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Consider the following identity: $$\delta(\vec r_1-\vec r_2)=\frac{1}{r_1^2}\delta (r_1-r_2)\delta (\cos\theta_1-\cos\theta_2)\delta (\phi_1-\phi_2) $$

I was trying to understand this geometrically.

It seems that the left-hand side implies that when the line joining the vectors becomes zero, a singularity occurs. This would mean that when vector $r_1$ is exactly on top of $r_2$ (happens when $r's$ and the angles are equal simultaneously) then the left side blows up.

The problem then is the right-hand side blows up even when $r_1=r_2$ and the angles do not necessarily have to be equal.

How is this happening?

Wouldn't $\vec r_1-\vec r_2=0$ be when $r_1-r_2=0$ $\ and$ $\theta_1-\theta_2=0$ $\ and $ $\phi_1-\phi_2=0$ ?

Lost
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    Here are a few hints. Suppose I put a singularity at the origin of $\mathbb{R}^2$, i.e. $\delta (x - x_0) \delta(y - y_0)$. Does that blow up when $x=x_0$ but $y \neq y_0$? No, because while the $x$ part is "blowing up" the $y$ part is identically zero. Remember two things: 1) Delta functions are defined by what they do inside integrals, i.e. $\int dx , dy , f(x, y) \delta(x-x_0) \delta(y - y_0) = f(x_0, y_0)$. 2) Always think of a delta function as a limit of another smooth, well behaved function, i.e. $\delta(x) = \lim_{\sigma \rightarrow 0} \sin(x / \sigma) / (\pi x)$. – DanielSank Aug 15 '21 at 21:17
  • Oh..I see. Is the identity $x \delta (x)=0 $ used here? – Lost Aug 15 '21 at 21:21
  • I'm not sure what you're getting at with that equation. The delta function in spherical coordinates is pretty much the same as in Cartesian, just with that $1/r^2$ in front to cancel the $r^2$ that shows up when you integrate in 3D in spherical coordinates. The other weird part, $\delta(\cos(\theta_1) - \cos(\theta_2))$ is there for a similar reason... – DanielSank Aug 15 '21 at 21:26

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