Solving Phonon Excitation in Diatomic Lattices

[aabb009]

Homework Statement

Ok, I need to show that in an acoustic mode of vibration in a diatomic lattice, for small k, \omega \propto k, and find the constant of proportionality.

Homework Equations

A_1\left(\omega^2M-\frac{2T}{a}\right)+A_2\left(\frac{2T}{a}cos(ka)\right)=0
, and
A_1\left(\frac{2T}{a}cos(ka)\right)+A_2\left(\omega^2m-\frac{2T}{a}\right)=0
hence:
\omega^2 = \frac{T}{a}\left[\frac{1}{M} + \frac{1}{m}\right] - \frac{T}{a}\left[\left(\frac{1}{M}+\frac{1}{m}\right)^2-\frac{4sin^2(ka)}{Mm}\right]^{1/2}

The Attempt at a Solution

I work through it, but repeatedly find that \omega^2 \propto k, and I can't see anyway of getting a k^2 factor on the right.
\omega^2 = \frac{T}{a}\left[\left(\frac{1}{M} + \frac{1}{m}\right) - \sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2-\frac{4sin^2(ka)}{Mm}}\right]
\omega^2 = \frac{T}{a}\left[\left(\frac{1}{M} + \frac{1}{m}\right) - \sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2\left[1-\frac{Mm4sin^2(ka)}{(M+m)^2}\right]}\right]
\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left[1 - \sqrt{\left[1-\frac{Mm4sin^2(ka)}{(M+m)^2}\right]}\right]
with small angle approximation we get:
\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left[1 - \sqrt{1+\frac{4Mmk^2a^2}{(M+m)^2}}\right]
\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left(1-1+\frac{2\sqrt{Mm}ka}{m+M}\right)
hence
\omega^2 = \frac{2T}{\sqrt{Mm}}k

Where am I going wrong? I don't see any way to prove this.

 

[olgranpappy]

Homework Statement

Ok, I need to show that in an acoustic mode of vibration in a diatomic lattice, for small k, \omega \propto k, and find the constant of proportionality.

Homework Equations

A_1\left(\omega^2M-\frac{2T}{a}\right)+A_2\left(\frac{2T}{a}cos(ka)\right)=0
, and
A_1\left(\frac{2T}{a}cos(ka)\right)+A_2\left(\omega^2m-\frac{2T}{a}\right)=0
hence:
\omega^2 = \frac{T}{a}\left[\frac{1}{M} + \frac{1}{m}\right] - \frac{T}{a}\left[\left(\frac{1}{M}+\frac{1}{m}\right)^2-\frac{4sin^2(ka)}{Mm}\right]^{1/2}

The Attempt at a Solution

I work through it, but repeatedly find that \omega^2 \propto k, and I can't see anyway of getting a k^2 factor on the right.
\omega^2 = \frac{T}{a}\left[\left(\frac{1}{M} + \frac{1}{m}\right) - \sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2-\frac{4sin^2(ka)}{Mm}}\right]
\omega^2 = \frac{T}{a}\left[\left(\frac{1}{M} + \frac{1}{m}\right) - \sqrt{\left(\frac{1}{M}+\frac{1}{m}\right)^2\left[1-\frac{Mm4sin^2(ka)}{(M+m)^2}\right]}\right]
\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left[1 - \sqrt{\left[1-\frac{Mm4sin^2(ka)}{(M+m)^2}\right]}\right]

the next step is wrong, you changed a minus into a plus between the terms in the sqrt

with small angle approximation we get:
\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left[1 - \sqrt{1+\frac{4Mmk^2a^2}{(M+m)^2}}\right]

the next step is wrong. you didn't use the right expansion of sqrt(1-x). use
<br /> \sqrt(1-x)\approx 1-x/2<br />

\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left(1-1+\frac{2\sqrt{Mm}ka}{m+M}\right)
hence
\omega^2 = \frac{2T}{\sqrt{Mm}}k

Where am I going wrong? I don't see any way to prove this.