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Onamor
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Homework Statement
Consider phonons propagating on a one-dimensional chain of N identical atoms of mass M interacting by nearest-neighbour spring constants of magnitude C.
Show that the Debye frequency can be written as [tex]w_{D}=\pi \left(\frac{C}{M}\right)^{1/2}[/tex].
Homework Equations
The dispersion relation for a 1D chain of monovalent atoms:
[tex]w(k)=\left(\frac{4C}{M}\right)^{1/2}\left|sin(\frac{1}{2}ka)\right|[/tex]
gives the density of states in frequency:
[tex]g(w)=\frac{2N}{\pi w_{ZB} \left(1-\frac{w^{2}}{w_{ZB}^{2}}\right)^{1/2}}[/tex]
where [tex]w_{ZB}=\left(\frac{4C}{M}\right)^{1/2}[/tex] is the frequency at the Brillouin zone boundary.
The Attempt at a Solution
The Debye cut-off frequency is defined by [tex]N=\int^{w_{D}}_{0}g(w)dw[/tex] where [tex]N[/tex] is the number of unit cells.
Doing the math i get [tex]N=\frac{2N}{\pi}\int^{w_D}_{0}\frac{dw}{\left(w_{ZB}^{2} - w^{2}\right)^{1/2}}[/tex]
And the integral is equal to arctan of the integrand... at which point I assume I've gone wrong.
Am I using the right dispersion relation? I've tried using [tex]w=v_{sound}k[/tex] but that didn't work either.
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