# 2D Hard Sphere (Hard Disk) Problem¶

Doruk Efe GĂ¶kmen -- 26/07/2018 -- Ankara

## Event Driven Hard Disks Simulation (Molecular Dynamics Simulation)¶

Solves the Newtonian dynamics of a 4 hard disk system exactly. Spheres move in straight lines between collisions and the collision events are calculated by finding the minimum of 6 time values until the walls and the other disks.

Note that this algorithm is exact assuming that the numbers for the velocity and the times are calculated with infinite precision. However this is not the case for computers where the numbers are truncated at a finite value. By definition, chaotic systems are characterised by the vas differences in the final states despite having only marginal differences in their initial conditions. Therefore, although the numbers calculated by the molecular dynamics algorithm have millions of digits, the simulation is valid only for a limited number of iterations. (The chaoticity of the system is caused by the negative curvature of the surface of the disk.)

In [22]:
%pylab qt
import os, math, pylab

output_dir = "event_disks_box_movie"
colors = ['r', 'b', 'g', 'orange']

def wall_time(pos_a, vel_a, sigma):
if vel_a > 0.0:
del_t = (1.0 - sigma - pos_a) / vel_a
elif vel_a < 0.0:
del_t = (pos_a - sigma) / abs(vel_a)
else:
del_t = float('inf')
return del_t

def pair_time(pos_a, vel_a, pos_b, vel_b, sigma):
del_x = [pos_b[0] - pos_a[0], pos_b[1] - pos_a[1]]
del_x_sq = del_x[0] ** 2 + del_x[1] ** 2
del_v = [vel_b[0] - vel_a[0], vel_b[1] - vel_a[1]]
del_v_sq = del_v[0] ** 2 + del_v[1] ** 2
scal = del_v[0] * del_x[0] + del_v[1] * del_x[1]
Upsilon = scal ** 2 - del_v_sq * (del_x_sq - 4.0 * sigma ** 2)
if Upsilon > 0.0 and scal < 0.0:
del_t = - (scal + math.sqrt(Upsilon)) / del_v_sq
else:
del_t = float('inf')
return del_t

def min_arg(l):
return min(zip(l, range(len(l))))

def compute_next_event(pos, vel):
wall_times = [wall_time(pos[k][l], vel[k][l], sigma) for k, l in singles]
pair_times = [pair_time(pos[k], vel[k], pos[l], vel[l], sigma) for k, l in pairs]
return min_arg(wall_times + pair_times)

def compute_new_velocities(pos, vel, next_event_arg):
if next_event_arg < len(singles):
collision_disk, direction = singles[next_event_arg]
vel[collision_disk][direction] *= -1.0
else:
a, b = pairs[next_event_arg - len(singles)]
del_x = [pos[b][0] - pos[a][0], pos[b][1] - pos[a][1]]
abs_x = math.sqrt(del_x[0] ** 2 + del_x[1] ** 2)
e_perp = [c / abs_x for c in del_x]
del_v = [vel[b][0] - vel[a][0], vel[b][1] - vel[a][1]]
scal = del_v[0] * e_perp[0] + del_v[1] * e_perp[1]
for k in range(2):
vel[a][k] += e_perp[k] * scal
vel[b][k] -= e_perp[k] * scal

pylab.gcf().set_size_inches(6, 6)
img = 0
if not os.path.exists(output_dir): os.makedirs(output_dir)
def snapshot(t, pos, vel, colors, arrow_scale=.2):
global img
pylab.cla()
pylab.axis([0, 1, 0, 1])
pylab.setp(pylab.gca(), xticks=[0, 1], yticks=[0, 1])
for (x, y), (dx, dy), c in zip(pos, vel, colors):
dx *= arrow_scale
dy *= arrow_scale
circle = pylab.Circle((x, y), radius=sigma, fc=c)
pylab.text(.5, 1.03, 't = %.2f' % t, ha='center')
pylab.savefig(os.path.join(output_dir, '%04i.png' % img))
pylab.pause(0.00001)
pylab.show()
img += 1

pos = [[0.25, 0.25], [0.75, 0.25], [0.25, 0.75], [0.75, 0.75]]
vel = [[0.21, 0.12], [0.71, 0.18], [-0.23, -0.79], [0.78, 0.1177]]
singles = [(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)]
pairs = [(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
sigma = 0.15
t = 0.0
dt = 0.02     # dt=0 corresponds to event-to-event animation
n_steps = 100
next_event, next_event_arg = compute_next_event(pos, vel)
snapshot(t, pos, vel, colors)
for step in range(n_steps):
pylab.clf()
if dt:
next_t = t + dt
else:
next_t = t + next_event
while t + next_event <= next_t:
t += next_event
for k, l in singles: pos[k][l] += vel[k][l] * next_event
compute_new_velocities(pos, vel, next_event_arg)
next_event, next_event_arg = compute_next_event(pos, vel)
remain_t = next_t - t
for k, l in singles: pos[k][l] += vel[k][l] * remain_t
t += remain_t
next_event -= remain_t
snapshot(t, pos, vel, colors)
#print 'time',t

print('Producing animation.gif using ImageMagick...')
os.system("convert -delay 1 -dispose Background +page " + str(output_dir)
+ "/*.png -loop 0 " + str(output_dir) + "/animation.gif")

Populating the interactive namespace from numpy and matplotlib

/anaconda3/envs/python2/lib/python2.7/site-packages/IPython/core/magics/pylab.py:161: UserWarning: pylab import has clobbered these variables: ['random']
%matplotlib prevents importing * from pylab and numpy
"\n%matplotlib prevents importing * from pylab and numpy"

Producing animation.gif using ImageMagick...

