def findDeviation(name):
lattice = np.zeros((3, 3))
original_pb = np.zeros((8, 3))
original_cl = np.zeros((24, 3))
original_c = np.zeros((8, 3))
original_n = np.zeros((8, 3))
original_h = np.zeros((48, 3))
op.in_coord(name, lattice, original_pb, original_cl, original_c,
original_n, original_h)
#enlarge the unit cell by 8 times
pb = np.zeros((64, 3))
cl = np.zeros((192, 3))
c = np.zeros((64, 3))
n = np.zeros((64, 3))
h = np.zeros((384, 3))
op.enlarge(original_pb, pb, lattice, 8)
op.enlarge(original_cl, cl, lattice, 24)
op.enlarge(original_c, c, lattice, 8)
op.enlarge(original_n, n, lattice, 8)
op.enlarge(original_h, h, lattice, 48)
#find the center of C-N
CNCenter = np.zeros((64, 3))
for i in range(64):
temp = np.zeros((3, 2))
op.gothrough(n[i], c, temp)
CNCenter[i] = 1 / 2 * (n[i] + c[int(temp[0, 0])])
#Find the 4 molecules that surround the Cl atoms. Take N atoms as indicator. Store the index
cl_n = np.zeros((24, 4))
for i in range(24):
temp = np.zeros((4, 2))
op.gothrough(original_cl[i], CNCenter, temp)
if temp[-1, 1] > 6:
print("There is something wrong my friend.")
sys.exit()
cl_n[i, :] = temp[:, 0]
#Assume the ideal ideal hydrogen bond number is 1 for each Cl atom
NumHBond = np.zeros((24, 1))
r0 = 2.2
B = 0.3
for i in range(24):
for j in range(4):
tempH = np.zeros((3, 2))
indexN = int(cl_n[i, j])
op.gothrough(n[indexN], h, tempH)
for k in range(3):
dij = np.linalg.norm(original_cl[i, :] - h[int(tempH[k, 0])])
NumHBond[i, 0] += countFD(r0, B, dij)
deviation = np.zeros((8, 1))
deviation = np.square(NumHBond - 1)
totDevi = np.sum(deviation)
return totDevi
def findNHBond(name):
#import the coordinates
lattice = np.zeros((3,3));
original_pb = np.zeros((8,3));
original_cl = np.zeros((24,3));
original_c = np.zeros((8,3));
original_n = np.zeros((8,3));
original_h = np.zeros((48,3));
op.in_coord(name, lattice, original_pb, original_cl, original_c, original_n, original_h);
#enlarge the unit cell by 8 times
pb = np.zeros((64,3));
cl = np.zeros((192,3));
c = np.zeros((64,3));
n = np.zeros((64,3));
h = np.zeros((384,3));
op.enlarge(original_pb,pb,lattice,8);
op.enlarge(original_cl,cl,lattice,24);
op.enlarge(original_c,c,lattice,8);
op.enlarge(original_n,n,lattice,8);
op.enlarge(original_h,h,lattice,48);
the_n = np.zeros((8,3));
index_n = np.zeros((8,1));
the_c = np.zeros((8,3));
index_c = np.zeros((8,1));
the_cl = np.zeros((8,12,3));
the_pb = np.zeros((8,8,3));
distance_o_cl = np.zeros((8,12));
distance_h_cl = np.zeros((8,3));
#first get one small cell
count = 0;
for i in range(64):
temp_n = n[i];
temp_center = np.zeros((3,1));
temp_cn = np.zeros((64,2));
op.gothrough(temp_n,c,temp_cn);
temp_c = c[int(temp_cn[0][0])];
op.midpoint(temp_c,temp_n,temp_center);#determine the center coordinate
