The Debye model is a solid-state equivalent of Planck's law of black body photon radiation, where one treats electromagnetic photonic radiation as a photon gas. The Debye model treats atomic vibrations as phonons in a box (the box being the solid). Most of the calculation steps are identical as both are examples of a massless Bose gas with linear dispersion relation.
Consider a cube of side L{\displaystyle L}
where n{\displaystyle n}
where h{\displaystyle h}
in which cs{\displaystyle c_{s}}
in which pn{\displaystyle p_{n}}
The approximation that the frequency is inversely proportional to the wavelength (giving a constant speed of sound) is good for low-energy phonons but not for high-energy phonons (see the article on phonons). This disagreement is one of the limitations of the Debye model. It produces incorrect results at intermediate temperatures, whereas the results are exact at the low and high temperatures limits.
Let's now compute the total energy in the box,
E=∑nEnN¯(En),{\displaystyle E=\sum _{n}E_{n}\,{\bar {N}}(E_{n})\,,}where N¯(En){\displaystyle {\bar {N}}(E_{n})}
Here, the Debye model and Planck's law of black body photon radiation differ. Unlike electromagnetic photon radiation in a box, there is a finite number of phonon energy states because a phonon cannot have arbitrarily high frequencies. Its frequency is bounded by the medium of its propagation—the atomic lattice of the solid. Consider an illustration of a transverse phonon below.
It is reasonable to assume that the minimum wavelength of a phonon is twice the atom separation, as shown in the lower figure. There are N{\displaystyle N}
making the maximum mode number n{\displaystyle n} (infinite for photons)
nmax=N3.{\displaystyle n_{\rm {max}}={\sqrt[{3}]{N}}\,.}This number bounds the upper limit of the triple energy sum
U=∑nxN3∑nyN3∑nzN3EnN¯(En).{\displaystyle U=\sum _{n_{x}}^{\sqrt[{3}]{N}}\sum _{n_{y}}^{\sqrt[{3}]{N}}\sum _{n_{z}}^{\sqrt[{3}]{N}}E_{n}\,{\bar {N}}(E_{n})\,.}For slowly varying, well-behaved functions, a sum can be replaced with an integral (also known as Thomas–Fermi approximation)
U≈∫0N3∫0N3∫0N3E(n)N¯(E(n))dnxdnydnz.{\displaystyle U\approx \int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}E(n)\,{\bar {N}}\left(E(n)\right)\,dn_{x}\,dn_{y}\,dn_{z}\,.}So far, there has been no mention of N¯(E){\displaystyle {\bar {N}}(E)}
Because a phonon has three possible polarization states (one longitudinal, and two transverse which approximately do not affect its energy) the formula above must be multiplied by 3,
N¯(E)=3eE/kT−1.{\displaystyle {\bar {N}}(E)={3 \over e^{E/kT}-1}\,.}Actually one uses an effective sonic velocity cs:=ceff{\displaystyle c_{s}:=c_{\rm {eff}}}
Substituting into the energy integral yields
U=∫0N3∫0N3∫0N3E(n)3eE(n)/kT−1dnxdnydnz.{\displaystyle U=\int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}\int _{0}^{\sqrt[{3}]{N}}E(n)\,{3 \over e^{E(n)/kT}-1}\,dn_{x}\,dn_{y}\,dn_{z}\,.}The ease with which these integrals are evaluated for photons is due to the fact that light's frequency, at least semi-classically, is unbound. As the figure above illustrates, this is not true for phonons. In order to approximate this triple integral, Debye used spherical coordinates.
