# One dimensional spring chain and density of states

1. Oct 28, 2012

### FillBill

I have a classic infinite, linear chain of atoms, each of mass m, each separated by a spring with spring constant $b$ and equilibrium distance $a$ between each adjacent one. I know from my textbook that the dispersion relationship you get for this is:

$$\Omega(k) = 2\sqrt{\frac{b}{m}} |sin(ka/2)|$$

Now I want to find the density of states (per unit length of the chain). I know that for 1 dimension, the equation is

$$g(\omega) = \frac{1}{2\pi}\int \delta(\omega - \Omega(k))dk$$

Now my general understanding of delta functions was such that if you have one function of the integration variable inside the delta function and another one outside, the delta function is 'triggered' at the value it hits zero, and then the integrand is just evaluated at this value, so this is true:

$$\int_{-\infty}^\infty f(x)\delta(g(x) - x_0)dx = f(g^{-1}(x_0))$$

So, in my case, it seems like f is just 1 (or 1/2π), and the inverse function is $k(\omega) = (2/a)arcsin(\frac{\omega}{2}\sqrt{\frac{m}{b}})$. But it doesn't seem like we need the inverse here, because it seems like the integral should just equal 2 (because the delta function is triggered at two points).

But this isn't the answer given. Can anyone help me out?