Calculation of coordination number

Hello LAMMPS users, respectivamente

I have a question about the coordination number. Would you please answer it and help me?

Yes, with the commands “all c1 compute coord / atom 3.2”, “c2 Nickel compute coord / atom 3.2” and “Aluminum c3 compute coord / atom 3.6” calculate the total coordination numbers of the cell, Aluminum and Nickel respectively.

What would be the command to calculate the number of coordination between the Nickel-Aluminum atoms?
I appreciate the attention given and any help you can give me.

regards

coord/atom does not distinguish between different groups or atom
types. If you would like to get info on different atom types,
consider using compute rdf

http://lammps.sandia.gov/doc/compute_rdf.html

Aidan

Hello
Aidan, thanks for answering my query.

I see the rdf compute calculates the coordination between different atoms, but I get an average value. As I can do to get a value atom by atom.

I am studying a nonhomogeneous system, and I calculate coordination as an order parameter that tells me whether or not intermetallic formed between Al and Ni

I appreciate any information you can give me

regards

2012/7/12 Aidan Thompson <[email protected]>

I see. You want to know the coordination number of each atom
separately, and also separately for each atom type. This would not be
hard to do in compute_coord/atom, if you can find someone to do it.

The hardest part of modifying the code would be extending the compute
from a per-atom vector to a per-atom array. You could this by creating
your own renamed copies of the compute coord/atom as new classes (you
could have them derive from ComputeCoordAtom) and hard code the
desired types in the function compute_peratom().

Aidan

Extended the compute coord/atom command to allow computation
of multiple coord numbers, each with respect to neighbors of specified
atom types.

See the 11 Aug patch ...

Steve

Dear,
Dr Steve Plimpton and Dr Aidan Thompson
Thanks for the information.

regards

2012/8/8 Steve Plimpton <[email protected]>