Subject: Re: CCL:MD/MM combination

From: Konrad Hinsen <[email-masked]>
Date: Wed, 6 Nov 96 10:28:17 +0100

> Now I have a question.  What "temperature" is this type of MD going to
> simulate?  Temperature normally just reflects the average kinetic energy of
> the system, so it can be calculated from the velocities.  But if part of the
> system is frozen, should those atoms be included in the "average"? 

Strictly speaking, no. Equipartition of energy assigns an equal
amount of kinetic energy to each degree of freedom in the system,
but fixed coordinates are not degrees of freedom.

> Obviously, if all atoms are included, it will add a bunch of zero
> velocities, so the calculated temperature will be lower than what is
> reflected by the movement of molecules actually allowed to move.  But would

If the fixed atoms ever constitute a significant part of your system,
then you are too far away from the thermodynamic limit anyway. So
it shouldn't make a difference in practice how you calculate the
temperature.

-- 
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From: "Dr. Teerakiat Kerdcharoen" <[email-masked]>
Date: Wed, 6 Nov 1996 20:52:51 -0700 (GMT+7)

> the "temperature" calculated only by averaging the moving molecules have any
> more meaning?  I suppose that either way this is just an artificial
> situation, but if there is something that corresponds to temperature, I hope
> any answer to Rochus's question can also answer my question.  Thanks.

Hi!
	This point has confined me in a cycle of confusion for quite a
long time. Traditionally, we calculate "temperature" from the ensemble
average of kinetic energy. The kinetic energy is unambiguously defined
as a square of velocity. Well, we assign thermodynamic temperature with
a factor of boltzmann constant and the total degree of freedom of the
system. Things are clear if we consider system as a system of N atoms
moving only translationally and so there are 3N degrees of freedom.
	Things are more ambiguous when people try to devide motions
into 3 modes, translation (of the centre of mass), rotation (about the
C.O.M.) and vibration. Then they introduce the equipartition theorem 
to cope with this. They calculate translational and rotational energies
separately and they have translational and rotational temperature
that are believed to be equal due to the equipartition theorem (frequently
they are much different in some systems that two modes of motion are
not coupled then the macroscopic temperature is calculated by
averaging translational and rotational degrees of freedom). In a system
of rigid molecules, vibration is absent and then the degrees of freedom
>from vibration are not employed to calculate temperature.
	In a constraint method, system of rigid molecules can be
consider as a system of flexible molecules with a set of constraint.
For this, only translation motions of N atoms are involved
but the degree of freedom is not 3N, but substracted by an amount
of constraint.
	Your problem may be consider like that, the solute is fixed
with constraint and so not used to calculate temperature.

take care,
Teerakiat

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Teerakiat Kerdcharoen, Ph.D.
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From: Konrad Hinsen <[email-masked]>
Date: Wed, 6 Nov 96 17:14:32 +0100

> to cope with this. They calculate translational and rotational energies

And this division must be done very carefully. The equipartition
theorem is valid *only* in a set of canonical coordinates and momenta.
An important practical case of non-canonical variables are angular
velocities (there are even no corresponding coordinates). It is
therefore *not* allowed to simply assign equal kinetic energy to
rotation about each of the principal axes of a molecule. However,
you can use the equipartition theorem for rotation if you use some
canonical cooridinate set to describe orientation, such as three
Euler angles per molecule, plus the associated momenta.

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From: Walter Polkosnik <[email-masked]>
Date: Wed, 6 Nov 1996 13:38:51 -0500 (EST)

> but fixed coordinates are not degrees of freedom.

There seems to be some confusion here with understanding of the
equipartition theorm. The classical equipartition theorem assigns a total
energy (note not just kinetic energy) of kT/2 per each term in the system's
Hamiltonian (momentum or position) that is quadratic. Only for ideal gases
is the total energy kinetic, for other molecules this is not true. Also,
non quadratic terms will contibute to the energy of the system but not
necessarily in the simple kT/2 way. For empirical potential energy
functions used in MD how many quadratic terms you have depends on the
choice of force field. Another complication are the VDW and Coulomb terms
in the FF, which are not quadratic, so the classical equipartition theorem
can't be used. This is always something that has bothered me about MD, the
thermodynamics does not seem incorporated into simulations well.

The statistical temperature is defined as 1/T= dS/dE, where S is the
system's entropy. This really gives and unambiguous definition of
temperature to your system, but I'm not sure if the current state MD/MM
simulations allow S or dS/dE to be calculated easily. 

--
Walter Polkosnik                             [email-masked]  
Physics Department                           http://www.physics.qc.edu/~walt
Queens College, City University of New York  (718)-997-3364 voice
Credo quia absurdum est.                     (718)-997-3349 fax

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From: Walter Polkosnik <[email-masked]>
Date: Wed, 6 Nov 1996 14:33:30 -0500 (EST)

On Wed, 6 Nov 1996, Konrad Hinsen wrote:

> I don't see the problem. The equipartition theorem says that there
> will be kT/2 of (kinetic) energy for each degree of freedom whose
> momentum enters quadratically into the Hamiltonian, and kT/2 of
> (potential) energy for each degree of freedom whose coordinate enters
> quadratically into the Hamiltonian. The first part remains true even
> if the second part can't be applied. As far as I know, the equipartition
> theorem is used in MD only for the momenta, not for the coordinates.
> The form of the potential energy function is therefore irrelevant.
> 

What I was trying to point out (perhaps in a very roundabout way) is that
defining 'temperature' in this way for an MD situation is a problem (and it
seems like all MD simply uses ideal gas results). This 'temperature' is not
simply related to the real physical temperature.

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From: Konrad Hinsen <[email-masked]>
Date: Wed, 6 Nov 96 20:13:26 +0100

For those who want it precisely, here's the equipartition theorem
for the kinetic energy in its exact form:

   /      d H   \      kT
  <  p  -------  >  = ----
   \  i   d p   /       2
	     i

(Of course it's a partial derivative, but ASCII didn't provide for this!)