The Debye model is a solidstate 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. Show
Consider a cube of side L{\displaystyle L} . From the particle in a box article, the resonating modes of the sonic disturbances inside the box (considering for now only those aligned with one axis) have wavelengths given byλn=2Ln,{\displaystyle \lambda _{n}={2L \over n}\,,}where n{\displaystyle n} is an integer. The energy of a phonon isEn =hνn,{\displaystyle E_{n}\ =h\nu _{n}\,,}where h{\displaystyle h} is Planck's constant and νn{\displaystyle \nu _{n}} is the frequency of the phonon. Making the approximation that the frequency is inversely proportional to the wavelength, we haveEn=hνn=hcsλn=hcsn2L,{\displaystyle E_{n}=h\nu _{n}={hc_{\rm {s}} \over \lambda _{n}}={hc_{s}n \over 2L}\,,}in which cs{\displaystyle c_{s}} is the speed of sound inside the solid. In three dimensions we will useEn2=pn2cs2=(hcs2L)2(nx2+ny2+nz2),{\displaystyle E_{n}^{2}={p_{n}^{2}c_{\rm {s}}^{2}}=\left({hc_{\rm {s}} \over 2L}\right)^{2}\left(n_{x}^{2}+n_{y}^{2}+n_{z}^{2}\right)\,,}in which pn{\displaystyle p_{n}} is the magnitude of the threedimensional momentum of the phonon.The approximation that the frequency is inversely proportional to the wavelength (giving a constant speed of sound) is good for lowenergy phonons but not for highenergy 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})} is the number of phonons in the box with energy En{\displaystyle E_{n}}. In other words, the total energy is equal to the sum of energy multiplied by the number of phonons with that energy (in one dimension). In 3 dimensions we have:U=∑nx∑ny∑nzEnN¯(En).{\displaystyle U=\sum _{n_{x}}\sum _{n_{y}}\sum _{n_{z}}E_{n}\,{\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} atoms in a solid. Our solid is a cube, which means there are N3{\displaystyle {\sqrt[{3}]{N}}} atoms per edge. Atom separation is then given by L/N3{\displaystyle L/{\sqrt[{3}]{N}}}, and the minimum wavelength isλmin=2LN3,{\displaystyle \lambda _{\rm {min}}={2L \over {\sqrt[{3}]{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, wellbehaved 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)} , the number of phonons with energy E.{\displaystyle E\,.} Phonons obey Bose–Einstein statistics. Their distribution is given by the famous Bose–Einstein statistics formula⟨N⟩BE=1eE/kT−1.{\displaystyle \langle N\rangle _{BE}={1 \over e^{E/kT}1}\,.}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}}} , i.e. the Debye temperature TD{\displaystyle T_{\rm {D}}} (see below) is proportional to ceff{\displaystyle c_{\rm {eff}}}, more precisely TD−3∝ceff−3:=(1/3)clong−3+(2/3)ctrans−3{\displaystyle T_{\rm {D}}^{3}\propto c_{\rm {eff}}^{3}:=(1/3)c_{\rm {long}}^{3}+(2/3)c_{\rm {trans}}^{3}}, where one distinguishes longitudinal and transversal soundwave velocities (contributions 1/3 and 2/3, respectively). The Debye temperature or the effective sonic velocity is a measure of the hardness of the crystal.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 semiclassically, 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} is the radius of this sphere, which is found by conserving the number of particles in the cube and in the eighth of a sphere. The volume of the cube is N{\displaystyle N} unitcell volumes,N=1843πR3,{\displaystyle N={1 \over 8}{4 \over 3}\pi R^{3}\,,}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}} ,U=3π2kT(2LkThcs)3∫0hcsR/2LkTx3ex−1dx.{\displaystyle U={3\pi \over 2}kT\left({2LkT \over hc_{\rm {s}}}\right)^{3}\int _{0}^{hc_{\rm {s}}R/2LkT}{x^{3} \over e^{x}1}\,dx.}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} is the volume of the cubic box of side L{\displaystyle L}.Many references[2][3] describe the Debye temperature as merely shorthand for some constants and materialdependent variables. However, as shown below, kTD{\displaystyle kT_{\rm {D}}} is roughly equal to the phonon energy of the minimum wavelength mode, and so we can interpret the Debye temperature as the temperature at which the highestfrequency mode (and hence every mode) is excited.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)} is the (third) Debye function.Differentiating with respect to T{\displaystyle T} we get the dimensionless heat capacity: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 )} at low frequencies, and (ii) corresponds to the exact density of states (∫g(ν)dν≡3N){\textstyle (\int g(\nu )\,d\nu \equiv 3N)}, concerning the number of vibrations per frequency interval.