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I am trying to calculate with a general approach the gradient of an axis symmetric vector field in cylindrical coordinates and then express it in cartesian coordinates.

First I write my vector field in cylindircal coordinates as $$\mathbb{f} = [f_\rho(\rho,\phi,z),f_\phi(\rho,\phi,z),f_z(\rho,\phi,z)]^T$$ then I consider that it is axis symmetric so $$\mathbb{f} = [f_\rho(\rho,z),f_\phi(\rho,z),f_z(\rho,z)]^T$$

The gradient of a scalar function $g=g(\rho,\phi,z)$ in cylindrical coordinates is $$\small \nabla g = [\frac{\partial g}{\partial \rho} \mathbb, \frac{1}{\rho}\frac{\partial g}{\partial \phi}, \frac{\partial g}{\partial z}]$$ but the gradient of a vector field in cylindrical coordinates seems to be far more complicated:
$$ \small \frac{\partial ( f_\rho,f_\phi,f_z)}{\partial ( \rho,\phi,z)} =\left[ \begin{array}{clc} \frac{\partial f_\rho}{\partial \rho} & \frac{1}{\rho} \frac{\partial f_\rho}{\partial \phi} - \frac{f_\phi}{\rho} & \frac{\partial f_\rho}{\partial z} \\ \frac{\partial f_\phi}{\partial \rho} & \frac{1}{\rho} \frac{\partial f_\phi}{\partial \phi} + \frac{f_\rho}{\rho}& \frac{\partial f_\phi}{\partial z} \\ \frac{\partial f_z}{\partial \rho} & \frac{1}{\rho} \frac{\partial f_z}{\partial \phi} & \frac{\partial f_z}{\partial z} \\ \end{array} \right] $$ First Question: Where $\frac{f_\phi}{\rho}$ and $\frac{f_\rho}{\rho}$ additional terms come from?

OK, let's move on and consider the gradient of the transformation from cylindrical to cartesian as $$ \small \frac{\partial ( \rho,\phi,z)}{\partial ( X,Y,Z)} =\left[ \begin{array}{clc} cos(\phi) & sin(\phi) & 0 \\ -sin(\phi) & cos(\phi) & 0 \\ 0 & 0 & 1 \\ \end{array} \right] $$

Secon Question: My idea at the end is to multiply the two matrix in a sort of chain rule way in order to obtain the gradient in cartesian coordinates $$ \frac{\partial ( f_\rho,f_\phi,f_z)}{\partial ( X,Y,Z)} =\frac{\partial ( f_\rho,f_\phi,f_z)}{\partial ( \rho,\phi,z)} \frac{\partial ( \rho,\phi,z)}{\partial ( X,Y,Z)} $$ Is this procedure correct.? Thanks

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You need to carefully distinguish between gradients and derivatives (since the metric is non-trivial) and between tangent vectors expressed in the coordinate basis $\partial/\partial \rho, \partial/\partial \theta, \partial/\partial z$ and those expressed in the "unit" basis $\partial/\partial \hat\rho, \partial/\partial \hat\theta, \partial/\partial \hat z$. These two factors account for the extra terms. –  user7530 Mar 13 '13 at 22:50

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