- M₁M₂w¹ – 2C(M₁ + M₂)² + 2C²(1 — cos Ka) = 0. (22) We can solve this equation exactly for w², but it is simpler to examine the limiting cases Ka < 1 and Ka = ±π at the zone boundary. For small Ka we have cos Ka 1-K²a² + ..., and the two roots are 1 w² = 2C - ( 1₁ + M₂) C -K²a² M₁ + M₂ (optical branch); (acoustical branch). (23) (24)

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Given the (22) equation, prove that (23) and (24) are the frequencies of the optical and acoustic branch
The homogeneous linear equations have a solution only if the determinant of
the coefficients of the unknowns u, v vanishes:
or
2C-M₁w²
-C[1 + exp(iKa)]
² = 20 (₁ + M₂)
1
1
2C
w² =
-C[1 + exp(ika)]
2C-M₂w²
M₁M₂w¹ - 2C(M₁ + M₂)w² + 2C²(1 - cos Ka) = 0.
(22)
We can solve this equation exactly for w², but it is simpler to examine the
limiting cases Ka < 1 and Ka = ±π at the zone boundary. For small Ka we
have cos Ka 1-K²a² + ..., and the two roots are
C
--K²a²
M₁ + M₂
=0,
(optical branch);
(21)
(acoustical branch).
(23)
(24)
Transcribed Image Text:The homogeneous linear equations have a solution only if the determinant of the coefficients of the unknowns u, v vanishes: or 2C-M₁w² -C[1 + exp(iKa)] ² = 20 (₁ + M₂) 1 1 2C w² = -C[1 + exp(ika)] 2C-M₂w² M₁M₂w¹ - 2C(M₁ + M₂)w² + 2C²(1 - cos Ka) = 0. (22) We can solve this equation exactly for w², but it is simpler to examine the limiting cases Ka < 1 and Ka = ±π at the zone boundary. For small Ka we have cos Ka 1-K²a² + ..., and the two roots are C --K²a² M₁ + M₂ =0, (optical branch); (21) (acoustical branch). (23) (24)
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