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
$$\sqrt(1-x)\approx 1-x/2$$

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

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