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I often see the following:

$$\delta\left(\boldsymbol{r}-\boldsymbol{r_{0}}\right)=\frac{1}{r^{2}\sin\theta}\delta\left(r-r_{0}\right)\delta\left(\theta-\theta_{0}\right)\delta\left(\varphi-\varphi_{0}\right)\overset{?}{=}\frac{1}{2\pi r^{2}\sin\theta}\delta\left(r-r_{0}\right)\delta\left(\theta-\theta_{0}\right)$$ and I fail to see how the $1/2\pi$ factor got out of the $\varphi$ Dirac delta when we have azimuthal symmetry in a problem. How does it come about?

*Wasn't sure what to tag, but it is relevant in many areas in physics.

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  • $\begingroup$ Often see where? Which page? $\endgroup$
    – Qmechanic ♦
    Commented Jul 5, 2021 at 17:37
  • $\begingroup$ In many different textbooks in many different areas - it is used especially when solving electrostatic problems using Green function. $\endgroup$
    – Darkenin
    Commented Jul 5, 2021 at 17:49
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    $\begingroup$ A form which might make this more clear is $\int_0^{2\pi} \delta(\varphi - \varphi_0) d\varphi = \frac{1}{2\pi} \int_0^{2\pi} d\varphi = 1$. $\endgroup$
    – Connor Behan
    Commented Jul 5, 2021 at 21:26
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    $\begingroup$ What is your equal-? symbol meant to indicate? Surely not that the two expressions are equal? Three Dirac delta functions shouldn't be equal to two Dirac delta functions. $\endgroup$
    – David
    Commented Jul 5, 2021 at 21:34

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