What is Debye model theory?

In solid-state physics, debye theory is used to estimate the phonons contributing to the specific heat capacity in a solid. It explains that the specific heat is a consequence of the vibrations of the atomic lattice of the solid, which is in contrast to the Einstein model. In the Einstein model, solids are treated as independent, non-interacting quantum harmonic oscillators. This model clearly explains the low-temperature dependence of the heat capacity, which is proportional to T3. This is known as Debye Tlaw.

Introduction to Debye’s theory

In 1912, Peter Debye found the debye model of specific heat capacity. The Einstein model of specific heat of solid consider each atom as an individual atom and gave only the correct high temperatures limit. But debye model explains that there are a maximum number of modes of vibrations in solid and framed the vibrations as standing wave modes.

This model is similar to black body radiation, where electromagnetic radiation acts as a photon gas. The density of states of modes is called phonons–quanta of sound waves.

Debye found that the harmonic oscillation of atoms is the same as sound. As sound is also a wave, it should be quantized the same way as Planck quantized light. There is only one difference between the light and sound wave: for light, there are two polarizations for each wave vector k, whereas sound has three modes for each wave vector. One in longitudinal mode, atomic motion is in the same direction as, k, and the other two are transverse mode, where the wave motion is perpendicular to the wave vector k. Light has only transverse mode.

Derivation of Debye's specific heat capacity

To determine the low-temperatures capacity of the heat, it is important to go beyond the assumption of the Einstein model which states that all the modes have the same frequency. As an alternate, we have to find and count the number of modes present in the crystal structure based on its model. In the crystal, the modes are standing waves.

The number of possible wave numbers (k) for a significant crystallographic orientation is,


Here, L is the size of the grain and N is the number of atoms along with the orientation of the crystal, respectively.

Oscillation with wave vectors  |k|>NπLfalls outside the brillouin zone boundary. By counting the different wave numbers, we understand that for one dimension, one k value per 2πLand in 3- dimension, one k value per (2πL)3.

The total number of modes in the crystal is calculated by multiplying the numbers of the density of modes L2π by the reciprocal space value, which is taken by all wave vectors within the sphere whose radius is k. Therefore, the total number of modes N is,

N=L2π343πk3N=Vk36π2                 (1)

Here, V is the macroscopic volume of the crystal, which is given by V=L3.

D(ω)=dNdωD(ω)=ddωVk36π2=V6π2ddωk3D(ω)=V6π23k2dkdωD(ω)=Vk22π2dkdω            (2)

The debye model treats the crystal as an isotropic elastic medium, where the periodic pattern of atoms is the same. In this model, the wave vector is directly proportional to the frequency.

ω α kω=vskk=ωvs         (3)

Where, vs is the velocity of sound acts as the proportionality constant here.This is similar to the dispersion relation in optics, where the speed of sounds takes the role of speed of light.

By using this relation, we can derive debye’s density of states. Substitute the value of k in (2),

D(ω)=Vω22π2vs2ddωωvsD(ω)=Vω22π2vs3         (4)

The thermal vibrational energy is given by,


Where, kB is the Boltzmann distribution constant and h is the planck's constant.

Substitute the value of D(ω),

Evib=0ωD3Vω22π2vs3hωexphωkBT-1dω       (5)

The factor 3 represents the three different polarizations. One longitudinal and two transverse for each wave vector in the atomic lattice. The upper linmit is reduced from infinity to Debye frequency ωD, the maximum frequency in the crystal lattice. This observation is given on the basis of one k per unit cell.

Therefore, the total numbr of modes becomes,

N=Vk36π2=Vω36π2vs3        (6)

Hence, the cut off frequency is,

ωD=vs6π2NV3     and

The corresponding wave vector is,


From the equation (5), substitute hωkBT=x


Therefore, the vibrational energy is,

Evib=3Vh2π2vs3kB3T3h3kBTh0xDxex-1dx  Evib=3VkB4T42π2vs3h30xDxex-1dx  

The Debye temperature θD in terms of Debye frequency is expressed as,


The Debye model of heat capacity is calculated easily by differentiating the vibrational energy with respect to the temperature.

cv=EvibTcv=T0ωD3Vω22π2vs3hωexphωkBT-1dω      cv=3Vh2π2vs30ωDω3exphωkBT-1dω  cv=3Vh2π2vs30ωDω3-1exphωkBT-1hωkBexphωkBT-1T2dω  cv=3Vh22π2vs30ωDω4 exphωkBTexphωkBT-12dω

In terms of Debye temperature,


Low temperature limit

The Debye temperature is said to be low, if the upper limit (xD=θDT)tends to infinity. This infinite integral is reduced to constant (convergent series).


which is the Debye T3 approximation.

Therefore, the experimental observation of T3 dependence of heat capacity at low temperatures is correctly proved.

High temperature limit

The Debye temperature is said to be high if(T>>θD), it is given by,


This is the Dulong-Petit law.

Einstein-Debye specific heat

At very low temperatures, the T dependence of specific heat agrees for nonmetals. And for metals, the specific heat of the atoms at high temperatures is defined by the Einstein model. The temperature dependence of the einstein model is just on T. Both Einstein model and debye model gives major contribution to the high temperature. To explain the low-temperatures specific heats of metals, they include electron contribution to the specific heat. The combined expression is given as,


Context and Applications

This is an important topic in solid-state physics and thermodynamics, for all the graduates and postgraduates, especially for

  • Bachelors in science (physics and chemistry)
  • Bachelors in technology (electrical engineering)

Practice Problems

Question 1: Phonons are ____.

a. Quanta of sound

b. Quanta of light

c. Quanta of energy

d. Quanta of heat

Answer: The correct option is a.

Explanation: Phonons are the quanta of sound waves. When energy is given to the atomic lattice, the lattice absorbs energy and gets excited to a higher energy level. While returning to the ground state, it emits radiation in the sound-wave region, which is known as phonons.

Question 2: The Debye temperature for silver is 225 K. Calculate the maximum frequency for lattice vibrations in silver.

a. 3.8×1012Hzb.4.7×1012Hzc. 2.6×1013Hzd.4.5×1015Hz

Answer: The correct option is b.


Given data:

Debye temperature θD=225K

Boltzmann constant kB=1.38×10-23J/K

Planck's constant h=6.632×10-34 J.s

The maximum debye frequency is derived from the debye temperature,


Substitute all the values,

ωD=1.38×10-23×2256.632×10-34ωD=4.7×1012 Hz.

Question 3: Find the specific heat capacity of sodium at 20K. The Debye temperature of sodium is 150K.


Answer: The correct option is c.


Given data:

Debye temperature of sodium θD=150K.

The temperature of sodium T= 20K.

At low temperatures, Debye specific heat capacity is given by,


Question 4: The total number of polarizations in a sound wave in solid is______.

a. 1

b. 2

c. 3

d. 4

Answer: The correct option is c.

Explanation: There are 3 modes of polarisations in a sound wave. One is longitudinal mode and the other two are transverse mode.

Question 5: In the Debye model, the wave vector is proportional to______.

a. Frequency

b. Wavelength

c. Temperature

d. None of these

Answer: The correct option is a.

Explanation: In the debye model, the wave vector is directly proportional to frequency.

ω α kω=vs k

Where, ω is the frequency, k is the wave vector and vs is the proportionality constant (speed of sound).

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