Helium gas with Lennard-Jones potential in MC&MD 4100054004 Jung-Ping, Wang & 4100054101 Tzu-Ping, Chen

Department of Physics, National Chung Hsing University, Taichung, Taiwan

(Dated: January 22, 2015)

Introduction

First, we use the MC method to simulate many He atoms in order to find the equllibrium

temperature. And use the result as a reference of the starting point of Molecular Dynamics(MD)

But then we find the problem of convergence in MC leads to a dilemma on analyzing. So we use only

four particles to find the compressibility KT and deduce

the representation function of heat capacity Cv.

Many He particles in MC MD simulation

At first we assume that there’s ten [He] atoms

scattering randomly in a box which volume isV =

10−27m3 with temperature T = 200K. We use the

Lennard-jones potential to illustrate the interaction

between each atom. In here we use the sum of

potential energy between each particles as the system

total energy ( neglecting kinetic energy)But in room

temperature, the probability of one system jumping to

another bigger potential system is ,

obviously ,approaching to zero (As 𝛽~1022 ). In such

strict condition, almost every trial leads the system to a

smaller total potential energy. Only if the temperature

is high enough, where 𝛽~1 , there’s a chance that the

system evolving to higher potential energy system 【fig.2】.

【fig.1】[He] atom’s Lennard-Jones potential:

VLJሺrሻ = 4ε ൬ቀσ

rቁ

12

− ቀσ

rቁ

6

൰ ,which ε = 10.22、σ =

2.556 ∙ 10−10.While r>1nm, VLJ→0. So we choose L <

1nm. where L is the length of the box

r(nm)

V(r)

MC steps

VLJtot(J)

【fig.2】The figure of VLJtot against MC steps in different temperatureሺbetaሻ. Just as we

expected. It’s not stable while β < 1。(V = 10−27m3，N = 20)

So we change our plan. In every step, we randomly pick one of those atoms and testing it while

fixed others. This little trick speeds up the process of gathering atoms together 【fig.3】，Although it

certainly lose some of the randomness, the system would however, still , developing to a equilibrium

temperature while e−βΔV is bigger. But then we found, though increase the trials to 106, there’s no

sign of convergence.【fig.4】

It’s means, although the atoms adjusting their position continuously in order to find the lowest

potential energy spot, MC is not the ideal method to find the ,theoretically, best position. Although

we can’t acquire the theoretical minimum potential. If we use the result found in MC as the starting

point of molecular dynamics【fig.5】and using Boltzmann distribution to every particle. We will find

particles moving pretty stable as the figure down below. One of the reason is Lennard-Jones potential

energy is much higher than the kinetic energy of Boltzmann distribution. It’s like a potential well

which you can’t break through.

【fig.5】MD simulation ሺT = 200K，V = 10−27m3，N = 20，MC steps = 1Mሻ.

(331)XY projection. (332)XZ projection. (333)The impulse on the wall from particles

against steps. (334)Total Energy against steps. (335)Total potential against steps.

(336)Total kinetic energy against step. These plots is to secure the stability of

system’s total energy. Then we have(337)Boltzmann distribution of momentum.

X-axis is momentum. Y-axis is the accepted number between random 0~1.

【fig.3】Random starting point and equilibrium state of He

atoms In MC simulation.ሺ T = 200K，V = 10−27m3，N =

20，MC steps = 1Mሻ

【fig.4】The figure of potential VLJtot against MC

stepsሺ T = 200K，V = 10−20m3，N = 10ሻ

MC steps V

LJt

ot(

J)

Because atoms is restricted. So the impulse of the cube wall is rather small. It’s not a good timing to

calculate system pressure by impulse.

We doubt that enormous volume may leads to a bad convergence. In order to prove this .we

made MC simulation in different volume. And the result shows, no matter the volume is big or small,

it didn’t converge at all. And the total potential energy is not inversely proportional to volume as we

expected, too. 【fig.6】

But if we put this result in MD simulation, the volume is too small so it only repulse between

each atom. Every atoms stick against the wall to have a lower total potential energy. So it looks

like this【fig.7】, in which it’s hard to find the theoretical value of minimum system potential.

From experiment above, we know it’s hard to analysis while there’re so many atoms in

it. So we construct two atoms which in equilibrium distance. Then we can find the distance is

r0 = 21/6σ. And if there’re four atoms, the equilibrium state would be a tetrahedron.【fig.8】

【fig.6】VLJtot against MC steps in different volume ሺT = 200K，N = 10ሻ

MC steps

VL

Jto

t(J)

【fig.7】Using conditions in Fig.6 to MD simulate XY and XZ projection( V = 10−28m3,

steps=1000)

Four helium atoms in MC simulation

Calculate the compressibility κTሺVሻ.