Out[22]:
32512

## Direct Sampling Hard Disks¶

By Boltzmann's equiprobability hypothesis and ergodic hypothesis, given infinite time, the system explores all of its possible configurations with equal probability. This key idea leads to the following algorithm that samples possible configurations of 4 hard disk system through direct sampling. For an infinitely large sampling, the resulting frames of the system constitute a "time scrambled" version of the actual dynamics of the system.

In [1]:
%pylab qt
import random, math, os, pylab

output_dir = 'direct_disks_box_movie'

def direct_disks_box(N, sigma):
condition = False
while condition == False:
L = [(random.uniform(sigma, 1.0 - sigma), random.uniform(sigma, 1.0 - sigma))]
for k in range(1, N):
a = (random.uniform(sigma, 1.0 - sigma), random.uniform(sigma, 1.0 - sigma))
min_dist = min(math.sqrt((a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2) for b in L)
if min_dist < 2.0 * sigma:
condition = False
break
else:
L.append(a)
condition = True
return L

img = 0
if not os.path.exists(output_dir): os.makedirs(output_dir)
def snapshot(pos, colors):
global img
pylab.gcf().set_size_inches(6, 6)
pylab.axis([0, 1, 0, 1])
pylab.setp(pylab.gca(), xticks=[0, 1], yticks=[0, 1])
for (x, y), c in zip(pos, colors):
circle = pylab.Circle((x, y), radius=sigma, fc=c)
#pylab.savefig(os.path.join(output_dir, '%d.png' % img), transparent=True)
pylab.pause(0.01)
pylab.show()
img += 1

N = 4 #the number of disks
colors = ['r', 'b', 'g', 'orange']
sigma = 0.2 #the radii of the disks
n_runs = 8
for run in range(n_runs):
pos = direct_disks_box(N, sigma)
snapshot(pos, colors)
pylab.clf()

Populating the interactive namespace from numpy and matplotlib


Direct sampling with periodic boundary conditions. Notice that for $N=16$ disks of radii $\sigma=\sqrt{\frac{\eta}{\pi N}}$, where $\eta=0.26$ is the density of the disks, it is found that it usually occurred thousands of rejections before producing a legitimate configuration! It turns out that the acceptance ratio in the case of $\eta=0.3$ and $N=16$, the acceptance ratio $p_{\text{acceptance}}=5\times 10^{-6}$. The accepted cases strictly characterise the portion of the $32$-dimensional configuration space that corresponds to the $16$ hard disk system. In other words, the rejection probability can be used to calculate the partition function $Z(\eta)$ of the system.

For a $2$-D box of volume V, the $32$-dimensional configuration space has the volume $V^{16}$. If the radius of the disk is $\sigma=0 \implies \eta=0$, i.e. the density of the box is $0$, then all the points of the configuration space are legal configurations of the system. Hence the partition function of that system is $Z(N,\eta=0)=V^N$. On the other hand $Z(N,\eta)=Z(N,\eta=0)p_{\text{acceptance}}$.

In [4]:
%pylab inline
import random, math, pylab, os

#Introduce the periodic boundary conditions via the modular distance function between two 2D vectors x, y:
def dist(x,y):
d_x = abs(x[0] - y[0]) % 1.0 #distance between the first compononents of two vectors in modulo 1
d_x = min(d_x, 1.0 - d_x) #the modular distance is the minimum of cases due to periodicity
d_y = abs(x[1] - y[1]) % 1.0 #distance between the second compononents of two vectors in modulo 1
d_y = min(d_y, 1.0 - d_y)
return  math.sqrt(d_x**2 + d_y**2) #returns the modular distance

def direct_disks(N, sigma): #constructs a legal sample in the 2N-dimensional unit hypercube
n_iter = 0
condition = False
while condition == False:
n_iter += 1 #increases in the case of failure to find non-overlapping disk coordinates
L = [(random.random(), random.random())] #random vector in 2N-dimensional hypercube
for k in range(1, N):
a = (random.random(), random.random()) #random vector in the 2D square
min_dist = min(dist(a, b) for b in L) #minimum of the distances between any two disks
#if the modular distance between any two disks is less than their radii, then breaks and tries again
if min_dist < 2.0 * sigma:
condition = False
break
else:
L.append(a) #if no overlap occurs, then adds a new disk with coordinates a to the system
condition = True
return n_iter, L #the number of iterations required to find a legal configuration and its coordinates

