Editing Debye model (section)

===Debye's derivation===
====Three-dimensional crystal====
In Debye's derivation of the [[heat capacity]], he sums over all possible modes of the system, accounting for different directions and polarisations. He assumed the total number of modes per polarization to be <math>N</math>, the amount of masses in the system, and the total to be<ref name=":0" />

:<math>\sum_{\rm modes}3=3 N,</math>

with three polarizations per mode. The sum runs over all modes without differentiating between different polarizations, and then counts the total number of polarization-mode combinations. Debye made this assumption based on an assumption from [[classical mechanics]] that the number of modes per polarization in a chain of masses should always be equal to the number of masses in the chain.

The left hand side can be made explicit to show how it depends on the Debye frequency, introduced first as a cut-off frequency beyond which no frequencies exist. By relating the cut-off frequency to the maximum number of modes, an expression for the cut-off frequency can be derived.

First of all, by assuming <math>L</math> to be very large (<math>L</math> ≫ 1, with <math>L</math> the size of the system in any of the three directions) the smallest wave vector in any direction could be approximated by: <math> d k_i = 2 \pi / L </math>, with <math>i = x, y, z</math>. Smaller wave vectors cannot exist because of the [[periodic boundary conditions]]. Thus the summation would become<ref>{{Cite web |title=The Oxford Solid State Basics |url=https://podcasts.ox.ac.uk/series/oxford-solid-state-basics |access-date=2024-01-12 |website=podcasts.ox.ac.uk |language=en}}</ref> 
:<math>\sum_{\rm modes}3=\frac {3 V}{(2 \pi)^3} \iiint d \mathbf k,</math>
where <math> \mathbf k \equiv (k_x, k_y, k_z) </math>; <math> V \equiv L^3 </math> is the size of the system; and the integral is (as the summation) over all possible modes, which is assumed to be a finite region (bounded by the cut-off frequency).

The triple integral could be rewritten as a single integral over all possible values of the absolute value of <math> \mathbf k </math> (see [[Jacobian matrix and determinant|Jacobian for spherical coordinates]]). The result is
:<math>\frac {3 V}{(2 \pi)^3} \iiint d \mathbf k = \frac {3 V}{2 \pi^2} \int_0^{k_{\rm D}} |\mathbf k|^2 d \mathbf k ,</math>
with <math>k_{\rm D}</math> the absolute value of the wave vector corresponding with the Debye frequency, so <math>k_{\rm D} = \omega_{\rm D}/v_{\rm s}</math>.

Since the dispersion relation is <math>\omega =v_{\rm s}|\mathbf k|</math>, it can be written as an integral over all possible <math> \omega </math>:
:<math> \frac {3 V}{2 \pi^2} \int_0^{k_{\rm D}} |\mathbf k|^2 d \mathbf k = \frac {3 V}{2 \pi^2 v_{\rm s}^3} \int_0^{\omega_{\rm D}} \omega^2 d \omega ,</math>

After solving the integral it is again equated to <math>3 N</math> to find
:<math> \frac {V}{2 \pi^2 v_{\rm s}^3} \omega_{\rm D}^3 = 3 N .</math>

It can be rearranged into
:<math>  \omega_{\rm D}^3 =\frac {6 \pi^2 N}{V} v_{\rm s}^3 .</math>

==== One-dimensional chain in 3D space ====
The same derivation could be done for a one-dimensional chain of atoms. The number of modes remains unchanged, because there are still three polarizations, so
:<math>\sum_{\rm modes}3=3 N.</math>

The rest of the derivation is analogous to the previous, so the left hand side is rewritten with respect to the Debye frequency:
:<math>\sum_{\rm modes}3=\frac {3 L}{2 \pi} \int_{-k_{\rm D}}^{k_{\rm D}}d k = \frac {3 L}{\pi v_{\rm s}} \int_{0}^{\omega_{\rm D}}d \omega.</math>

The last step is multiplied by two is because the integrand in the first integral is even and the bounds of integration are symmetric about the origin, so the integral can be rewritten as from 0 to <math>k_D</math> after scaling by a factor of 2. This is also equivalent to the statement that the volume of a one-dimensional ball is twice its radius. Applying a change a substitution of <math>k=\frac{\omega}{v_s}</math> , our bounds are now 0 to <math>\omega_D = k_Dv_s</math>, which gives us our rightmost integral. We continue;
:<math> \frac {3 L}{\pi v_{\rm s}} \int_{0}^{\omega_{\rm D}}d \omega = \frac {3 L}{\pi v_{\rm s}} \omega_{\rm D} = 3 N .</math>

Conclusion:
:<math> \omega_{\rm D} = \frac {\pi v_{\rm s} N}{L} .</math>

==== Two-dimensional crystal ====
The same derivation could be done for a two-dimensional crystal. The number of modes remains unchanged, because there are still three polarizations. The derivation is analogous to the previous two. We start with the same equation,
:<math>\sum_{\rm modes}3=3 N.</math>

And then the left hand side is rewritten and equated to <math>3N</math>
:<math> \sum_{\rm modes}3=\frac {3 A}{(2 \pi)^2} \iint d \mathbf k = \frac {3 A}{2 \pi v_{\rm s}^2} \int_{0}^{\omega_{\rm D}} \omega d \omega = \frac {3 A \omega_{\rm D}^2}{4 \pi v_{\rm s}^2} = 3 N ,</math>
where <math> A \equiv L^2</math> is the size of the system.

It can be rewritten as
:<math> \omega_{\rm D}^2 = \frac {4 \pi N}{A} v_{\rm s}^2 .</math>