# Debye model and reciprocal space

• I
• WWCY
In summary, the Debye model for solids assumes a linear dispersion relation and treats the solid as a box with phonons inside. The mean number of phonons excited at a given temperature is given by Planck's distribution with a chemical potential of 0. The calculation for the mean energy involves integrating over the phonon density of states, which is given by ## D(\omega) = L \pi /c ## in one dimension. This model is not applicable to small numbers of atoms as it is based on quantum statistical mechanics and requires a large number of atoms for statistical calculations.

#### WWCY

Hi everyone, I need a little help understanding how periodic reciprocal space applies to the Debye model for solids. Many thanks in advance!

If we start with the general derivation of a dispersion relation for a 1D system, with atoms coupled by springs, one gets the following relation
$$\omega = 2\sqrt{\frac{k}{m}} |\sin \big( \frac{ka}{2} \big)|$$
whereby it is possible to show that the function is periodic with ##\omega (k + 2\pi/a) = \omega (k)##.

However if we consider Debye's model, which comes as a result of taking the long wavelength (and small k) limit, we get
$$\omega = v_s k$$
where ##v_s## is the speed of sound.

##\textbf{Question}##: Does the concept of a periodic ##k## still apply here? Also, is the number of allowed modes for a chain of N masses still equal to N?

For example if I were to calculate energy contributions by individual frequencies ##\omega (k)## using the following equation
$$\langle E \rangle _k= \hbar \omega(k) \big( n_b (\beta \hbar \omega (k)) + 1/2 \big)$$
clearly energy for the ##k=0## mode differs from that of the ##k = 2\pi /a## mode. Should I pick ##k = 0## for my calculations or ##k = 2 \pi /a##?

Hi,

the Debye model assumes a linear dispersion relation, as you say. This model "forgets" the oscillators (like coupled springs); it treats the problem as if the solid was a box with phonons inside. The mean number of phonons excited at temperature T is given by Planck's distribution (Bose-Einstein with ## \mu = 0## ). So, since there is not a lattice of atoms now (only phonons in a box), you don't have the periodicity.

Mr rabbit said:
Hi,

the Debye model assumes a linear dispersion relation, as you say. This model "forgets" the oscillators (like coupled springs); it treats the problem as if the solid was a box with phonons inside. The mean number of phonons excited at temperature T is given by Planck's distribution (Bose-Einstein with ## \mu = 0## ). So, since there is not a lattice of atoms now (only phonons in a box), you don't have the periodicity.

Debye also postulated that a chain of N atoms (with motion in 1d) should only have N modes. Should I decide to find the expectation value of energy for this chain using
$$\langle E \rangle = \sum_{k} \hbar \omega(k) \big( n_b (\beta \hbar \omega (k)) + 1/2 \big)$$
how do I choose the N wavenumbers? I assume that the range to "pick" from is ##0<k\leq 2\pi/a## with a ##k##-space interval of ##\frac{2\pi}{Na}##? I'm also assuming that we ignore ##k=0## since by ##\omega = vk## it simply means that all the masses don't vibrate anyway. Is this right?

Thanks for assisting!

If you want to know the mean energy of your crystal (or chain, in this case), you have to sum over all k's which are accesibles given the temperature ##T## of the chain. For each mode (##\vec{k},## p) (p is the branch) the mean energy is

## \langle E_p (k) \rangle = [ \langle n_p (k) \rangle+1/2 ] \hbar \omega_p (k) ##

so the total mean energy is

## \displaystyle \langle E \rangle = \sum_{k, p} \langle E_p (k) \rangle ##

Since the wavenumber here is a continuous variable, you have to replace the sum by an integral weighted with the density of states. In 1D this density is ## D (\omega) = L \pi /c ##, where c is the speed of sound (##\omega = c k##) and L the length of the chain. Finally, the calculation you have to do is

## \displaystyle \langle E \rangle = \int _0^{\omega_D} (\langle n \left( \omega) \rangle+1/2 \right) \hbar \omega \:D(\omega) d\omega ##

where ## \displaystyle \langle n (\omega) \rangle = \frac{1}{e^{\frac{\hbar \omega}{k T}}-1} ## is the ocupation number for phonons (Planck's distribution).

WWCY
Hi, thanks for the reply!

Mr rabbit said:
Since the wavenumber here is a continuous variable, you have to replace the sum by an integral weighted with the density of states. In 1D this density is ## D (\omega) = L \pi /c ##, where c is the speed of sound (##\omega = c k##) and L the length of the chain. Finally, the calculation you have to do is

## \displaystyle \langle E \rangle = \int _0^{\omega_D} (\langle n \left( \omega) \rangle+1/2 \right) \hbar \omega \:D(\omega) d\omega ##

where ## \displaystyle \langle n (\omega) \rangle = \frac{1}{e^{\frac{\hbar \omega}{k T}}-1} ## is the ocupation number for phonons (Planck's distribution).

I understand that as the number of atoms (N) becomes large, the wavenumber becomes continuous and we can do integrals. However if N is not large, (perhaps a chain of 2 atoms for example), does what I said here still apply?
WWCY said:
I assume that the range to "pick" from is ##0<k\leq 2\pi/a## with a ##k##-space interval of ##\frac{2\pi}{Na}##? I'm also assuming that we ignore ##k=0## since by ##\omega = vk## it simply means that all the masses don't vibrate anyway.

I think not. Remember that in the Debye model phonons can have any wavenumber from 0 to ##k_D = \omega _D /c##. Any state can be occupied if the temperature is high enough. Maybe you are mixing concepts: classical treatment of a chain of atoms and the Debye model, which is based on quantum statistical mechanics.

I'm not sure you can apply the Debye model to a chain of 2 atoms, or 5 or 20. It is a macroscopic model and uses statistical mechanics, and you can't make statistics with 2 atoms.

## 1. What is the Debye model?

The Debye model is a theoretical model used to describe the behavior of a solid at low temperatures. It assumes that the atoms in a solid are arranged in a lattice structure and can vibrate around their equilibrium positions. This model is based on the concept of phonons, which are quantized lattice vibrations.

## 2. How does the Debye model differ from the Einstein model?

The Debye model takes into account the fact that different atoms in a solid can have different vibrational frequencies, while the Einstein model assumes that all atoms in a solid have the same vibrational frequency. Additionally, the Debye model accounts for the effects of temperature on the vibrations, while the Einstein model does not.

## 3. What is reciprocal space?

Reciprocal space is a mathematical construct used to describe the structure of crystals. It is a way of representing the periodicity of a crystal lattice in terms of reciprocal vectors, which are perpendicular to the crystal lattice planes. This allows for the analysis of diffraction patterns and the determination of crystal structures.

## 4. How is reciprocal space related to the Debye model?

The Debye model can be used to predict the vibrational frequencies of a solid, which can then be used to calculate the diffraction pattern of the solid in reciprocal space. This allows for the comparison of experimental diffraction patterns with predicted patterns, providing insight into the structure and properties of the solid.

## 5. What are some limitations of the Debye model?

The Debye model is only accurate at low temperatures and does not take into account the effects of temperature on the vibrations. It also assumes that the atoms in a solid are arranged in a perfect lattice, which is not always the case. Additionally, the Debye model does not account for the anharmonicity of vibrations, which can become significant at higher temperatures.