img = 0
output_dir = 'direct_disks_multirun_movie'
if not os.path.exists(output_dir): os.makedirs(output_dir)
def snapshot(pos, colors, border_color = 'k'):
global img
pylab.figure()
pylab.axis([0, 1, 0, 1])
[i.set_linewidth(2) for i in pylab.gca().spines.itervalues()]
[i.set_color(border_color) for i in pylab.gca().spines.itervalues()]
pylab.setp(pylab.gca(), xticks = [0, 1], yticks = [0, 1], aspect = 'equal')
for (x, y), c in zip(pos, colors):
circle = pylab.Circle((x, y), radius = sigma, fc = c)
pylab.savefig(output_dir+'/snapshot_%03i.png'%img)
pylab.pause(0.001)
pylab.close()
img += 1

def periodicize(config):
images = [-1.0, 0.0, 1.0]
return [(x + dx, y + dy) for (x,y) in config for dx in images for dy in images]

N = 16 #the number of the disks
eta = 0.26 #the disk density of the system (fraction of space occupied by the disks)
sigma = math.sqrt(eta / N / math.pi) #the radii of the disks
n_runs = 5 #number of runs
colors = ['r' for i in range(8 * N)]
for run in range(n_runs):
pylab.clf()
iterations, config =  direct_disks(N, sigma)
print 'run',run
print iterations - 1, 'tabula rasa wipe-outs before producing the following configuration'
print config
print
config_per = periodicize(config)
snapshot(config_per, colors, border_color = 'k')

Populating the interactive namespace from numpy and matplotlib
run 0
30036 tabula rasa wipe-outs before producing the following configuration
[(0.5290863088869059, 0.026602165942082534), (0.7068861913834033, 0.841765426057141), (0.350246769636395, 0.5691655942222656), (0.5645476848710024, 0.25386832218349087), (0.04798050921429031, 0.35387875591009954), (0.7908119383793827, 0.016254686601902435), (0.9999214172719869, 0.4952435341395388), (0.14324225234241617, 0.7544997761130654), (0.790845238067139, 0.6926598186347003), (0.6603863557473032, 0.424173085826433), (0.5231276402387884, 0.7180344838877797), (0.3431909675284315, 0.025476194822818043), (0.34948072471303104, 0.3231388243348168), (0.958008437682202, 0.7639663257620449), (0.8479799683835139, 0.35298926296288435), (0.2538594432886999, 0.18606330998718656)]


<Figure size 432x288 with 0 Axes>
run 1
16547 tabula rasa wipe-outs before producing the following configuration
[(0.820775802536151, 0.6382203875836991), (0.4994876147259021, 0.6944488152203134), (0.8262359021855195, 0.015628632683440946), (0.9371208419822434, 0.8968165824312339), (0.17126203609104784, 0.5847827680049164), (0.14538249802057335, 0.8095048871852456), (0.6514025090157548, 0.24049746233092262), (0.8444344838268607, 0.3206571424503102), (0.4309482077730372, 0.4158203150383223), (0.6118328659936254, 0.4489655140103994), (0.26674751156518983, 0.07280573592629969), (0.3367265167647999, 0.5685097793003173), (0.44624884853100755, 0.8332232581634768), (0.6291172175885347, 0.903243985050751), (0.30286450459953096, 0.29800677302308276), (0.7924321388459232, 0.7967320275977108)]


<Figure size 432x288 with 0 Axes>
run 2
7516 tabula rasa wipe-outs before producing the following configuration
[(0.997011476912311, 0.23823009873908463), (0.20621709483185124, 0.5891035784239245), (0.10118350467709525, 0.07907699087981945), (0.4716465280346849, 0.417454687405741), (0.4678218946283592, 0.7141723043275596), (0.3147159813855247, 0.7210526639527041), (0.9952703671930787, 0.6922178099114037), (0.35585010258529315, 0.5649069148664847), (0.6609343694723954, 0.5163412127065151), (0.5608111707904527, 0.2627446089358014), (0.24357655441213055, 0.9462511646654779), (0.28150850754722967, 0.3405060330435056), (0.9018256207654266, 0.10565325251846991), (0.6224147643203051, 0.6670379720594759), (0.5510108953783537, 0.0866724260029228), (0.7678616786480671, 0.9479731188795825)]


<Figure size 432x288 with 0 Axes>
run 3
26138 tabula rasa wipe-outs before producing the following configuration
[(0.9434199654999524, 0.15218440298230407), (0.09745669212185293, 0.6460152193667532), (0.27098153940391645, 0.2922309691987671), (0.3188707740561313, 0.5554249964405475), (0.7312006374701823, 0.7166516747159463), (0.15511726616813903, 0.14901876055716923), (0.9306327850845793, 0.5145610011161166), (0.3417698978206465, 0.7565287669956946), (0.5230768300029479, 0.6176798382811663), (0.5923745591850242, 0.3794954882057897), (0.501353747848843, 0.7822898246983823), (0.37901538349632535, 0.12374150475647527), (0.002423264837273287, 0.3230211686176111), (0.20415136567818648, 0.8777440359462632), (0.8520738693427051, 0.8339903634866587), (0.8022041186441675, 0.05433996145526898)]


