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I am trying to understand how to write the differential operator $\vec{r}\cdot\nabla$ in cylindrical and spherical coordinates.

Say, we have this equation (an illustrative example):

$$(\vec{r}\cdot\nabla) f=a$$

In Cartesian coordinates, this equation is

$$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+z\frac{\partial f}{\partial z}=a$$

How does one write this same equation in cylindrical and spherical coordinates?

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Sphericals

By the chain rule, $$ \frac{\partial f}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial f}{\partial y} + \frac{\partial z}{\partial r} \frac{\partial f}{\partial z}, $$ and $\partial x/\partial r = \cos{\theta}\cos{\phi} = x/r$, and similarly $\partial y/\partial r = y/r$, $\partial z/\partial r = z/r$, so $$ \frac{\partial f}{\partial r} = \frac{x}{r} \frac{\partial f}{\partial x} + \frac{y}{r} \frac{\partial f}{\partial y} + \frac{z}{r} \frac{\partial f}{\partial z}. $$ Hence $$ (\vec{r} \cdot \nabla) f = r\frac{\partial f}{\partial r}. $$

Cylindricals

In cylindricals $(\rho,\phi,z)$, the $z$ part does not change, while a similar argument shows that $\frac{\partial f}{\partial \rho} = \frac{x}{\rho} \frac{\partial f}{\partial x} + \frac{y}{\rho} \frac{\partial f}{\partial y}$, and so $$ (\vec{r} \cdot \nabla)f = \rho\frac{\partial f}{\partial \rho} + z\frac{\partial f}{\partial z}. $$

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  • $\begingroup$ Thank you for your answer! $\endgroup$ – user142523 Sep 8 '17 at 22:02

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