# Calculate density of states for 1D system

The density of states for a 1D system can be written as:

$$D\left(\omega\right)=\frac{L}{\pi}\frac{1}{d\omega\left(k\right)/dk}$$

I have some expressions for $$\omega\left(k\right)$$:

w1[k_] = Sqrt[(k1 (m1 + m2)/(m1*m2))*(1 -
Sqrt[1 - (2*(1 - Cos[k*a]) m1*m2)/((m1 + m2)^2)])];
w2[k_] = Sqrt[(k1 (m1 + m2)/(m1*m2))*(1 +
Sqrt[1 - (2*(1 - Cos[k*a]) m1*m2)/((m1 + m2)^2)])];


I need to calculate $$D\left(\omega\right)$$ analytically eliminating $$k$$ but I can only do that if I simplify the expression for w1,2[k] by assigning values to k1,m1,m2,a:

k1 = 1; a = 1; m1 = 1; m2 = 2;
w1[k_] = Sqrt[(k1 (m1 + m2)/(m1*m2))*(1 -
Sqrt[1 - (2*(1 - Cos[k*a]) m1*m2)/((m1 + m2)^2)])];
w2[k_] = Sqrt[(k1 (m1 + m2)/(m1*m2))*(1 +
Sqrt[1 - (2*(1 - Cos[k*a]) m1*m2)/((m1 + m2)^2)])];
sol = Solve[p == D[w2[k], k] && w == w2[k], {p}, {k}];
Simplify[(p /. sol[[2]])^-1]


Output:

(2 Sqrt[(-3 + 2 w^2)^2])/Sqrt[6 - 11 w^2 + 6 w^4 - w^6]


Is it possible to do it for the general case?

• Why not just solve for $k$ first and then take the derivative with respect to $\omega$ instead? Since you can write $D\left(\omega\right)=\frac{L}{\pi}\frac{d\omega}{dk}$, do D[k /. Solve[w == w1[k], k] // Evaluate, w]. YOu don't need to worry about domains and such because both $\omega$ and $k$ can be taken to be positive. – march Jun 30 at 22:18