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Debye Model of Crystals


We[*] represented each atom as a harmonic oscillator. It did not work.

Let us represent our system as a set of 3N oscillators, but choose them differently, including collective motion.

One-Dimensional Crystal

We have N particles with masses m on springs k:


Equation of motion:  
 m\ddot{u_i} = k(u_{i+1}-u_i) - k(u_i-u_{i-1})\end{displaymath} (3)
 \ddot{u}(x) = \omega_0^2\bigl(u(x+a)+u(x-a)-2u(x)\bigr)\end{displaymath} (4)
Periodic boundary conditions: u1=uN+1. Sound velocity:  
 c=a\omega_0,\qquad \omega_0^2=k/m\end{displaymath} (5)


All ui are entangled! To disentangle them--use a trick[*]:  
 u(x) = \sum_q \exp(iqx)u(q),\qquad u(q) = u(-q)\end{displaymath} (6)
What can we say about q? We want u(Na)=u(0), or $\exp(iqNa)+\exp(-iqNa)=1$ for all q. This means that $\cos(qNa)=1$,or

q = 0,\,\pm\frac{\pi}{Na},\,\pm\frac{2\pi}{Na},\ldots,\pm\frac{\pi}{a}\end{displaymath}

Physics: we represented deviation as sum of waves:
These waves are called modes

Dispersion Equation. Meaning of Phonons

Substitute (6) into (4):



\omega^2(q) = \omega_0^2\bigl(2-\exp(iqa)-\exp(-iqa)\bigr) =
 2\omega_0^2(1-\cos qa)\end{displaymath}

For small q

 q^2a^2\omega_0^2=c^2q^2 \end{displaymath}


u(q,t)= \mathit{const}\, e^{i(\omega t)}\end{displaymath}

In real space we obtain sum of terms proportional to

u(x,t) = e^{i(\omega t - qx)} = e^{iq(ct-x)}\end{displaymath}

This describes sound waves propagating with velocity c (that is why we claimed that c is sound velocity)

 \omega^2(q)=2(1-\cos qa)\end{displaymath} (7)
is called dispersion equation. Each q corresponds to one normal mode or one kind of phonons--in other words, to a sound wave.

Why Debye model is better than Einstein model: since $\omega(q)\to0$at $q\to0$, some phonons are excited even at low temperatures!

Heat Capacity for 1D Crystal

We obtained N independent oscillators (modes!) with frequencies

\omega^2(q)=c^2 q^2, \quad q = \frac{\pi}{Na}n, \quad n
 =0,\,1,\dots, N\end{displaymath}

Each gives a contribution $C_{\text{Einstein}}$ to the heat capacity. Total contribution:

C = \sum_q
 ... \omega(q)/k_BT)}{\bigl(1 -\exp(-\hbar\omega(q)/k_BT)\bigr)^2} \end{displaymath}

If $N\to\infty$--integration instead of summation. We have N values for q in the interval $[0,\pi/a]$--so

\sum_q \mapsto 2N\int_0^{\pi/a}\frac{dk}{2\pi}\end{displaymath}

3D Crystals

What's new for 3D?

$\mathbf{q}$ is now vector--instead of dq we must integrate over $d\mathbf{q}=4\pi q^2\,dq$, q2=qx2+qy2+qz2.
There are three waves for every $\mathbf{q}$: two transversal and one longitudinal

Instead of integrating by $d\mathbf{q}$ we integrate by $d\omega$. Number of modes:

g(\omega)\,d\omega= 3V 4\pi q^2\frac{dq}{(2\pi)^3} \end{displaymath}

Limits: lower limit $\omega=0$, upper limit should be $\Theta_E$, but we used $\omega^2=c^2 q^2$ instead of exact equation (7), so we correct this by taking $\Theta_D$instead:


Result (for simple lattice!):


 \Theta_D \propto\frac{\hbar\omega_0}{k_B}&...
 ...e^x}{(e^x-1)^2}dx& \quad \text{Debye
 \end{aligned}\end{displaymath} (8)

High temperatures:
for $u\gg1$ upper limit in the integral in (8) is small (1/u),

 = \frac{1}{3u}\end{displaymath}

and C=3NkB
Low temperatures:
for $u\ll1$ integral is constant, and $C\propto T^3$
. This is an example of a scaling, made by a classic of physics as early as 1912!

next up previous
Next: Quiz Up: Solid State Previous: Einstein Model of Crystals

© 1997 Boris Veytsman and Michael Kotelyanskii
Wed Oct 1 00:45:35 EDT 1997