*

*> Ph.D. Graduate Student
*

*> 352-392-4032
*

*>
*

*> On Jul 14, 2010, at 12:50 PM, Lekpa Duukori <duukori.gmail.com> wrote:
*

*>
*

*> > Hello,
*

*> >
*

*> > I am having some trouble with some analysis on mass weighted covariance
*

*> > matrix from ptraj
*

*> >
*

*> > I am trying to get entropy estimates using the Schlitter formula (Chem
*

*> Phys
*

*> > Lett, 1993,215,617). I first do an rms fit to the first frame of my data
*

*> > using the following ptraj commands ( I am using AmberTools 1.2)
*

*> >
*

*> > trajin model.mdcrd.nowaters
*

*> > trajout model.mdcrd.fit
*

*> > rms first mass out rms.out.fit
*

*> >
*

*> > Then do
*

*> >
*

*> > trajin model.mdcrd.fit
*

*> > matrix mwcovar out mwcovar.out
*

*> >
*

*> > I then use the following python script to calculate the entropy
*

*> >
*

*> > #!/usr/bin/python
*

*> > import numpy
*

*> > sigma = numpy.loadtxt('mwcovar.out')
*

*> > hbar = 1.054571628e-34 #Joules seconds
*

*> > T = 300.0 # Kelvin
*

*> > Kb = 1.3806504e-23 #Joules/Kelvin
*

*> > Euler = numpy.exp(1)
*

*> > constant = 10e-23 #Angstroms squared multiplied by g --> Kg (10^-3)
*

*> >
*

*> > unit_matrix = numpy.eye(len(sigma[0]),len(sigma[0]),dtype=float)
*

*> >
*

*> > matrix = unit_matrix + ((Kb*T*Euler*Euler)/(hbar*hbar))*sigma*constant
*

*> > Entropy = 0.5*Kb*numpy.log(numpy.linalg.det(matrix))
*

*> >
*

*> > But I get an infinite determinant and a NaN entropy, the elements matrix
*

*> are
*

*> > very large (10^26 or so, see below)
*

*> >
*

*> > I have done some dimensional analysis and think my units should be ok,
*

*> but
*

*> > the values in the matrix indicates otherwise so I don't know exactly what
*

*> is
*

*> > going wrong here. Any comments welcome.
*

*> >
*

*> > Thanks.
*

*> >
*

*> > Lekpa
*

*> >
*

*> > mwcovar matrix
*

*> > [[ 0.75 -0.541 0.269 ..., 0.232 -0.571 0.065]
*

*> > [-0.541 0.894 -0.4 ..., -0.597 0.793 -0.113]
*

*> > [ 0.269 -0.4 1.099 ..., 0.354 -0.537 0.167]
*

*> > ...,
*

*> > [ 0.232 -0.597 0.354 ..., 1.102 -0.755 -0.031]
*

*> > [-0.571 0.793 -0.537 ..., -0.755 1.762 -0.103]
*

*> > [ 0.065 -0.113 0.167 ..., -0.031 -0.103 0.491]]
*

*> >
*

*> >
*

*> > Output of python script
*

*> >
*

*> > [[ 2.06396838e+26 -1.48880919e+26 7.40276658e+25 ...,
*

*> 6.38454218e+25
*

*> > -1.57136792e+26 1.78877259e+25]
*

*> > [ -1.48880919e+26 2.46025031e+26 -1.10078313e+26 ..., -1.64291883e+26
*

*> > 2.18230256e+26 -3.10971236e+25]
*

*> > [ 7.40276658e+25 -1.10078313e+26 3.02440166e+26 ..., 9.74193074e+25
*

*> > -1.47780136e+26 4.59576959e+25]
*

*> > ...,
*

*> > [ 6.38454218e+25 -1.64291883e+26 9.74193074e+25 ..., 3.03265754e+26
*

*> > -2.07772817e+26 -8.53106929e+24]
*

*> > [ -1.57136792e+26 2.18230256e+26 -1.47780136e+26 ..., -2.07772817e+26
*

*> > 4.84894971e+26 -2.83451657e+25]
*

*> > [ 1.78877259e+25 -3.10971236e+25 4.59576959e+25 ..., -8.53106929e+24
*

*> > -2.83451657e+25 1.35121130e+26]]
*

*> > -inf
*

*> > Entropy = nan
*

*> > _______________________________________________
*

*> > AMBER mailing list
*

*> > AMBER.ambermd.org
*

*> > http://lists.ambermd.org/mailman/listinfo/amber
*

*>
*

*> _______________________________________________
*

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*

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*

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*

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*

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Received on Wed Jul 14 2010 - 14:30:03 PDT

Date: Wed, 14 Jul 2010 15:29:18 -0600

Thanks Jason. I have used the quasi-harmonic facility with ptraj but I

wanted to compare that to what I would get from the Schlitter formula.

Lepka

On Wed, Jul 14, 2010 at 1:30 PM, Jason Swails <jason.swails.gmail.com>wrote:

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Received on Wed Jul 14 2010 - 14:30:03 PDT

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