Delta function with different variables

Question

How can I deal with something like that:

$\int\int\int dx dydz \delta(E-E_0+x^2+y^2+z^2)$

I could substitue $x^2\rightarrow a$ and do the first integral, but the the delta function vanishes?

Best

Answer 1

With the exception of $E=E_0$, this just means you integrate over the subspace, where $E-E_0+x^2+y^2+z^2=0$. This can be done with various methods: you can find coordinates where this is trivially satisfied (spherical coordinates, for instance), or you could just express one of the variables with the rest (accounting for multiple branches) and integrate that way.

Answer 2

First, we note that if $E>E_0$, then $E-E_0+x^2+y^2+z^2>0$ and thus the integral is zero.

Now, assume that $E<E_0$. Use of spherical coordinates will facilitate analysis. So, let $r^2=x^2+y^2+z^2$. Then, we have

$$\begin{align} \iiint\delta (E-E_0+x^2+y^2+z^2)dx\,dy\,dz&=4\pi\int_0^{\infty}r^2\delta \left(r^2-(E_0-E)\right)dr\\\\ &=4\pi\int_0^{\infty}\,\left(u+(E_0-E)\right)\frac{\delta \left(u\right)}{2\sqrt{u+(E_0-E)}}du\\\\ &=2\pi\sqrt{E_0-E} \end{align}$$

Answer 3

Warning about the fact that $\delta$ is not a function! So one should be careful by using the notation $\delta(r)\,\mathrm{d}r$ and remember that it means $\delta(\mathrm{d}r)$, or just use the definition in the sense of distributions. In particular, the rules for the change of variable are different from the Lebesgue measure: one directly has $\delta(r^2-a^2) = \delta_a(\mathrm{d}r)$ and so $$ \begin{align*} \int_{\mathbb{R}^3}\delta_0 (x^2+y^2+z^2 - a^2)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z &=4\pi\int_0^{\infty}r^2\delta_a(r)\,\mathrm{d}r = 4π\,a^2. \end{align*} $$ Hence, we find the size of the sphere of radius $a$, which is coherent with the interpretation of the distribution $\delta_0 (x^2+y^2+z^2 - a^2)$ being the uniform distribution on such a sphere.

In this particular example, i.e. when $a^2 = E_0-E ≥0$, one gets $$ \begin{align*} \int_{\mathbb{R}^3}\delta_0 (E-E_0+x^2+y^2+z^2)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z &=4\pi\int_0^{\infty}r^2\delta_a(r)\,\mathrm{d}r = 4π\,(E_0-E)_+. \end{align*} $$