ex1_1_lennardjones

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Example 1.1: Lennard-Jones Potential

A NumFys Example

http://www.numfys.net

Fall 2012

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p.1 Introduction to the Lennard-Jones Potential

The Lennard-Jones potential (also referred to as theL-J potential or 12-6 potential) is a simple physical model that

approximates the interaction between a pair of neutral atoms or

molecules.

The most common expressions for the L-J potential are:

V LJ = 4

σ

r

12

−

σ

r

6 =

r m

r

12

− 2

r m

r

6 , (1)

where r is the distance between the particles.Here, is the depth (i.e. the minimum) of the potential well,

reached at r = r m , and σ is the distance at which the

inter-particle potential is zero. It can be shown that r m = 21/6σ.



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p.2 The L-J Potential.

For

= σ

= 1, we obtain the following graph:

As we can see, we find a global minimum and expect

oscillations about this equilibrium as long as the total energy,

E total = E kin + E pot , does not exceed zero.



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p.3 Parameters for our Particular Example

In this example, we will consider the Lennard-Jones Potential

for two H2O molecules. The parameters are:

σ = 0.32 × 10−9 m = 1.08 × 10−21 J

Note that the orientation-dependent interaction between the

water dipoles is not included here explicitly.



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p.4 Plot

We obtain this plot:



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p.5 The Problem

We will now analyze oscillations about the equilibrium in a

classical manner.

First, we want to examine the frequency of small

oscillations near r = r m .

To do this, we approximate the force near r = r m .

Thereafter, we want to compute the period of oscillations

for larger perturbations when the linearization is a poor

approximation.



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p.6 Deriving the Force from the Potential

Let us calculate the force from the potential by taking the

derivative of the potential:

F (r ) = −V (r ) (2)

= −

−

12r m

r m

r

13

+ 12r m

r m

r

7 (3)

= 12

r m r m

r 13

−r m

r 7

(4)

Note that F (r m ) = 0. Hence, r = r m is an equilibrium point.

Now, we Taylor-expand the force about this equilibrium point.



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p.7 Approximating the Force near r m

We write

F (r m + ∆r ) ≈ F (r m ) + F (r m )∆r = 0 + F (r m )∆r (5)

Now, we need to determine F (r m ):

F

(r ) = 12

r m 2− 13

r m

r 14

+ 7 r m

r 8

(6)

Substituting r = r m yields

F (r m ) = −72

r m

2 (7)

Using this result in equation (5), we find the force near r m

F (r m + ∆r ) ≈−72

r m 2

∆r (8)



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p.8 The Frequency f of Small OscillationsWe can now compute the frequency of small oscillations about

the equilibrium by writing

m d 2∆r

dt 2 = F (r m + ∆r ) ≈

−72

r m 2

∆r , (9)

where m is the effective mass of the water-water molecule

system. This describes harmonic oscillations with a frequency

ω2 = (2πf )2 =

72

mr m 2. (10)

Using the values for and r m = 21/6σ from above, and

m ≈ 9m proton , we obtain a frequency

f = ω

2π=

1

2π

72

mr 2m

= 1.01× 1012

Hz (11)



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p.9 Expanding the Problem

So far, we have examined the frequency near r = r m where

the dynamics can be approximated by that of a harmonic

oscillator.

This was simple and we did not need any computational

tools. However, what happens if we look at oscillations further

away from r = r m ?

If we look at our plot of the L-J potential, we can see that

the graph is not symmetric about the equilibrium. During

oscillations, we would expect the molecules to spend more

time to the right of the equilibrium (r > r m ) then to the left

of it. Let us study this phenomenon numerically.



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p.10 Nonlinear Oscillations

Now, we will solve the exact dynamic equation

m r = F (r ), (12)

where dots indicate time derivatives.

We transform this ODE into a set of first-order ODEs andemploy Euler’s method to solve it. We do this by introducing

new variables v and w , with v = r and w = r . Substitution into

the equation above gives

w = r = F (v )

m =

12

mr m

r m

v

13

−

r m

v

7. (13)



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p.11 Euler Method: Set of EquationsThis gives us the following set of equations

v = w , (14)

w = 12

mr m

r m

v

13

−

r m

v

7. (15)

For our initial conditions, we choose v (0) = r m and w (0) > 0.

Following our notation in Module 4.1, we can now write the

Euler method as (v 1 = v (0), w 1 = w (0))

v n +1 = v n + h ·w n , (16)

w n +1 = w n + h 12

mr m

r m

v n

13

−

r m

v n

7. (17)



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p.12 Implementation in Matlab

We can implement this in a simple for-loop in Matlab

(see corresponding script):

The step size is chosen to be h = 10−15 s.



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p.13 Oscillations near r = r m

First, we set 1

2mw 2

0 = 0.001. This corresponds to very small

oscillations because the kinetic energy is much smaller than

and, therefore, we are deep down in the potential well where

the oscillations are harmonic. We plot the vector v which

represents the position of the molecule vs. time steps.



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p.14 Oscillations near r = r m

We find the following features.

The oscillations around the equilibrium r = r m seem

harmonic. Matlab uses around 1000 time steps of magnitude 10−15 s.

Hence, the period is about T = 1000 ∗ 10−15s = 10−12 s,

in agreement with our linear analysis.

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p.15 Oscillations further up the Potential Well

Now let’s see what the oscillations look like when we addkinetic energy. We set 1

2mw 20 = 0.3 and n = 1270. We have to

increase the number of time steps n because we expect the

period to be longer.


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p.16 Oscillations further up the Potential Well

We can clearly see how the molecule spends more time on

the right side (r > r m ) of the potential well which we

expected from the L-J plot.

In this case, the period is about T = 1270 ∗ 10−15 s. Since

T increases with the amplitude, we have no longerharmonic oscillations.

Note that we treat this microscopic system classically and

not quantum mechanically. However, it gives us some

ideas about how a full quantum mechanical treatmentwould differ from that of a quantum mechanical harmonic

oscillator.


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p.17 Repeating Oscillations

So far, we have only looked at oscillations for one period.

If we increase the number of oscillations, we would expect

neither diminishing nor increasing amplitudes since energyis conserved.

Unfortunately, Euler’s method contains errors which can be

large enough to clearly violate energy conservation in our

solution. Hence, the amplitude may not be constant.


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p.18 Plot: Repeating Oscillations

As the number of oscillations increases, the molecules move

further away from equilibrium with each oscillation which would

correspond to an increase in energy.


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p.19 Error in the Euler Method

Remember that this method is an approximation of thesolution.

Each time we calculate a new step, we make an error.

This error results in an increase in the amplitude of each

new oscillation. The error can be reduced by using better but more

complicated approximations, or by reducing the step size h .

In the next plot, we have reduced the step size to

h = 10

−16

s. We have to increase the number ofcalculations to n = 100 000 so as to obtain the same

amount of oscillations. This demands more computational

time but reduces the error.


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p.20 A Better But More Demanding Plot


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