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Finding the force in a simple harmonic lattice

  1. Mar 22, 2015 #1

    Fyj

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    1. The problem statement, all variables and given/known data
    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.

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

    3. 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?
     
  2. jcsd
  3. Mar 25, 2015 #2

    ehild

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

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

    The force on the n-th particle is the negative gradient with respect to un, 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?
     
    Last edited: Mar 25, 2015
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