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} is a factor which he calculated from elasticity coefficients and density. Combining this formula with the expected energy of a harmonic oscillator at temperature T (already used by Einstein in his model) would give an energy ofU=∫0∞hν3VFehν/kT−1dν,{\displaystyle U=\int _{0}^{\infty }\,{h\nu ^{3}VF \over e^{h\nu /kT}1}\,d\nu \,,}if the vibrational frequencies continued to infinity. This form gives the T3{\displaystyle T^{3}} behaviour which is correct at low temperatures. But Debye realized that there could not be more than 3N{\displaystyle 3N} vibrational states for N atoms. He made the assumption that in an atomic solid, the spectrum of frequencies of the vibrational states would continue to follow the above rule, up to a maximum frequency νm{\displaystyle \nu _{m}}chosen so that the total number of states is3N=13νm3VF.{\displaystyle 3N={1 \over 3}\nu _{m}^{3}VF\,.}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} ,U=9NkT(T/TD)3∫0TD/Tx3ex−1dx,=3NkTD3(TD/T),{\displaystyle {\begin{aligned}U&=9NkT(T/T_{\rm {D}})^{3}\int _{0}^{T_{\rm {D}}/T}\,{x^{3} \over e^{x}1}\,dx\,,\\&=3NkTD_{3}(T_{\rm {D}}/T)\,,\end{aligned}}}where D3{\displaystyle D_{3}} is the function later given the name of thirdorder Debye function.Another derivation[edit]First we derive the vibrational frequency distribution; the following derivation is based on Appendix VI from.[4] Consider a threedimensional isotropic elastic solid with N atoms in the shape of a rectangular parallelepiped with sidelengths Lx,Ly,Lz{\displaystyle L_{x},L_{y},L_{z}} . The elastic wave will obey the wave equation and will be plane waves; consider the wave vector k=(kx,ky,kz){\displaystyle \mathbf {k} =(k_{x},k_{y},k_{z})} and define lx=kxk,ly=kyk,lz=kzk{\displaystyle l_{x}={\frac {k_{x}}{\mathbf {k} }},l_{y}={\frac {k_{y}}{\mathbf {k} }},l_{z}={\frac {k_{z}}{\mathbf {k} }}}. Note that we havelx2+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} at x,y,z=0,x=Lx,y=Ly,z=Lz{\displaystyle x,y,z=0,x=L_{x},y=L_{y},z=L_{z}}, we have2lxLxλ=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}} are positive integers. Substituting (2) into (1) and also using the dispersion relation cs=λν{\displaystyle c_{s}=\lambda \nu }, we havenx2(2νLx/cs)2+ny2(2νLy/cs)2+nz2(2νLz/cs)2=1.{\displaystyle {\frac {n_{x}^{2}}{(2\nu L_{x}/c_{s})^{2}}}+{\frac {n_{y}^{2}}{(2\nu L_{y}/c_{s})^{2}}}+{\frac {n_{z}^{2}}{(2\nu L_{z}/c_{s})^{2}}}=1.}The above equation, for fixed frequency ν{\displaystyle \nu } , describes an eighth of an ellipse in "mode space" (an eighth because nx,ny,nz{\displaystyle n_{x},n_{y},n_{z}} are positive). The number of modes with frequency less than ν{\displaystyle \nu } is thus the number of integral points inside the ellipse, which, in the limit of Lx,Ly,Lz→∞{\displaystyle L_{x},L_{y},L_{z}\to \infty } (i.e. for a very large parallelepiped) can be approximated to the volume of the ellipse. Hence, the number of modes N(ν){\displaystyle N(\nu )} with frequency in the range [0,ν]{\displaystyle [0,\nu ]} isN(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}} is the volume of the parallelepiped. Note that the wave speed in the longitudinal direction is different from the transverse direction and that the waves can be polarised one way in the longitudinal direction and two ways in the transverse direction; thus we define 3cs3=1clong3+2ctrans3{\displaystyle {\frac {3}{c_{s}^{3}}}={\frac {1}{c_{\text{long}}^{3}}}+{\frac {2}{c_{\text{trans}}^{3}}}}.Following the derivation from,[5] we define an upper limit to the frequency of vibration νD{\displaystyle \nu _{D}} ; since there are N atoms in the solid, there are 3N quantum harmonic oscillators (3 for each x, y, z direction) oscillating over the range of frequencies [0,νD]{\displaystyle [0,\nu _{D}]}. Hence we can determine νD{\displaystyle \nu _{D}} using3N=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}}} , where k is Boltzmann's constant and h is Planck's constant, and substituting (4) into (3), we getN(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 } where i=0,1,2,…{\displaystyle i=0,1,2,\dotsc } and using MaxwellBoltzmann statistics, the number of particles with energy Ei{\displaystyle E_{i}} isni=1Ae−Ei/(kT)=1Ae−(i+1/2)hν/(kT).