I am trying to obtain the density of states of the Debye model in one dimension. I know that the correct expression is $$\frac{L}{\pi c_s}$$ where $c_s$ is the speed of sound in the material. However, when I try to work it outmyself, I get $$\sum_k{f(c_s|k|)} \rightarrow \frac{L}{2\pi}\int{f(c_s|k|)~dk} \, .$$ Using $\omega=c_s|k|$ we get $$dk=\frac{d\omega}{c_s}$$ which gives $$\rightarrow\frac{L}{2\pi{c_s}} \int{f(\omega) \, d\omega} \, .$$ Therefore, the density of states is $$\frac{L}{2\pi c_s} \, .$$ I know I need to multiply by two. Why? Is it like with the 2 transverse one longitudinal modes of vibration? I have done a lot of searching for help on page 7 of some online notes (pdf) where they show the calculation just not why the 2 is placed in front.

I have also found on Wikipedia that

"The 1, 2 and 3-dimensional density of wave vector states for a line, disk, or sphere are explicitly written as \begin{align} N_1(k) =& 2 \\ N_2(k) =& 2\pi{k} \\ N_3(k) =& 4\pi{k^2} \end{align}

I understand the last two which is recognizing the independence on the direction and using spherical integration (because the density of states in two and three dimensions would have $dk^N$ for the integration). It's the last one I don't understand.


2 Answers 2


It's because you have two $\omega$ for each $\vert \vec{k} \vert$, $\vec{k}$ ''to the left'' and ''to the right''. You can also see this as the boundary of an one-dimensional disk. Looking at $\vert \vec{r} - \vec{r}_{0} \vert < R$ as a ball in one dimension, the boundary are two points, just as boundary in two dimensional space would be $2\pi R$ and so on.


First response was a misunderstanding of the question I apologise for this.

The reason for the introduction of the factor of 2 is to account for the 2 independent directions along the 1-D line, i.e acounts for right and left moving waves

  • $\begingroup$ The OP is referring to density of states for phonon modes. $\endgroup$ Commented Dec 5, 2014 at 1:16
  • $\begingroup$ I believe the answer should be correct now after my latest edit. $\endgroup$
    – PhotonBoom
    Commented Dec 5, 2014 at 1:34
  • $\begingroup$ I considered it a few times but I am not entirely sure about it. My major reason is that $w=c_s|k|$, but considering it now to include the negative direction I could multiply by two (before is only half the number of states). Can someone confirm this sort of thinking? $\endgroup$
    – Jesse Ross
    Commented Dec 5, 2014 at 1:54

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