Dear users,
Many recent papers on the mechanical strength of graphene and CNTs have used the AIREBO potential (with the cutoff function removed) to simulate various tests. But, can this potential (with cutoff removed [i.e. rc = 2]) capture BOND FORMING situations?
Many thanks,
Daniel
No AIREBO expert has answered, so I’ll just note
that bond-order potentials like (AI)REBO
both create and break bonds dynamically depending
on the current local conformation of atoms.
Steve
Thanks Steve,
But can bonds be created if rcmin (the bond distance at which we BEGIN to ramp down the potential to zero at bond distance r=2) is put equal to 1.95 or 2 (instead of the default value of 1.7).
Most papers that deal with mechanical testing of CNTs or graphene have had to alter the rcmin close to, or equal to 2 in order to avoid the spurious (and considerable) overestimation of strength caused by the discontinuity at rcmin = 1.7.
Many Thanks,
Daniel
Thanks Steve,
But can bonds be created if rcmin (the bond distance at which we BEGIN to ramp down the potential to zero at bond distance r=2) is put equal to 1.95 or 2 (instead of the default value of 1.7).
Most papers that deal with mechanical testing of CNTs or graphene have had to alter the rcmin close to, or equal to 2 in order to avoid the spurious (and considerable) overestimation of strength caused by the discontinuity at rcmin = 1.7.
Many Thanks,
Daniel
Do you wish to completely remove the cutoff function (as suggested by your title and your first post) or do you want to just change the inner cutoff?
If it is the later, then any pair of atoms that are separated by 2A or smaller will be considered in the BO calculation - if a bond is considered favorable, a bond will form. Then the bond strength is just scaled by the cutoff function determined by the inner/outer cutoffs.
Ray
Many thanks Ray
Changing the inner cutoff (rcmin) to a value of 2 removes the cutoff function right?Thats what I mean. Would bond forming occur here?
Daniel
Many thanks Ray
Changing the inner cutoff (rcmin) to a value of 2 removes the cutoff
function right?
No, the cutoff function is still there but you just change it from a cosine
function to a step function.
Thats what I mean. Would bond forming occur here?
It should, but you should reach such a conclusion after you conduct a
series of tests.
Ray
Thanks Ray
Yes, in certain trial simulations to test this I seem to see that new bonds can form even though I have cut off the potential by a step at r=2
I was wondering is this still a correct description of the formation of new bonds?
Many thanks
Daniel
What if I say it is not correct and all your simulations will not be useful? What if I say it is absolutely correct and you should not worry about anything? What if I say this and then someone else says that? Who should you trust?
You should trust your own results after conducting a series of rigorous tests. You should compare them with the original cutoff and be convinced that changing the cutoff does or does not have effects on chemical bond formation. Personally I would never use a step function for a cutoff function.
Please keep in mind that these are empirical force fields and are fitted primarily to QM calculations and to some limited experimental data. Once you alter anything from the original force field parameterization (i.e. changing the cutoffs), you are deviating from what the original authors created and it is entirely on your own to verify that it generates desirable results. No one on this mailing list can answer that or do the work for you.
Ray
OK, Thanks for your help and discussion Ray,
Regards,
Daniel