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* Math: Density of states

A math theorem tells that for a function $ f(x)$,

$\displaystyle \lim_{L\to \infty}\frac{1}{L}\sum_{n=0}^{\infty}f(n/L)=\int_0^{\infty}f(x)dx.$ (8)

Assume standing waves in a box of size $ L^3$ with wave vectors

$\displaystyle {\bf k}=(\pi n_x/L,\pi n_y/L,\pi n_z/L).$

Assume we wish to calculate a function $ f({\bf k})=f(k)$ that only depends on the modulus of $ {\bf k}$. We use
$\displaystyle \lim_{L\to \infty}\frac{1}{L^3}\sum_{n_x,n_y,n_z=0}^{\infty}f\left(\pi n_x/L,\pi n_y/L,\pi n_z/L)\right)$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^3}\int_0^{\infty}f({\bf k})dk_xdk_ydk_z,$ (9)

i.e.
$\displaystyle \lim_{L\to \infty}\frac{1}{L^3}\sum_{k_x,k_y,k_z>0}f(k)$ $\displaystyle =$ $\displaystyle \frac{1}{\pi^3}
\int_0^{\infty}f(k)dk_xdk_ydk_z
=\frac{1}{\pi^3 2^3}\int_{-\infty}^{\infty}f(k)dk_xdk_ydk_z$  
  $\displaystyle =$ $\displaystyle \frac{1}{(2\pi)^3}\int_{0}^{\infty}dk 4\pi k^2f(k)
=\int_0^{\infty}dk\frac{k^2}{2\pi^2}f(k)$  
  $\displaystyle =$ $\displaystyle \int_0^{\infty}d\omega\frac{\omega^2}{2\pi^2c^3}f(\omega)
=:\int_0^{\infty}d\omega\rho(\omega)f(\omega)$  
$\displaystyle \rho(\omega)$ $\displaystyle :=$ $\displaystyle \frac{\omega^2}{2\pi^2c^3}.$ (10)

Here, $ \rho(\omega)$ is the density of states if there was only one polarization direction per possible `state' (wave vector $ {\bf k}$). If there are two polarization directions, like for unpolarized light, the density of states is twice as large, i.e. $ \rho(\omega):=\frac{\omega^2}{\pi^2c^3}$.


next up previous contents
Next: Waves, particles, and wave Up: The Radiation Laws and Previous: Remarks   Contents
Tobias Brandes 2004-02-04