xn+1 - 2xn + xn-1 = - h2 xn

    where h is step size, xn = x(t=nh) is the value at time nh. Solve the
    difference equation exactly and compare its solution with that of the
    differential equation. Hint: look for solution of the form xn=Re{A exp(i nh w)}.
    Due to discretization, the energy is not exactly conserved. Can you construct
    an exactly conserved quantity which differs from total energy only by higher
    orders in h?
 

Solution:
 

Assume:

xn = A cos(nhw)

Then:

xn-1 = A cos((n-1)hw)
xn+1 = A cos((n+1)hw)

The equation becomes:

cos((n+1)hw) + cos((n-1)hw) + (h2-2)cos(nhw) =  0

The solution of this equation gives w:

2cos(hw)-2+h2=0

cos(hw)=1-h2/2

If h -> 0:  cos(hw) -> 1- (hw)2/2
Then: w -> 1

Energy:

E = K + V =  sum n [ (xn+1 - xn)² + 1/2 xn² ]
 

    Give a detailed proof of the second order and forth order Runge-Kuttta
    method.
 

solutions:

2nd order:
 

substiute k1 and k2 into the equation:

x(t + h) = x(t) + k2 + O(h3)

to show that it is a higher order Tayler expansion of the equation:

x(t+h) = x(t) + h f(t, y(t)) + O(h2)

4th order:
 

substiute k1, k2 , k3 into the equaton:

x(t + h) = x(t) + k1/6 + k2/3 + k3/3 + k4/6 + O(h5)

to show that it is a higher order Tayler expansion of the equation:

x(t+h) = x(t) + h f(t, y(t)) + O(h2)

    The particles in a Lennard-Jones fluid have the parameters: r0=3.6 A (1 A =
    10-10 m) and V0 = 0.2 Kcal/mol. What is the energy of interaction between
    two particles at seperation of (a) 2.7 A, (b) 3.6 A, (c) 5.5 A, and (d) 10.0 A
    respectively?
 
 
 

Lennard-Jones potential:

V(r) = V0 [ (r0/r)12 - 2(r0/r)6 ]

Substitution of the parameters in (a), (b), and (c) gives you the energy.

    Why do we need to study the  statistical mechanics of a fluid?

       To derive average macrosopic physical quantities such as the relationship between
        pressure, volume and temperature from microscopic random motions.
 

    Show that the unit of V0 /kB is temperature, and that of sqrt{ m r02/ 48 V0}
    is time.
 

solution:
 

Since:

kB ~ [energy / T ]

V0 ~ [energy]

Then:

V0/kB ~ [T]

 

Solution:
 

Use dp/dt = - dV(r)/dr r/r and V(r'ij) = (1/r'ij)12 - 2(1/r'ij)6 ]

Here r'ij = (ri -rj)/r0