# why is the laplacian of 1/r equal zero outside the origin?

In spherical coordinates, the laplacian can be written as:

$\nabla^2 = \frac{1}{r}\frac{\partial^2}{\partial r^2}r + \frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial f}{\partial \theta}) + \frac{1}{r^2 \sin^2 \theta}\frac{\partial^2 f}{\partial \phi^2}$

Suppose the function I look at is $f(r,\theta,\phi) = \frac{1}{r}$, then

$\nabla^2f = \frac{1}{r}\frac{\partial^2}{\partial r^2}r(\frac{1}{r}) = 0 .$ (outside the origin)

My question is if the laplacian is a kind of second derivative in higher dimensions how is the laplacian of this function equal to zero? I mean if we consider $1/r$ in 1 - dimension hitting it with two derivatives makes it go like $1/r^3$. What's different in three dimensions. I understand the calculation; I am looking for a more intuitive explanation.

Also, according to the view that the laplacian measures how the function deviates from the average value around a point, I don't see how this comes to be zero. $\frac{1}{r}$ has curvature, especially near $r=1$. I am probably thinking about this the wrong way.

• In three dimensions $\nabla ^{2}\frac{1}{r}$ is proportional to the Dirac delta-function $\delta (\mathbf{x})$, a distribution. Although it vanishes outside the origin it is not identically $0$. Nov 17, 2015 at 15:37
• @Urgje The OP clearly understands that their analysis is only correct away from 0, since the both title and the equation in the post states "outside the origin". Nov 17, 2015 at 15:47
• @Urgje How does one show that $\nabla^2 1/r$ is a Dirac delta ? Is there a reference you can point me to ? Apr 16, 2016 at 21:28

Let's answer the analogous question in 2D because it's a lot easier to visualize! Consider $f = - \log r$, which has $\nabla^2 f = 0$ in 2D.

In image 1, I've plotted a (rescaled) $f$ with the lines of constant $r,\theta$ drawn on. What you are imagining is that as you go down the $r$ curve, you have non-zero second derivative because it's bending, and that you can ignore the $\theta$ direction because it's completely flat that way: But this isn't really what the Laplacian derivatives are measuring. In image two I've plotted the same curve but with lines of constant $x,y$ drawn on: Now it's clear that whilst in the $x$ direction (along the nearer side), the function is bending upwards gradually, we can see by the contours that in the $y$ direction (away from us) it is currently bending downwards. And on average, as you say, nothing is happening. Hence $\nabla^2 f = 0$ can make sense: $\partial_x^2 > 0,\partial_y^2 < 0$.

The mathematical way that we capture the relationship between these two diagrams is by converting between derivatives in one coordinate system ($x,y$) to those in another one ($r,\theta$). In this way, we capture all the subtleties of measuring things in different directions in simple formulae, like

$$\nabla^2 f = \frac{\partial^2 f}{\partial r^2} + \boxed{\frac 1 r \frac{\partial f}{\partial r}} + \frac 1 {r^2} \frac{\partial^2 f}{\partial \theta^2}$$

The highlighted term here tells you that if a function has a non-zero slope in $r$, then there's secretly some extra curvature other than that from looking in the increasing $r$ direction: this is due to motion in the direction orthogonal to $r$, and it is because if you look orthogonal to increasing $r$ you're actually still looking at points where $r$ is larger! This is just saying that the tangent to a circle passes through points of increasing distance from the centre of the circle. As a result, because the function $f$ is smaller at larger $r$, we measure a negative curvature as we look in that orthogonal direction, and this contributes to $\nabla^2$.

Hope that helps!

Edit: Notice that if we write $r= \sqrt{x^2+y^2}$ and $\theta = \tan^{-1}(y/x)$ and take $f$ to be only a function of $r$ then

$$\frac{\partial f}{\partial x} = \frac{\partial r}{\partial x}\frac{\partial f}{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial f}{\partial \theta} = \frac x r f'(r)$$

so $$\frac{\partial^2 f}{\partial x^2} = \frac{x^2}{r^2} f''(r) + \frac{y^2}{r^3} f'(r)$$

Hence if we are moving along the $x$ axis ($y=0$) then

$$\left[\frac{\partial^2 f}{\partial x^2} \right]_{y=0} = f''(r)$$

But similarly

$$\frac{\partial^2 f}{\partial y^2} = \frac{y^2}{r^2} f''(r) + \frac{x^2}{r^3} f'(r)$$

$$\left[\frac{\partial^2 f}{\partial y^2} \right]_{y=0} = \frac{1}{r} f'(r)$$

This shows that it is the bending in the $y$ direction (which is what is locally orthogonal to the $r$ direction) which gives the extra highlighted term. Meanwhile, the bending in the $x$ direction only gives the naively expected two $r$ derivative term.

Edit: To address Ian's question in the comments - one reason why you actually get exactly zero for these choices of $f$ ($\log$ in 2D, $1/r$ in 3D, ...) is the following, based around the divergence theorem which says that $$\int_V \nabla^2 f \ \mathrm{d} V = \int_{\partial V} \nabla f \cdot \ \mathrm{d} S$$ or in more intuitive words, the Laplacian represents a measurement of whether or not on average $\nabla f$ points in or out of the region $V$.

Suppose you are given a point $\mathbf{x}_0$. Consider a very thin spherical shell at $|\mathbf{x}_0| \pm \delta$ from the origin. Clearly the Laplacian should be the same everywhere in that shell, so it would be sufficient to show that the integral over the shell of the Laplacian vanishes for all sufficiently small $\delta$.

But by the above theorem, this is the same as making the flux of $\nabla f$ through the shell vanish. Since $$\nabla f \cdot \pm \hat{\mathbf{r}} = \pm f'(r)$$ and since the surface areas of each boundary of the shell are proportional to $r^{d-1}$ in $d$ dimensions, we need $f' \sim r^{1-d}$ in order to obtain a vanishing result.

I.e. in two dimensions, $f' \sim 1/r$ and $f \sim \log r$ or in at least three dimensions, $f \sim r^{2-d}$.

• +1, though I have some concerns. In trying to write up an answer I struggled with explaining something that I don't think you've encapsulated here, either. You can see that there are cancellations between the terms in various ways. For instance I was thinking of explaining it with the divergence theorem: since the gradient of $1/r$ points toward the origin, provided the origin is not in the domain, the dot product between the gradient and the surface normal will always take on both signs. But it's not geometrically obvious to me that there is exact cancellation.
– Ian
Nov 17, 2015 at 15:11
• I suppose you could rephrase it this way: suppose $f$ is smooth, convex, and decreasing, then $\Delta f(r)$ exhibits the cancellations you've mentioned. What is geometrically special about $\log(r)$ in 2D or $1/r$ in 3D to make the cancellations exact?
– Ian
Nov 17, 2015 at 15:16
• @Ian Thanks. as I understand the question, what is desired is some intuition as to how to count up the curvature in such a way that it is possible to get 0 from a non-trivial radial function. Of course there is no way my answer explains why it is specifically zero, I didn't use any properties of $f$ at at all! I can add a geometric justification, though. Nov 17, 2015 at 15:16
• A geometric justification would be great. I've always had this taught in a very analytic manner (set up the ODE, solve it, then manage the singularities that the ODE can't handle classically).
– Ian
Nov 17, 2015 at 15:18
• @Ian I've added a quick justification based around the divergence theorem and the scaling of volumes of spherical shell surfaces in $d$ dimensions. Nov 17, 2015 at 15:41