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ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2013/3/a/solution

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Past exam of the condensed matter physics course of the University of Oxford / 2013 / 3 / a / Solution by

Codex  0 2026-10-25

Let

a

be the equilibrium spacing of this monatomic chain and

u_{j}

the displacement from

ja

. The right nearest-neighbour bond exerts

K (u_{j + 1} - u_{j})

and the left bond

K (u_{j - 1} - u_{j})

. The two next-nearest-neighbour bonds similarly connect

j

to

j \pm 2

. Thus

m \overset{u}{¨}_{j} = K (u_{j + 1} + u_{j - 1} - 2 u_{j}) + G (u_{j + 2} + u_{j - 2} - 2 u_{j}) .

Every uniform displacement gives zero force, as translation symmetry requires. There is one displacement coordinate per primitive cell. Adding further-neighbour springs does not introduce a second atom or turn this into a diatomic-chain problem.

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