Phonon density of states and density of states of free electrons

In summary, the conversation discusses the calculation of density of states for free electrons and phonons. The free electron DOS is found to be proportional to (energy)^(-1/2) in 1D and constant in 2D. The question arises whether the phonon DOS can also be expressed in terms of energy or if it should be found using the equation D(w)dw. The suggested resources for the phonon part are Kittel and Ashcroft and Mermin. It is mentioned that the dispersion relation for phonons can resemble that of photons. The DOS of a phonon in Kittel's book is given as 1/pi*vg in 1D and k/2pivg in 2D, both of
  • #1
chikchok
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Homework Statement
Compare phonon DOS and DOS of free electrons in 1D and 2D
Relevant Equations
D(E)1d=(1/L)dN/dE
D(E)2d=1/A)dN/dE for electrons
In the following pdf I tried to calculate the density of states of free electrons and phonons. First, I found the free electron DOS in 1D, it turns to be proportional to (energy)^(-1/2) and in 2D it is constant. However, I am not sure I found the DOS for phonons in the second part of the solution. Because the homework said to compare two DOS, I thought phonon DOS needs to be in terms of energy D(E) and not frequency w D(w). But I suspect it is wrong. Can phonon density of states be in terms of energy? If so, how to find it? And if not, should I find it trough equation D(w)dw?
 

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  • #2
This is a pretty broad question and I would suggest looking at Kittel or Ashcroft and Mermin for the phonon part. I believe the question wants you to see that the dispersion relation for phonons can resemble that of photons. There is a lot of good physics in this question and it is worth some effort.
 
  • #3
I`ve looked up the DOS of a phonon in Kittel`s book and in 1D DOS is 1/pi*vg (vg as a group velocity dw/dk) and in 2D it`s k/2pivg . Both of them have no relationship with energy. That is why I was wondering if there is a way to calculate the DOS as a function of energy.
 
  • #4
$$E=\hbar \omega$$
 
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FAQ: Phonon density of states and density of states of free electrons

1. What is the phonon density of states?

The phonon density of states is a measure of the number of phonon states per unit energy interval in a solid material. It describes the distribution of phonon energies in a material, and is an important factor in understanding the thermal and mechanical properties of solids.

2. How is the phonon density of states calculated?

The phonon density of states is calculated using the Debye model, which assumes that the vibrations of atoms in a solid can be treated as harmonic oscillators. The Debye model takes into account the speed of sound, the mass of atoms, and the density of the material to calculate the phonon density of states.

3. What is the density of states of free electrons?

The density of states of free electrons is a measure of the number of energy states available to electrons in a solid material. It describes the distribution of electron energies in a material, and is an important factor in understanding the electrical conductivity and other electronic properties of solids.

4. How is the density of states of free electrons related to the Fermi energy?

The density of states of free electrons is directly related to the Fermi energy, which is the highest energy level occupied by electrons at absolute zero temperature. The density of states at the Fermi energy determines the number of electrons available for conduction and is a crucial factor in determining the electrical conductivity of a material.

5. Can the density of states of free electrons be affected by external factors?

Yes, the density of states of free electrons can be affected by external factors such as temperature, pressure, and impurities in the material. Changes in these factors can alter the energy distribution of electrons, leading to changes in the material's electrical and thermal properties.

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