Editing Debye model (section)

==Debye frequency==
The '''Debye frequency''' (Symbol: <math> \omega_{\rm Debye} </math> or <math> \omega_{\rm D} </math>) is a parameter in the Debye model that refers to a cut-off [[angular frequency]] for [[wave]]s of a harmonic chain of masses, used to describe the movement of [[ion]]s in a [[Crystal structure|crystal lattice]] and more specifically, to correctly predict that the [[heat capacity]] in such crystals is constant at high temperatures (Dulong–Petit law). The concept was first introduced by Peter Debye in 1912.<ref>{{Cite journal|last=Debye|first=P.|author-link=Peter Debye|title=Zur Theorie der spezifischen Wärmen|journal=Annalen der Physik|language=en|volume=344|issue=14|pages=789–839| doi=10.1002/andp.19123441404| issn=1521-3889 |year=1912 |bibcode=1912AnP...344..789D |url=https://zenodo.org/record/1424256}}</ref>

Throughout this section, [[periodic boundary conditions]] are assumed.

=== Definition ===
Assuming the [[dispersion relation]] is
:<math> \omega = v_{\rm s} |\mathbf k| ,</math>
with <math>v_{\rm s}</math> the [[speed of sound]] in the crystal and '''k''' the wave vector, the value of the Debye frequency is as follows:

For a one-dimensional monatomic chain, the Debye frequency is equal to<ref>{{Cite web| url=https://openphysicslums.files.wordpress.com/2012/08/latticevibrations.pdf| title=The one dimensional monatomic solid| access-date=2018-04-27}}</ref>
:<math> \omega_{\rm D} = v_{\rm s} \pi / a = v_{\rm s} \pi N / L = v_{\rm s} \pi \lambda ,</math>
with <math>a</math> as the distance between two neighbouring atoms in the chain when the system is in its [[ground state]] of energy, here being that none of the atoms are moving with respect to one another; <math>N</math> the total number of atoms in the chain; <math>L</math> the size of the system, which is the length of the chain; and <math> \lambda </math> the [[linear density|linear number density]]. For <math>L</math>, <math>N</math>, and <math>a</math>, the relation <math>L = N a</math> holds.

For a two-dimensional monatomic square lattice, the Debye frequency is equal to
:<math> \omega_{\rm D}^2 = \frac {4 \pi}{a^2} v_{\rm s}^2 = \frac {4 \pi N}{A} v_{\rm s}^2 \equiv 4 \pi \sigma v_{\rm s}^2 ,</math>
with <math> A \equiv L^{2} = N a^{2} </math> is the size (area) of the surface, and <math>\sigma</math> the [[surface density|surface number density]].

For a three-dimensional monatomic [[cubic crystal system|primitive cubic crystal]], the Debye frequency is equal to<ref>{{Cite web|url=http://farside.ph.utexas.edu/teaching/sm1/lectures/node71.html|title=Specific heats of solids| last=Fitzpatrick|first=Richard|date=2006|website=Richard Fitzpatrick [[University of Texas at Austin]]|access-date=2018-04-27}}</ref>
:<math> \omega_{\rm D}^3 = \frac {6 \pi^2}{a^3} v_{\rm s}^3 = \frac {6 \pi^2 N}{V} v_{\rm s}^3 \equiv 6 \pi^2 \rho v_{\rm s}^3 ,</math>
with <math> V \equiv L^3 = N a^3 </math> the size of the system, and <math>\rho</math> the [[density|volume number density]].

The general formula for the Debye frequency as a function of <math>n</math>, the number of dimensions for a (hyper)cubic lattice is
:<math> \omega_{\rm D}^n = 2^n \pi^{n/2} \Gamma\left(1+\tfrac{n}{2}\right) \frac {N}{L^n} v_{\rm s}^n ,</math>
with <math>\Gamma</math> being the [[gamma function]].

The speed of sound in the crystal depends on the mass of the atoms, the strength of their interaction, the [[pressure]] on the system, and the [[polarization (waves)|polarisation]] of the spin wave (longitudinal or transverse), among others. For the following, the speed of sound is assumed to be the same for any polarisation, although this limits the applicability of the result.<ref name=":0">{{Cite book|title=The Oxford Solid State Basics|last=Simon|first=Steven H.| publisher=Oxford University Press|isbn=9780199680764|edition= First|location=Oxford|oclc=859577633|author-link=Steven H. Simon | date = 2013-06-20}}</ref>

The assumed [[dispersion relation]] is easily proven inaccurate for a one-dimensional chain of masses, but in Debye's model, this does not prove to be problematic.{{Citation needed|date=January 2024}}

=== Relation to Debye's temperature ===
The Debye temperature <math> \theta_{\rm D} </math>, another parameter in Debye model, is related to the Debye frequency by the relation <math display="block"> \theta_{\rm D}=\frac{\hbar}{k_{\rm B}}\omega_{\rm D}, </math> where <math> \hbar </math> is the reduced Planck constant and <math> k_{\rm B} </math> is the [[Boltzmann constant]].

