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.
Best Answer
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.