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How to calculate $$∇\bigg(\frac{(ρ^2)}{2z(1+\frac{a}{z^2})}\bigg)$$ where the function is in cylindrical coordinates $$ρ^2=x^2+z^2$$

$$∇\bigg(\frac{x^2+z^2}{2z(1+\frac{a}{z^2})}\bigg)$$

Is the answer in Cartesian coordinates a vector

$$[x/(z(1+a/z^2))),y/(z(1+a/z^2))),1/((a-z^2)/(a+z^2)^2)]$$

In cylindrical coordinates a vector ?

$$[ρ/(z(1+a/z^2))),0,1/((a-z^2)/(a+z^2)^2)]$$

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  • $\begingroup$ Is it not $\rho^2 = x^2 + y^2$? $\endgroup$
    – mvw
    Jun 24 '16 at 16:56
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You need the gradient operator for cylindrical coordinates, e.g. see here.

$$ \nabla f = \frac{\partial f}{\partial \rho} e_\rho + \frac{1}{\rho}\frac{\partial f}{\partial \varphi} e_\varphi + \frac{\partial f}{\partial z} e_z $$

and apply it to $$ f(\rho, \phi, z) = \frac{\rho^2}{2z \left( 1 + \frac{a}{z^2} \right)} $$ so \begin{align} \nabla f = \frac{\rho}{z \left( 1 + \frac{a}{z^2} \right)} e_\rho - \frac{\rho^2}{\left( 2z \left( 1 + \frac{a}{z^2} \right) \right)^2} \left( \left( 1 + \frac{a}{z^2} \right) - z \left(2\frac{a}{z^3} \right) \right) e_z \\ = \frac{\rho}{z \left( 1 + \frac{a}{z^2} \right)} e_\rho - \rho^2 \frac{1 - \frac{a}{z^2}}{\left( 2z \left( 1 + \frac{a}{z^2} \right) \right)^2} e_z \\ \end{align} In Cartesian coordinates it would be $$ \nabla f = (\partial_x, \partial_y, \partial_z) f(x,y,z) = (\partial_x, \partial_y, \partial_z) \frac{x^2 + y^2}{2z \left( 1 + \frac{a}{z^2} \right)} $$

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  • $\begingroup$ I would prefer an answer in Cartesian coordinates, would it work just using cartesian nabla ? $\endgroup$
    – Anonymous
    Jun 24 '16 at 16:55
  • $\begingroup$ Why should it not. You need your $f$ in Cartesian coordinates as well. $\endgroup$
    – mvw
    Jun 24 '16 at 16:56

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