<Figure size 432x288 with 0 Axes>
run 4
6084 tabula rasa wipe-outs before producing the following configuration
[(0.7590878221547396, 0.38609759080933814), (0.7785486579710811, 0.5892174114517533), (0.5732064268778967, 0.3005173238565446), (0.5168950312242604, 0.7537428959811848), (0.4195615209396405, 0.9178446574429164), (0.20762995655358862, 0.5829254689199423), (0.765030883264898, 0.8155794809216419), (0.33782792836027997, 0.33804887534482286), (0.16593128400739898, 0.9556577252452482), (0.1015162527233131, 0.3432614494975966), (0.20573498888013464, 0.7501051776901527), (0.9916406419116316, 0.5596533897780842), (0.4981527878862717, 0.4537827964280835), (0.6059255756849841, 0.10226042014195891), (0.13771319471837562, 0.15787588170708022), (0.38774080885250817, 0.16451894627990304)]


<Figure size 432x288 with 0 Axes>

The acceptance probability $p_{\text{acceptance}}(\eta)$ is calculated by the following code.

In [10]:
%pylab inline
import random, math, pylab

def dist(x, y): #periodic boundary conditions as before
d_x = abs(x[0] - y[0]) % 1.0
d_x = min(d_x, 1.0 - d_x)
d_y = abs(x[1] - y[1]) % 1.0
d_y = min(d_y, 1.0 - d_y)
return  math.sqrt(d_x**2 + d_y**2)

N = 16 #number of disks
n_confs = 10 ** 5 #number of configurations
pairs = [(i, j) for i in range(N - 1) for j in range(i + 1, N)]
eta_max_list = [] #initialise the allowed maximum densities
for conf in xrange(n_confs):
#sample a random configuration -- overlapping/non-overlapping
#i.e. pick a random vector in 2N-D unit hypercube
L = [(random.random(), random.random()) for k in range(N)]
#determine the maximum possible radius so that any of the two disks in the configuration overlap
sigma_max = min(dist(L[i], L[j]) for i, j in pairs) / 2.0
eta_max = N * math.pi * sigma_max ** 2 #calculate the corresponding maximum density
eta_max_list.append(eta_max)
#The histogram of these maximum densities corresponds to the acceptance probability!

# Begin of graphics output
pylab.figure()
n, bins, patches = pylab.hist(eta_max_list, 100, histtype='step', cumulative=-1,
log=True, normed=True, label="numerical evaluation of $p_{accept}$")
explaw = [math.exp( - 2.0 * (N - 1) * eta) for eta in bins] #first term in Virial expansion
pylab.plot(bins, explaw, 'r--', linewidth=1.5, label="1st order virial expansion")
pylab.xlabel('density \eta')
pylab.ylabel('$p_{accept}(\eta)$')
pylab.legend()
pylab.show()

Populating the interactive namespace from numpy and matplotlib


## Markov-Chain Sampling Hard Disks¶

Direct sampling for hard disks works only at low densities and small particle numbers, and we thus switch to a more general Markov-chain Monte Carlo algorithm. Similar approach as direct sampling but this time the sampling is done through the Markov-chains, i.e. the location of a particular disk is updated from where it was at the previous frame. The rejection cases are the cases where any two of the disks overlap. The necessary conditions are aperiodicity and irreducibility. Aperiodicity is trivial in this case. Here, reducibility corresponds to the case where the radii of the disks are so large, so that they stuck and vibrate at their initial locations. There, the system can be reduced into four independent systems. Hence, irreducibility is a condition on the size of the radii of the disks.

There is a link between the acceptance probability of the system and the partition function as seen in the previous section.

In [5]:
%pylab qt
import random, os, pylab

output_dir = 'markov_disks_box_movie'

img = 0
if not os.path.exists(output_dir): os.makedirs(output_dir)
def snapshot(pos, colors):
global img
pylab.gcf().set_size_inches(6, 6)
pylab.axis([0, 1, 0, 1])
pylab.setp(pylab.gca(), xticks=[0, 1], yticks=[0, 1])
for (x, y), c in zip(pos, colors):
circle = pylab.Circle((x, y), radius=sigma, fc=c)
#pylab.savefig(os.path.join(output_dir, '%d.png' % img), transparent=True)
pylab.pause(0.0001)
pylab.show()
img += 1

L = [[0.25, 0.25], [0.75, 0.25], [0.25, 0.75], [0.75, 0.75]] #initial positions of the disks
sigma = 0.15 #the radii of disks
sigma_sq = sigma ** 2
delta = 0.1
colors = ['r', 'b', 'g', 'orange']
n_steps = 50
for step in range(n_steps):
pylab.clf()
snapshot(L, colors)
a = random.choice(L)
b = [a[0] + random.uniform(-delta, delta), a[1] + random.uniform(-delta, delta)]
min_dist = min((b[0] - c[0]) ** 2 + (b[1] - c[1]) ** 2 for c in L if c != a)
box_cond = min(b[0], b[1]) < sigma or max(b[0], b[1]) > 1.0 - sigma
if not (box_cond or min_dist < 4.0 * sigma ** 2):
a[:] = b

