# Transform derivatives from 2D Cartesian to axisymmetric cylindrical coordinates

Consider the 1st and 2nd derivatives (differential operations) of a function $$z=f(x)$$ with respect to horizontal coordinate $$x$$ in 2D Cartesian coordinates $$(x,z)$$,

$$\frac{df}{dx} \quad \text{and} \quad \frac{d^2f}{dx^2}.$$

If $$f(x)$$ is axisymmetric about the vertical $$z$$ axis and we want to transform the derivatives into cylindrical coordinates $$(r,z)$$, can they be written straightforward as follows:

$$\frac{1}{r}\frac{d(rf)}{dr} \quad \text{and} \quad \frac{1}{r}\frac{d}{dr}(r\frac{df}{dr}),$$ where $$z=f(r)$$ and $$r$$ is the radial coordinate.

Update:

For the 1st derivative, I believe the answer of @David K is correct. However, I don't understand the 2nd derivative. Actually, I got the transformation of the 2nd derivative by comparing the Laplace operators in Cartesian coordinates ($$z=f(x)$$) and in axisymmetric cylindrical coordinates ($$z=f(r)$$):

In Cartesian coordinates:

$$\nabla^2 f=\frac{d^2f}{dx^2}$$

In cylinderical coordiantes:

$$\nabla^2 f=\frac{1}{r}\frac{d}{d r}(r\frac{d f}{d r})$$

You have no angular component in your "cylindrical" coordinates, so all you have actually is Cartesian coordinates in which one coordinate is named $$r$$ instead of $$x.$$ The transformation of the derivative is indeed extremely straightforward:

\begin{align} \frac{\mathrm df}{\mathrm dx} &\to \frac{\mathrm df}{\mathrm dr}, \\ \frac{\mathrm d^2f}{\mathrm dx^2} &\to \frac{\mathrm d^2f}{\mathrm dr^2}. \end{align}

Why would it be anything else?

You could extend your function $$f$$ from a function on the $$x$$ axis to a function $$\bar f$$ over a plane $$\pi_0$$ through the origin, perpendicular to the $$z$$ axis, by setting up cylindrical coordinates around the $$z$$ axis, thereby assigning polar coordinates $$r,\theta$$ to the plane $$\pi_0$$, and defining the function

$$\bar f(r, \theta) = f(r).$$

You can then take partial derivatives with respect to $$r$$ and $$\theta$$:

\begin{align} \frac{\partial \bar f}{\partial r} &= \frac{\mathrm df}{\mathrm dr}, \\ \frac{\partial \bar f}{\partial \theta} &= 0. \\ \end{align}

The gradient of $$\bar f$$ is then

$$\nabla \bar f = \hat{\mathbf r} \frac{\partial \bar f}{\partial r} + \hat{\mathbf \theta} \frac 1r \frac{\partial \bar f}{\partial \theta}.$$

(See How to obtain the gradient in polar coordinates and its answers.) Note that there is a factor of $$\frac1r$$ outside one of the partial derivatives, but it's on the partial derivative with respect to $$\theta,$$ not $$r,$$ and there is no factor $$r$$ inside any partial derivative.

• Thank you so much! Did you mean $df/dx\rightarrow df/dr$ and $d^2f/dx^2\rightarrow d^2f/dr^2$? BTW, please see my update. Commented Nov 18, 2020 at 13:48
• You were right, I was lazily cutting and pasting and forget to change the $x$s. What you found with the Laplacian is that the Laplacian of a function on the plane whose level curves are all parallel to the $y$ axis is different from the Laplacian of a function whose level curves are all circles around the origin, even if the functions agree on the $x$ axis. The connection to derivatives of a single-variable function is unclear. (Even the applicability of the gradient is unclear.) Commented Nov 19, 2020 at 0:21
• what did you mean by the $y$ axis? Is it the vertical axis, which is the $z$ axis in my original post? Commented Nov 19, 2020 at 3:06
• I mean the $y$ axis of cartesian coordinates of the plane over which your function is defined, perpendicular to the $z$ axis. Notice that you are using $z$ as an output of the function, not an input, so the domain of the function has only one dimension in the case of $f(x)$ or $f(r)$ and only two dimensions when you go to cylindrical coordinates. You never define a function on a 3D domain. Commented Nov 19, 2020 at 5:13