(nx,ny,nz)=(nsinθcosϕ,nsinθsinϕ,ncosθ){\displaystyle \ (n_{x},n_{y},n_{z})=(n\sin \theta \cos \phi ,n\sin \theta \sin \phi ,n\cos \theta )}and approximated the cube by an eighth of a sphere
U≈∫0π/2∫0π/2∫0RE(n)3eE(n)/kT−1n2sinθdndθdϕ,{\displaystyle U\approx \int _{0}^{\pi /2}\int _{0}^{\pi /2}\int _{0}^{R}E(n)\,{3 \over e^{E(n)/kT}-1}n^{2}\sin \theta \,dn\,d\theta \,d\phi \,,}where R{\displaystyle R}
so we get
R=6Nπ3.{\displaystyle R={\sqrt[{3}]{6N \over \pi }}\,.}The substitution of integration over a sphere for the correct integral introduces another source of inaccuracy into the model.
The energy integral becomes
U=3π2∫0Rhcsn2Ln2ehcsn/2LkT−1dn{\displaystyle U={3\pi \over 2}\int _{0}^{R}\,{hc_{s}n \over 2L}{n^{2} \over e^{hc_{\rm {s}}n/2LkT}-1}\,dn}.Changing the integration variable to x=hcsn2LkT{\displaystyle x={hc_{\rm {s}}n \over 2LkT}}
To simplify the appearance of this expression, define the Debye temperature TD{\displaystyle T_{\rm {D}}}
TD =def hcsR2Lk=hcs2Lk6Nπ3=hcs2k6πNV3{\displaystyle T_{\rm {D}}\ {\stackrel {\mathrm {def} }{=}}\ {hc_{\rm {s}}R \over 2Lk}={hc_{\rm {s}} \over 2Lk}{\sqrt[{3}]{6N \over \pi }}={hc_{\rm {s}} \over 2k}{\sqrt[{3}]{{6 \over \pi }{N \over V}}}}where V{\displaystyle V}
Many references[2][3] describe the Debye temperature as merely shorthand for some constants and material-dependent variables. However, as shown below, kTD{\displaystyle kT_{\rm {D}}}
Continuing, we then have the specific internal energy:
UNk=9T(TTD)3∫0TD/Tx3ex−1dx=3TD3(TDT),{\displaystyle {\frac {U}{Nk}}=9T\left({T \over T_{\rm {D}}}\right)^{3}\int _{0}^{T_{\rm {D}}/T}{x^{3} \over e^{x}-1}\,dx=3TD_{3}\left({T_{\rm {D}} \over T}\right)\,,}where D3(x){\displaystyle D_{3}(x)}
Differentiating with respect to T{\displaystyle T}
These formulae treat the Debye model at all temperatures. The more elementary formulae given further down give the asymptotic behavior in the limit of low and high temperatures. As already mentioned, this behavior is exact, in contrast to the intermediate behavior. The essential reason for the exactness at low and high energies, respectively, is that the Debye model gives (i) the exact dispersion relation E(ν){\displaystyle E(\nu )}
Debye's derivation[edit]
Debye derived his equation somewhat differently and more simply. Using continuum mechanics, he found that the number of vibrational states with a frequency less than a particular value was asymptotic to
n∼13ν3VF,{\displaystyle n\sim {1 \over 3}\nu ^{3}VF\,,}in which V{\displaystyle V} is the volume and F{\displaystyle F}
if the vibrational frequencies continued to infinity. This form gives the T3{\displaystyle T^{3}}
Debye knew that this assumption was not really correct (the higher frequencies are more closely spaced than assumed), but it guarantees the proper behaviour at high temperature (the Dulong–Petit law). The energy is then given by
U=∫0νmhν3VFehν/kT−1dν,=VFkT(kT/h)3∫0TD/Tx3ex−1dx.{\displaystyle {\begin{aligned}U&=\int _{0}^{\nu _{m}}\,{h\nu ^{3}VF \over e^{h\nu /kT}-1}\,d\nu \,,\\&=VFkT(kT/h)^{3}\int _{0}^{T_{\rm {D}}/T}\,{x^{3} \over e^{x}-1}\,dx\,.\end{aligned}}}Substituting TD{\displaystyle T_{\rm {D}}} for hνm/k{\displaystyle h\nu _{m}/k}
where D3{\displaystyle D_{3}}
Another derivation[edit]
