How to prove $\frac{\theta(t)}{4\pi a^2 t}\delta_{Sat}(x) = \frac{\theta(t)}{2\pi a}\delta( a^2t^2 - |x|^2)$?

Question

How to prove $\frac{\theta(t)}{4\pi a^2 t}\delta_{Sat}(x) = \frac{\theta(t)}{2\pi a}\delta( a^2t^2 - |x|^2)$

This 'formula' is essentially the Green function for the wave equation in $R^3$.

In most textbooks I found, this equality is given as obvious, but I somehow don't really find it as so.

$\frac{\theta(t)}{2\pi a}\delta( a^2t^2 - |x|^2) = \frac{\theta(t)}{2\pi a}\delta( (at - |x|)(at + |x|)) = \frac{\theta(t)}{2\pi a}(\delta( (at - |x|) + \delta(at + |x|))$.

Since $t>0$, the second 'delta' is always zero, whereas the first one is a 'sphere of deltas', so, essentially $\delta_{Sat}(x)$. Where does the $\frac{1}{2 at}$ appear from?

Answer 1

By definition, we have \begin{align} \int_{\mathbb{R}^3} f(x)\frac{\theta(t)}{2\pi a}\delta(a^2t^2-|x|^2)\ dx =&\ \int_{|x|^2 = a^2t^2}f(x)\frac{\theta(t)}{4\pi a} \frac{d\sigma(x)}{|x|} \\ =&\ \int_{|x|=at} f(x) \frac{\theta(t)}{4\pi a^2t} d\sigma(x)\\ =&\ \int_{\mathbb{R}^3} f(x)\frac{\theta(t)}{4\pi a^2 t} \delta(at-|x|)\ dx \end{align} for all $f$. Hence it follows \begin{align} \frac{\theta(t)}{2\pi a}\delta(a^2t^2-|x|^2) = \frac{\theta(t)}{4\pi a^2 t} \delta(at-|x|). \end{align} Note, I used the fact that \begin{align} \int_{\mathbb{R}^3} f(x)\delta(g(x))\ dx = \int_{g^{-1}(0)} \frac{f(x)}{|\nabla g|}\ d\sigma(x). \end{align}