# Calculating bond orientational order parameters¶

This example illustrates the calculation of bond orientational order parameters. Bond order parameters, $$q_l$$ and their averaged versions, $$\bar{q}_l$$ are widely used to identify atoms belong to different crystal structures. In this example, we will consider bcc, fcc, and hcp, and calculate the $$q_4$$ and $$q_6$$ parameters and their averaged versions which are widely used in literature. More details can be found here.

import pyscal.core as pc
import pyscal.crystal_structures as pcs
import numpy as np
import matplotlib.pyplot as plt


In this example, we analyse MD configurations, first a set of perfect bcc, fcc and hcp structures and another set with thermal vibrations.

## Perfect structures¶

To create atoms and box for perfect structures, the :mod:~pyscal.crystal_structures module is used. The created atoms and boxes are then assigned to System objects.

bcc_atoms, bcc_box = pcs.make_crystal('bcc', lattice_constant=3.147, repetitions=[4,4,4])
bcc = pc.System()
bcc.atoms = bcc_atoms
bcc.box = bcc_box

fcc_atoms, fcc_box = pcs.make_crystal('fcc', lattice_constant=3.147, repetitions=[4,4,4])
fcc = pc.System()
fcc.atoms = fcc_atoms
fcc.box = fcc_box

hcp_atoms, hcp_box = pcs.make_crystal('hcp', lattice_constant=3.147, repetitions=[4,4,4])
hcp = pc.System()
hcp.atoms = hcp_atoms
hcp.box = hcp_box


Next step is calculation of nearest neighbors. There are two ways to calculate neighbors, by using a cutoff distance or by using the voronoi cells. In this example, we will use the cutoff method and provide a cutoff distance for each structure.

### Finding the cutoff distance¶

The cutoff distance is normally calculated in a such a way that the atoms within the first shell is incorporated in this distance. The :func:pyscal.core.System.calculate_rdf function can be used to find this cutoff distance.

bccrdf = bcc.calculate_rdf()
fccrdf = fcc.calculate_rdf()
hcprdf = hcp.calculate_rdf()


Now the calculated rdf is plotted

fig, (ax1, ax2, ax3) = plt.subplots(1, 3, figsize=(11,4))
ax1.plot(bccrdf[1], bccrdf[0])
ax2.plot(fccrdf[1], fccrdf[0])
ax3.plot(hcprdf[1], hcprdf[0])
ax1.set_xlim(0,5)
ax2.set_xlim(0,5)
ax3.set_xlim(0,5)
ax1.set_title('bcc')
ax2.set_title('fcc')
ax3.set_title('hcp')
ax2.set_xlabel("distance")
ax1.axvline(3.6, color='red')
ax2.axvline(2.7, color='red')
ax3.axvline(3.6, color='red')

<matplotlib.lines.Line2D at 0x7f5cfc5f11d0>


The selected cutoff distances are marked in red in the above plot. For bcc, since the first two shells are close to each other, for this example, we will take the cutoff in such a way that both shells are included.

### Steinhardt’s parameters - cutoff neighbor method¶

bcc.find_neighbors(method='cutoff', cutoff=3.6)
fcc.find_neighbors(method='cutoff', cutoff=2.7)
hcp.find_neighbors(method='cutoff', cutoff=3.6)


We have used a cutoff of 3 here, but this is a parameter that has to be tuned. Using a different cutoff for each structure is possible, but it would complicate the method if the system has a mix of structures. Now we can calculate the $$q_4$$ and $$q_6$$ distributions

bcc.calculate_q([4,6])
fcc.calculate_q([4,6])
hcp.calculate_q([4,6])


Thats it! Now lets gather the results and plot them.

bccq = bcc.get_qvals([4, 6])
fccq = fcc.get_qvals([4, 6])
hcpq = hcp.get_qvals([4, 6])

plt.scatter(bccq[0], bccq[1], s=60, label='bcc', color='#C62828')
plt.scatter(fccq[0], fccq[1], s=60, label='fcc', color='#FFB300')
plt.scatter(hcpq[0], hcpq[1], s=60, label='hcp', color='#388E3C')
plt.xlabel("$q_4$", fontsize=20)
plt.ylabel("$q_6$", fontsize=20)
plt.legend(loc=4, fontsize=15)

<matplotlib.legend.Legend at 0x7f5cfc4edc10>


Firstly, we can see that Steinhardt parameter values of all the atoms fall on one specific point which is due to the absence of thermal vibrations. Next, all the points are well separated and show good distinction. However, at finite temperatures, the atomic positions are affected by thermal vibrations and hence show a spread in the distribution. We will show the effect of thermal vibrations in the next example.

