# Density of states for the diffusion

+ 4 like - 0 dislike
57 views

For the wave equation, the propagator in the Fourier domain is written as $$G(\mathbf{k},\omega)=-\frac{1}{\frac{\omega^2}{c^2}-\mathbf{k}^2+\mathrm i\epsilon}.$$ When $\omega/c$ is close to $\|\mathbf{k}\|=k$, we have the following approximation (Stokholski-Plemelj theorem) $$\frac{1}{\pi}\operatorname{Im} G(\mathbf{k},\omega)=-\delta\left(\frac{\omega^2}{c^2}-\mathbf{k}^2\right)=-\frac{c^2}{2\omega}\delta\left(\omega-ck\right),$$ such that, if we integrate over $\mathbf{k}$ we get the famous expression of the density of states $$\rho(\omega)=-\frac{2\omega}{\pi c^2}\operatorname{Im}G(\mathbf{r}=0,\omega).\tag{1}$$

Can one do the same computation with the diffusive propagator (diffuson) $$G(\mathbf{k},\omega)=\frac{1}{\frac{\mathrm i\omega}{D}-\mathbf{k}^2} ?$$

If $\omega$ is close to zero we have $$\frac{1}{\pi}\operatorname{Im}G(\mathbf{k},\omega=0)=\delta(\mathbf k^2)$$ or taking the real part $$\frac{1}{\pi}\operatorname{Re}G(\mathbf{k}=0,\omega)=D\delta(\omega),$$ but I can't really see how one could obtain a relation such as (1).

EDIT

Considering that $$\operatorname{Im}G(\mathbf{r}=0,\omega)=-\frac{1}{4\pi}\sqrt{\frac{\omega}{2D}}$$ in three dimensions, it is tempting to use a similar relation to get

$$\rho^{\text{diff}}_3(\omega)=\frac{1}{4\pi^2}\sqrt{\frac{\omega}{2D^3}}.$$

But is this really justified ?

This post imported from StackExchange Physics at 2015-06-17 16:39 (UTC), posted by SE-user Tom-Tom

asked Jun 17, 2015
edited Jun 18, 2015

## Your answer

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.