How do we determine the gradient and curl of a scalar/vector field in polar coordinates? For instance, if we have the following potential energy function for a force,

$$U = \frac{kx}{(x^2+y^2)^{3/2}}$$

it makes much more sense to compute the force in polar coordinates


But what is $\vec{\nabla}\cdot U$ in this case? The first thing that comes to mind is

$$\vec{F} = \frac{\partial U}{\partial r}\hat{r}+\frac{\partial U}{\partial\theta}\hat{\theta}$$

But that can't true, since the dimensions don't work out. Similarly, how do we calculate the curl for a vector field in polar coordinates? For instance, a central force like

$$F = -\frac{A}{r^2}\hat{r}$$

is much easier to deal with in polar.

I am a beginner in calculus, vector calculus in particular is a completely alien field to me. I would appreciate it if the answer could be kept as simple as possible.

Moreover, apart from the mathematical definition, it would be nice if the answer could include some reasoning (physical as well as mathematical) on why the curl and gradient are what they are in polar coordinates and how they tie to their counterparts in Cartesian coordinates.


1 Answer 1


You were pretty close already. There is a handy table on Wikipedia for a variety of coordinate systems. But for the polar system:

$$ \vec{\nabla} \cdot \vec{U} = \frac{\partial U_r}{\partial r} + \frac{1}{r} \frac{\partial U_\theta}{\partial \theta} $$

and you can look up the curl in the same table.

These can be derived from the Cartesian definitions by considering the total differentials:

$$ dr = \frac{\partial r}{\partial x} dx + \frac{\partial r}{\partial y} dy$$


$$ d\theta = \frac{\partial \theta}{\partial x} dx + \frac{\partial \theta}{\partial y} dy$$

Or to put it another way, you end up with:

$$ \frac{\partial \phi}{\partial r} = \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial r}$$

which is just the chain-rule for derivatives. $\phi$ is any variable. A similar expression can be written for the $\theta$ coordinate. You can use this to go between any coordinate systems you would like to go between and relate it all back to the Cartesian system by successive chaining.

As for a physical insight, that's a little bit harder. But the way I like to look at it for the polar (aka cylindrical) system is that the unit length of $\theta$ changes as $r$ changes. If you draw a big circle vs a small circle, the arc length of the wedge defined by an angle $\theta$ is certainly different. And so it is natural that the derivative in the $\theta$ direction will have some radial dependence.

  • $\begingroup$ Thanks. However, that solves only part of my problem. I am still looking for intuition on how these link to Cartesian coordinates. $\endgroup$
    – Gerard
    Mar 27, 2015 at 6:40
  • 1
    $\begingroup$ Check out Arfken and Webber "Mathematical Methods fro Physicists" if you have a copy, this is explained in there. $\endgroup$
    – hft
    Mar 27, 2015 at 6:44
  • $\begingroup$ @Gerard It's a work in progress -- it's late! Or early, I guess, depending on perspective :) $\endgroup$
    – tpg2114
    Mar 27, 2015 at 6:49

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