why PotEng is equal to E_vdwl ?

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1

Dear All,

I have two CNTs in parallel, and then pull one CNT away from another CNT with thermostat.

The thing is, as shown in the log file, the potential energy is always equal to energy due to van der waals. Could any one of you explain why? Should the potential energy include not only the van der waals, but also the short range rebo?

Many thanks in advance.

---------------- Step 0 ----- CPU = 0.0000 (sec) ----------------
TotEng = -41371.5973 KinEng = 217.1183 Temp = 292.6817
PotEng = -41588.7156 E_bond = 0.0000 E_angle = 0.0000
E_dihed = 0.0000 E_impro = 0.0000 E_vdwl = -41588.7156
E_coul = 0.0000 E_long = 0.0000 Press = 6.7531
---------------- Step 20 ----- CPU = 0.8452 (sec) ----------------
TotEng = -41367.5043 KinEng = 90.3261 Temp = 121.7622
PotEng = -41457.8304 E_bond = 0.0000 E_angle = 0.0000
E_dihed = 0.0000 E_impro = 0.0000 E_vdwl = -41457.8304
E_coul = 0.0000 E_long = 0.0000 Press = 6.8752
---------------- Step 40 ----- CPU = 1.6899 (sec) ----------------
TotEng = -41365.4953 KinEng = 124.9776 Temp = 168.4734
PotEng = -41490.4730 E_bond = 0.0000 E_angle = 0.0000
E_dihed = 0.0000 E_impro = 0.0000 E_vdwl = -41490.4730
E_coul = 0.0000 E_long = 0.0000 Press = 2.8685

2

e_vdwl is just a quantity in LAMMPS to track all short range, charge-independent energy contributions. For REBO, both rebo and LJ are lumped into this sum.

Ray

3

Dear Ray,

Thank you very much for the reply.
Would you mind explaining a little bit more?

It is said in the manual that

evdwl = VanderWaal pairwise energy

Then, how come rebo is included?

4

That is assuming vdw is the only short range. Run some tests and you will see e_vdwl includes all charge independent terms.

Ray

5

See the compute pair doc page for a discussion of this.

Unless the pair style tallies extra values (few do),

everything that is not Couombic is “vdwl”.

Steve