In spherical coordinates, it can be proved that the laplacian of $r = \sqrt{x^2+y^2+z^2}$ at the origin is

\begin{align} \nabla^2 \dfrac{1}{r} = -4\pi\delta(\vec{r}) \end{align}

as demonstrated in this answer:


I am trying to prove a the same thing except in cylindrical coordinates $(\rho,\theta,z)$ at $(\rho = 0, \theta, z = 0)$ and $r = \sqrt{x^2+y^2}$, ie

\begin{align} \nabla^2 \dfrac{1}{r} =? \end{align}

Following the proof in spherical coordinates, I consider the circle, $C$, enclosing the area around the origin, $D$, and use the 2D divergence theorem

\begin{align} \iint_D (\nabla \cdot \nabla\dfrac{1}{r})dA =\oint_C \nabla\dfrac{1}{r} \cdot \hat{n} ds= \oint_C \nabla\dfrac{1}{r} \cdot \hat{e}_\rho ds \end{align}

$\hat{n} = \hat{e}_\rho$ since the contour is a circle. The gradient of $1/r$ is

\begin{align} \nabla\dfrac{1}{r} = \nabla\dfrac{1}{\sqrt{x^2+y^2}} = -\dfrac{x}{(x^2+y^2)^{3/2}}\hat{e}_x-\dfrac{y}{(x^2+y^2)^{3/2}}\hat{e}_y = \dfrac{\vec{\rho}}{r^3} \end{align}

Therefore I get

\begin{align} \oint_C \dfrac{\vec{\rho}}{r^3} \cdot \hat{e}_\rho ds = \int_0^{2\pi}\dfrac{r}{r^3}rd\theta = \dfrac{2\pi}{r} \end{align}

This seems wrong to me as I expect that it should be some constant similar to the spherical case of $-4\pi$. Am I approaching this problem correctly or have I made a mistake somewhere?


You lost a negative sign, but overall your work holds together. In any number of dimensions, say $n\geq 2$, $$\vec{\nabla}\left(\frac{1}{r}\right) = \frac{-r\hat{r}}{r^{3}}$$ has zero divergence for $r\neq 0$: $$\Delta\left(\frac{1}{r}\right)=0 \mbox{ for } r\neq 0.$$ Physically this means that no electric field is being generated away from the origin, for example. The only question, then, is what is happening at $r=0$, where you envision a point charge being located at the origin that is generating an electric field. How strong is that source?

We consider this problem by letting our domain $\Omega$ be an $n$-dimensional sphere of radius $R$ and letting $R\to 0^{+}$. The term in our integral that will work, if any such term exists, gives the strength of our point source, which in your context is the value of $\Delta \left(\frac{1}{r}\right)$ at the origin.

This is the context into which your computations fit. Your computations are correct: in determining the Laplacian of $\frac{1}{r}$ at the origin you look at a volume integral and turn it into a surface integral over the boundary of a ball of radius $R>0$, where again your focus is on what is happening as $R\to 0^{+}$.

You determine this by looking at $$\int_{\Omega}\Delta\left(\frac{1}{r}\right)dV = \int_{\partial \Omega}\vec{\nabla}\left(\frac{1}{r}\right)\cdot \hat{r}dA = \int_{\partial \Omega}\frac{-R\hat{r}}{R^{3}}\cdot \hat{r}dA = \int_{\partial \Omega}\frac{-1}{R^{2}}dA.$$ The spherical area element in $n$ dimensions has a factor of $R^{n-1}$ with additional angular terms that, big-picture, we don't care about. The important point is that when $n=3$ the radial factors cancel out perfectly and we are left with a point source that has an intrinsic strength that is easily defined as it can be measured as the flux of electric field through a sphere of any radius about the point charge, since there is no $R$ is our answer.

For $n\neq 3$ the easiest, most accurate, thing to say is simply that we don't have a particularly meaningful way of defining the strength of a point source as we can longer meaningfully connect it to the flux out through a sphere of arbitrary radius: the flux now depends upon $R$.

If you attempt to define the "intrinsic strength" of a point source as the flux through a sphere of radius $R$ as $R\to 0^{+}$, then we find that:

When $n=2$ there is an overall factor of $\frac{1}{R}$ that goes to $\infty$ as $R\to 0^{+}$, so the "intrinsic strength" of the point source is infinite.

When $n\geq 4$ there is an overall factor of $R^{n-3}$ that goes to $0$ as $R\to 0^{+}$, so the "intrinsic strength" of the point source is $0$.

These results make intuitive sense in that with fewer dimensions there is less "space" for the electric field to dissipate, so that $n=2$ is too cramped and the "intrinsic strength" is infinite, $n\geq 4$ is too spacious and the "intrinsic strength" is $0$, and $n=3$ is just right, which makes a proper definition of the strength of a point charge possible.

  • $\begingroup$ At the beginning you state "zero divergence for $r \neq 0$". but your math statement is for the laplacian of $1/r$. Shouldn't it be the laplacian is zero for $r \neq 0$? Also shouldn't this only be valid for $ n = 3$? In the cylindrical 2D case, $\nabla^2 \dfrac{1}{r} = \dfrac{1}{r^3}$ for $r \neq 0$. Thanks for the detailed answer $\endgroup$ – Ragnar May 24 '15 at 17:37

$$\frac{1}{r}$$ function is harmonic only in 3 dimensions. Take a look at Laplace operator radial component in N dimensions: $$\frac{1}{r^{N-1}}\frac{\partial}{\partial r}(r^{N-1}\frac{\partial}{\partial r})\frac{1}{r}$$ If you use cylindrical coordinates, make sure you express distance from the origin in cylindrical coordinates as well: i.e. $\sqrt{r^2 + z^2}$. Then laplasian would be: $$\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial}{\partial r})\frac{1}{\sqrt{r^2 + z^2}} + \frac{\partial}{\partial z}\frac{\partial}{\partial z}\frac{1}{\sqrt{r^2 + z^2}}$$ You came up with strange results because you used polar 2-D representation of coordinates.

  • $\begingroup$ I am defining $r$ specifically like this because I am working with the biharmonic equation in cylindrical coordinates where $\nabla^2 \dfrac{1}{r} = \nabla^2 \dfrac{1}{\sqrt{x^2+y^2}}$ specifically appears. there should still be a singularity at $\rho = 0$ for all $z$ I was trying to see if it is measurable in the same way as the spherical version $\endgroup$ – Ragnar May 24 '15 at 17:09

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