## Structures with thermal vibrations¶

Once again, we create the reqd structures using the pyscal.crystal_structures module. Noise can be applied to atomic positions using the noise keyword as shown below.

bcc_atoms, bcc_box = pcs.make_crystal('bcc', lattice_constant=3.147, repetitions=[10,10,10], noise=0.1)
bcc = pc.System()
bcc.atoms = bcc_atoms
bcc.box = bcc_box

fcc_atoms, fcc_box = pcs.make_crystal('fcc', lattice_constant=3.147, repetitions=[10,10,10], noise=0.1)
fcc = pc.System()
fcc.atoms = fcc_atoms
fcc.box = fcc_box

hcp_atoms, hcp_box = pcs.make_crystal('hcp', lattice_constant=3.147, repetitions=[10,10,10], noise=0.1)
hcp = pc.System()
hcp.atoms = hcp_atoms
hcp.box = hcp_box


### cutoff method¶

bcc.find_neighbors(method='cutoff', cutoff=3.6)
fcc.find_neighbors(method='cutoff', cutoff=2.7)
hcp.find_neighbors(method='cutoff', cutoff=3.6)


And now, calculate $$q_4$$, $$q_6$$ parameters

bcc.calculate_q([4,6])
fcc.calculate_q([4,6])
hcp.calculate_q([4,6])


Gather the q vales and plot them

bccq = bcc.get_qvals([4, 6])
fccq = fcc.get_qvals([4, 6])
hcpq = hcp.get_qvals([4, 6])

plt.scatter(fccq[0], fccq[1], s=10, label='fcc', color='#FFB300')
plt.scatter(hcpq[0], hcpq[1], s=10, label='hcp', color='#388E3C')
plt.scatter(bccq[0], bccq[1], s=10, label='bcc', color='#C62828')
plt.xlabel("$q_4$", fontsize=20)
plt.ylabel("$q_6$", fontsize=20)
plt.legend(loc=4, fontsize=15)

<matplotlib.legend.Legend at 0x7f5cfc743a50>


The thermal vibrations cause the distributions to spread, but it still very good. Lechner and Dellago proposed using the averaged distributions, $$\bar{q}_4-\bar{q}_6$$ to better distinguish the distributions. Lets try that.

bcc.calculate_q([4,6], averaged=True)
fcc.calculate_q([4,6], averaged=True)
hcp.calculate_q([4,6], averaged=True)

bccaq = bcc.get_qvals([4, 6], averaged=True)
fccaq = fcc.get_qvals([4, 6], averaged=True)
hcpaq = hcp.get_qvals([4, 6], averaged=True)


Lets see if these distributions are better..

plt.scatter(fccaq[0], fccaq[1], s=10, label='fcc', color='#FFB300')
plt.scatter(hcpaq[0], hcpaq[1], s=10, label='hcp', color='#388E3C')
plt.scatter(bccaq[0], bccaq[1], s=10, label='bcc', color='#C62828')
plt.xlabel("$q_4$", fontsize=20)
plt.ylabel("$q_6$", fontsize=20)
plt.legend(loc=4, fontsize=15)

<matplotlib.legend.Legend at 0x7f5cfc125650>


This looks much better! We can see that the resolution is much better than the non averaged versions.

There is also the possibility to calculate structures using Voronoi based neighbor identification too. Let’s try that now.

bcc.find_neighbors(method='voronoi')
fcc.find_neighbors(method='voronoi')
hcp.find_neighbors(method='voronoi')

bcc.calculate_q([4,6], averaged=True)
fcc.calculate_q([4,6], averaged=True)
hcp.calculate_q([4,6], averaged=True)

bccaq = bcc.get_qvals([4, 6], averaged=True)
fccaq = fcc.get_qvals([4, 6], averaged=True)
hcpaq = hcp.get_qvals([4, 6], averaged=True)


Plot the calculated points..

plt.scatter(fccaq[0], fccaq[1], s=10, label='fcc', color='#FFB300')
plt.scatter(hcpaq[0], hcpaq[1], s=10, label='hcp', color='#388E3C')
plt.scatter(bccaq[0], bccaq[1], s=10, label='bcc', color='#C62828')
plt.xlabel("$q_4$", fontsize=20)
plt.ylabel("$q_6$", fontsize=20)
plt.legend(loc=4, fontsize=15)

<matplotlib.legend.Legend at 0x7f5cfbe5a510>


Voronoi based method also provides good resolution,the major difference being that the location of bcc distribution is different.