#determine the neighbor Cl atoms
temp_o_cl = np.zeros((12,2));
op.gothrough(temp_center, cl, temp_o_cl);
#determine the neighbor Pb atoms
temp_o_pb = np.zeros((8,2));
op.gothrough(temp_center,pb,temp_o_pb);
if temp_o_cl[11][1] > 5.5:
continue;
else:
#determine the H(N) atoms in this unit cell
temp_h_n = np.zeros((3,2));
op.gothrough(temp_n, h, temp_h_n);
temp_nh_cl = np.zeros((3,2));
for j in range(3):
op.gothrough(h[int(temp_h_n[j][0])],cl,temp_nh_cl[j,:]);
#decide whether we choose this unit cell or not, trying to avoid overlap
count2 = 0;
for j in range(count+1):
indicator1 = np.linalg.norm(temp_o_cl[:,1] - distance_o_cl[j,:]);
indicator2 = np.linalg.norm(temp_nh_cl[:,1] - distance_h_cl[j,:]);
# print(indicator1)
# print(indicator2)
if indicator1 > 0.0000001 and indicator2 > 0.0000001:
count2 = count2 + 1;
continue;
else:
break;
if count2 == count+1:
distance_o_cl[count,:] = temp_o_cl[:,1];
distance_h_cl[count,:] = temp_nh_cl[:,1];
the_n[count] = temp_n;
the_c[count] = temp_c;
index_n[count] = i;
index_c[count] = temp_cn[0][0];
for k in range(12):
the_cl[count][k] = cl[int(temp_o_cl[k][0])];
for k in range(8):
the_pb[count][k] = pb[int(temp_o_pb[k][0])];
count = count + 1;
#rearrange the chlorine atoms
for i in range(8):
op.rearrange(the_cl[i]);
#After spliting into small cell, let's calculate the hydrogen bonds length.
#Every Hydrogen atom keep its first two shortest distances between Cl
the_nh = np.zeros((8,3,3));
the_ch = np.zeros((8,3,3));
for i in range(8):
temp_h_n = np.zeros((3,2));
temp_h_c = np.zeros((3,2));
op.gothrough(the_n[i],h,temp_h_n);
op.gothrough(the_c[i],h,temp_h_c);
for j in range(3):
the_nh[i][j] = h[int(temp_h_n[j][0])];
the_ch[i][j] = h[int(temp_h_c[j][0])];
#3rd method
r0 = 2.2;
B = 0.3;
numHBond = np.zeros((8,1));
for i in range(8):
for j in range(3):
for k in range(12):
dij = np.linalg.norm(the_nh[i][j] - the_cl[i][k]);
numHBond[i,0] += countFD(r0,B,dij);
#print(numHBond)
aveBond = np.sum(numHBond)/24;
return aveBond;
original_cl = np.zeros((24, 3))
original_c = np.zeros((8, 3))
original_n = np.zeros((8, 3))
original_h = np.zeros((48, 3))
op.in_coord(name, lattice, original_pb, original_cl, original_c,
original_n, original_h)
#enlarge the unit cell by 8 times
pb = np.zeros((64, 3))
cl = np.zeros((192, 3))
c = np.zeros((64, 3))
n = np.zeros((64, 3))
h = np.zeros((384, 3))
op.enlarge(original_pb, pb, lattice, 8)
op.enlarge(original_cl, cl, lattice, 24)
op.enlarge(original_c, c, lattice, 8)
op.enlarge(original_n, n, lattice, 8)
op.enlarge(original_h, h, lattice, 48)
#According to our own system, we can just choose the original Pb atoms
#We take the Pb atom that has the least x,y,z coordinate as our central Pb atoms, i.e. the reference octahedron
