herz wrote:
Hi,

I have done exactly what you suggest in your first paragraph, seperating
translation and rotation for each water molecule.
If you want I can send you the new ekin_vs_t_and_erot_vs_t.dat where you
can see that the ekin is actually equidistributed
but the erot isn't!

So in this rigid body perspective each water molecule has 6 dof (3
translat 3 rot) and the 3 rot dof are not
equipartitioned! Now the rigid body is pretty much the text book example
for the equipartition theorem.

So you are right in so far as that the first measurements of ekin we did
while ago only are insufficient for the rigid body.
Now we measure ekin and erot seperatly for all 3 components per rigid
body and they are still not equipartitioned!

Have you tested the equipartitioning per atom? This must be correct.

To make water move as a rigid body one would probably need to determine the force and torque on the center of mass and use that to integrate the equations of motion. A quaternion description of rigid water would do this I presume. Did you analyze the EkRot around the center of mass of the molecule? Whether or not a quaternion is equivalent to a constrained atomic system I do not know, but Berk's answer below seems to indicate it is not.

Alex

Berk Hess schrieb:
If you would consider a water molecule a generic rigid body,
you would have to calculate the kinetic energy in a different way,
you would have to separate the rotational and translational kinetic energy.

I'll try to explain it with an example:
say you have a water molecule with all three atoms in the x-y plane.
Non-constrained there will be 3*2=6 degrees of freedom in the x-y plane.
If you constrain the two bonds and the angle you remove 3 degrees
of freedom in the x-y plane. You are left with 3: x and y translation
of the center of mass and rotation in the x-y plane.
Out of this plane (the z direction) you still have 3 degrees of freedom.
So in z you have the full Ekin, whereas in x and y you each have half
the degrees of freedom and half the Ekin of the unconstrained water.

Berk

Date: Mon, 14 Sep 2009 19:58:20 +0200
From: alexander.h...@mytum.de
To: gmx-users@gromacs.org
Subject: Re: [gmx-users] non isotropic kinetic energy

Hey,

I thought a little more about this. Eventually, the water molecules are
rigid bodies.
The rigidity constraint is implemented via constraints (settle or so)
but could also (in theory) be implemented
using some euler angles or what ever scheme you prefer. It should not
matter how you implement
the rigidity, the result of a rigid body in some potential should be
identical (if implemented correctly).

The equipartition theorem must hold for a rigid body (it has 3
translational and 3 (for water) rotational degrees
of freedom which must contribute 1/2*k_b*T per molecule, irrespective
how you implement rigidity.
Also I wouldn't see how you could explain the energy difference if the
molecules were actually implemented
as rigid bodies with e.g. euler angles. Why should the euler angles care
about the water orientation at the interface boundary?

I ran some more analysis assuming the water molecules are rigid bodies.
This shows that the translational energy of the molecules is actually
correct (so 1/2k T) but the average rotational energy is incorrect (with
settle the z part is too high and x/y ok,with lincs x/y is too low and z
too high).

Summarizing, I fail to see why choosing one method over the other for
performing rigid body dynamics should have an influence over
equipartition of energy in the system. I'm still not convinced the
ensemble is correct.

Alex


Berk Hess schrieb:
Equipartitioning still holds.
The issue here is that constraints remove degrees of freedom.
If you remove a degree of freedom, there is not potential or kinetic
energy
anymore in that generalized coordinate.
This does not introduce any problems with ensembles or something like
that.

I did not have time yet to implement pull code with pbc.
I will do it this week.

Berk

Date: Mon, 14 Sep 2009 16:08:55 +0200
From: alexander.h...@mytum.de
To: gmx-users@gromacs.org
Subject: Re: [gmx-users] non isotropic kinetic energy

Hm,

apparently there are constraint algorithms which don't change the
ensemble
(e.g.
http://www.informaworld.com/smpp/content~db=all~content=a750934359)
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