# Homework Help: Equation of motion of magnetic dipole chain

1. Dec 2, 2015

### throneoo

1. The problem statement, all variables and given/known data
Find the equation of motion of a chain of atoms in 1D with alternating magnetic dipoles
At stationary equilibrium the atoms of mass m are separated by d , all displacements are small compared to d
2. Relevant equations
U=μBx=2μ20/4π)(1/x^3)
F(x)=-dU/dx

3. The attempt at a solution
The net force on particle n due to the dipole interactions
=F(xn-1-xn)-F(xn-xn+1)
However, I've found that if I fix the positions of the adjacent particles, the net force on particle n is a linear restoring force (with small displacement from eq. position). Can I assume that the atomic chain behaves as though they are connected by springs with identical fixed spring constant k? If so, the resultant equation would be

m*d2(xn)/dt2=-k*(2xn-xn-1-xn+1)

2. Dec 3, 2015

bump

3. Dec 3, 2015

### haruspex

I could use a fuller description of the set-up. Are the dipoles end-to-end, or (more likely) parallel?
Are we after end-to-end oscillations, lateral oscillations, or both?
For an equilibrium position to exist, don't there need to be opposing forces? Are the ends fixed, perhaps? Or some short-range repulsion?

4. Dec 3, 2015

### throneoo

In a 1d infinite lattice with lattice constant d
the dipoles are assumed to be fixed in direction such that they are either end to end or head to head, so the nth particle will be repelled by the n-1 th particle and n+1 th particle in opposite directions. The particles are also assumed to be only affected by adjacent particles and oscillate longitudinally.

It turns out that I don't need to mentioned assumption in OP as I could directly obtain the form
m*d2(xn)/dt2=-k*(2xn-xn-1-xn+1)
from FNet ; n=F(xn-1-xn)-F(xn-xn+1) where xm=m*d+hm(t) and hm is the displacement from the eq. position of the mth particle

5. Dec 3, 2015

### haruspex

Are you using 'particle' interchangeably with 'dipole', or is one dipole two particles?
If two dipoles are end to end in the same direction, won't they attract? Always?
I still don't see how you get from dipoles to springs. The force in a spring is proportional to the spring extension. Force between electric charges goes as the inverse square of distance and doesn't switch between attraction and repulsion.

6. Dec 4, 2015

### throneoo

the particles are individual magnetic dipoles. they are arranged such they have alternating spins so they behave like bar magnets with same poles facing each other. The linearity of the force is only an approximation at small displacements from their eq. position

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7. Dec 4, 2015

### haruspex

Ok, now I understand the arrangement, I see how you get the first order linearity.
Of course, each dipole has that sort of relationship to all the other dipoles, but it looks like this falls off as the inverse fifth power of distance, so maybe you only need consider adjacent ones.
Yes, that leads to the relationship you wrote in the OP, but I feel it would be more useful to write the right hand side as though n is a continuous variable. That should produce the usual wave equation.
What do you get for k in terms of the underlying constants (dipole strengths, spacing...)?

The physical embodiment of this arrangement still bothers me. What stops alternate dipoles flipping so that they all line up? What stops the whole chain expanding as the alternating dipoles repel each other? These considerations imply other forces, which would surely affect the wave behaviour.

8. Dec 4, 2015

### throneoo

I got k=6μ2μ0/πd5

and yeah I also feel the situation would be more complicated realistically

9. Dec 4, 2015

### haruspex

Depending on what μ represents, I got the same.

10. Dec 4, 2015

### throneoo

The magnetic moment