Debye model and reciprocal space

  • #1
477
12

Main Question or Discussion Point

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##?
 

Answers and Replies

  • #2
26
3
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.
 
  • #3
477
12
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!
 
  • #4
26
3
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).
 
  • #5
477
12
Hi, thanks for the reply!

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?
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.
 
  • #6
26
3
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.
 

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