アルミニウム結晶の状態方程式（エネルギーの体積依存性）

def get_fcc_unitcell(p):
  c = [(  0.0, p/2.0, p/2.0),
       (p/2.0,   0.0, p/2.0),
       (p/2.0, p/2.0,   0.0)]
  return c
 
b = 4.05
 
from ase import Atom, Atoms
from ase.calculators.jacapo import Jacapo
bulk = Atoms([Atom('Al', (0, 0, 0))], pbc = True)
 
para = {}
para.update(kpts = (4, 4, 4))
para.update(nbands = 6)
para.update(ft = 0.01)
 
print 'Volume(A^3) Energy(eV)'
 
for fraction in [0.9, 0.95, 1.0, 1.05, 1.1]:
    ncfile = 'bulkAl_%1.2f.nc' % fraction
    new_cell = get_fcc_unitcell(b*(fraction**(1./3.)))
    bulk.set_cell(new_cell, scale_atoms = True)
    para.update(atoms = bulk)
    solver = Jacapo(ncfile, **para)
    vol = bulk.get_volume()
    eng = bulk.get_potential_energy()
    print vol, eng

$ python Al_equation_of_state.py 
Volume(A^3) Energy(eV)
14.946778125 -56.4090488622
15.7771546875 -56.4473195776
16.60753125 -56.4583266527
17.4379078125 -56.4483594784
18.268284375 -56.4233507908
$ 

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