[gmx-users] energy unit
Mark Abraham
mark.abraham at anu.edu.au
Mon Aug 30 01:46:26 CEST 2010
----- Original Message -----
From: Moeed <lecielll at googlemail.com>
Date: Monday, August 30, 2010 9:09
Subject: [gmx-users] energy unit
To: gmx-users at gromacs.org
> Dear Justin,
>
> Thanks for your explanation. I am sorry to bother you again. Like you said: "a mole of a given species in the given configuration would have this energy in kJ". This is absolutely clear but actually I think increasing the total energy with system size contradict this statement unless values given by g_energy are in KJ/system. If there are more interactions in a bigger system, as you said eventually values refer to mole number of particles (regardless of number of particles). If there are half Na (avogardo) particles, finally energy is multiplied by 2 to get KJ/mole and if there are 10*Na, g_energy divides total energy by 10. Please enlightem me on this issue. Many thanks
To say that "the energy of a single-component system is in kJ/mol" is to say that if (somehow) a mole of that component were present in the system in that configuration, then the energy would be that number in kJ.
If the system was twice as large, the magnitude of the energy in kJ/mol does increase, because there are more interactions. If a mole of that system were present, then its energy would be that number in kJ. This appears to generate a contradiction, however there is a size dependence in the energy that is not linear in the number of molecules, so that this comparison is between two different kinds of quantities.
To compare these two cases, you need to normalize not by the number of particles, but by the number of interactions.
Think about building up a LJ-only system of a noble gas in an infinite non-periodic system. With one atom, the energy is zero. With two atoms, the energy from the single interaction is normally non-zero, and has some value if you average it over all possible configurations. It doesn't matter whether you measure it in Hartree, or kJ, or kJ/mol - these are just multiplication by a conversion constant. With three atoms, there are now three interactions, the configurational-average energy has greater magnitude - but it makes no sense to compare these last two cases without normalizing by the number of interactions. When you do that normalization, then the numbers will almost agree - because the total energy was a simple sum of three nearly independent components (except that the atoms could not coincide). The same kind of thinking goes for the kind of systems the OP was originally considering, but there are more complications from the number and kind of interactions.
Mark
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