In thermodynamics and solid-state physics, the Debye model is a method developed by Peter Debye in 1912 to estimate phonon contribution to the specific heat (heat capacity) in a solid.[1] It treats the vibrations of the atomic lattice (heat) as phonons in a box in contrast to the Einstein photoelectron model, which treats the solid as many individual, non-interacting quantum harmonic oscillators. The Debye model correctly predicts the low-temperature dependence of the heat capacity of solids, which is proportional to
T3
The Debye model treats atomic vibrations as phonons confined in the solid's volume. It is analogous to Planck's law of black body radiation, which treats electromagnetic radiation as a photon gas confined in a vacuum space. Most of the calculation steps are identical, as both are examples of a massless Bose gas with a linear dispersion relation.
For a cube of side-length
L
λn={2L\overn},
n
En =h\nun,
h
\nun
En=h\nun={hc\rm\overλn}={hcsn\over2L},
cs
2 | |
E | |
n |
2}=\left({hc | |
c | |
\rms |
2\right), | |
\over2L}\right) | |
z |
pn
nx
ny
nz
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, which is a limitation of the Debye model. This approximation leads to incorrect results at intermediate temperatures, whereas the results are exact at the low and high temperature limits.
The total energy in the box,
U
U=\sumnEn\bar{N}(En),
\bar{N}(En)
En
U=
\sum | |
nx |
\sum | |
ny |
\sum | |
nz |
En\bar{N}(En).
The Debye model and Planck's law of black body radiation differ here with respect to this sum. Unlike electromagnetic photon radiation in a box, there are a finite number of phonon energy states because a phonon cannot have an arbitrarily high frequency. Its frequency is bounded by its propagation medium—the atomic lattice of the solid. The following illustration describes transverse phonons in a cubic solid at varying frequencies:
It is reasonable to assume that the minimum wavelength of a phonon is twice the atomic separation, as shown in the lowest example. With
N
\sqrt[3]{N}
L/\sqrt[3]{N}
λ\rm={2L\over\sqrt[3]{N}},
nmax
n\rm=\sqrt[3]{N}.
This contrasts with photons, for which the maximum mode number is infinite. This number bounds the upper limit of the triple energy sum
U=
\sqrt[3]{N | |
\sum | |
nx |
If
En
n
U
\sqrt[3]{N | |
≈ \int | |
0 |
To evaluate this integral, the function
\bar{N}(E)
E,
\langleN\rangleBE={1\overeE/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,
\bar{N}(E)={3\overeE/kT-1}.
Considering all three polarization states together also means that an effective sonic velocity
c{\rm
cs.
T\rm
c{\rm
-3 | |
T | |
\rmD |
\proptoc{\rm
Substituting
\bar{N}(E)
U=
\sqrt[3]{N | |
\int | |
0 |
These integrals are evaluated for photons easily because their frequency, at least semi-classically, is unbound. The same is not true for phonons, so in order to approximate this triple integral, Peter Debye used spherical coordinates,
(nx,ny,nz)=(n\sin\theta\cos\phi,n\sin\theta\sin\phi,n\cos\theta),
U
\pi/2 | |
≈ \int | |
0 |
\pi/2 | |
\int | |
0 |
R | |
\int | |
0 |
E(n){3\overeE(n)/kT-1}n2\sin\thetadnd\thetad\phi,
R
3
\pi/2 | |
\int | |
0 |
\pi/2 | |
\int | |
0 |
\sin\thetad\theta
R | ||
d\phi\int | E(n) | |
0 |
1 | |
eE(n)/kT-1 |
n2dn=
3\pi | |
2 |
R | ||
\int | E(n) | |
0 |
1 | |
eE(n)/kT-1 |
n2dn
N
N={1\over8}{4\over3}\piR3,
R=\sqrt[3]{6N\over\pi}.
The substitution of integration over a sphere for the correct integral over a cube introduces another source of inaccuracy into the resulting model.
