2 First-order surface-anisotropy energy $ \mathcal {E}_{1}$

In Ref. [74] the case of non-symmetrical particles was also discussed. More precisely, for small deviations from spheres and cubes, i.e., for weakly ellipsoidal or weakly rectangular particles, there is a corresponding weak first-order contribution $ \mathcal {E}_{1}$ that adds up to $ \mathcal {E}_{2}$. Hence, for an ellipsoid of revolution with axes $ a$ and $ b=a(1+\epsilon)$, $ \epsilon\ll1,$ cut out of an sc lattice so that the ellipsoid's axes are parallel to the crystallographic direction, one has that the first order anisotropy is given by [97]:

$\displaystyle \mathcal{E}_{1} = -k_{1} m_{z}^{2}\quad.$ (33)


$\displaystyle k_{1} \sim k_s {\cal N}^{-1/3}\epsilon ,$ (34)

i.e. the $ \mathcal {E}_{1}$ energy contribution scales with the surface of the system.

It is necessary to mention that for crystal shapes such as spheres and cubes the contribution $ \mathcal {E}_{1}$ vanishes by symmetry.

The ratio of the second to the first order surface contributions is:

$\displaystyle \frac{\mathcal{E}_{2}}{\mathcal{E}_{1}}\sim k_s\frac{{\cal N}^{1/3}}{\epsilon},$ (35)

It can be significant even for $ K_{s}\ll J$ due to the combined influence of the large particle size and the small deviation from symmetry, $ \epsilon \ll 1$ .
Rocio Yanes