Past exam of the condensed matter physics course of the University of Oxford / 2015 / 3 / b

Table of contents
- Solution b

Solution

Let

u_{p}

be the longitudinal displacement of atom

p

. Translational invariance and a symmetric harmonic pair force give the long-range harmonic-chain dispersion equations

M \overset{u}{¨}_{p} = \sum_{n \neq = 0} C_{∣ n ∣} (u_{p + n} - u_{p}) = \sum_{n > 0} C_{n} (u_{p + n} + u_{p - n} - 2 u_{p}) .

Use a traveling normal mode

u_{p} = u_{0} e^{i (k p a - ω t)}

. Dividing by its nonzero amplitude gives

- M ω^{2} = \sum_{n > 0} C_{n} (e^{inka} + e^{- inka} - 2) = - 4 \sum_{n > 0} C_{n} sin^{2} (nka /2) .

Hence

ω (k) = \frac{2}{M} [n > 0 \sum C_{n} sin^{2} (nka /2)]^{1/2} .

Positive pair constants give a nonnegative squared frequency and a uniform-translation zero mode. More general signed force constants require checking stability of the entire sum. The frequency is even in

k

and periodic with

2 π / a

, reflecting reciprocal-lattice equivalence. The force sum must converge in the model. A finite ordinary acoustic slope additionally requires convergence of

\sum n^{2} C_{n}

, in which case

v_{s} = a \sum n^{2} C_{n} / M

. Both finite-range examples below satisfy this condition.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version