After making the spherical substitution and substituting in the function
E(n)
U=
R | |
{3\pi\over2}\int | |
0 |
{hcsn\over2L}{n2\over
hc\rmn/2LkT | |
e |
-1}dn
Changing the integration variable to
x={hc\rmn\over2LkT}
U={3\pi\over2}kT\left({2LkT\overhc\rm
To simplify the appearance of this expression, define the Debye temperature
T\rm
T\rm \stackrel{def
V
L
Some authors[2] [3] describe the Debye temperature as shorthand for some constants and material-dependent variables. However,
kT\rm
From the total energy, the specific internal energy can be calculated:
U | |
Nk |
=9T\left({T\overT\rm
D3(x)
T
CV | |
Nk |
=9\left({T\overT\rm
E(\nu)
Debye derived his equation 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\sim{1\over3}\nu3VF,
V
F
T
U=
infty | |
\int | |
0 |
{h\nu3VF\overeh\nu/kT-1}d\nu,
T3
3N
\num
3N={1\over3}
3 | |
\nu | |
m |
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
\begin{align} U&=
\num | |
\int | |
0 |
{h\nu3VF\overeh\nu/kT-1}d\nu,\\ &=VFkT(kT/h)3
T\rm/T | |
\int | |
0 |
{x3\overex-1}dx. \end{align}
T\rm
h\num/k
\begin{align} U&=9NkT(T/T\rm)3
T\rm/T | |
\int | |
0 |
{x3\overex-1}dx,\\ &=3NkTD3(T\rm/T), \end{align}
D3
First the vibrational frequency distribution is derived from Appendix VI of Terrell L. Hill's An Introduction to Statistical Mechanics.[4] Consider a three-dimensional isotropic elastic solid with N atoms in the shape of a rectangular parallelepiped with side-lengths
Lx,Ly,Lz
k=(kx,ky,kz)
l | ||||
|
,
l | ||||
|
,
l | ||||
|
Solutions to the wave equation are
u(x,y,z,t)=\sin(2\pi\nut)\sin\left(
2\pilxx | \right)\sin\left( | |
λ |
2\pilyy | \right)\sin\left( | |
λ |
2\pilzz | |
λ |
\right)
u=0
x,y,z=0,x=Lx,y=Ly,z=Lz
nx,ny,nz
cs=λ\nu
| ||||||||
|
+
| ||||||||
|
+
| ||||||||
|
=1.
\nu
nx,ny,nz
\nu
Lx,Ly,Lz\toinfty
N(\nu)
[0,\nu]
V=LxLyLz
3 | ||||||
|
=
1 | ||||||
|
+
2 | ||||||
|
Following the derivation from A First Course in Thermodynamics,[5] an upper limit to the frequency of vibration is defined
\nuD
N
3N
[0,\nuD]
\nuD
\nu\rm=
kT\rm | |
h |
\nu
Ei=(i+1/2)h\nu
i=0,1,2,...c
Ei
n | ||||
|
-Ei/(kT) | ||
e | = |
1 | |
A |
e-(i+1/2)h\nu/(kT).
The energy contribution for oscillators with frequency
\nu
By noting that
infty | |
\sum | |
i=0 |
ni=dN(\nu)
dN(\nu)
\nu
1 | |
A |
e-1/2h\nu/(kT)
infty | |
\sum | |
i=0 |
e-ih\nu/(kT)=
1 | |
A |
e-1/2h\nu/(kT)
1 | |
1-e-h\nu/(kT) |
=dN(\nu).
From above, we can get an expression for 1/A; substituting it into,
\begin{align} dU&=dN(\nu)e1/2h\nu/(kT)(1-e-h\nu/(kT)
infty | |
)\sum | |
i=0 |
h\nu(i+1/2)e-h\nu(i+1/2)/(kT)\ \\ &=dN(\nu)(1-e-h\nu/(kT)
infty | |
)\sum | |
i=0 |
h\nu(i+1/2)e-h\nu\\ &=dN(\nu)h\nu\left(
1 | |
2 |
+(1-e-h\nu/(kT)
infty | |
)\sum | |
i=0 |
ie-h\nu\right)\\ &=dN(\nu)h\nu\left(
1 | + | |
2 |
1 | |
eh\nu/(kT)-1 |
\right). \end{align}
Integrating with respect to ν yields
U=
9Nh4 | ||||||
|
\nuD | ||
\int | \left( | |
0 |
1 | + | |
2 |
1 | |
eh\nu/(kT)-1 |
\right)\nu3d\nu.