re_pb = np.zeros((8, 3))
op.rearrange_pb(original_pb, re_pb)
#reference Pb is the first Pb atom
EightSets = np.zeros((8, 3))
for i in range(8):
EightSets[i, :] = euler(i, re_pb, cl)
def findNHBond(name):
#import the coordinates
lattice = np.zeros((3, 3))
original_pb = np.zeros((8, 3))
original_cl = np.zeros((24, 3))
original_c = np.zeros((8, 3))
original_n = np.zeros((8, 3))
original_h = np.zeros((48, 3))
op.in_coord(name, lattice, original_pb, original_cl, original_c,
original_n, original_h)
#enlarge the unit cell by 8 times
pb = np.zeros((64, 3))
cl = np.zeros((192, 3))
c = np.zeros((64, 3))
n = np.zeros((64, 3))
h = np.zeros((384, 3))
op.enlarge(original_pb, pb, lattice, 8)
op.enlarge(original_cl, cl, lattice, 24)
op.enlarge(original_c, c, lattice, 8)
op.enlarge(original_n, n, lattice, 8)
op.enlarge(original_h, h, lattice, 48)
the_n = np.zeros((8, 3))
index_n = np.zeros((8, 1))
the_c = np.zeros((8, 3))
index_c = np.zeros((8, 1))
the_cl = np.zeros((8, 12, 3))
the_pb = np.zeros((8, 8, 3))
distance_o_cl = np.zeros((8, 12))
distance_h_cl = np.zeros((8, 3))
#first get one small cell
count = 0
for i in range(64):
temp_n = n[i]
temp_center = np.zeros((3, 1))
temp_cn = np.zeros((64, 2))
op.gothrough(temp_n, c, temp_cn)
temp_c = c[int(temp_cn[0][0])]
op.midpoint(temp_c, temp_n, temp_center)
#determine the center coordinate
#determine the neighbor Cl atoms
temp_o_cl = np.zeros((12, 2))
op.gothrough(temp_center, cl, temp_o_cl)
#determine the neighbor Pb atoms
temp_o_pb = np.zeros((8, 2))
op.gothrough(temp_center, pb, temp_o_pb)
if temp_o_cl[11][1] > 5.5:
continue
else:
#determine the H(N) atoms in this unit cell
temp_h_n = np.zeros((3, 2))
op.gothrough(temp_n, h, temp_h_n)
temp_nh_cl = np.zeros((3, 2))
for j in range(3):
op.gothrough(h[int(temp_h_n[j][0])], cl, temp_nh_cl[j, :])
#decide whether we choose this unit cell or not, trying to avoid overlap
count2 = 0
for j in range(count + 1):
indicator1 = np.linalg.norm(temp_o_cl[:, 1] -
distance_o_cl[j, :])
indicator2 = np.linalg.norm(temp_nh_cl[:, 1] -
distance_h_cl[j, :])
# print(indicator1)
# print(indicator2)
if indicator1 > 0.0000001 and indicator2 > 0.0000001:
count2 = count2 + 1
continue
else:
break
if count2 == count + 1:
distance_o_cl[count, :] = temp_o_cl[:, 1]
distance_h_cl[count, :] = temp_nh_cl[:, 1]
the_n[count] = temp_n
the_c[count] = temp_c
index_n[count] = i
index_c[count] = temp_cn[0][0]
for k in range(12):
the_cl[count][k] = cl[int(temp_o_cl[k][0])]
for k in range(8):
the_pb[count][k] = pb[int(temp_o_pb[k][0])]
count = count + 1
#rearrange the chlorine atoms
for i in range(8):
op.rearrange(the_cl[i])
#After spliting into small cell, let's calculate the hydrogen bonds length.