First we derive the vibrational frequency distribution; the following derivation is based on Appendix VI from.[4] Consider a three-dimensional isotropic elastic solid with N atoms in the shape of a rectangular parallelepiped with side-lengths Lx,Ly,Lz{\displaystyle L_{x},L_{y},L_{z}}
lx2+ly2+lz2=1.{\displaystyle l_{x}^{2}+l_{y}^{2}+l_{z}^{2}=1.}
(1)
Solutions to the wave equation are
u(x,y,z,t)=sin(2πνt)sin(2πlxxλ)sin(2πlyyλ)sin(2πlzzλ){\displaystyle u(x,y,z,t)=\sin(2\pi \nu t)\sin \left({\frac {2\pi l_{x}x}{\lambda }}\right)\sin \left({\frac {2\pi l_{y}y}{\lambda }}\right)\sin \left({\frac {2\pi l_{z}z}{\lambda }}\right)}and with the boundary conditions u=0{\displaystyle u=0}
2lxLxλ=nx;2lyLyλ=ny;2lzLzλ=nz{\displaystyle {\frac {2l_{x}L_{x}}{\lambda }}=n_{x};{\frac {2l_{y}L_{y}}{\lambda }}=n_{y};{\frac {2l_{z}L_{z}}{\lambda }}=n_{z}}
(2)
where nx,ny,nz{\displaystyle n_{x},n_{y},n_{z}}
The above equation, for fixed frequency ν{\displaystyle \nu }
N(ν)=184π3(2νcs)3LxLyLz=4πν3V3cs3,{\displaystyle N(\nu )={\frac {1}{8}}{\frac {4\pi }{3}}\left({\frac {2\nu }{c_{\mathrm {s} }}}\right)^{3}L_{x}L_{y}L_{z}={\frac {4\pi \nu ^{3}V}{3c_{\mathrm {s} }^{3}}},}
(3)
where V=LxLyLz{\displaystyle V=L_{x}L_{y}L_{z}}
Following the derivation from,[5] we define an upper limit to the frequency of vibration νD{\displaystyle \nu _{D}}
3N=N(νD)=4πνD3V3cs3{\displaystyle 3N=N(\nu _{\rm {D}})={\frac {4\pi \nu _{\rm {D}}^{3}V}{3c_{\rm {s}}^{3}}}}
(4)
By defining νD=kTDh{\displaystyle \nu _{\rm {D}}={\frac {kT_{\rm {D}}}{h}}}
N(ν)=3Nh3ν3k3TD3,{\displaystyle N(\nu )={\frac {3Nh^{3}\nu ^{3}}{k^{3}T_{\rm {D}}^{3}}},}
(5)
this definition is more standard. We can find the energy contribution for all oscillators oscillating at frequency ν{\displaystyle \nu }. Quantum harmonic oscillators can have energies Ei=(i+1/2)hν{\displaystyle E_{i}=(i+1/2)h\nu }
The energy contribution for oscillators with frequency ν{\displaystyle \nu } is then
dU(ν)=∑i=0∞Ei1Ae−Ei/(kT){\displaystyle dU(\nu )=\sum _{i=0}^{\infty }E_{i}{\frac {1}{A}}e^{-E_{i}/(kT)}}
(6)
By noting that ∑i=0∞ni=dN(ν){\displaystyle \sum _{i=0}^{\infty }n_{i}=dN(\nu )}
From above, we can get an expression for 1/A; substituting it into (6), we have
dU=dN(ν)e1/2hν/(kT)(1−e−hν/(kT))∑i=0∞hν(i+1/2)e−hν(i+1/2)/(kT)=dN(ν)(1−e−hν/(kT))∑i=0∞hν(i+1/2)e−hνi/(kT)=dN(ν)hν(12+(1−e−hν/(kT))∑i=0∞ie−hνi/(kT))=dN(ν)hν(12+1ehν/(kT)−1).{\displaystyle {\begin{aligned}dU&=dN(\nu )e^{1/2h\nu /(kT)}(1-e^{-h\nu /(kT)})\sum _{i=0}^{\infty }h\nu (i+1/2)e^{-h\nu (i+1/2)/(kT)}\\\\&=dN(\nu )(1-e^{-h\nu /(kT)})\sum _{i=0}^{\infty }h\nu (i+1/2)e^{-h\nu i/(kT)}\\&=dN(\nu )h\nu \left({\frac {1}{2}}+(1-e^{-h\nu /(kT)})\sum _{i=0}^{\infty }ie^{-h\nu i/(kT)}\right)\\&=dN(\nu )h\nu \left({\frac {1}{2}}+{\frac {1}{e^{h\nu /(kT)}-1}}\right).\end{aligned}}}Integrating with respect to ν yields
U=9Nh4k3TD3∫0νD(12+1ehν/(kT)−1)ν3dν.{\displaystyle U={\frac {9Nh^{4}}{k^{3}T_{\rm {D}}^{3}}}\int _{0}^{\nu _{D}}\left({\frac {1}{2}}+{\frac {1}{e^{h\nu /(kT)}-1}}\right)\nu ^{3}d\nu .}Low-temperature limit[edit]
The temperature of a Debye solid is said to be low if T≪TD{\displaystyle T\ll T_{\rm {D}}}
This definite integral can be evaluated exactly:
CVNk∼12π45(TTD)3.{\displaystyle {\frac {C_{V}}{Nk}}\sim {12\pi ^{4} \over 5}\left({T \over T_{\rm {D}}}\right)^{3}.}In the low-temperature limit, the limitations of the Debye model mentioned above do not apply, and it gives a correct relationship between (phononic) heat capacity, temperature, the elastic coefficients, and the volume per atom (the latter quantities being contained in the Debye temperature).