Populating the interactive namespace from numpy and matplotlib


The "tabula rasa" strategy explained:

In [6]:
import random

N = 4
sigma = 0.2 #the radii of disks
pairs = [(i, j) for i in range(N - 1) for j in range(i + 1, N)]
while True:
#place four disks at randomly chosen positions: uniform vector in an 8D hypercube
L = [(random.uniform(sigma, 1.0 - sigma), random.uniform(sigma, 1.0 - sigma)) for k in range(N)]
#check for an overlap between the disks by calculating the distance in between all disks
if  min((L[i][0] - L[j][0]) ** 2 + (L[i][1] - L[j][1]) ** 2 for i, j in pairs) > 4.0 * sigma ** 2:
break #if overlap occurs, the algorithm breaks
print L #Sample coordinates of the four disks are outputted

[(0.3029469955941781, 0.7670627214525076), (0.7129377088143414, 0.7693773660570689), (0.32291343374765724, 0.2737530107568247), (0.7867778045636022, 0.25558665751782333)]


On the other hand, it is crucial to note that random sequential deposition is forbidden since it yields inequal probabilities. The following example illustrates this in a 1D discrete example where two rods are placed on a grid of size 5.

In [1]:
%pylab qt
import random, pylab, os

output_dir = 'random_sequential_discrete_movie'
if not os.path.exists(output_dir): os.makedirs(output_dir)
def show_rods(red_rod, blue_rod, run, trial, frame):
fig, ax = pylab.subplots()
ax.set_xticks([0, 1, 2, 3, 4])
ax.set_yticks([])
height = 1.0
redrect = pylab.Rectangle((red_rod - 1.5, 0.0), 3.0, 1.1 * height,  fc = 'r')
bluerect = pylab.Rectangle((blue_rod-1.5,0.0), 3.0, height,  fc = 'b')
pylab.axis('scaled')
pylab.axis([-1.5, 5.5, 0.0, 2.5*height])
pylab.xlabel("x")
if abs(red_rod - blue_rod) > 2:
pylab.title('run %d, trial %d (ACCEPTED!)' % (run, trial))
else:
pylab.title('run %d, trial %d (REJECTED!)' % (run, trial))
pylab.savefig(output_dir+'/random_sequential_discrete_frame%04i.png' % (frame))
pylab.pause(0.0001)
pylab.close()

configurations = {(0, 3): 'a', (0, 4): 'b', (1, 4): 'c',
(3, 0): 'd', (4, 0): 'e', (4, 1): 'f'}
counts = {'a': 0, 'b': 0, 'c': 0, 'd': 0, 'e': 0, 'f': 0}
n_runs = 10
frame = 0
trial = 0
for run in range(n_runs):
pylab.clf()
red_rod = random.randint(0, 3)
if red_rod >= 2: red_rod += 1
trial = 0
while True:
blue_rod = random.randint(0, 4)
show_rods(red_rod, blue_rod, run, trial, frame)
trial += 1
frame += 1
if abs(red_rod - blue_rod) > 2: break
conf = configurations[(red_rod, blue_rod)]
counts[conf] += 1
for conf in counts:
print conf, counts[conf] / float(n_runs)

Populating the interactive namespace from numpy and matplotlib
a 0.1
c 0.2
b 0.0
e 0.0
d 0.4
f 0.3


## Equiprobability¶

Using small boxes [x - del_xy, x + del_xy], etc, we can show that the probability to sample the following "marked" configurations a, b, and c (given in the code) are the same (within the numerical precision), with the following codes for direct sampling, Markov-chain sampling, and simulation of Newtonian mechanics via the event driven algorithm.

### Direct sampling:¶

In [3]:
import random, math
def direct_disks_box(N, sigma): #same direct sampling as in the second section
condition = False
while condition == False:
L = [(random.uniform(sigma, 1.0 - sigma), random.uniform(sigma, 1.0 - sigma))]
for k in range(1, N):
a = (random.uniform(sigma, 1.0 - sigma), random.uniform(sigma, 1.0 - sigma))
min_dist = min(math.sqrt((a[0] - b[0]) ** 2 + (a[1] - b[1]) ** 2) for b in L)
if min_dist < 2.0 * sigma:
condition = False
break
else:
L.append(a)
condition = True
return L

del_xy = 0.05 #"uncertainty"
n_runs = 1000000

#Define the "marked" configurations:
conf_a = ((0.30, 0.30), (0.30, 0.70), (0.70, 0.30), (0.70,0.70))
conf_b = ((0.20, 0.20), (0.20, 0.80), (0.75, 0.25), (0.75,0.75))
conf_c = ((0.30, 0.20), (0.30, 0.80), (0.70, 0.20), (0.70,0.70))
configurations = [conf_a, conf_b, conf_c] #list the configurations

hits = {conf_a: 0, conf_b: 0, conf_c: 0} #initialise the number of times each marked configuration occurs