The temperature of a Debye solid is said to be low if
T\llT\rm
CV | |
Nk |
\sim9\left({T\overT\rm
This definite integral can be evaluated exactly:
CV | |
Nk |
\sim{12\pi4\over5}\left({T\overT\rm
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).
The temperature of a Debye solid is said to be high if
T\ggT\rm
ex-1 ≈ x
|x|\ll1
CV | |
Nk |
\sim9\left({T\overT\rm
CV | |
Nk |
\sim3.
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
\epsilon/k
CV=3Nk\left({\epsilon\overkT}\right)2{e\epsilon/kT\over\left(e\epsilon/kT-1\right)2}.
The scale of the Debye model is
T\rm
{\epsilon\overk}\neT\rm,
T\rm \stackrel{def
T\rm\neT\rm,
T\rm | |
T\rm |
The Einstein solid is composed of single-frequency quantum harmonic oscillators,
\epsilon=\hbar\omega=h\nu
λmin
\nu={c\rm\overλ}={c\rm\sqrt[3]{N}\over2L}={c\rm\over2}\sqrt[3]{N\overV}
T\rm={\epsilon\overk}={h\nu\overk}={hc\rm\over2k}\sqrt[3]{N\overV},
{T\rm\overT\rm
Using the ratio, both models can be plotted on the same graph. It is the cube root of the ratio of the volume of one octant of a three-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.
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
T3
|
|
|
|
For other bosonic quasi-particles, e.g., 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(\nu)\proptok2
E(\nu)\proptok
k=2\pi/λ
\intg(\nu){\rmd}\nu\equivN
\DeltaC{\rm
\DeltaC{\rm
\proptoT
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] More recently, it has been shown that instantaneous normal modes associated with relaxations from saddle points in the liquid energy landscape, which dominate the frequency spectrum of liquids at low frequencies, may determine the specific heat of liquids as a function of temperature over a broad range.[11]
The Debye frequency (Symbol:
\omega\rm
\omega\rm
Throughout this section, periodic boundary conditions are assumed.
Assuming the dispersion relation is
\omega=v\rm|k|,
v\rm
For a one-dimensional monatomic chain, the Debye frequency is equal to[13]
\omega\rm=v\rm\pi/a=v\rm\piN/L=v\rm\piλ,
a
N
L
λ
L
N
a
L=Na
For a two-dimensional monatomic square lattice, the Debye frequency is equal to
2 | |
\omega | |
\rmD |
=
4\pi | |
a2 |
2 | |
v | |
\rms |
=
4\piN | |
A |
2 | |
v | |
\rms |
\equiv4\pi\sigma
2 | |
v | |
\rms |
,
A\equivL2=Na2
\sigma
For a three-dimensional monatomic primitive cubic crystal, the Debye frequency is equal to[14]
3 | |
\omega | |
\rmD |
=
6\pi2 | |
a3 |
3 | |
v | |
\rms |
=
6\pi2N | |
V |
3 | |
v | |
\rms |
\equiv6\pi2\rho
3 | |
v | |
\rms |
,
V\equivL3=Na3
\rho
The general formula for the Debye frequency as a function of
n
n | |
\omega | |
\rmD |
=2n\pin/2\Gamma\left(1+\tfrac{n}{2}\right)
N | |
Ln |
n | |
v | |
\rms |
,
\Gamma
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 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.[15]
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.
The Debye temperature
\theta\rm
\hbar
k\rm
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
N
\sum\rm3=3N,
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
L
L
L
dki=2\pi/L
i=x,y,z
\sum\rm3=
3V | |
(2\pi)3 |
\iiintdk,
k\equiv(kx,ky,kz)
V\equivL3
The triple integral could be rewritten as a single integral over all possible values of the absolute value of
k
3V | |
(2\pi)3 |
\iiintdk=
3V | |
2\pi2 |
k\rm | |
\int | |
0 |
|k|2dk,
k\rm
k\rm=\omega\rm/v\rm
Since the dispersion relation is
\omega=v\rm|k|
\omega
3V | |
2\pi2 |
k\rm | |
\int | |
0 |
|k|2dk=
3V | ||||||||
|
\omega\rm | |
\int | |
0 |
\omega2d\omega,
After solving the integral it is again equated to
3N
V | ||||||||
|
3 | |
\omega | |
\rmD |
=3N.