High-temperature limit[edit]
The temperature of a Debye solid is said to be high if T≫TD{\displaystyle T\gg T_{\rm {D}}}
where
CVNk∼3.{\displaystyle {\frac {C_{V}}{Nk}}\sim 3\,.}This is the Dulong–Petit law, and is fairly accurate although it does not take into account anharmonicity, which causes the heat capacity to rise further. The total heat capacity of the solid, if it is a conductor or semiconductor, may also contain a non-negligible contribution from the electrons.
Debye versus Einstein[edit]
Debye vs. Einstein. Predicted heat capacity as a function of temperature.
The Debye and Einstein models correspond closely to experimental data, but the Debye model is correct at low temperatures whereas the Einstein model is not. To visualize the difference between the models, one would naturally plot the two on the same set of axes, but this is not immediately possible as both the Einstein model and the Debye model provide a functional form for the heat capacity. As models, they require scales to relate them to their real-world counterparts. One can see that the scale of the Einstein model is given by ϵ/k{\displaystyle \epsilon /k}
The scale of the Debye model is TD{\displaystyle T_{\rm {D}}}, the Debye temperature. Both are usually found by fitting the models to the experimental data. (The Debye temperature can theoretically be calculated from the speed of sound and crystal dimensions.) Because the two methods approach the problem from different directions and different geometries, Einstein and Debye scales are not the same, that is to say
ϵk≠TD,{\displaystyle {\epsilon \over k}\neq T_{\rm {D}}\,,}which means that plotting them on the same set of axes makes no sense. They are two models of the same thing, but of different scales. If one defines the Einstein condensation temperature as
TE =def ϵk,{\displaystyle T_{\rm {E}}\ {\stackrel {\mathrm {def} }{=}}\ {\epsilon \over k}\,,}then one can say
TE≠TD,{\displaystyle T_{\rm {E}}\neq T_{\rm {D}}\,,}and, to relate the two, we must seek the ratio
TETD.{\displaystyle {\frac {T_{\rm {E}}}{T_{\rm {D}}}}\,.}The Einstein solid is composed of single-frequency quantum harmonic oscillators, ϵ=ℏω=hν{\displaystyle \epsilon =\hbar \omega =h\nu }
which makes the Einstein temperature
TE=ϵk=hνk=hcs2kNV3,{\displaystyle T_{\rm {E}}={\epsilon \over k}={h\nu \over k}={hc_{\rm {s}} \over 2k}{\sqrt[{3}]{N \over V}}\,,}
and the sought ratio is therefore
TETD=π63 =0.805995977...{\displaystyle {T_{\rm {E}} \over T_{\rm {D}}}={\sqrt[{3}]{\pi \over 6}}\ =0.805995977...}Using the ratio, both models can be plotted on the same graph. This ratio is the cube root of the ratio of the volume of one octant of a 3-dimensional sphere to the volume of the cube that contains it, which is just the correction factor used by Debye when approximating the energy integral above. Alternatively, the ratio of the two temperatures can be seen to be the ratio of Einstein's single frequency at which all oscillators oscillate and Debye's maximum frequency. Einstein's single frequency can then be seen to be a mean of the frequencies available to the Debye model.
