[gmx-users] non isotropic kinetic energy

herz alexander.herz at mytum.de
Tue Sep 15 10:51:05 CEST 2009


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!


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.herz at mytum.de
>> To: gmx-users at 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.herz at mytum.de
>>>> To: gmx-users at 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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