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why does the radial dependence of Laplacian in spherical and polar coordinate vary? ie, in polar coordinates if there is no $\theta$ dependence the laplacian goes as $\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial}{\partial r}) $ and in spherical coordinates it is $\frac{1}{\rho^2} \frac{\partial}{\partial \rho} (\rho^2 \frac{\partial}{\partial \rho})$. if my function only depends on $r$ shouldn't the laplacian be the same in both the coordinate system so that when $z\rightarrow 0$ in polar cordinate, $\theta\rightarrow \pi/2$ in spherical cordinate and $\phi_{polar}=\phi_{spherical}$ such that on $z=0 (\theta=\pi/2)$ plane as $\rho \rightarrow r$ the value of laplacian remains invariant since it is a scalar quantity?

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    $\begingroup$ Might Mathematics be better suited for this math question? $\endgroup$ – Kyle Kanos Nov 24 '17 at 14:58
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The meaning of the number $r$ is different in both cases. Imagine you want to locate a point $P$, whereas in spherical coordinates $r$ means the distance from the origin to the point $P$, in cylindrical coordinates $r$ is the distance from the origin to the projection of $P$ onto the $z=0$ plane

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To avoid confusion, label your coordinates differently, i.e. $r$ in the spherical case, $R$ in the cylindrical/polar case

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In polar coordinates, there are only two dimensions $x$ and $y$ that contribute to $r$, as opposed to three dimensions for the spherical coordinate version of $r$. You would have a different form yet if you work out the Laplacian for the radial coordinate in spherical coordinates in higher dimensions.

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  • $\begingroup$ Is there any intuitive way of understanding this? @Chris $\endgroup$ – Prince M S Nov 24 '17 at 6:45

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