# Deriving equation to describe lattice vibrations of a one dimensional crystal

1. Nov 9, 2008

### peterjaybee

When setting up this derivation one assumes a chain of identical atoms. The interatomic interaction between atoms is short ranged and so only affects neighbouring atoms (see Hook and Hall, "Solid State Physics" chapter 2.3.1).

The potential V(r) is expanded as a taylor series about r = a to give,

V(r) = V(a) + (1/2)(d^2V/dr^2)(r-a)^2 + ...

1) I am just taking it as read that this is true but i dont understand why V(r) can actually be expanded it like this (I just cant get my head around the problem, and the lecture notes and formentioned book do not go into detail or show any steps leading up to this equation). Can anyone explain or show why you can use this expansion?

2) The dV/dr expression in the expansion is dropped "because the first derivative must vanish at the equilibrium spacing where V(r) is a minimum" - Hook/Hall again. Can anyone explain this in a different way because once again it has gone over my head?

Many thanks

Peter

2. Nov 9, 2008

### thuong

We can derive the fomula
V(r) = V(a) + (1/2)(d^2V/dr^2)(r-a)^2 + ...
as follow:

We write
V(r) = V(r - a + a)
where a is the equilibrium position of a particle. We assume that the displacement of the particle with respect to the equilibrium position is very small, so we can expand:
V(r-a+a) = V(a) + (dV/dr)(where r = a)* (r-a) + second order + ...
you see a is the equilibrium position of particle, so it corresponds to a minimum V(r) --> derivative = 0

that means the coefficient of (r-a) is zero.

cheers,

3. Nov 10, 2008

Thankyou