Debye temperature table[edit]
Even though the Debye model is not completely correct, it gives a good approximation for the low temperature heat capacity of insulating, crystalline solids where other contributions (such as highly mobile conduction electrons) are negligible. For metals, the electron contribution to the heat is proportional to T{\displaystyle T}, which at low temperatures dominates the Debye T3{\displaystyle T^{3}} result for lattice vibrations. In this case, the Debye model can only be said to approximate the lattice contribution to the specific heat. The following table lists Debye temperatures for several pure elements[2] and sapphire:
The Debye model's fit to experimental data is often phenomenologically improved by allowing the Debye temperature to become temperature dependent;[6] for example, the value for water ice increases from about 222 K[7] to 300 K[8] as the temperature goes from absolute zero to about 100 K.
Extension to other quasi-particles[edit]
For other bosonic quasi-particles, e.g., for magnons (quantized spin waves) in ferromagnets instead of the phonons (quantized sound waves) one can derive analogous results. In this case at low frequencies one has different dispersion relations of momentum and energy, e.g., E(ν)∝k2{\displaystyle E(\nu )\propto k^{2}}
Extension to liquids[edit]
It was long thought that phonon theory is not able to explain the heat capacity of liquids, since liquids only sustain longitudinal, but not transverse phonons, which in solids are responsible for 2/3 of the heat capacity. However, Brillouin scattering experiments with neutrons and with X-rays, confirming an intuition of Yakov Frenkel,[9] have shown that transverse phonons do exist in liquids, albeit restricted to frequencies above a threshold called the Frenkel frequency. Since most energy is contained in these high-frequency modes, a simple modification of the Debye model is sufficient to yield a good approximation to experimental heat capacities of simple liquids.[10]
Debye frequency[edit]
The Debye frequency (Symbol: ωDebye{\displaystyle \omega _{\rm {Debye}}}
Throughout this whole article periodic boundary conditions are assumed.
Definition[edit]
Assuming the dispersion relation is
ω=vs|k|,{\displaystyle \omega =v_{\rm {s}}|\mathbf {k} |,}with vs{\displaystyle v_{\rm {s}}}
For a one dimensional monatomic chain the Debye frequency is equal to[12]
ωD=vsπ/a=vsπN/L=vsπλ,{\displaystyle \omega _{\rm {D}}=v_{\rm {s}}\pi /a=v_{\rm {s}}\pi N/L=v_{\rm {s}}\pi \lambda ,}with a{\displaystyle a}
For a two dimensional monatomic square lattice the Debye frequency is equal to
ωD2=4πa2vs2=4πNAvs2≡4πσvs2,{\displaystyle \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},}where a{\displaystyle a} and N{\displaystyle N} are the same as before; A≡L2=Na2{\displaystyle A\equiv L^{2}=Na^{2}}
For a three dimensional monatomic primitive cubic crystal, the Debye frequency is equal to[13]
ωD3=6π2a3vs3=6π2NVvs3≡6π2ρvs3,{\displaystyle \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},}where a{\displaystyle a} and N{\displaystyle N} are the same as before; V≡L3=Na3{\displaystyle V\equiv L^{3}=Na^{3}}
The speed of sound in the crystal could depend on (among others) the mass of the atoms, the strength of their interaction, the pressure on the system, and/or the polarization of the spin wave (longitudinal or transverse), but in the following we will first assume the speed of sound to be the same for any polarization (this assumption however does not render far-reaching implications).[14]
The assumed dispersion relation is easily proven wrong for a one-dimensional chain of masses, but in Debye's model this did not prove to be problematic.
