This is more of a maths question, but several sources point at different expressions for the gradient in cylindrical coordiantes.

Sometimes I see the radial component for the gradient of a scalar function $\nabla f$ written as $\frac{1}{r} \frac{ \partial}{\partial r}(rf)$ while sometimes I see just $\frac{ \partial}{\partial r}$. Which version is correct?

I'm applying it to a scalar potential $\phi$ that involves Bessel functions. Fore reference, $\phi = Ae^{-kz} ( B_1 \cos{n \theta} + B_2 \sin{n \theta}) J_n(kr)$ where $J_n(kr)$ is the Bessel function of the first kind.

• The first term is from the divergence of a vector. Commented Jul 10, 2015 at 17:11

Given a function in cylindrical coordinates $f(r, \phi, z)$, the gradient of $f$ is
$$\nabla f = \frac{\partial f}{\partial r} \textbf{e}_r + \frac{1}{r}\frac{\partial f}{\partial \phi} \textbf{e}_\phi + \frac{\partial f}{\partial z} \textbf{e}_z,$$
where $\{\textbf{e}_i\}_\text{cyl}$ is the standard orthonormal basis in cylindrical coordinates. One can obtain this formula simply by finding the directional derivatives of $f$ in Cartesian coordinates with respect to the elements of $\{\textbf{e}_i\}_\text{cyl}$.
• Then why does T. Conway in equation (5) of ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=947050 write it as $\frac{1}{r} \frac{\partial}{\partial r} (r A_{\theta})$ then? I realise he uses $\textbf{B} = \nabla \times A_{\theta}$ instead of $\textbf{B} = \nabla f$, but I'm not aware of a major difference between the two when it comes to the radial term besides the addition of the $\frac{1}{r} \frac{\partial}{\partial \theta} A_r$ term that is zero in the case of the article. Commented Jul 10, 2015 at 17:39
• @Tetradic: Conway is writing the curl of a vector field, $\nabla\times\mathbf A$, not the gradient. The two are easily derived from Cartesian, but you can also see them on Wikipedia. Commented Jul 10, 2015 at 17:55
• I agree with what @KyleKanos said. Also, in Cartesian coordinates, one can pretend that $\nabla$ is a vector with components $(\partial_x, \partial_y, \partial_z)$. This invalid assumption wrecks havoc on curvilinear coordinate systems. You cannot treat $\nabla \times \textbf{v}$ as the cross product, because it isn't. The curl is a linear operator on $\mathbb{R}^3$. Commented Jul 10, 2015 at 17:55