Consider there’s four He atoms, each atoms has a distance r0 and sit diagonally in four corners,

lying in a cube of length L0. So L0 :

L0√2

= r0

L0 =r0

√2

Minimum volume of the box V0 ∶

V0

= L03

= r03ሺ2ሻ−

32

=1

2σ3

So we can have the minimum potential while

atoms’ tetrahedron Volume is bigger thanV0 :

VLJmin

= −C24 ∙ Vሺr0ሻ

= −C24 ∙ ε

= −6ε

≅ −61.32ሺJሻ

When the volume <V0, Four atoms will stay where they are (at four corner forming a tetrahedron)

and shrink depends on the volume to maintain the minimum of system potential.(assumeβ ≫ 1) :

Vሺrሻ = r3ሺ2ሻ−32r = √2V1/3

Utotalሺrሻ = C24 ∙ VLJሺrሻ

= 24εσ6ሺσ6r−12 − r−6ሻ

UtotalሺVሻ = 24εσ6 (σ6 ൬√2V13൰

−12

− ൬√2V13൰

−6

)

= 3εσ6 ൬1

8σ6V−4 − V−2൰

Then we can calculate pressure as a function of volume:

PሺVሻ = − ൬∂U

∂V൰

S,N

= 3εσ6 ൬1

2σ6V−5 − 2V−3൰

【fig.8】The position of atoms doesn’t change in

minimum potential tetrahedron using MC ሺβ =

10 , V = 10−27m3, steps 1Mሻ

And κT:

κTሺVሻ = −1

V൬

∂V

∂P൰

T,N

=1

V(

−1

ቀ∂P∂V

ቁ)

=1

V(

−1

18εσ6 ቀV−4 −5

12 σ6V−6ቁ)

=1

18εσ6 ቀ5

12 σ6V−5−V−3ቁ

Caculate the capacity of heat (Cv).

Only in high temperature we can tear the

tetrahedron model apart( β < 1). On the other

hand, high temperature also makes kinetic energy

exceed the potential well. BecauseUKper = 3/2βሺJሻ,

we need β < 0.1 to escape the well where

【fig.9】The figure of UtotalሺVሻ、PሺVሻ、κT ሺVሻ in

tetrahedron model (low temperature). Shows κT

declining while volume shrinks. Means it becomes

hard to compress.

【fig.10】The atoms finally moved in high

temperature(β = 1 , V = 10−27m3, steps 1Mሻ)

VLJper = VLJmin/4 = −15.33ሺJሻ. What we’re interested in is the ratio of potential and kinetic energy

contribute to total energy in this high temperature region.

We draw a line of UKperሺβሻ = 3/2β. Launch 50 times of MC simulation (steps = 10000) in

each different β and take notes of the increase of potential VLJ+ሺβሻ = (VLJtot − VLJper)/4 of

each He atom. Then take average of VLJ+ሺβሻ over 50 times .Where < VLJ+ሺβሻ > is the average

contribution of He atom’s potential in system’s total energy (Volume = V0)【fig.11】

We guess the contribution of potential energy to total energy have an exponential decay to β. So

we do a curve fitting of< VLJ+ሺβሻ > :

< VLJ+ሺβሻ >= 0.248𝛽−1.486

0 2 4 6 8 10

-1

0

1

2

3

4

5

6

7

8

V (

J)

beta (1/K)

Model Allometric

Equation y = a*x^b

Adj. R-Sq 0.99925

Value

V a 0.2487

V b -1.4866

【fig.11】Red line indicates the contribution of potential energy from each particles, UKperሺβሻ = 3/2β，The small dots is the

increase of potential VLJ+perሺβሻ in each MC steps while big dots represent the average< VLJperሺβሻ >. We find the chance is rather

small where potential surpass kinetic in contribution of energy. And < VLJperሺβሻ > is smaller too. (volume = V0)

UKperሺβሻ VLJ+(J)

β

β

< VLJ+ሺβሻ >

ሺJሻ

From this we can obtain the energy of each particle :

Utotal

𝑁=

3

2β+

VLJmin

𝑁+< VLJperሺβሻ >

=3

2β− 15.33 + 0.248𝛽−1.486

The deduce Cv :

CV =1

N൬

∂Utotal

∂T൰

= ൬∂β

∂T൰ ൬

∂

∂β൰ ൬

Utotal

N൰

= ሺ−kBβ2ሻ ൬−3

2β2− 1.734𝛽−2.486൰

= kB ൬3

2+ 1.734𝛽−0.486൰

Conclusion :

After these experiments ,We know it’s hard to have a conclusion of too many He

particles in MC simulation. Then we change our plan, discussing the minimum potential

energy of four He particles (tetrahedron model). And find that in low temperature, the

relationship of compressed tetrahedron ሺV ≤1

2σ3ሻ and system total energy is UtotalሺVሻ =

3εσ6 ቀ1

8σ6V−4 − V−2ቁ, Pressure PሺVሻ = 3εσ6 ቀ

1

2σ6V−5 − 2V−3ቁ, compressibility κTሺVሻ =

1

18εσ6ቀ5

12σ6V−5−V−3ቁ

. While V =1

2σ3. We can approach heat capacity to CV = kB ቀ

3

2+

【fig.12】< VLJ+ሺβሻ > Fitting in exponential

【fig.13】result of < VLJ+ሺβሻ > in exponential fitting

1.734𝛽−0.486ቁ

Need for Further research

We use lots of time to calculate the partition function Z of many He particles to make up

to the previous possibility we neglected. But it didn’t work either hand writing or computer

coding. If there’s any breakthrough, the more realistic P-V plot will emerge. And be able to

draw part of the phase transition plot of He atom.

Reference

Lennard-Jones potential - wiki

http://en.wikipedia.org/wiki/Lennard-Jones_potential

An Introduction to Statistical Mechanics and Thermodynamics(by Robert H. Swendsen)