Re: SMD Energy

From: Sterling Paramore (
Date: Wed May 31 2006 - 11:41:57 CDT

The problem is that the total energy you calculate in MD doesn't include
the "energy" of the thermostat. You must keep in mind that heat
exchange is also a source of energy change. One of the first things we
learn in a thermodynamics class is that the change in total energy of a
system is equal to the work done on it PLUS the energy due to heat
exchange with the environment. If you're doing NVE dynamics, then there
is no "environment" to speak of and yes, the work done on the system
will be equal to the energy change. However, in this case the
temperature of the system will change (usually it will increase due to
viscous heating). When you have a thermostat on the system (either NVT
or NPT), the temperature is kept constant by removing the heat produced
during the process. Therefore, the total energy change of the MD system
is equal to the work done on it minus the heat removed from it. While
it's very easy to calculate the energy removed as heat with a
Nose-Hoover thermostat, I am not aware of how (or if) this can be done
with a Langevin thermostat.

Again, what do you hope to learn about your system by calculating the
energy difference?


Pijush Ghosh wrote:

>Dear Sterling and Li
> Sorry to bother you again. Well, fantastic discussion so far. I completely
>agree with what Sterling said about the fluctuation of total energy in NPT
>ensemble particularly for a large biosystem. I have had the experience of
>such in the past. But what I want to insist on is, purely from the of
>mechanics viewpoint whats the flaw in going with the theory that "change in
>total energy is equal to the work done"
> Where:
>change in energy is the difference in initial and final energy of the whole
>system, and,
>wrok done is the mechanical work done in displacing the pulling SMD atom.
>If they don't match, where the remaining energy dissipates to? Does all of
>it goes to the thermostat system.
>For example in one of my organic-inorganic system:
>At vel= 0.50A/ps
> Change in total energy: -4745 kcal/mol
> Area under force-disp. Curve: 8142 Kcal/mol
>Whereas at vel= 0.25A/ps
> Change in total energy: -4626 kcal/mol
> Area under force-disp. Curve: 2143 Kcal/mol
>I am not quite sure how to explain this....or if at all it makes sense to
>draw any conclusion from these numbers.
> I will look forward for some discussion from your end.
>I appreciate your time so much.
>Pijush Ghosh
>PhD Student
>Department of Civil Engineering
>North Dakota State University
>Fargo. ND. 58105. USA
> 701-231-6491(Lab)
> 701-231-4341(Res)

This archive was generated by hypermail 2.1.6 : Wed Feb 29 2012 - 15:43:40 CST