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Phonon excitation

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Homework Statement


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

Homework Equations


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


The Attempt at a Solution


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

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

Answers and Replies

  • #2
olgranpappy
Homework Helper
1,271
3

Homework Statement


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

Homework Equations


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


The Attempt at a Solution


I work through it, but repeatedly find that [tex]\omega^2 \propto k[/tex], and I can't see anyway of getting a [tex]k^2[/tex] factor on the right.
[tex]\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][/tex]
[tex]\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][/tex]
[tex]\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][/tex]
the next step is wrong, you changed a minus into a plus between the terms in the sqrt
with small angle approximation we get:
[tex]\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left[1 - \sqrt{1+\frac{4Mmk^2a^2}{(M+m)^2}}\right][/tex]
the next step is wrong. you didn't use the right expansion of sqrt(1-x). use
[tex]
\sqrt(1-x)\approx 1-x/2
[/tex]
[tex]\omega^2 = \frac{T}{a}\left(\frac{M+m}{Mm}\right)\left(1-1+\frac{2\sqrt{Mm}ka}{m+M}\right)[/tex]
hence
[tex]\omega^2 = \frac{2T}{\sqrt{Mm}}k[/tex]

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

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