what is a vector dot del in cylindrical coordinates? $\vec{U}\cdot \nabla$

Question

what should be the product of :

$\vec{U}\cdot \nabla$ in a cylindrical coordinate ? for example with a scalar following the product such as $(\vec{U}\cdot \nabla) \Omega$ ? Thank you

Answer 1

$\renewcommand\phi\varphi$

The question already has answers here and here. However, those answers only treat the case when one is using an orthonormal basis $\{e_r,e_\phi,e_z\}$. In that case the gradient vector is $$\nabla f = \partial_rf \,e_r + \frac 1 r \partial_\phi f \,e_\phi +\partial_zf \,e_z$$ So for a vector $u=u_re_r+u_\phi e_\phi+u_ze_z$ you have $$\begin{align*} u\cdot \nabla f &= u_r\partial_rf \, e_r\cdot e_r + u_\phi\frac 1 r\partial_\phi f \, e_\phi\cdot e_\phi + u_z\partial_zf \, e_z\cdot e_z \\ &= u_r\partial_rf + u_\phi\frac 1 r\partial_\phi f+u_z\partial_zf \\ &= \left(u_r\partial_r + u_\phi\frac 1 r\partial_\phi +u_z\partial_z\right)f \end{align*}$$ Hence $$u\cdot \nabla =u_r\partial_r + u_\phi\frac 1 r\partial_\phi +u_z\partial_z.$$

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On the other hand, if you are using the holonomic basis $\{v_r,v_\phi,v_z\}$ (that is, $v_r=e_r, v_\phi=re_\phi, v_z=e_z$), then the gradient vector is $$\nabla f = \partial_rf \, v_r + \frac 1 {r^2}\partial_\phi f \, v_\phi +\partial_zf \, v_z$$ So for a vector $u=u_rv_r+u_\phi v_\phi+u_zv_z$ you have $$\begin{align*} u\cdot \nabla f &= u_r\partial_rf \, v_r\cdot v_r + u_\phi\frac 1 {r^2}\partial_\phi f \, v_\phi\cdot v_\phi + u_z\partial_zf \, v_z\cdot v_z \\ &= u_r\partial_rf + u_\phi\partial_\phi f+u_z\partial_zf \\ &= \left(u_r\partial_r + u_\phi\partial_\phi +u_z\partial_z\right)f \end{align*}$$ (the middle factor goes away because $v_\phi\cdot v_\phi=r^2$). Hence $$u\cdot \nabla =u_r\partial_r + u_\phi\partial_\phi +u_z\partial_z.$$

Again please note that these formulas only hold when the directional derivative operator $u\cdot \nabla$ is applied to a scalar.