{\displaystyle n_{i}={\frac {1}{A}}e^{E_{i}/(kT)}={\frac {1}{A}}e^{(i+1/2)h\nu /(kT)}.}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 )} (because there are dN(ν){\displaystyle dN(\nu )} modes oscillating with frequency ν{\displaystyle \nu }), we have1Ae−1/2hν/(kT)∑i=0∞e−ihν/(kT)=1Ae−1/2hν/(kT)11−e−hν/(kT)=dN(ν).{\displaystyle {\frac {1}{A}}e^{1/2h\nu /(kT)}\sum _{i=0}^{\infty }e^{ih\nu /(kT)}={\frac {1}{A}}e^{1/2h\nu /(kT)}{\frac {1}{1e^{h\nu /(kT)}}}=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)}(1e^{h\nu /(kT)})\sum _{i=0}^{\infty }h\nu (i+1/2)e^{h\nu (i+1/2)/(kT)}\\\\&=dN(\nu )(1e^{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}}+(1e^{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 Lowtemperature limit[edit]The temperature of a Debye solid is said to be low if T≪TD{\displaystyle T\ll T_{\rm {D}}} , leading toCVNk∼9(TTD)3∫0∞x4ex(ex−1)2dx.{\displaystyle {\frac {C_{V}}{Nk}}\sim 9\left({T \over T_{\rm {D}}}\right)^{3}\int _{0}^{\infty }{x^{4}e^{x} \over \left(e^{x}1\right)^{2}}\,dx.}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 lowtemperature 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). Hightemperature limit[edit]The temperature of a Debye solid is said to be high if T≫TD{\displaystyle T\gg T_{\rm {D}}} . Using ex−1≈x{\displaystyle e^{x}1\approx x} if x≪1{\displaystyle x\ll 1} leads toCVNk∼9(TTD)3∫0TD/Tx4x2dx{\displaystyle {\frac {C_{V}}{Nk}}\sim 9\left({T \over T_{\rm {D}}}\right)^{3}\int _{0}^{T_{\rm {D}}/T}{x^{4} \over x^{2}}\,dx}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 nonnegligible 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 realworld counterparts. One can see that the scale of the Einstein model is given by ϵ/k{\displaystyle \epsilon /k} :CV=3Nk(ϵkT)2eϵ/kT(eϵ/kT−1)2.{\displaystyle C_{V}=3Nk\left({\epsilon \over kT}\right)^{2}{e^{\epsilon /kT} \over \left(e^{\epsilon /kT}1\right)^{2}}.}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 singlefrequency quantum harmonic oscillators, ϵ=ℏω=hν{\displaystyle \epsilon =\hbar \omega =h\nu } . That frequency, if it indeed existed, would be related to the speed of sound in the solid. If one imagines the propagation of sound as a sequence of atoms hitting one another, then it becomes obvious that the frequency of oscillation must correspond to the minimum wavelength sustainable by the atomic lattice, λmin{\displaystyle \lambda _{min}}, whereν=csλ=csN32L=cs2NV3{\displaystyle \nu ={c_{\rm {s}} \over \lambda }={c_{\rm {s}}{\sqrt[{3}]{N}} \over 2L}={c_{\rm {s}} \over 2}{\sqrt[{3}]{N \over V}}},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 3dimensional 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 quasiparticles[edit]For other bosonic quasiparticles, 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}} in the case of magnons, instead of E(ν)∝k{\displaystyle E(\nu )\propto k} for phonons (with k=2π/λ{\displaystyle k=2\pi /\lambda }). One also has different density of states (e.g., ∫g(ν)dν≡N{\displaystyle \int g(\nu ){\rm {d}}\nu \equiv N\,}). As a consequence, in ferromagnets one gets a magnon contribution to the heat capacity, ΔCVmagnon∝T3/2{\displaystyle \Delta C_{\,{\rm {V\,magnon}}}\,\propto T^{3/2}}, which dominates at sufficiently low temperatures the phonon contribution, ΔCVphonon∝T3{\displaystyle \,\Delta C_{\,{\rm {V\,phonon}}}\propto T^{3}}. In metals, in contrast, the main lowtemperature contribution to the heat capacity, ∝T{\displaystyle \propto T}, comes from the electrons. It is fermionic, and is calculated by different methods going back to Sommerfeld's free electron model.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 Xrays, 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 highfrequency 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}}} or ωD{\displaystyle \omega _{\rm {D}}}) is a parameter in the Debye model. It refers to a cutoff angular frequency for waves of a harmonic chain of masses, used to describe the movement of ions in a crystal lattice and more specifically, to correctly predict the heat capacity in such crystals to be constant for high temperatures (Dulong–Petit law). The term was first introduced by Peter Debye in 1912.