for run in range(n_runs):
x_vec = direct_disks_box(4, sigma) #generates a random sample by direct sampling
for conf in configurations: #run a loop iterating over the given 3 configurations
#condition that a randomly generated configuration L is the same as a, b or c up to uncertainty of del_xy
condition_hit = True
for b in conf: #run a loop iterating over the 4 disk coordinates in a specific configuration
#If the max(x distance and y distance between a disk in L and a disk in conf_a,b,c)
#is less than the given uncertainty del_xy, then we treat the two disks as in the same location.
#Note that the "any two disks" condition is realised by minimising over all 4 disks in a
#randomly sampled configuration L.
condition_b = min(max(abs(a[0] - b[0]), abs(a[1] - b[1])) for a in x_vec) < del_xy
#The following logical variable is 1 only if there exists 4 disk pairs are within del_xy range.
#If at least any one of the disks does not have a pair within del_xy, then it is 0.
condition_hit *= condition_b #multiplies condition_b's (for all 4 disks)
#If the current L and a, b or c are the same up to uncertainty del_xy, then increase:
if condition_hit:
hits[conf] += 1

for conf in configurations:
print conf, hits[conf] #Print the configurations and the number of times they occured.

((0.3, 0.3), (0.3, 0.7), (0.7, 0.3), (0.7, 0.7)) 110
((0.2, 0.2), (0.2, 0.8), (0.75, 0.25), (0.75, 0.75)) 114
((0.3, 0.2), (0.3, 0.8), (0.7, 0.2), (0.7, 0.7)) 115


### Markov-chain sampling:¶

In [13]:
import random

def markov_disks_box(L, delta, sigma):
condition = True
while condition == True:
a = random.choice(L)
b = [a[0] + random.uniform(-delta, delta), a[1] + random.uniform(-delta, delta)]
min_dist = min((b[0] - c[0]) ** 2 + (b[1] - c[1]) ** 2 for c in L if c != a)
box_cond = min(b[0], b[1]) < sigma or max(b[0], b[1]) > 1.0 - sigma
if not (box_cond or min_dist < 4.0 * sigma ** 2):
a[:] = b
condition = False
break
return L

#inputs of the markov_disks_box function:
#initial positions of the disks to startup the Markov-chain
L = [[0.25, 0.25], [0.75, 0.25], [0.25, 0.75], [0.75, 0.75]]
delta = 0.1

n_steps = 10000
del_xy = 0.05 #"uncertainty"

#Define the "marked" configurations:
conf_a = ((0.30, 0.30), (0.30, 0.70), (0.70, 0.30), (0.70,0.70))
conf_b = ((0.20, 0.20), (0.20, 0.80), (0.75, 0.25), (0.75,0.75))
conf_c = ((0.30, 0.20), (0.30, 0.80), (0.70, 0.20), (0.70,0.70))
configurations = [conf_a, conf_b, conf_c] #list the configurations

hits = {conf_a: 0, conf_b: 0, conf_c: 0} #initialise the number of times each marked configuration occurs

for run in range(n_steps):
x_vec = markov_disks_box(L, delta, sigma) #generates a random sample by direct sampling
for conf in configurations: #run a loop iterating over the given 3 configurations
#condition that a randomly generated configuration L is the same as a, b or c up to uncertainty of del_xy
condition_hit = True
for b in conf: #run a loop iterating over the 4 disk coordinates in a specific configuration
#If the max(x distance and y distance between a disk in L and a disk in conf_a,b,c)
#is less than the given uncertainty del_xy, then we treat the two disks as in the same location.
#Note that the "any two disks" condition is realised by minimising over all 4 disks in a
#randomly sampled configuration L.
condition_b = min(max(abs(a[0] - b[0]), abs(a[1] - b[1])) for a in x_vec) < del_xy
#The following logical variable is 1 only if there exists 4 disk pairs are within del_xy range.
#If at least any one of the disks does not have a pair within del_xy, then it is 0.
condition_hit *= condition_b #multiplies condition_b's (for all 4 disks)
#If the current L and a, b or c are the same up to uncertainty del_xy, then increase:
if condition_hit:
hits[conf] += 1

for conf in configurations:
print conf, hits[conf] #Print the configurations and the number of times they occured.

((0.3, 0.3), (0.3, 0.7), (0.7, 0.3), (0.7, 0.7)) 3
((0.2, 0.2), (0.2, 0.8), (0.75, 0.25), (0.75, 0.75)) 1
((0.3, 0.2), (0.3, 0.8), (0.7, 0.2), (0.7, 0.7)) 0


