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Chemistry 380.37 Fall 2015 Dr. Jean M. Standard November 2, 2015 Introduction to Molecular Dynamics What is Molecular Dynamics? Molecular dynamics is a computational method that generally employs Newton's equations of motion to determine the time dependence of a molecular system. In most typical molecular dynamics methods, the atoms in each molecule are treated classically. For example, for particle motion in a single x dimension, Newton's equation of motion is (1) Fx = ma x , where Fx is the force in the x direction, m is the mass of the particle, and a x is the acceleration in the x direction. Recall that acceleration is the first derivative of velocity, and velocity is the first derivative of position, € dv x € = v˙ x (t), dt dx v x (t) = = x˙ (t). dt € a x (t) = (2) Therefore, acceleration is the second derivative of position, € a x (t) = d 2x dt 2 = ˙x˙(t). (3) For N particles that are able to move in three dimensions, Newton's equations of motion are € ! ! Fi = mi a i , i = 1,2, …N . (4) These equations are a set of N coupled vector equations. Each vector equation has three components representing motion in the x, y, and z directions for € the ith particle. As an example, consider the motion of two particles. Newton's equations of motion consists of a set of six coupled equations for the motion of the two particles, F1x = m1 a1x F1y = m1 a1y F1z = m1 a1z F2x = m2 a 2x F2y = m2 a 2y F2z = m2 a 2z . It is often easier to use the definitions for position and velocity above and solve an equivalent set of 12 first-order differential equations instead. € (5) Determination of Forces In the traditional molecular dynamics method, a molecular mechanics empirical force field is utilized to determine the forces. A typical molecular mechanics force field is U = ∑U s + bonds ∑U b ∑U + angles t + ∑ nonbond interactions torsions ∑ U nb + U el . (6) charged pairs This force field depends on all 3N cartesian coordinates of the atoms in the molecular system. This is the reason that the set of Newton's equations is coupled. € The force is by definition calculated as the derivative of the force field; for example, the force in the x direction for particle i is determined as Fix = − ∂U . ∂x i (7) As an example, consider a two-particle system interacting through a stretching force field, € U = 1 2 ( k s r − req ) 2 (8) , where r is the instantaneous bond length and req is the equilibrium bond length. The bond length depends upon the Cartesian coordinates of each atom, € € r = [( x − x ) 1 2 2 2 + ( y1 − y 2 ) + ( z1 − z 2 ) 2 1/ 2 ] (9) . The forces that appear in Newton's equations can be computed directly from the force field expression. For example, for the force in the € x direction on atom 1, we can use the chain rule for derivatives to obtain F1x = − ∂U ∂U ∂r = − ⋅ ∂x1 ∂r ∂x1 ( )[( x1 − x 2 ) 2 k s ( r − req ) x1 − . = − k s r − req F1x = r 2 + ( y1 − y 2 ) + ( z1 − z 2 ) 2 −1/ 2 ] ⋅ ( x1) (10) Similar expressions can be developed for the other forces. € Applications of Molecular Dynamics Molecular dynamics (MD) provides insight into the important atomic motions in a molecular system. For example, it allows us to determine whether or not a molecular system undergoes large conformational changes at room temperature. MD may also be used to search for low energy conformations of a molecular system. Solvent effects may be studied by explicit inclusion of solvent molecules. And finally, thermodynamic properties may be predicted from MD simulations. 2 Example 1: Dynamics of Bond Stretching Consider a simple example of the dynamics of bond stretching in a diatomic molecule. In this example, only the vibrational motion of the bond will be considered; rotational and translation motion will be ignored. A bond involving atoms A and B is represented using a harmonic force field of the form U = 1 2 k x2 , (11) where the bond displacement coordinate x is defined as € x = rAB − rAB,eq . (12) For this system, Newton's equation of motion is € (13) Fx = µa x , where µ is the reduced mass associated wit h the A-B bond, € mA mB . mA + mb µ = (14) The force is obtained from the derivative of the force field, € ∂U ∂x = − kx . Fx = − (15) Substituting into Newton’s equation of motion (Eq. 13) yields € Fx = µa x −kx = µa x , or − kx = µx˙˙ . (16) Solving for the acceleration, € ˙x˙( t ) = − k x( t ) . µ (17) This equation has a simple analytic solution. The exact solution of Newton's equation for a harmonic oscillator with an initial position given by x(0) and€an initial velocity of v(0) are x(t) = x(0) cos(ω t ) + v( 0) ω sin(ω t ) (18) v(t) = − x(0)ω sin(ω t ) + v( 0) cos(ω t ) . The parameter ω is the angular velocity and is defined by the relation € ω = € k . µ (19) 3 4 The angular velocity ω is related to the harmonic frequency of vibration ν o , νo = ω 1 = 2π €2π k . µ (20) For a sample bond with a force constant k=500 kcal mol–1 Å–2 and a reduced mass µ=6.0 g/mol, the harmonic frequency ν o is 2.97×1013 s–1 (or 991€cm–1). Plots of the position and velocity as functions of time for this bond vibration starting from the initial conditions x ( 0) = 0.3 Å and v( 0) = 0 are shown in Figure 1. € € € (a) (b) Figure 1. Dynamics of bond stretching for a bond with force constant k=500 kcal mol–1 Å–2, reduced mass µ=6.0 g/mol, and equilibrium bond length req=1.30 Å. The initial conditions are x(0)=0.,3 Å and v(0)=0. (a) Bond length vs. time; (b) velocity vs. time. The period of oscillation is 2π / ω ; for this example, the period is 0.034 ps (34 fs), which is fairly typical for the magnitude of the period of oscillation of a bond vibration. As a result, the time step employed in molecular dynamics simulations must be ~ 1 femtosecond in order to capture the motion of bond stretching properly. €