Cohesive energy and Solubility parameter

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1

Hello LAMMPS users,

I am interested in the cohesive energy between the polymer chains and later use it to determine the Hildebrand solubility parameters as given below.

Solubility = sqrt ( E_coh / Volume)
E_coh = summation (E_iso) - E_tot; E_iso is the potential energy of each chain in the system and E_tot is the energy.

Here are list of things I did:

Method-1) Generate an equilibrated polymer box with 5 chains. Use the last frame to determine the potential energy (run 0), which would be E_tot
Later, I have created 5 independent datafiles, for each polymer chain, using exact polymer configuration and simulation box to determined energy of the polymer chain (E_iso). I have determined E_coh as mentioned above.

Of course, this procedure is not ideal when there are several chains and several frames to determined the solubility parameter.

Method-2) To determine the solubility parameter, on the fly, while the simulation is running.
In the same example, I have defined five groups and determined as shown below.

2

I think the main problem here is the kspace contribution, which you probably handling differently in the three methods. I suggest turning off kspace and redoing the intramolecular energy calculations. You should be able to recover the same values for each molecule using all three methods.

3

Hello Aidan,

Thanks for your response.

My pair style is “lj/charmm/coul/long/soft 2.0 0.5 10 9.0 12.0”.

I tried to remove kspace using “lj/cut/coul/cut/soft” (lj/charmm/coul/cut/soft is not available), the individual energies are quite high.

Also, turning off Kspace for a polymer chain for 40-50 Ang, how does it evaluate the potential energy if the cutoff is just 12 Ang. Unless I consider a huge box, with cutoff slightly less than half-of-the box length. Sorry, If I missed something here.

Thanks,
Karteek