[11]Throughout this whole article periodic boundary conditions are assumed. Definition[edit]Assuming the dispersion relation is ω=vsk,{\displaystyle \omega =v_{\rm {s}}\mathbf {k} ,}with vs{\displaystyle v_{\rm {s}}} 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[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} the distance between two neighbouring atoms in the chain when the system is in its ground state of energy (in this case that means that none of the atoms are moving with respect to each other); N{\displaystyle N} the total number of atoms in the chain; and L{\displaystyle L} the size (volume) of the system (length of the chain); and λ{\displaystyle \lambda } is the linear number density. Where the following relation holds: L=Na{\displaystyle L=Na}.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}} is the size (area) of the surface; and σ{\displaystyle \sigma } the surface number density.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 size of the system; and ρ{\displaystyle \rho } the volume number density.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 farreaching implications).[14] The assumed dispersion relation is easily proven wrong for a onedimensional 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}}} , another parameter in Debye model, is related to the Debye frequency by the relationθD=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} on both sides is because of the three polarizations, so the sum runs over all modes for one specific polarization. Debye made this assumption because he knew from classical mechanics that the number of modes per polarization in a chain of masses should always be equal to the amount of masses in the chain.The left hand side is now to be made explicit to show how it depends on the Debye frequency (here simply introduced as a cutoff 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} , with i=x,y,z{\displaystyle i=x,y,z}. Smaller wave vectors cannot exist because of the periodic boundary conditions. Thus the summation would become 4∑modes3=3V(2π)3∭dk,{\displaystyle \sum _{\rm {modes}}3={\frac {3V}{(2\pi )^{3}}}\iiint d\mathbf {k} ,}where k≡(kx,ky,kz){\displaystyle \mathbf {k} \equiv (k_{x},k_{y},k_{z})} ; V≡L3{\displaystyle V\equiv L^{3}} 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 cutoff frequency).The triple integral could be rewritten as a single integral over all possible values of the absolute value of k{\displaystyle \mathbf {k} } (see: Jacobian for spherical coordinates). The result is3V(2π)3∭dk=3V2π2∫0kDk2dk,{\displaystyle {\frac {3V}{(2\pi )^{3}}}\iiint d\mathbf {k} ={\frac {3V}{2\pi ^{2}}}\int _{0}^{k_{\rm {D}}}\mathbf {k} ^{2}d\mathbf {k} ,}with kD{\displaystyle k_{\rm {D}}} the absolute value of the wave vector corresponding with the Debye frequency, so kD=ωD/vs{\displaystyle k_{\rm {D}}=\omega _{\rm {D}}/v_{\rm {s}}}.Since we know the dispersion relation to be ω=vsk{\displaystyle \omega =v_{\rm {s}}\mathbf {k} } , this can be written as an integral over all possible ω{\displaystyle \omega }3V2π2∫0kDk2dk=3V2π2vs3∫0ωDω2dω,{\displaystyle {\frac {3V}{2\pi ^{2}}}\int _{0}^{k_{\rm {D}}}\mathbf {k} ^{2}d\mathbf {k} ={\frac {3V}{2\pi ^{2}v_{\rm {s}}^{3}}}\int _{0}^{\omega _{\rm {D}}}\omega ^{2}d\omega ,}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}} after scaling by a factor of 2. Applying a change a substitution of k=ωvs{\displaystyle k={\frac {\omega }{v_{s}}}} , our bounds are now 0 to ωD=kDvs{\displaystyle \omega _{D}=k_{D}v_{s}}, which gives us our rightmost integral. We continue;3Lπvs∫0ωDdω=3LπvsωD=3N.{\displaystyle {\frac {3L}{\pi v_{\rm {s}}}}\int _{0}^{\omega _{\rm {D}}}d\omega ={\frac {3L}{\pi v_{\rm {s}}}}\omega _{\rm {D}}=3N.}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}} is the size of the system.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,ik{\displaystyle \omega _{i}=v_{s,i}\mathbf {k} } , where i=1,2,3{\displaystyle i=1,2,3}, which correspond to the three polarizations. The cutoff frequency (Debye frequency) however does not depend on i{\displaystyle i}. And we can write the total number of modes as ∑i∑modes1{\displaystyle \sum _{i}\sum _{\rm {modes}}1}, which is again equal to 3N{\displaystyle 3N}. Here the summation over the modes is (although not explicitly stated) dependent on i{\displaystyle i}.