**
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 -> 1Energy:

E = K + V = sum n [ (xn+1 - xn)

^{2}+ 1/2 xn^{2}]

**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)/drr/r and V(r'ij) = (1/r'ij)12 - 2(1/r'ij)6 ]Here

r'ij = (ri -rj)/r0