# Going from 2D dispersion relation to density of states

If one was to have a 2D dispersion say: $$\varepsilon(k)=k_x^2-k_y^2$$ We know the dispersion relation generally can be written as:$$D(\varepsilon)=\sum_{k_x}\sum_{k_y}\delta(E-\varepsilon(k_x,k_y)$$ which as can be seen in many different pieces of literature (e.g Ashcroft and Mermin), the density of states can be written as:$$D(\varepsilon)=\frac{1}{(2\pi)^2}\int_{S(\varepsilon)}\frac{dS}{|\nabla \varepsilon(k)|}$$ My question is how to actually compute this surface integral? As it seems in different literature that this step is not mentioned or completed in its entirety.

I am led to believe some transformation of $$k_x$$ and $$k_y$$ is required. However, I am stuck on the meaning of this as a requirement to get an actual answer for $$D(\varepsilon)$$. Any advice on this area would be greatly appreciated.

That integral denotes that you want to integrate $$1/|\nabla \epsilon(k)|$$ over all wavevectors that satisfy $$\epsilon(k) = \epsilon$$ for any given $$\epsilon$$ which is taken to be a constant. Unless you find some convenient parametrization of these constant-energy surfaces, this is a hard to do analytically and one needs to numerical techniques.
Now, for the problem at hand, I do not have a full solution nor a good intuition what the result should be. But here is an outline which should work: First note $$\nabla \epsilon(k) = (2 k_x,-2 k_y)$$ so $$|\nabla \varepsilon(k)| = 4 \sqrt{k_x^2+k_y^2}$$. We then have $$D(\epsilon) = \int_{\epsilon(k_x,k_y)=\epsilon} d S \frac{1}{4 \sqrt{k_x^2+ k_y^2}}$$. The form of the dispersion suggests to use a hyperbolic parametrization. Letting $$k_x = a \cosh u$$ and $$k_y = a \sinh u$$ which parametrizes the region, $$k_x > 0$$ and $$-k_x < k_y < k_x$$, we note that $$\varepsilon(k) = a^2 (\cosh^2 u - \sinh^2 u) = a^2$$ which implies we can set $$a=\sqrt{\epsilon}$$ and only integrate over $$u$$, where the surface element is given by $$a du$$ (this can be checked from the Jacobian). You then need the integral $$\int_{-\infty}^\infty 1/\sqrt{\cosh 2u} du$$ which can be done with WolframAlpha, yielding $$\approx 2.62$$. Note that this is only one contribution to the integral, by appropriate parametrizations of $$k_x$$ and $$k_y$$ you should be able to cover the entire plane. Basically we are integrating $$1/|k|$$ along a level set of $$k_x^2 - k_y^2$$ as illustrated by the contour lines in this plot.