Gradient in cylindrical coordinates

Question

I am a bit confused on why the vector gradient of a function $f$ in a cylindrical coordinate is equal to: $$\mathbf{e}_r\frac {\partial f}{\partial r} + \mathbf{e}_\theta\frac 1 r \frac {\partial f}{\partial\theta} + \mathbf{e}_z\frac {\partial f}{\partial z}.$$

From where did we get the $1/r$?

Answer 1

Conceptually, you can think of the factor of $1/r$ as a unit conversion. That is, because the angle $\theta$ is dimensionless, we need to multiply the $\partial f/\partial \theta$ term by something with units of inverse length to match the units of the other two terms.

That isn't very satisfying, so let's derive the form of the gradient in cylindrical coordinates explicitly. The crucial fact about $\nabla f$ is that, over a small displacement $d\mathbf{l}$ through space, the infinitesimal change in $f$ is $$df = \nabla f\cdot d\mathbf{l}. \tag{1}$$ In terms of the basis vectors in cylindrical coordinates, $$d\mathbf{l} = dr\,\hat{r} + r\,d\theta\,\hat{\theta} + dz\,\hat{z}. \tag{2}$$ Now, we are assuming that $f$ is differentiable, i.e., it is locally linear. So $$df = \frac{\partial f}{\partial r}\,dr + \frac{\partial f}{\partial \theta}\,d\theta + \frac{\partial f}{\partial z}\,dz. \tag{3}$$ Plugging (2) and (3) into (1), we get $$\frac{\partial f}{\partial r}\,dr + \frac{\partial f}{\partial \theta}\,d\theta + \frac{\partial f}{\partial z}\,dz = \left(\nabla f\right)_r\,dr + (\nabla f)_\theta\,r\,d\theta + (\nabla f)_z\,dz.$$ (I'm using the notation $(\nabla f)_r = \nabla f\cdot\hat{r}$.) Comparing term by term, we find that $$\nabla f = \frac{\partial f}{\partial r}\,\hat{r} + \frac{1}{r}\frac{\partial f}{\partial \theta}\,\hat{\theta} + \frac{\partial f}{\partial z}\,\hat{z},$$ as expected.

As an exercise, you might want to try using this same procedure to find the form of the gradient in spherical coordinates.

Answer 2

It's there because a change in $\theta$ is much more relevant if it happens close to the origin (low $r$) than far away from the origin (high $r$).

To derive the exact formula, you need to express the cylindrical coordinates in Cartesian coordinates and differentiate. This is done e.g. in the answers to Del operator in Cylindrical coordinates (problem in partial differentiation).