# Reason for Debye cutoff frequency?

Dulong and Petit came up with $C_v$ = $3R$ using classical equipartion theorem, which Einstein modified (because it could not hold up for objects other than metals) by using Planck's quantization and assumed that all the atoms in a solid vibrated with the same $\omega$ which made it act like $3N$ independent harmonic oscillators....and it gave the (incorrect) result that the heat capacity tends to $0$ as temperature tends to $0$ K. So Debye corrected this by saying

Long wavelength modes have lower frequencies than short wavelength modes, so the former are much harder to freeze out than the latter (because the spacing between quantum energy levels, $\hbar \,\omega$, is smaller in the former case). The molar heat capacity does not decrease with temperature as rapidly as suggested by Einstein's model because these long wavelength modes are able to make a significant contribution to the heat capacity even at very low temperatures.

as given in this link.

It also says that:

this approach is reasonable because the only modes which really matter at low temperatures are the long wavelength modes: i.e., those whose wavelengths greatly exceed the interatomic spacing. It is plausible that these modes are not particularly sensitive to the discrete nature of the solid: i.e., the fact that it is made up of atoms rather than being continuous.

## Question 1

I dont understand why only long wavelengths matter. Why should the vibrations care if the solid is discrete or continuous? Why do phonons need neighbouring atoms to be closer than the wavelength of their vibrational frequency?

He goes on to say:

The Debye approach consists in approximating the actual density of normal modes $\sigma(\omega)$ by the density in a continuous medium $\sigma_c(\omega)$, not only at low frequencies (long wavelengths) where these should be nearly the same, but also at higher frequencies where they may differ substantially. Suppose that we are dealing with a solid consisting of $N$ atoms. We know that there are only $3\,N$ independent normal modes. It follows that we must cut off the density of states above some critical frequency, $\omega_D$ say, otherwise we will have too many modes

## Question 2

Theoretically, why cant we have "too many modes" other than the fact that experimental evidence says otherwise? May be this is a naive question but I still need the intuition for the CUTOFF Debye decided to invoke.

He also says:

the Debye frequency depends only on the sound velocity in the solid and the number of atoms per unit volume. The wavelength corresponding to the Debye frequency is $2\pi\,c_s/\omega_D$ ($c_s$ is the speed of sound in solid), which is clearly on the order of the interatomic spacing $a\sim (V/N)^{1/3}$. It follows that the cut-off of normal modes whose frequencies exceed the Debye frequency is equivalent to a cut-off of normal modes whose wavelengths are less than the interatomic spacing. Of course, it makes physical sense that such modes should be absent

## Question 3

Again, why? Why does it make physical sense?

I looked around more and found this paper. On page 5 near the bottom, it says:

The temperature where the collective or acoustic vibration shifts to an independent thermal vibration is the Debye temperature.

And Debye temperature is directly proportional to the Debye cutoff frequency.

## Question 4

Why do we have to draw that line between acoustic and independent thermal vibration? Are thermal vibrations too not capable of contributing to the heat capacity of the solid?

• The cutoff is there because waves longer than the interatomic distance are sufficient to describe the motions of the atoms. Shorter waves just describe oscillation between the atoms - there are no masses oscillating there. – Pieter Nov 27 '16 at 15:16
• Shorter waves describe oscillations of WHAT between the atoms? And whatever they may be, aren't they capable of absorbing and storing heat without raising the temperature thereby contributing more to the heat capacity? – Prasad Mani Nov 27 '16 at 15:19

## 1 Answer

To answer this, we must understand how this cutoff appears.

It arises from the approximations made in the continuous theory. The material is definitely not continuous, but made of a finite number of atoms. Long wavelengths correspond to waves which comprise a lot of atoms in each period. This means there is little difference between the continuous and discrete theories. When you reduce the wavelength, you reduce the number of atoms in each period. You can imagine that a wave with only 5 atom in a period is not exactly a continuous wave: it is just a bunch of points which oscillate with some resemblance to a usual sine wave. If you have only 2 atoms per period, or less, are you sure you make a good approximation saying it is a wave? It is actually not.

In fact, with very few points, an oscillation of atoms might correspond to several types of sine waves, as depicted in this picture. In this case, one given oscillation can be described by several wavelengths. This means you count too many modes! Indeed, let us take the example of 2 atoms per wavelength. Divide the wavelength by 2 => it still matches the oscillation. Divide again the wavelength by 2 => it still matches the oscillation. And so on ... This means that, for this particular oscillation, you should not use all these multiples because they represent the same oscillation. Repeat this reasoning for all oscillations, and you will realize that you must cutoff the wavelengths below the interatomic distance.

I hope this answers all your questions.

• I understand the motivation behind that cutoff now, thank you. But the question still remains why dont we take into account those extra waves which represent the same oscillations since those extra waves are also capable of storing heat right? Which means they are contributing to the heat capacity which means we SHOULD count them. I mean if interatomic distance corresponds to oscillations of 1 or more than 1 wavelength, who cares? They are still oscillations and they are storing heat. – Prasad Mani Nov 30 '16 at 2:53
• The representation as waves is a mathematical construct. Waves don't exist here. Only atoms and their movement. Thus counting waves twice for the same atom movement is a mistake. – fffred Nov 30 '16 at 6:40
• Tell me if this is right. Sound or phonon is lattice vibrations; and if the vibrational frequency of the atoms exceed a particular value we cut it off, ie if the wavelength is lesser than interatomic distance then we dont include it. But the atom is still vibrating with a higher frequency (lower wavelength whose multiple could be the interatomic distance), its not as if it is a duplicate of the larger wavelength, so why does it matter if the distance is represented by 1 or more different wavelengths? the shorter wavelengths are still valid since they are also excitations right? – Prasad Mani Nov 30 '16 at 7:57
• The high frequencies do not satisfy the dispersion relation: they are not collective excitations. It is true that these vibrations exist (they are thermal) but they fade at low temperature. – fffred Nov 30 '16 at 8:08
• so is that the reason? their insignificant contribution at low temperature where debye's $T^3$ law was a success? as opposed to multiple representations of same distance? – Prasad Mani Nov 30 '16 at 8:11