1. 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 =
10−27
m3
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 ൬ቀ
σ
r
ቁ
12
− ቀ
σ
r
ቁ
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 = 10−27
m3
，N = 20)

2. 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 = 10−27
m3
，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 = 10−27
m3
，N =
20，MC steps = 1Mሻ
【fig.4】The figure of potential VLJtot against MC
stepsሺ T = 200K，V = 10−20
m3
，N = 10ሻ
MC steps
VLJtot(J)

3. 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 = 21/6
σ. 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
VLJtot(J)
【fig.7】Using conditions in Fig.6 to MD simulate XY and XZ projection( V = 10−28
m3
,
steps=1000)

4.  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
= L0
3
= r0
3ሺ2ሻ−
3
2
=
1
2
σ3
So we can have the minimum potential while
atoms’ tetrahedron Volume is bigger thanV0 :
VLJmin
= −C2
4
∙ Vሺr0ሻ
= −C2
4
∙ 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−
3
2r = √2V1/3
Utotalሺrሻ = C2
4
∙ VLJሺrሻ
= 24εσ6
ሺσ6
r−12
− r−6
ሻ
UtotalሺVሻ = 24εσ6
(σ6
൬√2V
1
3൰
−12
− ൬√2V
1
3൰
−6
)
= 3εσ6
൬
1
8
σ6
V−4
− V−2
൰
Then we can calculate pressure as a function of volume:
PሺVሻ = − ൬
∂U
∂V
൰
S,N
= 3εσ6
൬
1
2
σ6
V−5
− 2V−3
൰
【fig.8】The position of atoms doesn’t change in
minimum potential tetrahedron using MC ሺβ =
10 , V = 10−27
m3
, steps 1Mሻ

5. And κT:
κTሺVሻ = −
1
V
൬
∂V
∂P
൰
T,N
=
1
V
(
−1
ቀ
∂P
∂V
ቁ
)
=
1
V
(
−1
18εσ6 ቀV−4 −
5
12
σ6V−6ቁ
)
=
1
18εσ6 ቀ
5
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 = 10−27
m3
, steps 1Mሻ)

6. 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)
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
< VLJ+ሺβሻ >
ሺJሻ

7. 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
N
൬
∂Utotal
∂T
൰
= ൬
∂β
∂T
൰ ൬
∂
∂β
൰ ൬
Utotal
N
൰
= ሺ−kBβ2ሻ ൬−
3
2β2
− 1.734𝛽−2.486
൰
= kB ൬
3
2
+ 1.734𝛽−0.486
൰
 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
2
σ3
ሻ and system total energy is UtotalሺVሻ =
3εσ6
ቀ
1
8
σ6
V−4
− V−2
ቁ, Pressure PሺVሻ = 3εσ6
ቀ
1
2
σ6
V−5
− 2V−3
ቁ, compressibility κTሺVሻ =
1
18εσ6ቀ
5
12
σ6V−5−V−3ቁ
. While V =
1
2
σ3
. 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

8. 1.734𝛽−0.486
ቁ
 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)