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Debye model

From Biocrawler, the free encyclopedia.

In thermodynamics and solid state physics, the Debye model is a method of calculating the phonon contribution to the specific heat (heat capacity) in a solid. Debye model treats the vibrations of the atomic lattice (heat) as phonons in a box. In contrast to Einstein model, which treats the solid as many individual, non-interacting quantum harmonic oscillators. This model correctly predicts the low temperature dependence of the heat capacity, which is proportional to $T 3$ . Just like the Einstein model, it also recovers the Dulong-Petit law at high temperatures.

Contents

5 Debye temperature table

6 References

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Derivation

Debye model is a solid-state equivalent of Planck's law of black body radiation, where one treats electromagnetic radiation as photons in a box. Debye model treats atomic vibrations as phonons in a box (box is the solid). Most of the calculation steps are identical.

Consider a cube of side $L$ . From the particle in a box article, the resonating modes of the sonic disturbances inside the box have wavelengths given by

$\lambda_n = {2L\over n}$

where $n$ is an integer, and the energy formulae for a phonon

$E_n=p_nc_s=h\nu_n={hc_s\over\lambda_n}={hc_sn\over 2L}$

where $c s$ is the speed of sound inside the solid. This is in one dimension. In three dimensions the energy is

$E_n^2=E_{nx}^2+E_{ny}^2+E_{nz}^2=\left({hc_s\over2L}\right)^2\left(n_x^2+n_y^2+n_z^2\right)$

Let's now compute the total energy in the box

$U = \sum_n E_n\,\bar{N}(E_n)$

where $\bar{N}(E_n)$ is the number of particles in the box with energy $E n$ . In other words, the total energy is equal to the sum of energy multiplied by the number of particles with that energy (in one dimension). In 3 dimensions we have:

$U = \sum_{n_x}\sum_{n_y}\sum_{n_z}E_n\,\bar{N}(E_n)$

Now, this is where Debye model and Planck's law of black body radiation differ. Unlike electromagnetic radiation in a box, there is a finite number of phonon energy states because a phonon cannot have infinite frequency. Its frequency is bound 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 $N$ atoms in a solid. Our solid is a cube, which means there are $\sqrt[3]{N}$ atoms per side. Atom separation is then given by $L/\sqrt[3]{N}$ , and minimum wavelength is

$\lambda_{min} = {2L \over \sqrt[3]{N}}$

making the maximum mode $n$ , which is infinity for photons

$n_{max} = \sqrt[3]{N}$

This is the upper limit of the triple energy sum

$U = \sum_{n_x}^{\sqrt[3]{N}}\sum_{n_y}^{\sqrt[3]{N}}\sum_{n_z}^{\sqrt[3]{N}}E_n\,\bar{N}(E_n)$

For slowly-varying, well-behaved functions, a sum can be replaced with an integral (a.k.a Thomas Fermi approximation)

$U \approx\int_0^{\sqrt[3]{N}}\int_0^{\sqrt[3]{N}}\int_0^{\sqrt[3]{N}} E(n)\,\bar{N}\left(E(n)\right)dn_x dn_y dn_z$

So far, there has been no mention of $\bar{N}(E)$ , the number of particles with energy $E$ . The unit of sonic radiation is a phonon, and phonons obey Bose-Einstein statistics. Their distribution is given by the famous Bose-Einstein formula

$\langle N\rangle_{BE} = {1\over e^{E/kT}-1}$

Because a phonon has three possible polarization states (one longitudinal and two transverse) which do not affect its energy, the formula above must be multiplied by 3

$\bar{N}(E) = {3\over e^{E/kT}-1}$

Substituting this into the energy integral yields

$U = \int_0^{\sqrt[3]{N}}\int_0^{\sqrt[3]{N}}\int_0^{\sqrt[3]{N}} E(n)\,{3\over e^{E(n)/kT}-1}dn_x dn_y dn_z$

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. To compute this triple integral, Peter Debye made a bold assumption: he approximated the cube by an eighth of a sphere thus allowing the integral, and the limits, to be represented in spherical coordinates