Relation to Debye's temperature[edit]
The Debye temperature θD{\displaystyle \theta _{\rm {D}}}
θD=ℏkBωD,{\displaystyle \theta _{\rm {D}}={\frac {\hbar }{k_{\rm {B}}}}\omega _{\rm {D}},}
where ℏ{\displaystyle \hbar } is the reduced Planck constant and kB{\displaystyle k_{\rm {B}}}is the Boltzmann constant.Debye's derivation[edit]
Three dimensional crystal[edit]
In Debye's derivation of the heat capacity he sums over all possible modes of the system. That is: including different directions and polarizations. He assumed the total number of modes per polarization to be 3N{\displaystyle 3N} (with N{\displaystyle N} the amount of masses in the system), or in mathematical language[14]
∑modes3=3N,{\displaystyle \sum _{\rm {modes}}3=3N,}where the 3{\displaystyle 3}
The left hand side is now to be made explicit to show how it depends on the Debye frequency (here simply introduced as a cut-off frequency, that is: higher frequencies than the Debye frequency cannot exist), so that an expression for it could be found.
First of all, by assuming L{\displaystyle L} to be very large (L{\displaystyle L}>>1, with L{\displaystyle L} the size of the system in any of the three directions) the smallest wave vector in any direction could be approximated by: dki=2π/L{\displaystyle dk_{i}=2\pi /L}
where k≡(kx,ky,kz){\displaystyle \mathbf {k} \equiv (k_{x},k_{y},k_{z})}
The triple integral could be rewritten as a single integral over all possible values of the absolute value of k{\displaystyle \mathbf {k} }
with kD{\displaystyle k_{\rm {D}}}
Since we know the dispersion relation to be ω=vs|k|{\displaystyle \omega =v_{\rm {s}}|\mathbf {k} |}
After solving the integral it is again equated to 3N{\displaystyle 3N} to find
V2π2vs3ωD3=3N.{\displaystyle {\frac {V}{2\pi ^{2}v_{\rm {s}}^{3}}}\omega _{\rm {D}}^{3}=3N.}Conclusion:
ωD3=6π2NVvs3.{\displaystyle \omega _{\rm {D}}^{3}={\frac {6\pi ^{2}N}{V}}v_{\rm {s}}^{3}.}One dimensional chain in 3D space[edit]
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
∑modes3=3N.{\displaystyle \sum _{\rm {modes}}3=3N.}The rest of the derivation is analogous to the previous, so again the left hand side is rewritten;
∑modes3=3L2π∫−kDkDdk=3Lπvs∫0ωDdω.{\displaystyle \sum _{\rm {modes}}3={\frac {3L}{2\pi }}\int _{-k_{\rm {D}}}^{k_{\rm {D}}}dk={\frac {3L}{\pi v_{\rm {s}}}}\int _{0}^{\omega _{\rm {D}}}d\omega .}In the last step the multiplication 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 kD{\displaystyle k_{D}}
Conclusion:
ωD=πvsNL.{\displaystyle \omega _{\rm {D}}={\frac {\pi v_{\rm {s}}N}{L}}.}Two dimensional crystal[edit]
