Strain energy function

Dear LAMMPS users,

I was trying to derive the strain energy function by deforming my polymer model for continuum level modeling. But I am getting abnormally high values and couldn't correct by myself. I am using "real" unit in lammps which gives my potential energy (the sum of pairs,bonds,angles, and dihedrals) in kcal/mol which needs to be multiplied by the number of mole I have in the model. For example, I got 13000 kcal/mole of potential energy. My model has 10,000 atoms so the mole is 10,000/6.023e23=1.66e-20 mole.
P.E=13000*1.66e-20 kcal
=2.16e-16 kcal
=2.16e-16*4.18 kjoul
=9.02e-16 kjoul

My simulation box size is 180X180X180 angstrom.
So, the strain energy density (per unit volume) =9.02e-16 / (180e-10)^3 kjoul/m^3=1.55e8 kjoul/m^3.

But this value is too high for my polymer (approximately 10,000 times larger). I thought my potential energy values are wrong and tried to minimize/relax the structure by different ensemble npt, nvt where it is reducing insignificantly. the entropic effect (considering the free energy) might be a solution but subtracting the roughly estimated statisical configurational entropy from the potential energy is not good enough to reduce it to 10,000 times smaller. I thing I did something wrong in my calculation during unit conversion. I would highly appreciate if somebody helps me to be corrected.



isn't it suggestive that you multiplied your numbers by 10000 atoms and obtained the value _10000_ times higher than it should be? The mistake is that in this case kcal/mole is an absolute value, not the per-atom one. Consider it as kcal/Na, Na = 6e23. It will increase with system size, you can check this.


"Uddin, Md Salah" <[email protected]> 14 апреля 2014 г. 23:25:53 написал:

Thanks Dear Oleg. I was confused at the same point but not sure as the value is given in kcal/mole, I did the normal calculation of mole dividing the number of atoms by Avogadro number. Thanks again !

Best Regards