 Coupled-chain equations and dispersion relation

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2017/paper-2/section-c/c10/coupled-chain-equations-and-dispersion-relation

Put type-A atoms at

2 ja

and type-B atoms at

(2 j + 1) a

u_{j}, v_{j}

m_{A} \overset{u}{¨}_{j} = α (v_{j} + v_{j - 1} - 2 u_{j}), m_{B} \overset{v}{¨}_{j} = α (u_{j} + u_{j + 1} - 2 v_{j}) .

u_{j} = U e^{i (2 ja q - ω t)}

and

v_{j} = V e^{i ((2 j + 1) a q - ω t)}

. Substitution gives

(2 α - m_{A} ω^{2} - 2 α cos q a - 2 α cos q a 2 α - m_{B} ω^{2}) (U V) = 0.

m_{A} m_{B} ω^{4} - 2 α (m_{A} + m_{B}) ω^{2} + 4 α^{2} sin^{2} q a = 0.

Solving this quadratic in

ω^{2}

ω_{\pm}^{2} = \frac{α}{m _{A} m _{B}} [(m_{A} + m_{B}) \pm (m_{A} + m_{B})^{2} - 4 m_{A} m_{B} sin^{2} q a] .

2 a

∣ q ∣ \leq π / (2 a)

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