### Newtonian dynamics¶

In [12]:
import math, pylab

def wall_time(pos_a, vel_a, sigma):
if vel_a > 0.0:
del_t = (1.0 - sigma - pos_a) / vel_a
elif vel_a < 0.0:
del_t = (pos_a - sigma) / abs(vel_a)
else:
del_t = float('inf')
return del_t

def pair_time(pos_a, vel_a, pos_b, vel_b, sigma):
del_x = [pos_b[0] - pos_a[0], pos_b[1] - pos_a[1]]
del_x_sq = del_x[0] ** 2 + del_x[1] ** 2
del_v = [vel_b[0] - vel_a[0], vel_b[1] - vel_a[1]]
del_v_sq = del_v[0] ** 2 + del_v[1] ** 2
scal = del_v[0] * del_x[0] + del_v[1] * del_x[1]
Upsilon = scal ** 2 - del_v_sq * ( del_x_sq - 4.0 * sigma **2)
if Upsilon > 0.0 and scal < 0.0:
del_t = - (scal + math.sqrt(Upsilon)) / del_v_sq
else:
del_t = float('inf')
return del_t

conf_a = ((0.30, 0.30), (0.30, 0.70), (0.70, 0.30), (0.70,0.70))
conf_b = ((0.20, 0.20), (0.20, 0.80), (0.75, 0.25), (0.75,0.75))
conf_c = ((0.30, 0.20), (0.30, 0.80), (0.70, 0.20), (0.70,0.70))
configurations = [conf_a, conf_b, conf_c]
hits = {conf_a: 0, conf_b: 0, conf_c: 0}
del_xy = 0.10
pos = [[0.25, 0.25], [0.75, 0.25], [0.25, 0.75], [0.75, 0.75]]
vel = [[0.21, 0.12], [0.71, 0.18], [-0.23, -0.79], [0.78, 0.1177]]
singles = [(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)]
pairs = [(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]
sigma = 0.10
t = 0.0
n_events = 5000000
for event in range(n_events):
if event % 100000 == 0:
print event
wall_times = [wall_time(pos[k][l], vel[k][l], sigma) for k, l  in singles]
pair_times = [pair_time(pos[k], vel[k], pos[l], vel[l], sigma) for k, l in pairs]
next_event = min(wall_times + pair_times)
t_previous = t
for inter_times in range(int(t + 1), int(t + next_event + 1)):
del_t = inter_times - t_previous
for k, l in singles:
pos[k][l] += vel[k][l] * del_t
t_previous = inter_times
#print t "Configuration analysis is done"
for conf in configurations:
condition_hit = True
for b in conf:
condition_b = min(max(abs(a[0] - b[0]), abs(a[1] - b[1])) for a in pos) < del_xy
condition_hit *= condition_b
if condition_hit:
hits[conf] += 1
t += next_event
del_t = t - t_previous
for k, l in singles:
pos[k][l] += vel[k][l] * del_t
if min(wall_times) < min(pair_times):
collision_disk, direction = singles[wall_times.index(next_event)]
vel[collision_disk][direction] *= -1.0
else:
a, b = pairs[pair_times.index(next_event)]
del_x = [pos[b][0] - pos[a][0], pos[b][1] - pos[a][1]]
abs_x = math.sqrt(del_x[0] ** 2 + del_x[1] ** 2)
e_perp = [c / abs_x for c in del_x]
del_v = [vel[b][0] - vel[a][0], vel[b][1] - vel[a][1]]
scal = del_v[0] * e_perp[0] + del_v[1] * e_perp[1]
for k in range(2):
vel[a][k] += e_perp[k] * scal
vel[b][k] -= e_perp[k] * scal

print t, "The total running time of the program"

for conf in configurations:
print conf, hits[conf]

0
100000
200000
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500000
600000
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800000
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1000000
1100000
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2000000
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3000000
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3600000
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3900000
4000000
4100000
4200000
4300000
4400000
4500000
4600000
4700000
4800000
4900000
611527.898476
((0.3, 0.3), (0.3, 0.7), (0.7, 0.3), (0.7, 0.7)) 668
((0.2, 0.2), (0.2, 0.8), (0.75, 0.25), (0.75, 0.75)) 690
((0.3, 0.2), (0.3, 0.8), (0.7, 0.2), (0.7, 0.7)) 655


### Calculations in terms of observables¶

Instead of the configurations themselves, and their probability distribution, we now consider an observable, in fact a particularly simple one, the position x: the x-coordinate of the center of a disk. We will compute its probability distribution, as the normed histogram of x-positions. This histogram is the same for all disks, so we can collect data for one disk or for all of them.

Direct sampling:

In [14]:
%pylab inline
import random, pylab

N = 4
sigma = 0.1197
n_runs = 1000000

histo_data = []
for run in range(n_runs):
pos = direct_disks_box(N, sigma) #this function was defined in the previous section
for k in range(N):
histo_data.append(pos[k][0])
pylab.hist(histo_data, bins=100, normed=True)
pylab.xlabel('x')
pylab.ylabel('frequency')
pylab.title('Direct sampling: x coordinate histogram (density eta=0.18)')
pylab.grid()
pylab.savefig('direct_disks_histo.png')
pylab.show()

Populating the interactive namespace from numpy and matplotlib

/anaconda3/envs/python2/lib/python2.7/site-packages/IPython/core/magics/pylab.py:161: UserWarning: pylab import has clobbered these variables: ['pylab', 'random']
%matplotlib prevents importing * from pylab and numpy
"\n%matplotlib prevents importing * from pylab and numpy"


Markov-chain sampling:

