A simple example of MD simulation

This is a homework of computational nano mechanics. The basic requirement is to use Lagrangian function to describe the motion of particles. In homework, it requires 5 particles. As an enhancement, I rewrote the code with Qt and use GNU Scientific Library(GSL) to finish the task.

Concept

The Lagrangian equation of motion describe the particle motion with energy method, the equation is:

\displaystyle \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_k}-\frac{\partial L}{\partial x_k}=0

L is the Lagrangian function of the system, k denotes the degree of freedom.

\displaystyle L = T - U

T is the kinetic energy and U is the potential energy.

Since the derivative of U with respect to x is the force

\displaystyle f = -\frac{dU(r)}{dr}

the Lagrangian equation of motion uses another way to illustrate the Newton’s second law.

Single particle example

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For a single particle in a chamber, two degree of freedom x_1 = x, x_2 = y, the kinetic energy is:

\displaystyle T =\sum  \frac{1}{2} m \left({\dot{x}}^2 + {\dot{y}}^2\right)

k is the degree of freedom. m denotes the mass of the particle.

The potential energy of the particle in a chamber is expressed as:

\displaystyle U = \alpha e^{- \beta(R-r)}

Here the exponent function is used to describe the repulsion from the chamber wall. when the distance R-r close to zero, the repulsion potential will be much large and prevent the particle penetrating the chamber wall. The Lagrangian function and equations are:

\displaystyle L = \frac{m}{2} (\dot{x}^2+\dot{y}^2)-\alpha e^{-\beta (R-\sqrt{x^2+y^2})}

and for each degree of freedom:

\displaystyle \left\{\begin{matrix} m \ddot{x}(t) = -\alpha \beta e^{-\beta (R-\sqrt{x^2+y^2})} x(t) / \sqrt{x^2+y^2} \\ m \ddot{y}(t) = -\alpha \beta e^{-\beta (R-\sqrt{x^2+y^2})} y(t) / \sqrt{x^2+y^2} \end{matrix}\right.

For this example, we take the following parameters:

\displaystyle \alpha = 10^{-27}, \beta = 4nm^-1, R = 10nm, m = 10^{-9}kg

The initial conditions are:

\displaystyle x(0) = -3nm, \dot{x}(0) = 10nm/s, y(0) = 5nm, \dot{y}(0) = 20nm/s

Using Mathematica (FF01Y5.nb) to plot the trajectory or use this script (YR29Q8.nb) to plot the animation:

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My94c2.gif

Lennard-Jones potential function

Lennard-Jones potential function is almost the world-famous potential function in nano mechanics. It describes the interaction between atoms. The function was proposed through quantum perturbation theory in which the attraction should follow a r^{6} relation. It combines the attraction and repulsion together.

\displaystyle U(r_{ij})=4 \varepsilon \left[ \left(\frac{\sigma}{r_{ij}} \right)^{12} - \left( \frac{\sigma}{r_{ij}}\right)^{6} \right]

where: \varepsilon denotes the depth of the energy well, and \sigma denotes the equilibrium spacing of atoms.

Take \sigma = 3, \varepsilon = 0.2, the result is shown below:

ZY62q9.png

As mentioned before, the derivative of U with respect to r is the force. The force from Lennard-Jones interaction is simply get the derivative of U:

\displaystyle F=-U^{\prime}(r_{ij}) = -4 \varepsilon \left(\frac{6 \sigma ^6}{r^7}-\frac{12 \sigma ^{12}}{r^{13}}\right)

BS79E0.png

The equilibrium distance between two atoms is the position F=0.

\displaystyle \sigma \sqrt[6]{2} = r_0

\displaystyle r_0 = 1.225 \sigma

Two particles example

If two particles are put into the chamber, Lennard-Jones potential should be considered in calculation. Here, I try to use GNU Scientific Library(GSL) to solve the problem in three dimension.

The Explicit 4th order (classical) Runge-Kutta(RK4) method is chosen for this problem. Two particles are denoted as i and j respectively. The Lagrangian function is (k denotes the degree of  freedom):

\displaystyle L=T-U

\displaystyle T = \sum \frac{1}{2}m_i (\dot{x}_{ik}^2)+\sum \frac{1}{2}m_j (\dot{x}_{jk}^2)

\displaystyle U = E_{pi} + E_{pj} + W_{LJ}

E_{pi} and E_{pj} denotes the repulsive potential from the chamber wall, the equations of them are similar to Single particle example. Here, I just analyze the Lanner-Jones potential:

\displaystyle W_{LJ-ijk} = 4 \varepsilon \left[ \frac{\sigma ^ {12}}{\left(\sum_{k=1}^3 \left( x_{ik} - x_{jk}\right)^2 \right)^6} - \frac{\sigma ^ 6}{\left(\sum_{k=1}^3 \left( x_{ik} - x_{jk}\right)^2 \right)^3} \right]

For the derivative W_{LJ-ijk}  with respect to x_{ik} and x_{jk}:

\displaystyle \frac{\partial W_{LJ-ijk}}{\partial x_{ik}} = 4 \varepsilon \left[ \frac{6 (x_{ik}-x_{jk})\sigma ^ {6}}{\left(\sum_{k=1}^3 \left( x_{ik} - x_{jk}\right)^2 \right)^4} - \frac{12 (x_{ik}-x_{jk}) \sigma ^ {12}}{\left(\sum_{k=1}^3 \left( x_{ik} - x_{jk}\right)^2 \right)^7} \right]

\displaystyle \frac{\partial W_{LJ-ijk}}{\partial x_{jk}} = 4 \varepsilon \left[ -\frac{6 (x_{ik}-x_{jk})\sigma ^ {6}}{\left(\sum_{k=1}^3 \left( x_{ik} - x_{jk}\right)^2 \right)^4} + \frac{12 (x_{ik}-x_{jk}) \sigma ^ {12}}{\left(\sum_{k=1}^3 \left( x_{ik} - x_{jk}\right)^2 \right)^7} \right]

With these two equations, we can simply calculate the motion of these two particles:

Cn16H1.gif

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