Hard superball model: Difference between revisions

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[[Image:shape.png|thumb|right|The shape of superballs interpolates between octahedra (''q'' = 0.5) and cubes (''q'' = ∞) via spheres (''q'' = 1).]]
[[Image:shape.png|thumb|right|The shape of superballs interpolates between octahedra (''q'' = 0.5) and cubes (''q'' = ∞) via spheres (''q'' = 1).]]
[[Image:phase_diagram_superball.png|thumb|right|Phase diagram for hard superballs in the <math>\phi<math> (packing fraction) versus 1/''q'' (bottom axis) and q (top axis) representation where ''q'' is the deformation parameter [2].]]


A superball is defined by the inequality
A superball is defined by the inequality

Revision as of 21:06, 16 September 2012

The shape of superballs interpolates between octahedra (q = 0.5) and cubes (q = ∞) via spheres (q = 1).

[[Image:phase_diagram_superball.png|thumb|right|Phase diagram for hard superballs in the

where x, y and z are scaled Cartesian coordinates with q the deformation parameter and radius a. The shape of the superball interpolates smoothly between two Platonic solids, namely the octahedron (q = 0.5) and the cube (q = ∞) via the sphere (q = 1) as shown in the left figure.

Particle Volume

The volume of a superball with the shape parameter q and radius a is given by

Failed to parse (unknown function "\begin{eqnarray}"): {\displaystyle \begin{eqnarray} v(q,a) & = & 8 a^3 \int_{0}^1 \int_{0}^{(1-x^{2q})^{1/2q}} (1-x^{2q}-y^{2q})^{1/2q} \mathrm{d}\, y \, \mathrm{d}\, x \nonumber\\ & = & \frac{8a^3\left[ \Gamma\left(1+1/2q\right) \right]^3}{\Gamma\left(1+ 3/2q\right)}, \end{eqnarray} }

where is the Gamma function.

Overlap algorithm

The most widely used overlap algorithm is on the basis of Perram and Wertheim method[1] [2].

Phase diagram

The full phase diagram of hard superballs whose shape interpolates from cubes to octahedra was in Ref.[2].

References