===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>

=== Polarization dependence ===
In reality, longitudinal waves often have a different wave velocity from that of transverse waves. Making the assumption that the velocities are equal simplified the final result, but reintroducing the distinction improves the accuracy of the final result.

The dispersion relation becomes <math> \omega_i = v_{s,i}|\mathbf k|</math>, with <math> i = 1, 2, 3 </math>, each corresponding to one of the three polarizations. The cut-off frequency <math> \omega_{\rm D} </math>, however, does not depend on <math>i</math>. We can write the total number of modes as <math> \sum_{i}\sum_{\rm modes} 1 </math>, which is again equal to <math>3 N</math>. Here the summation over the modes is now dependent on <math>i</math>.

==== One-dimensional chain in 3D space ====
The summation over the modes is rewritten
:<math> \sum_{i}\sum_{\rm modes} 1 = \sum_i \frac {L}{\pi v_{s,i}} \int_0^{\omega_{\rm D}} d \omega_i = 3 N .</math>

The result is
:<math> \frac {L \omega_{\rm D}}{\pi} (\frac {1}{v_{s,1}} + \frac {1}{v_{s,2}} + \frac {1}{v_{s,3}}) = 3 N .</math>

Thus the Debye frequency is found
:<math> \omega_{\rm D} = \frac{ \pi N}{L} \frac{3}{\frac {1}{v_{s,1}} + \frac {1}{v_{s,2}} + \frac {1}{v_{s,3}}} = \frac {3 \pi N}{L} \frac {v_{s,1} v_{s,2} v_{s,3}}{v_{s,2} v_{s,3} + v_{s,1} v_{s,3} + v_{s,1} v_{s,2}} = \frac{\pi N}{L} v_{\mathrm{eff}}\,. </math>

The calculated effective velocity <math> v_{\mathrm{eff}} </math> is the harmonic mean of the velocities for each polarization. By assuming the two transverse polarizations to have the same phase speed and frequency,
:<math> \omega_{\rm D} = \frac {3 \pi N}{L} \frac {v_{s,t}v_{s,l}}{2v_{s,l} + v_{s,t}} .</math>

Setting <math> v_{s,t} = v_{s,l} </math> recovers the expression previously derived under the assumption that velocity is the same for all polarization modes.

==== Two-dimensional crystal ====
The same derivation can be done for a two-dimensional crystal to find
:<math> \omega_{\rm D}^2 = \frac {4 \pi N}{A} \frac{3}{\frac {1}{v_{s,1}^2} + \frac {1}{v_{s,2}^2} + \frac {1}{v_{s,3}^2}} = \frac {12 \pi N}{A} \frac {(v_{s,1} v_{s,2} v_{s,3})^2}{(v_{s,2} v_{s,3})^2 + (v_{s,1} v_{s,3})^2 + (v_{s,1} v_{s,2})^2} = \frac{4 \pi N}{A} v_{\mathrm{eff}}^2\,.</math>

The calculated effective velocity <math> v_{\mathrm{eff}}</math> is the square root of the harmonic mean of the squares of velocities. By assuming the two transverse polarizations to be the same,
:<math> \omega_{\rm D}^2 = \frac {12 \pi N}{A} \frac {(v_{s,t} v_{s,l})^2}{2 v_{s,l}^2 + v_{s,t}^2} .</math>

Setting <math> v_{s,t} = v_{s,l} </math> recovers the expression previously derived under the assumption that velocity is the same for all polarization modes.

====Three-dimensional crystal====
The same derivation can be done for a three-dimensional crystal to find (the derivation is analogous to previous derivations)
:<math> \omega_{\rm D}^2 = \frac{6 \pi^2 N}{V} \frac{3}{\frac {1}{v_{s,1}^3} + \frac {1}{v_{s,2}^3} + \frac {1}{v_{s,3}^3}} = \frac {18 \pi^2 N}{V} \frac {(v_{s,1} v_{s,2} v_{s,3})^3}{(v_{s,2} v_{s,3})^3 + (v_{s,1} v_{s,3})^3 + (v_{s,1} v_{s,2})^3 } = \frac{6 \pi^2 N}{V} v_{\mathrm{eff}}^3\,.</math>

The calculated effective velocity <math> v_{\mathrm{eff}}</math> is the cube root of the harmonic mean of the cubes of velocities. By assuming the two transverse polarizations to be the same,
:<math> \omega_{\rm D}^3 = \frac {18 \pi^2 N}{V} \frac {(v_{s,t} v_{s,l})^3}{2 v_{s,l}^3 + v_{s,t}^3} .</math>

Setting <math> v_{s,t} = v_{s,l} </math> recovers the expression previously derived under the assumption that velocity is the same for all polarization modes.

