Debye Frequency for 1D atomic chain

[Onamor]

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 w_{D}=\pi \left(\frac{C}{M}\right)^{1/2}.

Homework Equations

The dispersion relation for a 1D chain of monovalent atoms:
w(k)=\left(\frac{4C}{M}\right)^{1/2}\left|sin(\frac{1}{2}ka)\right|

gives the density of states in frequency:
g(w)=\frac{2N}{\pi w_{ZB} \left(1-\frac{w^{2}}{w_{ZB}^{2}}\right)^{1/2}}

where w_{ZB}=\left(\frac{4C}{M}\right)^{1/2} is the frequency at the Brillouin zone boundary.

The Attempt at a Solution

The Debye cut-off frequency is defined by N=\int^{w_{D}}_{0}g(w)dw where N is the number of unit cells.

Doing the math i get N=\frac{2N}{\pi}\int^{w_D}_{0}\frac{dw}{\left(w_{ZB}^{2} - w^{2}\right)^{1/2}}
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 w=v_{sound}k but that didn't work either.

 

[Dickfore]

The integral is not expressible in terms of arctan, but arcsin.

However, I think you are doing unnecessary calculations twice. You first convert the density of states from k to frequency domain using the dispresion relation \omega = \omega(k) and the number conservation principle:

<br /> g(\omega) \, d\omega = g(k) \, dk<br />

and then integrate this over \omega. However, you could easily convert the integral to k domain again:

<br /> N = \int_{0}^{\omega_{D}}{g(\omega) \, d\omega} = \int_{0}^{k_{D}}{g(k) \, dk}, \; \omega_{D} = \omega(k_{D})<br />

However, it is much easier to do the k integral because:

<br /> g(k) \, dk = \frac{L}{2\pi} \times 2 \, dk \Rightarrow g(k) = \frac{L}{\pi}<br />

where L is the total length of the 1-D crystal. So, obtain the Debye wavevector k_{D} and then use the dispersion relation to express the DEbye frequency.

Finally, what is:

<br /> \frac{N}{L} = ?<br />

 

[Onamor]

Hi, thanks so much for your help - very much appreciated.

I get the Debye wavevector from N=\int^{k_{D}}_{0}g(k)dk=\int^{k_{D}}_{0}\frac{L}{\pi}dk=\frac{k_{D}L}{\pi}
ie k_{D}=\frac{N\pi}{L}

putting this into the dispersion relation gives w(k_{D})=\left(\frac{4C}{M}\right)^{1/2}\left|sin\left(\frac{N\pi a}{2L}\right)\right|

and since \frac{N}{L}=\frac{1}{a} we have

w_{D}=w(k_{D})=\left(\frac{4C}{M}\right)^{1/2}\left|sin\left(\frac{\pi}{2}\right)\right|=2\left(\frac{C}{M}\right)^{1/2}

which isn't quite right according to the question, but there may be a mistake in the question (it's not from a past paper). sorry if I'm being dim...

 

[blueyellow]

I have this homework problem too, and I have no idea how to do it. How do you get rid of the '4' and the 'sin' in the equation?

The equation to be derived looks very similar to the equation for a spring, so do I just have to derive the equation of the spring? I'd appreciate it much if anyone helped.

 

[Level21]

Solution

the density of states for a linear atomic chain is (as previously stated)

D( \omega )=\frac{ L }{ \pi } \frac{ dk }{d \omega}

that in the Debye's assumption of \omega = v k, becomes

D( \omega )=\frac{ L }{ \pi } \frac{ dk }{d \omega} ≈ \frac{ L }{v \pi }

the total number of states in a 1D atomic chain (assuming a single degree of freedom per particle)
is N=L/a with "L" the total chain length and "a" the spacing.

hence by definition the Debye's frequency must fullfill

N= \frac {L}{a} = \int ^{ \omega _D} _{0} D( \omega ) d \omega = \omega _D \frac{ L }{v \pi }<br /> <br />

\omega _D = \frac{ \pi v }{a}

Now for a linear atomic chain \omega (k) = 2 (C/M)^{1/2} |Sin (ka/2)|

Hence in the long wavelength approximation (i.e. linearized around k->0),
\omega= k a (C/M)^{1/2}

hence v= a (C/M)^{1/2}


that by substitution provides: \omega _D= \pi (C/M)^{1/2} (q.e.d.)