November 08 2002

Average radius of the surface atoms for a liquid cluster of 603 atoms

1. Average radius from simulation

The surface atoms have been carried out (cutoff=3.7Angstrom, criterion of neighbors<=10 atoms) from one hundred configurations of the liquid cluster at 1500K. The average radius is just the average distance from the surface atoms to the center of mass:

<r> = sum( sqrt( x_i*x_i + y_i*y_i + z_i*z_i) ) / N = 12.1963 (Angstrom)

The corresponding average curvature is:

<c> = 1 / <r> = 0.081992 (1/Angstrom)

2. Radius of sphere from evaluation

Assume 603 atoms form an ideal sphere, the linear number density on the radius n_r can be evaluated as:

4/3 * Pi * n_r³ = 603

Thus n_r = 5.2409.

The average minimal distance between two atoms have been calculated by loop all the atoms and find the minimal distance two other atoms for a given atom. The mentioned 100 configurations at 1500K have been looped over and the average minimal distance d_m = 2.35787(Angstrom). So the evaluated radius for an ideal sphere is:

<r> = nr * dm = 12.3574 (Angstrom)

The corresponding ideal curvature is:

<c> = 1 / <r> = 0.080923 (1/Angstrom)

3. Peak values of the curvatures

Read from Yesterday's histograms of the curvatures we have the approximate peak values as:

Temperature (K)	Peak1 (1/Angstrom)	Peak2 (1/Angstrom)
100	0.030	0.098
200	0.030	0.098
300	0.035	0.100
400	0.035	0.100
500	0.035	0.100
600	0.035	0.085
700	0.060	---
800	0.060	---
900	0.065	---
1000	0.070	---
1500	0.080	---

Note there are 100 bins in the region of [ 0, 0.5 ], so the reslolution is no better than 0.005 1/Angstrom. The irregular shapes of the peaks make it worse.