Finding the force in a simple harmonic lattice

In summary, the conversation discusses the use of the chain rule to derive the force equation in a system where a large number of ions are oscillating in a straight line. The potential energy of the lattice is represented by U = 0.5*K[u(an)-u([n+1]a)]^2 summed over all n. The force equation is then derived using the negative gradient of U with respect to the displacement of the n-th particle. The correct derivative of (x-y)^2 with respect to x is 2(x-y), and with respect to y is -2(x-y).
  • #1
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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?
 
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  • #2
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?
 
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Related to Finding the force in a simple harmonic lattice

1. What is a simple harmonic lattice?

A simple harmonic lattice is a model used in physics to describe the motion of particles in a periodic potential. It is made up of a series of masses connected by springs and can be used to study the behavior of systems such as crystals, polymers, and molecules.

2. How is force calculated in a simple harmonic lattice?

In a simple harmonic lattice, the force is calculated using Hooke's law, which states that the force exerted by a spring is proportional to the displacement of the mass from its equilibrium position. This means that the force is directly proportional to the distance the spring is stretched or compressed.

3. What is the relationship between force and displacement in a simple harmonic lattice?

The relationship between force and displacement in a simple harmonic lattice is linear. This means that as the displacement of a mass increases, the force exerted on it also increases, but in the opposite direction. This relationship is described by Hooke's law.

4. How does the spring constant affect the force in a simple harmonic lattice?

The spring constant, also known as the stiffness coefficient, determines how much force is required to stretch or compress a spring in a simple harmonic lattice. A higher spring constant means that the spring is stiffer and will require a greater force to move it a certain distance.

5. What is the significance of finding the force in a simple harmonic lattice?

Finding the force in a simple harmonic lattice is important for understanding the dynamics and behavior of systems that can be represented by this model. It allows scientists to study the effects of different parameters, such as spring constant and mass, on the forces acting within the lattice and how these forces contribute to the overall motion of the system.

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