In [7]:
%pylab inline
import random, pylab

L = [[0.25, 0.25], [0.75, 0.25], [0.25, 0.75], [0.75, 0.75]]
delta = 0.2 #may need to be varied
sigma = 0.1197
N=4

n_steps = 2000000

histo_data = []
for steps in range(n_steps):
pos = markov_disks_box(L, delta, sigma) #this function was defined in the previous section
for k in range(N):
histo_data.append(pos[k][0])
pylab.hist(histo_data, bins=100, normed=True)
pylab.xlabel('x')
pylab.ylabel('frequency')
pylab.title('Markov sampling: x coordinate histogram (density eta=0.18)')
pylab.grid()
pylab.savefig('markov_disks_histo.png')
pylab.show()

Populating the interactive namespace from numpy and matplotlib

/anaconda3/envs/python2/lib/python2.7/site-packages/matplotlib/axes/_axes.py:6462: UserWarning: The 'normed' kwarg is deprecated, and has been replaced by the 'density' kwarg.
warnings.warn("The 'normed' kwarg is deprecated, and has been "


Event drivent Newtonian dynamics:

In [9]:
import math, pylab

def wall_time(pos_a, vel_a, sigma):
if vel_a > 0.0:
del_t = (1.0 - sigma - pos_a) / vel_a
elif vel_a < 0.0:
del_t = (pos_a - sigma) / abs(vel_a)
else:
del_t = float('inf')
return del_t

def pair_time(pos_a, vel_a, pos_b, vel_b, sigma):
del_x = [pos_b[0] - pos_a[0], pos_b[1] - pos_a[1]]
del_x_sq = del_x[0] ** 2 + del_x[1] ** 2
del_v = [vel_b[0] - vel_a[0], vel_b[1] - vel_a[1]]
del_v_sq = del_v[0] ** 2 + del_v[1] ** 2
scal = del_v[0] * del_x[0] + del_v[1] * del_x[1]
Upsilon = scal ** 2 - del_v_sq * ( del_x_sq - 4.0 * sigma **2)
if Upsilon > 0.0 and scal < 0.0:
del_t = - (scal + math.sqrt(Upsilon)) / del_v_sq
else:
del_t = float('inf')
return del_t

#define the marked conditions
conf_a = ((0.30, 0.30), (0.30, 0.70), (0.70, 0.30), (0.70,0.70))
conf_b = ((0.20, 0.20), (0.20, 0.80), (0.75, 0.25), (0.75,0.75))
conf_c = ((0.30, 0.20), (0.30, 0.80), (0.70, 0.20), (0.70,0.70))
configurations = [conf_a, conf_b, conf_c]

#initial conditions
pos = [[0.25, 0.25], [0.75, 0.25], [0.25, 0.75], [0.75, 0.75]]
vel = [[0.21, 0.12], [0.71, 0.18], [-0.23, -0.79], [0.78, 0.1177]]

singles = [(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)]
pairs = [(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)]

n_events = 5000000

t = 0.0 #initialise time
histo_data = [] #initialise histogram
for event in range(n_events):
if event % 100000 == 0:
print event
wall_times = [wall_time(pos[k][l], vel[k][l], sigma) for k, l  in singles]
pair_times = [pair_time(pos[k], vel[k], pos[l], vel[l], sigma) for k, l in pairs]
next_event = min(wall_times + pair_times)
t_previous = t
for inter_times in range(int(t + 1), int(t + next_event + 1)):
del_t = inter_times - t_previous
for k, l in singles:
pos[k][l] += vel[k][l] * del_t
t_previous = inter_times

#histogram update
for k in range(N):
histo_data.append(pos[k][0]) #take the histogram of the x (0th) coordinate

t += next_event
del_t = t - t_previous
for k, l in singles:
pos[k][l] += vel[k][l] * del_t
if min(wall_times) < min(pair_times):
collision_disk, direction = singles[wall_times.index(next_event)]
vel[collision_disk][direction] *= -1.0
else:
a, b = pairs[pair_times.index(next_event)]
del_x = [pos[b][0] - pos[a][0], pos[b][1] - pos[a][1]]
abs_x = math.sqrt(del_x[0] ** 2 + del_x[1] ** 2)
e_perp = [c / abs_x for c in del_x]
del_v = [vel[b][0] - vel[a][0], vel[b][1] - vel[a][1]]
scal = del_v[0] * e_perp[0] + del_v[1] * e_perp[1]
for k in range(2):
vel[a][k] += e_perp[k] * scal
vel[b][k] -= e_perp[k] * scal

print t, "The total running time of the program"

#figure output
pylab.hist(histo_data, bins=100, normed=True)
pylab.xlabel('x')
pylab.ylabel('frequency')
pylab.title('Event driven Newtonian simulation: x coordinate histogram (density eta=0.18)')
pylab.grid()
pylab.savefig('event_disks_histo.png')
pylab.show()

0
100000
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1000000
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3900000
4000000
4100000
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4800000
4900000
485918.792042 The total running time of the program


We see that event driven Newtonian dynamics, direct and Markov-chain sampling give the same histogram for the x-position of the disks. This is a hint for the equivalence between statistical mechanics and Newtonian mechanics.