It can be rearranged into
3 | ||
\omega | = | |
\rmD |
6\pi2N | |
V |
3 | |
v | |
\rms |
.
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
\sum\rm3=3N.
The rest of the derivation is analogous to the previous, so the left hand side is rewritten with respect to the Debye frequency:
\sum\rm3=
3L | |
2\pi |
k\rm | |
\int | |
-k\rm |
dk=
3L | |
\piv\rm |
\omega\rm | |
\int | |
0 |
d\omega.
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
kD
k= | \omega |
vs |
\omegaD=kDvs
3L | |
\piv\rm |
\omega\rm | |
\int | |
0 |
d\omega=
3L | |
\piv\rm |
\omega\rm=3N.
Conclusion:
\omega\rm=
\piv\rmN | |
L |
.
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,
\sum\rm3=3N.
And then the left hand side is rewritten and equated to
3N
\sum\rm3=
3A | |
(2\pi)2 |
\iintdk=
3A | ||||||||
|
\omega\rm | |
\int | |
0 |
\omegad\omega=
| |||||||||
|
=3N,
A\equivL2
It can be rewritten as
2 | |
\omega | |
\rmD |
=
4\piN | |
A |
2 | |
v | |
\rms |
.
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
\omegai=vs,i|k|
i=1,2,3
\omega\rm
i
\sumi\sum\rm1
3N
i
The summation over the modes is rewritten
\sumi\sum\rm1=\sumi
L | |
\pivs,i |
\omega\rm | |
\int | |
0 |
d\omegai=3N.
The result is
L\omega\rm | ( | |
\pi |
1 | |
vs,1 |
+
1 | |
vs,2 |
+
1 | |
vs,3 |
)=3N.
Thus the Debye frequency is found
\omega\rm=
\piN | |
L |
3 | ||||||||||
|
=
3\piN | |
L |
vs,1vs,2vs,3 | |
vs,2vs,3+vs,1vs,3+vs,1vs,2 |
=
\piN | |
L |
veff.
The calculated effective velocity
veff
\omega\rm=
3\piN | |
L |
vs,tvs,l | |
2vs,l+vs,t |
.
Setting
vs,t=vs,l
The same derivation can be done for a two-dimensional crystal to find
2 | |
\omega | |
\rmD |
=
4\piN | |
A |
3 | ||||||||||||||||||||||||||||
|
=
12\piN | |
A |
(vs,1vs,2vs,3)2 | |
(vs,2vs,3)2+(vs,1vs,3)2+(vs,1vs,2)2 |
=
4\piN | |
A |
2. | |
v | |
eff |
The calculated effective velocity
veff
2 | |
\omega | |
\rmD |
=
12\piN | |
A |
(vs,tvs,l)2 | ||||||||||||||
|
.
Setting
vs,t=vs,l
The same derivation can be done for a three-dimensional crystal to find (the derivation is analogous to previous derivations)
2 | |
\omega | |
\rmD |
=
6\pi2N | |
V |
3 | ||||||||||||||||||||||||||||
|
=
18\pi2N | |
V |
(vs,1vs,2vs,3)3 | |
(vs,2vs,3)3+(vs,1vs,3)3+(vs,1vs,2)3 |
=
6\pi2N | |
V |
3. | |
v | |
eff |
The calculated effective velocity
veff
3 | |
\omega | |
\rmD |
=
18\pi2N | |
V |
(vs,tvs,l)3 | ||||||||||||||
|
.
Setting
vs,t=vs,l
This problem could be made more applicable by relaxing the assumption of linearity of the dispersion relation. Instead of using the dispersion relation
\omega=v\rmk
with
m
\kappa
a
\pi/a
λ
2a
\pi/a
k
\pi/a
k
k=\pi/a
By simply inserting
k=\pi/a
Combining these results the same result is once again found
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.
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
2a
2a
λ\rm=2a
k\rm=
2\pi | |
λD |
=\pi/a
It does not matter which dispersion relation is used, as the same cut-off frequency would be calculated.