Second derivative in polar coordinates

Posted by David Zaslavsky on November 16, 2011 9:54 PM

— Comments

Here’s an interesting, and perhaps occasionally useful, identity: suppose you have a function defined on a two-dimensional space, $f(\vec{r})$. And suppose that this function is independent of the angle of $\vec{r}$ but rather is only a function of the magnitude $r = \norm{\vec{r}}$. Then

$$\lapl f(r) = \frac{1}{r}\pd{}{r}r\pd{}{r}f(r) = \frac{1}{r^2} r\pd{}{r}r\pd{}{r} = \frac{1}{r^2}\pdd{}{(\ln r)}f(r)$$

(where $\ln r$ is shorthand for $\ln\frac{r}{r_0}$ where $r_0$ is some constant) In other words, the radial term of the polar Laplacian is equal to the second logarithmic derivative divided by $r^2$ for rotationally invariant functions.

At this point you might be wondering, why the heck is this interesting? Well, for me, it’s because my current research project happens to involve logarithmic derivatives of rotationally invariant functions in 2D space. But for (almost) everyone else, this is related to the way electric potential drops off with distance in 2D space.

Consider, for example, a wire that carries a charge density $\lambda$ (but no current). Since this system is translationally symmetric (nothing changes if you slide it along the length of the wire) and rotationally symmetric (nothing changes if you rotate around the axis of the wire), the potential won’t depend on either of the coordinates $z$ or $\phi$; it only depends on $r$. This means you can use it as an analogy for a point charge in 2D space. (Imagine taking a “slice” perpendicular to the wire) Once you know that, you can almost instantly compute the potential using Poisson’s equation:

$$\lapl\Phi(r) = -\frac{\rho(r)}{\epsilon_0}$$

Since $\rho = 0$ everywhere except on the wire itself, you get

$$\begin{align}0 &= \frac{1}{r^2}\pdd{}{(\ln r)}\Phi(r) \\ 0 &= \pdd{}{(\ln r)}\Phi(r)\end{align}$$

And when the second derivative of a function is zero, you know that the function is linear:

$$\Phi(r) = \Phi_0\ln\frac{r}{r_0}$$

All that’s left is to choose the constants of integration, $\Phi_0$ and $r_0$.

Based on this relation between electrical potential and derivatives you might look for higher-dimensional analogues. For example, instead of a function on a 2D space, let’s say you have a 3D space. Electrical potential of a point charge in 3D drops off as $\frac{1}{r}$, and instead of polar coordinates, we use 3D spherical coordinates. So I’d expect to find a relationship like this:

$$\lapl f(r) = \frac{1}{r^4}\pdd{}{(1/r)}f(r)$$

(It has to be $r^4$ to make the units work out)

Is it true? The left side, the radial part of the Laplacian in 3D spherical coordinates, is

$$\lapl f(r) = \frac{1}{r^2}\pd{}{r}\biggl[r^2 \pd{}{r}f(r)\biggr] = \frac{1}{r^2}\biggl(2r\pd{}{r}f(r) + r^2\pdd{}{r}f(r)\biggr)$$

On the right side, let’s first figure out what $\pd{}{(1/r)}$ is:

$$\pd{}{(r^{-1})} = -\frac{1}{r^{-2}}\pd{}{r} = -r^2\pd{}{r}$$

That means

$$\frac{1}{r^4}\pdd{}{(1/r)}f(r) = \frac{1}{r^4}\times r^2\pd{}{r}r^2\pd{}{r}f(r) = \frac{1}{r^2}\biggl(2r\pd{}{r}f(r) + r^2\pdd{}{r}f(r)\biggr)$$

Yep, it works! Based on this, we can extrapolate a general rule for the radial part of the Laplacian in $n$ dimensions:

$$\lapl f(r) = \frac{1}{r^{2(n-1)}}\pdd{}{(r^{n-2})}f(r)$$

where $r^{n-2}$ gets replaced by $\ln r$ for $n = 2$ (just like integration of $\int r^{n-3}\udc r$). You can check that this does indeed match the usual form written in Wikipedia and elsewhere,

$$\lapl f(r) = \frac{1}{r^{n-1}}\pd{}{r}r^{n-1}\pd{}{r}f(r)$$