the_nh = np.zeros((8, 3, 3))
the_ch = np.zeros((8, 3, 3))
for i in range(8):
temp_h_n = np.zeros((3, 2))
temp_h_c = np.zeros((3, 2))
op.gothrough(the_n[i], h, temp_h_n)
op.gothrough(the_c[i], h, temp_h_c)
for j in range(3):
the_nh[i][j] = h[int(temp_h_n[j][0])]
the_ch[i][j] = h[int(temp_h_c[j][0])]
#Now starting to calculate the LJ terms
#For all Cl atoms in one small cell
HCl_6LJ = np.zeros((8, 1))
HCl_12LJ = np.zeros((8, 1))
CHCl_6LJ = np.zeros((8, 1))
CHCl_12LJ = np.zeros((8, 1))
NCl_6LJ = np.zeros((8, 1))
NCl_12LJ = np.zeros((8, 1))
CCl_6LJ = np.zeros((8, 1))
CCl_12LJ = np.zeros((8, 1))
for i in range(8):
for j in range(12):
dij_NCl = np.linalg.norm(the_n[i] - the_cl[i, j])
dij_CCl = np.linalg.norm(the_c[i] - the_cl[i, j])
NCl_6LJ[i, 0] += count6LJ(dij_NCl)
NCl_12LJ[i, 0] += count12LJ(dij_NCl)
CCl_6LJ[i, 0] += count6LJ(dij_CCl)
CCl_12LJ[i, 0] += count12LJ(dij_CCl)
for j in range(3):
for k in range(12):
dij = np.linalg.norm(the_nh[i][j] - the_cl[i][k])
dij_CH = np.linalg.norm(the_ch[i][j] - the_cl[i][k])
HCl_6LJ[i, 0] += count6LJ(dij)
HCl_12LJ[i, 0] += count12LJ(dij)
CHCl_6LJ[i, 0] += count6LJ(dij_CH)
CHCl_12LJ[i, 0] += count12LJ(dij_CH)
#For the nearest 3 Cl atoms for H
# for i in range(8):
# for j in range(12):
# dij_NCl = np.linalg.norm(the_n[i] - the_cl[i,j]);
# NCl_6LJ[i,0] += count6LJ(dij_NCl);
# NCl_12LJ[i,0] += count12LJ(dij_NCl);
#
# for j in range(3):
# tempCl = np.zeros((3,2));
# op.gothrough(the_nh[i,j],the_cl[i],tempCl);
# for k in range(3):
# dij = tempCl[k,1];
# HCl_6LJ[i,0] += count6LJ(dij);
# HCl_12LJ[i,0] += count12LJ(dij);
return np.sum(HCl_6LJ), np.sum(HCl_12LJ), np.sum(CHCl_6LJ),\
np.sum(CHCl_12LJ),np.sum(NCl_6LJ), np.sum(NCl_12LJ),np.sum(CCl_6LJ), np.sum(CCl_12LJ)
def findChargeCenter(name):
'''Atoms and Charge should be one-to-one correspondance'''
#import the coordinates
lattice = np.zeros((3, 3))
original_pb = np.zeros((8, 3))
original_cl = np.zeros((24, 3))
original_c = np.zeros((8, 3))
original_n = np.zeros((8, 3))
original_h = np.zeros((48, 3))
op.in_coord(name, lattice, original_pb, original_cl, original_c,
original_n, original_h)
#enlarge the unit cell by 8 times
pb = np.zeros((64, 3))
cl = np.zeros((192, 3))
c = np.zeros((64, 3))
n = np.zeros((64, 3))
h = np.zeros((384, 3))
op.enlarge(original_pb, pb, lattice, 8)
op.enlarge(original_cl, cl, lattice, 24)
op.enlarge(original_c, c, lattice, 8)
op.enlarge(original_n, n, lattice, 8)
op.enlarge(original_h, h, lattice, 48)
the_n = np.zeros((8, 3))
index_n = np.zeros((8, 1))
the_c = np.zeros((8, 3))
index_c = np.zeros((8, 1))
the_cl = np.zeros((8, 12, 3))
the_pb = np.zeros((8, 8, 3))
distance_o_cl = np.zeros((8, 12))
distance_h_cl = np.zeros((8, 3))
# mapToOrigin = np.zeros((8,2));# Tell which original molecule they actually correspond to
# mapToOrigin[:,0] = np.arange(0,8,1);
mapToOrigin = np.zeros((8, 1))
#first get one small cell
count = 0
for i in range(64):
temp_n = n[i]
temp_center = np.zeros((3, 1))
temp_cn = np.zeros((64, 2))
op.gothrough(temp_n, c, temp_cn)