$U \approx\int_0^{\pi/2}\int_0^{\pi/2}\int_0^R E(n)\,{3\over e^{E(n)/kT}-1}n^2 \sin\theta dn d\theta d\phi$

where $R$ 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. Volume of the cube is $N$ particles

$N = {1\over8}{4\over3}\pi R^3$

$R = \sqrt[3]{6N\over\pi}$

Energy integral becomes

$U = {3\pi\over2}\int_0^R \,{hc_sn\over 2L}{n^2\over e^{hc_sn/2LkT}-1} dn$

changing the integration variable to $x = {hc_sn\over 2LkT}$

$U = {3\pi\over2} kT \left({2LkT\over hc_s}\right)^3\int_0^{hc_sR/2LkT} {x^3\over e^x-1} dx$

To simplify the look of this expression, one can define

$T_D\equiv {hc_sR\over2Lk} = {hc_s\over2Lk}\sqrt[3]{6N\over\pi} = {hc_s\over2k}\sqrt[3]{{6\over\pi}{N\over V}}$

$T D$ is called the Debye temperature -- nothing more than a shorthand for some constants and material-dependent variables.

$U = 9NkT \left({T\over T_D}\right)^3\int_0^{T_D/T} {x^3\over e^x-1} dx$

differentiating with respect to $T$ to get heat capacity

$C_V = 9Nk \left({T\over T_D}\right)^3\int_0^{T_D/T} {x^4 e^x\over\left(e^x-1\right)^2} dx$

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Low temperature

The temperature of a Debye solid is said to be low if

T < < T D

that is, $x = T D / T > > 1$ , or $x = T_D/T\rightarrow\infty$ , leading to

$C_V(T\rightarrow0) \approx 9Nk \left({T\over T_D}\right)^3\int_0^{\infty} {x^4 e^x\over \left(e^x-1\right)^2} dx$

This approximately definite integral can be evaluated exactly

$C_V(T\rightarrow0) = {12\pi^4\over5} N k \left({T\over T_D}\right)^3$

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High temperature

The temperature of a Debye solid is said to be high if

T > > T D

that is, $x = T D / T < < 1$ . Recall that $e^x \approx 1 + x$ if $| x | < < 1$ , leading to

$C_V(T\rightarrow\infty) \approx 9Nk \left({T\over T_D}\right)^3\int_0^{T_D/T} {x^4 (1+x)\over \left((1+x)-1\right)^2} dx \approx 9Nk \left({T\over T_D}\right)^3\int_0^{T_D/T} x^2 dx$

$C_V(T\rightarrow\infty) = 3Nk$

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Debye vs. Einstein

Debye vs. Einstein. Predicted heat capacity as a function of temperature.

So how closely do the Debye and Einstein models correspond to experiment? -- Surprisingly close (does anybody have a good overlay of Debye, Einstein, and experimental data?). Debye does better at lower temperatures.

How different are the models? To answer that question one would naturally plot the two on the same set of axes... except one can't. Both, the Einstein model and the Debye model, provide the functional form of the heat capacity. They are models, and no model is without a scale. A scale relates the model to its real-world counterpart. One can see that the scale of the Einstein model, which is given by

$C_V = 3Nk\left({\epsilon\over k T}\right)^2{e^{\epsilon/kT}\over \left(e^{\epsilon/kT}-1\right)^2}$

is $ε / k$ . And the scale of the Debye model is $T D$ , the Debye temperature. Both have to be found by fitting the models to the experimental data. 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

${\epsilon\over k} \ne T_D$

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 Einstein temperature as

$T_E \equiv {\epsilon\over k}$

then one can say

$T_E \ne T_D$

and, to relate the two, seek the ratio

${T_E\over T_D} = ?$

Einstein solid is composed of single-frequency quantum harmonic oscillators, $\epsilon = \hbar\omega = h\nu$ . That frequency, if it indeed existed, would be related to the speed of sound in the solid... even though there is no sound in Einstein 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, $λ m i n$ .