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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.
€