The same derivation could be done for a two dimensional crystal. Again, 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,
∑modes3=3N.{\displaystyle \sum _{\rm {modes}}3=3N.}And then the left hand side is rewritten and equated to 3N{\displaystyle 3N}
∑modes3=3A(2π)2∬dk=3A2πvs2∫0ωDωdω=3AωD24πvs2=3N,{\displaystyle \sum _{\rm {modes}}3={\frac {3A}{(2\pi )^{2}}}\iint d\mathbf {k} ={\frac {3A}{2\pi v_{\rm {s}}^{2}}}\int _{0}^{\omega _{\rm {D}}}\omega d\omega ={\frac {3A\omega _{\rm {D}}^{2}}{4\pi v_{\rm {s}}^{2}}}=3N,}where A≡L2{\displaystyle A\equiv L^{2}}
Conclusion
ωD2=4πNAvs2.{\displaystyle \omega _{\rm {D}}^{2}={\frac {4\pi N}{A}}v_{\rm {s}}^{2}.}Allowing polarization to make a difference[edit]
As mentioned in the introduction: in general, longitudinal waves have a different wave velocity than transverse waves. For clarity they were first assumed to be equal, but now we drop that assumption.
The dispersion relation becomes ωi=vs,i|k|{\displaystyle \omega _{i}=v_{s,i}|\mathbf {k} |}
One dimension[edit]
Once again the summation over the modes is rewritten
∑i∑modes1=∑iLπvs,i∫0ωDdωi=3N.{\displaystyle \sum _{i}\sum _{\rm {modes}}1=\sum _{i}{\frac {L}{\pi v_{s,i}}}\int _{0}^{\omega _{\rm {D}}}d\omega _{i}=3N.}The result is
LωDπ(1vs,1+1vs,2+1vs,3)=3N.{\displaystyle {\frac {L\omega _{\rm {D}}}{\pi }}({\frac {1}{v_{s,1}}}+{\frac {1}{v_{s,2}}}+{\frac {1}{v_{s,3}}})=3N.}Thus the Debye frequency is found
ωD=3πNLvs,1vs,2vs,3vs,2vs,3+vs,1vs,3+vs,1vs,2.{\displaystyle \omega _{\rm {D}}={\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}}}.}Or by assuming the two transverse polarizations to be the same (to have the same phase speed and frequency)
ωD=3πNLvs,tvs,l2vs,l+vs,t.{\displaystyle \omega _{\rm {D}}={\frac {3\pi N}{L}}{\frac {v_{s,t}v_{s,l}}{2v_{s,l}+v_{s,t}}}.}One can check this relation is equivalent to the one found earlier (when polarization did not make a difference) by setting vs,t=vs,l{\displaystyle v_{s,t}=v_{s,l}}
Two dimensions[edit]
The same derivation can be done for a two dimensional crystal to find (the derivation is analogous to previous derivations)
ωD2=12πNA(vs,1vs,2vs,3)2(vs,2vs,3)2+(vs,1vs,3)2+(vs,1vs,2)2.{\displaystyle \omega _{\rm {D}}^{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}}}.}Or by assuming the two transverse polarizations are equal (although for two dimensions it would be more logical if all polarizations would be different):
ωD2=12πNA(vs,tvs,l)22vs,l2+vs,t2.{\displaystyle \omega _{\rm {D}}^{2}={\frac {12\pi N}{A}}{\frac {(v_{s,t}v_{s,l})^{2}}{2v_{s,l}^{2}+v_{s,t}^{2}}}.}Again, one can check this relation is equivalent to the one found earlier by setting vs,t=vs,l{\displaystyle v_{s,t}=v_{s,l}}.
