Dear Friends and Colleagues,
I have encountered a question regarding the partial radial distribution functions (RDFs) computed using OVITO.
My molecular dynamics model contains three atom types: O, Si, and Al. When I compute the partial RDFs in OVITO, it outputs the following pairs: O–O, O–Si, O–Al, Si–Si, Si–Al, and Al–Al (as shown in the attached figure).
However, I would like to obtain the RDFs with Si or Al as the central atom, specifically the Si–O and Al–O RDFs. Is there a way to specify the center atom type in OVITO to reverse the pair direction?
Thank you very much for your time and suggestions.
Best regards,
Tao
I am sorry if I have misunderstood your question. But I would say that the order of the two involved elements does not matter in a partial RDF. It’s symmetric.
The RDF for the pair Si-O is identical to the one for O-SI. The radial distances between Si and O atoms are statistically distributed in the same way as the distances between O and Si atoms.
This is also what the OVITO documentation says: g_{⍺β}(r) \equiv g_{β⍺}(r) for any pair of atom types ⍺ and β. That’s why OVITO outputs only one of the two.
Hi Stukowski
Thank you for your kind explanation.
I agure that gAB should not idential gBA, when calculating the coordination number. The atom concentration should be considered?
Tao
By the way, in this forum you can type g_{\alpha\beta}(r) as
$g_{\alpha\beta}(r)$
thanks to the Discourse math plugin.
I agure that gAB should not idential gBA, when calculating the coordination number. The atom concentration should be considered?
Right, g_{\alpha\beta}(r) has been normalized by the atom counts of species \alpha and of species \beta. It’s basically a histogram of the distances between all \alpha atoms and all \beta atoms, divided by the product of atom counts, N_\alpha * N_\beta, and the volume of the spherical shells at each r.
Depending on what you want to compute, you may have to multiply g_{\alpha\beta}(r) with just a single concentration (or absolute atom count), and then the symmetry between \alpha and \beta gets broken.
Hi Stukowski,
Thank you so much for your timely reponse. It has cleared my confusion.
Tao