temp_c = c[int(temp_cn[0][0])]
op.midpoint(temp_c, temp_n, temp_center)
#determine the center coordinate
#determine the neighbor Cl atoms
temp_o_cl = np.zeros((12, 2))
op.gothrough(temp_center, cl, temp_o_cl)
#determine the neighbor Pb atoms
temp_o_pb = np.zeros((8, 2))
op.gothrough(temp_center, pb, temp_o_pb)
if temp_o_cl[11][1] > 5.5:
continue
else:
#determine the H(N) atoms in this unit cell
temp_h_n = np.zeros((3, 2))
op.gothrough(temp_n, h, temp_h_n)
temp_nh_cl = np.zeros((3, 2))
for j in range(3):
op.gothrough(h[int(temp_h_n[j][0])], cl, temp_nh_cl[j, :])
#decide whether we choose this unit cell or not, trying to avoid overlap
count2 = 0
for j in range(count + 1):
indicator1 = np.linalg.norm(temp_o_cl[:, 1] -
distance_o_cl[j, :])
indicator2 = np.linalg.norm(temp_nh_cl[:, 1] -
distance_h_cl[j, :])
if indicator1 > 0.0000001 and indicator2 > 0.0000001:
count2 = count2 + 1
continue
else:
break
if count2 == count + 1:
distance_o_cl[count, :] = temp_o_cl[:, 1]
distance_h_cl[count, :] = temp_nh_cl[:, 1]
the_n[count] = temp_n
the_c[count] = temp_c
index_n[count] = i
mapToOrigin[count] = i % 8
index_c[count] = temp_cn[0][0]
for k in range(12):
the_cl[count][k] = cl[int(temp_o_cl[k][0])]
for k in range(8):
the_pb[count][k] = pb[int(temp_o_pb[k][0])]
count = count + 1
#rearrange the chlorine atoms
for i in range(8):
op.rearrange(the_cl[i])
#Find the hydrogen atoms in each of the cell
the_nh = np.zeros((8, 3, 3))
the_ch = np.zeros((8, 3, 3))
for i in range(8):
temp_h_n = np.zeros((3, 2))
temp_h_c = np.zeros((3, 2))
op.gothrough(the_n[i], h, temp_h_n)
op.gothrough(the_c[i], h, temp_h_c)
for j in range(3):
the_nh[i][j] = h[int(temp_h_n[j][0])]
the_ch[i][j] = h[int(temp_h_c[j][0])]
'''Charges. Could be replaced with Bader Charge'''
qCl = -1
qPb = 2
qC = -2
qN = -3
qH = 1
orgoCenter = np.zeros((8, 3))
inorgoCenter = np.zeros((8, 3))
for i in range(8):
tempCenter = np.zeros((1, 3))
for j in range(8):
tempCenter += abs(qPb) * the_pb[i][j][:]
for j in range(12):
tempCenter += abs(qCl) * the_cl[i][j][:]
inorgoCenter[i, :] = tempCenter / (8 * abs(qPb) + 12 * abs(qCl))
pullBackFactor = np.round(
np.matmul(np.linalg.inv(lattice),
n[int(index_n[i])] - n[int(mapToOrigin[i])]))
inorgoCenter[i, :] = inorgoCenter[i, :] - np.matmul(
lattice, pullBackFactor)
for i in range(8):
tempCenter = np.zeros((1, 3))
for j in range(3):
tempCenter += abs(qH) * the_nh[i][j][:]
for j in range(3):
tempCenter += abs(qH) * the_ch[i][j][:]
tempCenter += abs(qC) * the_c[i, :] + abs(qN) * the_n[i, :]
orgoCenter[i, :] = tempCenter / (6 * abs(qH) + abs(qC) + abs(qN))
pullBackFactor = np.round(
np.matmul(np.linalg.inv(lattice),
n[int(index_n[i])] - n[int(mapToOrigin[i])]))
orgoCenter[i, :] = orgoCenter[i, :] - np.matmul(
lattice, pullBackFactor)
mapToOrigin = np.argsort(mapToOrigin, axis=0)
orgo = orgoCenter[np.int64(mapToOrigin[:, 0]).reshape([8, 1]), [0, 1, 2]]
inorgo = inorgoCenter[np.int64(mapToOrigin[:, 0]).reshape([8, 1]),
[0, 1, 2]]
return orgo, inorgo