Three dimensions[edit]
The same derivation can be done for a three dimensional crystal to find (the derivation is analogous to previous derivations)
ωD3=18π2NV(vs,1vs,2vs,3)3(vs,2vs,3)3+(vs,1vs,3)3+(vs,1vs,2)3.{\displaystyle \omega _{\rm {D}}^{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}}}.}Or by assuming the two transverse polarizations are equal (although for three dimensions it would be more logical when all polarizations would be the same):
ωD3=18π2NV(vs,tvs,l)32vs,l3+vs,t3.{\displaystyle \omega _{\rm {D}}^{3}={\frac {18\pi ^{2}N}{V}}{\frac {(v_{s,t}v_{s,l})^{3}}{2v_{s,l}^{3}+v_{s,t}^{3}}}.}Again, one can check this relation is equivalent to the one found earlier by setting vs,t=vs,l{\displaystyle v_{s,t}=v_{s,l}}.
Derivation with the actual dispersion relation[edit]
Because only the discretized points matter, two different waves could render the same physical manifestation (see Phonon).
This problem could be made more insightful by making it more complex. Instead of using the dispersion relation ω=vsk{\displaystyle \omega =v_{\rm {s}}k}
ω(k)=2κm|sin(ka2)|.{\displaystyle \omega (k)=2{\sqrt {\frac {\kappa }{m}}}\left|\sin \left({\frac {ka}{2}}\right)\right|.}
After plotting this relation, it is clear that Debye's estimation of the cut-off wavelength was right after all. Because for every wavenumber bigger than π/a{\displaystyle \pi /a}
vs(k=π/a)=2aπκm.{\displaystyle v_{\rm {s}}(k=\pi /a)={\frac {2a}{\pi }}{\sqrt {\frac {\kappa }{m}}}.}
By simply inserting k=π/a{\displaystyle k=\pi /a} in the original dispersion relation we find
ω(k=π/a)=2κm=ωD.{\displaystyle \omega (k=\pi /a)=2{\sqrt {\frac {\kappa }{m}}}=\omega _{\rm {D}}.}
Combining these results the same result is once again found
ωD=πvsa.{\displaystyle \omega _{\rm {D}}={\frac {\pi v_{\rm {s}}}{a}}.}
However, for diatomic chains (and more complex chains) the associated cut-off frequency (and wavelength) is not very accurate, since the cut-off wavelength is twice as big and the dispersion relation consists of two branches (for a diatomic chain). It is also not certain from this whether for more dimensional systems the cut-off frequency was accurately predicted by Debye.
Alternative derivation[edit]
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 (taken from Nyquist–Shannon sampling theorem).
For a one dimensional chain this result could also be reproduced using theory on aliasing. The Nyquist–Shannon sampling theorem is used in the following derivation; the main difference being that in the following derivation the discretization is not in time, but in space. If we use the correct dispersion relation from last paragraph, it will be clear in another insightful way why the cut-off frequency has the value previously (twice) derived. So again,
ω(k)=2κm|sin(ka2)|{\displaystyle \omega (k)=2{\sqrt {\frac {\kappa }{m}}}\left|\sin \left({\frac {ka}{2}}\right)\right|}
is assumed.This derivation is completely equivalent to the previous one, that is: the same assumptions are made to retrieve the result. It is not more or less accurate, it is just a different approach.
To determine where the cut-off frequency should be, it is useful to first determine where the cut-off of the wavelength should be. From the dispersion relation we know that for k>π/a{\displaystyle k>\pi /a}
However, the dispersion relation from previous paragraph (the correct one) is not even necessary in reasoning as to why the cut-off should be at λ=2a{\displaystyle \lambda =2a}
This results again in kD=π/a{\displaystyle k_{\rm {D}}=\pi /a}
ωD=πvsa.{\displaystyle \omega _{\rm {D}}={\frac {\pi v_{\rm {s}}}{a}}.}
Also here it does not matter which dispersion relation is used (the correct one or the one Debye used), the same cut-off frequency would be found.
Unfortunately, the same method could not be used (as easily) for a two- or three-dimensional crystal, because diagonal waves would have a larger cut-off wavelength, which are also difficult to predict.