=== Derivation with the actual dispersion relation ===
[[File:phonon k 3k.gif|right|thumb|250px|Because only the [[Discretization|discretized]] points matter, two different waves could render the same physical manifestation (see [[Phonon]]).]]
This problem could be made more applicable by relaxing the assumption of linearity of the dispersion relation. Instead of using the dispersion relation <math> \omega = v_{\rm s} k </math>, a more accurate dispersion relation can be used. In classical mechanics, it is known that for an equidistant chain of masses which interact harmonically with each other, the dispersion relation is<ref name=":0" />

<math display="block"> \omega (k) = 2 \sqrt {\frac {\kappa}{m}}\left|\sin\left(\frac {k a}{2}\right)\right| ,</math>

with <math> m</math> being the mass of each atom, <math> \kappa</math> the spring constant for the [[harmonic oscillator]], and <math> a</math> still being the spacing between atoms in the ground state. After plotting this relation, Debye's estimation of the cut-off wavelength based on the linear assumption remains accurate, because for every wavenumber bigger than <math>\pi / a </math> (that is, for <math> \lambda </math> is smaller than <math>2 a</math>), a wavenumber that is smaller than <math>\pi / a</math> could be found with the same angular frequency. This means the resulting physical manifestation for the mode with the larger wavenumber is indistinguishable from the one with the smaller wavenumber. Therefore, the study of the dispersion relation can be limited to the first [[Brillouin zone]] <math display="inline"> k \in \left[-\frac{\pi}{a},\frac{\pi}{a}\right] </math> without any loss of accuracy or information.<ref>{{Cite book|last=Srivastava|first=G. P. |url=https://books.google.com/books?id=XiCmDwAAQBAJ&q=brillouin+zone+debye+model+book&pg=PA44|title=The Physics of Phonons |date=2019-07-16|publisher=Routledge|isbn=978-1-351-40955-1|language=en}}</ref> This is possible because the system consists of [[Discretization|discretized]] points, as is demonstrated in the animated picture. Dividing the dispersion relation by <math>k</math> and inserting <math>\pi / a</math> for <math>k</math>, we find the speed of a wave with <math>k = \pi / a</math> to be
<math display="block"> v_{\rm s}(k = \pi / a) = \frac {2 a}{\pi} \sqrt {\frac {\kappa}{m}} .</math>

By simply inserting <math>k = \pi/a</math> in the original dispersion relation we find
<math display="block"> \omega(k = \pi / a) = 2 \sqrt {\frac {\kappa}{m}} = \omega_{\rm D} .</math>

Combining these results the same result is once again found
<math display="block"> \omega_{\rm D} = \frac {\pi v_{\rm s}}{a} .</math>

However, for any chain with greater complexity, including diatomic chains, the associated cut-off frequency and wavelength are not very accurate, since the cut-off wavelength is twice as big and the dispersion relation consists of additional branches, two total for a diatomic chain. It is also not certain from this result whether for higher-dimensional systems the cut-off frequency was accurately predicted by Debye when taking into account the more accurate dispersion relation.

===Alternative derivation===
[[File:CPT-sound-nyquist-thereom-1.5percycle.svg|thumb|250px|The physical result of two waves can be identical when at least one of them has a wavelength that is bigger than twice the initial distance between the masses.]]

For a one-dimensional chain, the formula for the Debye frequency can also be reproduced using a theorem for describing [[aliasing]]. The [[Nyquist–Shannon sampling theorem]] is used for this derivation, the main difference being that in the case of a one-dimensional chain, the discretization is not in time, but in space.

The cut-off frequency can be determined from the cut-off wavelength. From the sampling theorem, we know that for wavelengths smaller than <math> 2a </math>, or twice the sampling distance, every mode is a repeat of a mode with wavelength larger than <math> 2a </math>, so the cut-off wavelength should be at <math> \lambda_{\rm D} = 2 a </math>. This results again in <math> k_{\rm D} = \frac{2 \pi}{\lambda_D} = \pi / a </math>, rendering
<math display="block"> \omega_{\rm D} = \frac {\pi v_{\rm s}}{a} .</math>

It does not matter which dispersion relation is used, as the same cut-off frequency would be calculated.