How to derive delta function δ(r₁(vector)-r₂(vector)) in spherical polar coordinate?

Question

I was trying to derive the relation: $$δ(r_1(\text{vector})-r_2(\text{vector}))={δ(r_1-r_2)δ(\cosθ_1-\cosθ_2)δ(φ_1-φ_2)}/r_1^2$$ where $r(\text{vector})=(r,θ,φ)$ in spherical polar coordinate. but I failed to derive it. Please answer if possible. I don't know whether this is the right place of asking such question or not. If it is not, then Sorry.

Answer 1

Let $\psi \in C_c^\infty(\mathbb R^3, \mathbb R)$, i.e. a smooth realvalued function that vanishes outside a bounded region.

In a spherical coordinate system where we instead of $\theta$ have $t = \cos\theta$ with $-1 \leq t \leq 1$, we have $$ \psi(\vec r_2) = \psi(r_2, t_2, \varphi_2) = \int_0^\infty dr_1 \int_{-1}^{1} dt_1 \int_{0}^{2\pi} d\varphi \, \delta(r_1-r_2) \, \delta(t_1-t_2) \, \delta(\varphi_1-\varphi_2) \, \psi(r_1, t_1, \varphi_1) \\ = \int_0^\infty dr_1 \int_{0}^{\pi} d\theta_1 \, \sin\theta_1 \int_{0}^{2\pi} d\varphi \, \delta(r_1-r_2) \, \delta(\cos\theta_1-\cos\theta_2) \, \delta(\varphi_1-\varphi_2) \, \psi(r_1, t_1, \varphi_1) \\ = \int_0^\infty r_1^2 \, dr_1 \int_{0}^{\pi} d\theta_1 \, \sin\theta_1 \int_{0}^{2\pi} d\varphi \, r_1^{-2} \, \delta(r_1-r_2) \, \delta(\cos\theta_1-\cos\theta_2) \, \delta(\varphi_1-\varphi_2) \, \psi(r_1, t_1, \varphi_1) \\ = \iiint_{\mathbb R^3} d^3r_1 \, r_1^{-2} \, \delta(r_1-r_2) \, \delta(\cos\theta_1-\cos\theta_2) \, \delta(\varphi_1-\varphi_2) \, \psi(r_1, t_1, \varphi_1) \\ $$ Here I had $dr_1 \, d\theta \, \sin\theta_1 \, d\varphi$, but since the volume element is $dr_1 \, r_1^2 \, d\theta \, \sin\theta_1 \, d\varphi$ so I had to compensate by dividing the integrand by $r_1^2$.

Comparing with $ \psi(\vec r_2) = \int_{\mathbb R^3} d^3r_1 \, \delta(\vec r_1 - \vec r_2) \, \psi(\vec r_1) $ that is valid by definition, we see that $$\delta(\vec r_1 - \vec r_2) = r_1^{-2} \, \delta(r_1-r_2) \, \delta(\cos\theta_1-\cos\theta_2) \, \delta(\varphi_1-\varphi_2).$$