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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:

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

user55777
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1 Answers1

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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.

David K
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  • 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. – user55777 Nov 18 '20 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.) – David K Nov 19 '20 at 00:21
  • what did you mean by the $y$ axis? Is it the vertical axis, which is the $z$ axis in my original post? – user55777 Nov 19 '20 at 03:06
  • 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. – David K Nov 19 '20 at 05:13