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It would be wonderful if someone could direct me to a proof or explanation on why we can express the delta function of $\delta(x-\xi)=\delta(x_{1}-\xi_{1})\delta(x_{2}-\xi_{2})$ as $\delta(x-\xi)=r^{-1}\delta(r-r')\delta(\theta-\phi)$ for $$x=(r\hspace{0.05cm}\cos\hspace{0.05cm}\theta,r\hspace{0.05cm}\sin\hspace{0.05cm}\theta)$$ and $$\xi=(r'\hspace{0.05cm}\cos\hspace{0.05cm}\phi,r'\hspace{0.05cm}\sin\hspace{0.05cm}\phi)$$

RobPratt
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curiousboi
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1 Answers1

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In short: For any $\varphi \in C_c^\infty(\mathbb R^2)$ you want $$ \varphi^{\text{rect}}(x_0, y_0) = \int_{x=-\infty}^{\infty} \int_{y=-\infty}^{\infty} \delta_{(x_0,y_0)}^{\text{rect}}(x-x_0, y-y_0) \, \varphi^{\text{rect}}(x, y) \, dx \, dy \\ = \int_{r=0}^{\infty} \int_{\theta=0}^{2\pi} \delta_{(r_0,\theta_0)}^{\text{polar}}(r,\theta) \, \varphi^{\text{polar}}(r,\theta)\,r\,dr\,d\theta, $$ where subperscript $\textit{rect}$ and $\textit{polar}$ denotes rectangular and polar representations.

Since $$ \varphi^{\text{rect}}(x_0, y_0) = \varphi^{\text{polar}}(r_0, \theta_0) = \int_{r=0}^{\infty} \int_{\theta=0}^{2\pi} \delta(r-r_0)\,\delta(\theta-\theta_0)\,\varphi^{\text{polar}}(r,\theta)\,dr\,d\theta $$ it follows that you should have $$ r\,\delta_{(r_0,\theta_0)}^{\text{polar}}(r,\theta) = \delta(r-r_0)\,\delta(\theta-\theta_0), $$ i.e. $$ \delta_{(r_0,\theta_0)}^{\text{polar}}(r,\theta) = r^{-1} \delta(r-r_0)\,\delta(\theta-\theta_0) = r_0^{-1} \delta(r-r_0)\,\delta(\theta-\theta_0) . $$

md2perpe
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