I am curious what the value that VASP 4.6 and 5.2 spit out for the distance between the images in a NEB calculation corresponds to. From chain.f it appears to be the sum of the displacements of each atom in the system.
However, when I compared the value for the distance to the previous image in the OUTCAR for the first image to the same sum of the displacements calculated by hand - the values didn't match up at all (an order of magnitude difference in fact).
distance between images in NEB?
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d-farrell2
distance between images in NEB?
Last edited by d-farrell2 on Mon Feb 08, 2010 4:03 pm, edited 1 time in total.
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forsdan
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distance between images in NEB?
Given N atoms, the position of the n:th image, R_n, is a 3N-vector. The distance d between two images n and n+1 should then be
d = | R_(n+1) - R_(n) | = sqrt( sum_(i = 1)^(3N) ( R_(n+1, i) - R_(n,i) )^2 )
where R_(n,i) denotes the i:th element of R_(n). This of course assumes no atoms have crossed the cell boundary between the two images. If so, translations must been performed so the correct displacement is taken into account. All this is in agreement with the implementation in chain.F.
Have you checked that you are in fact adding up Cartesian coordinates and not the direct coordinates?
/Dan
<span class='smallblacktext'>[ Edited Mon Feb 08 2010, 08:14PM ]</span>
d = | R_(n+1) - R_(n) | = sqrt( sum_(i = 1)^(3N) ( R_(n+1, i) - R_(n,i) )^2 )
where R_(n,i) denotes the i:th element of R_(n). This of course assumes no atoms have crossed the cell boundary between the two images. If so, translations must been performed so the correct displacement is taken into account. All this is in agreement with the implementation in chain.F.
Have you checked that you are in fact adding up Cartesian coordinates and not the direct coordinates?
/Dan
<span class='smallblacktext'>[ Edited Mon Feb 08 2010, 08:14PM ]</span>
Last edited by forsdan on Mon Feb 08, 2010 7:01 pm, edited 1 time in total.