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): 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} , the correct dispersion relation is now going to be assumed. From classical mechanics it is known that for an equidistant chain of masses which interact harmonically with each other the dispersion relation reads as follows[14]ω(k)=2κmsin(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 cutoff wavelength was right after all. Because for every wavenumber bigger than π/a{\displaystyle \pi /a} (that is: λ{\displaystyle \lambda } is smaller than 2a{\displaystyle 2a}) a wavenumber that is smaller than π/a{\displaystyle \pi /a} 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. Thereby, the study of the dispersion relation can be limited to the first brillouin zone[15] i.e. for k∈[−πa,πa]{\textstyle k\in \left[{\frac {\pi }{a}},{\frac {\pi }{a}}\right]}.This is possible because the system consists of discretized points, as is demonstrated in the animated picture. Dividing the dispersion relation by k{\displaystyle k} and inserting π/a{\displaystyle \pi /a} for k{\displaystyle k}, we find the speed of a wave with k=π/a{\displaystyle k=\pi /a} to bevs(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 cutoff frequency (and wavelength) is not very accurate, since the cutoff 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 cutoff 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 cutoff frequency has the value previously (twice) derived. So again, ω(k)=2κmsin(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 cutoff frequency should be, it is useful to first determine where the cutoff of the wavelength should be. From the dispersion relation we know that for k>π/a{\displaystyle k>\pi /a} every mode is repeated, so the cutoff wavelength would be at λD=2a{\displaystyle \lambda _{\rm {D}}=2a}. From this and the periodic boundary conditions you can immediately see that the total number of modes per polarization would be N{\displaystyle N}. As seen in the gif of the previous paragraph this is because every wave with a wavelength shorter than 2a{\displaystyle 2a} could be replaced by a wave with a wavelength longer than 2a{\displaystyle 2a} to regain the same physical result.However, the dispersion relation from previous paragraph (the correct one) is not even necessary in reasoning as to why the cutoff should be at λ=2a{\displaystyle \lambda =2a} . Because, as is depicted, only waves with a longer wavelength than 2a{\displaystyle 2a} could render the same physical result as another one. So this is another way to correctly predict the cutoff wavelength of phonons without using the correct dispersion relation (or even knowledge from classical mechanics as Debye did). However, using the wrong dispersion relation which Debye assumed, waves with a smaller wavelength would have a higher frequency, but the relative movement of the masses would be the same, so this does not render new modes.This results again in kD=π/a{\displaystyle k_{\rm {D}}=\pi /a} , renderingωD=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 cutoff frequency would be found. Unfortunately, the same method could not be used (as easily) for a two or threedimensional crystal, because diagonal waves would have a larger cutoff wavelength, which are also difficult to predict. What is true about electromagnetic radiation?Electromagnetic radiation is an electric and magnetic disturbance traveling through space at the speed of light (2.998 × 108 m/s). It contains neither mass nor charge but travels in packets of radiant energy called photons, or quanta.
Which of the following statements is true about electromagnetic waves?EXPLANATION: Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. So statement 1 is correct. The electric field and magnetic field of an electromagnetic wave are perpendiculars (at right angles) to each other.
Which of the following statements about electromagnetic wave is or are correct?The oscillating electric and magnetic fields produced by oscillating charged particles are perpendicular to each other and both are perpendicular to the direction of propagation of the wave.
