Spherical decomposition for a radial Dirac delta?

Question

Is this a spherical coordinate decomposition for a radial Dirac delta?

$\delta(\overrightarrow r-\overrightarrow r_0)=\frac{\delta(r-r_0)\delta(\theta -\theta_0)\delta(\phi-\phi_0)}{r^2sin(\theta)}$

Answer 1

Correct. Notation such as $\delta(u,\,v):=\delta(u-v)$, whether the variables be $1$- or $3$-dimensional, facilitates our equating two expressions for $f(\vec{r}_0)=\int_{\Bbb R^3}\delta(\vec{r},\,\vec{r}_0)f(\vec{r})d^3\vec{r}$:$$\begin{align}&\color{white}{=}\underbrace{\int_0^{2\pi}\int_0^\pi\int_0^\infty}_{\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty}\delta(\vec{r},\,\vec{r}_0)f(\vec{r})\underbrace{r^2\sin\theta drd\theta d\phi}_{d^3\vec{r}}\\&=\int_0^{2\pi}\int_0^\pi\int_0^\infty\delta(r,\,r_0)\delta(\theta,\,\theta_0)\delta(\phi,\,\phi_0)f(\vec{r})drd\theta d\phi.\end{align}$$Hence$$\delta(\vec{r},\,\vec{r}_0)r^2\sin\theta=\delta(r,\,r_0)\delta(\theta,\,\theta_0)\delta(\phi,\,\phi_0),$$as you suggested.