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I'm preparing for my condensed matter exam and I'm trying to solve problem 3a) of chapter 22 in Ashcroft & Mermin. The problem is basically to prove that the dispersion relation of a diatomic linear chain will reduce to the monoatomic one when the coupling constants are equal, [itex]K=G=K_0[/itex]. Starting off with equation (22.37)

[tex]

\omega^2 = \frac{K+G}{M} \pm \frac{1}{M}\sqrt{K^2 + G^2 + 2KG \cos k a}

[/tex]

I get

[tex]

\omega^2 = \frac{2 K_0}{M}(1 \pm | \cos \[(k a)/2 \] | )

[/tex]

if we take the minus sign (and ignore the absolute value) we get back to the monoatomic result with a lattice constant of [itex]a/2[/itex]. But why would prefer the minus sign and remove the absolute value?

Thanks

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# Homework Help: Normal modes of diatomic linear chain

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