Finding the force in a simple harmonic lattice

[Fyj]

Homework Statement

If there are a large number of ions oscillating in a straight line, we can pick the nth one oscillating about its equilibrium a*n. The potential of the entire lattice is then U = 0.5*K[u(an)-u([n+1]a)]^2 - summed over all n. How do I use Force = -dU/du(an) to derive that Force = -K[2u(an)-u([n-1]a)-u([n+1]a)? Where K is some constant.

Homework Equations

The chain rule is f'(g(x))= df/dg * dg/dx

The Attempt at a Solution

I tried using the chain rule. Then if g = [u(an)-u([n+1]a)], then dU/dg = k*[u(an)-u([n+1]a)]. dg/du(an) can't be done analytically, however because we have a set of discrete points, dg = [(u(an) - (u[n-1]a) - (u[n+1]a)-u(an))].

I get dU/du(an) = K*[u(an)-u([n+1]a)]*[2u(an)-u([n-1]a)-u([n+1]a)]/du(an).
I am guessing I could write that du = [u([n+1]a)-u(an)] to get
dU/du(an) = -K*[2u(an)-u([n-1]a)]u([n+1]a)], but there's a minus sign that I can't get rid of.

Am I going about this all wrong?

 

[ehild]

To make the formulae simpler, denote the displacement from equilibrium of the n-th particle (that one which equilibrium position is at na) by u_n.
The potential energy of the whole lattice is $U=0.5K \sum {(u_{n+1}-u_n)^2}$. Writing it out, u_n appears in two terms:

U = ... K/2 [( u_n-u_n-1)² + ( u_n+1-u_n)² ]...

The force on the n-th particle is the negative gradient with respect to u_n, the displacement of the n-th particle. You get non-zero derivative from the terms above only.
Your derivation was wrong. What is the derivative of (x-y)² with respect to x? With respect to y?