We know that a point charge $q$ located at the origin $r=0$ produces a potential $\sim \frac{q}{r}$, and this is consistent with the fact that the Laplacian of $\frac{q}{r}$ is

$$\nabla^2\frac{q}{r}~=~-4\pi q~ \delta^3(\vec{r}).$$

My question is, what is the Laplacian of $\frac{1}{r^2}$ (at the origin!)? Is there a charge distribution that would cause this potential?


5 Answers 5


I) Problem. We put (four pi times) the permittivity $4\pi\varepsilon=1$ equal to one (in the SI unit system) from now on for simplicity. Let us first rephrase OP's question a bit. Instead of starting from the potential

$$ \Phi~=~\frac{1}{r} \qquad \mathrm{and} \qquad \Phi~=~\frac{1}{2r^2}, \qquad r\neq 0,\tag{1} $$

respectively, let us assume that the electric field has be given as

$$ \vec{E}~=~\frac{\vec{r}}{r^3} \qquad \mathrm{and} \qquad\vec{E}~=~ \frac{\vec{r}}{r^4},\qquad r\neq 0,\tag{2}$$

respectively, and we want to know the charge density

$$ \rho~=~\vec{\nabla} \cdot \vec{E},\tag{3}$$

in particular, at the origin $r=0$. For $r\neq 0$, the charge density (3) is

$$ \rho~=~0 \qquad \mathrm{and} \qquad \rho~=~-\frac{1}{r^4},\qquad r\neq 0,\tag{4}$$

respectively. This rephrasing is just so we only have to differentiate one time instead of two times, but the argument is in principle the same, cf. footnote $1$.

II) Distribution theory and integration theory. An issue is that the electric field $\vec{E}$ has not been specified at $r=0$ in eq. (2). A way to make sense of this is to use distributions and test functions $f\in C^{\infty}_c(\mathbb{R}^3)$, i.e., infinitely often differentiable functions $f$ with compact support. We now declare that a smeared electric field $E^i$ is

$$ E^i[f]~:=~\int_{\mathbb{R}^3} \!d^3r \ E^i(\vec{r})f(\vec{r}).\tag{5}$$

A Lebesgue majorant of the integrand (5) is

$$ \frac{|f(\vec{r})|}{r^2} \qquad \mathrm{and} \qquad \frac{|f(\vec{r})|}{r^3},\qquad r\neq 0,\tag{6}$$

respectively. Only the first case of (6) is Lebesgue integrable in $\mathbb{R}^3$. This is the heart of the problem. In the second case, even after smearing with a test function $f$, the electric field $E^i$ does not make sense as a distribution.

III) In distribution theory, the derivative of a distribution is always defined by applying the derivative to the test function with a minus sign$^{1}$. If $E^i$ is a distribution (5), we can carry out the next step to define the derivative of $E^i$,

$$-\vec{\nabla} \cdot \vec{E}[f] ~:=~ \int_{\mathbb{R}^3} \!d^3r \ \vec{E}(\vec{r})\cdot\vec{\nabla} f(\vec{r}) ~\stackrel{\begin{matrix}\text{Leb. dom.}\\ \text{conv. thm.}\end{matrix}}{=}~ \lim_{\varepsilon\to 0^+} \int_{\{r\geq\varepsilon\}} \!d^3r \ \vec{E}(\vec{r})\cdot\vec{\nabla} f(\vec{r}) $$ $$ ~\stackrel{\begin{matrix}\text{int. by}\\ \text{parts}\end{matrix}}{=}~ \lim_{\varepsilon\to 0^+} \int_{\{r\geq\varepsilon\}} \!d^3r \left[ \vec{\nabla} \cdot\left( \vec{E}(\vec{r})f(\vec{r})\right)-f(\vec{r}) \underbrace{\vec{\nabla} \cdot \vec{E}(\vec{r})}_{=0}\right] $$ $$ ~\stackrel{\begin{matrix}\text{div.}\\ \text{thm.}\end{matrix}}{=}~ -\lim_{\varepsilon\to 0^+} \int_{\{r=\varepsilon\}} \!d^2\vec{A} \cdot \vec{E}(\vec{r})f(\vec{r}) ~\stackrel{(2)}{=}~-\lim_{\varepsilon\to 0^+} \int_{\{r=\varepsilon\}} \!\frac{d^2A}{\varepsilon^2}f(\vec{r}) $$ $$~=~-4\pi f(0),\tag{7} $$

where we performed the well-known manipulations in the first case to show that

$$ \vec{\nabla} \cdot\frac{\vec{r}}{r^3} ~=~ 4\pi\delta^3(\vec{r}).\tag{8} $$

IV) Regularization. Well, so much for distribution theory and mathematical idealization in Sections II-III. In reality, the $\frac{-1}{r^4}$ charge density, $r\neq 0$, in eq. (4) would break down as we approach the singularity $r\to 0$, so that we never get to ask: What sits at $r=0$? This leads to the idea of regularization

$$ \Phi_{\varepsilon}~=~\frac{1}{\sqrt{r^2+\varepsilon}} \qquad \mathrm{and} \qquad \Phi_{\varepsilon}~=~\frac{1}{2(r^2+\varepsilon)}, \qquad \varepsilon>0,\tag{9}$$

respectively, of eq. (1). The regularized charge density

$$ \rho_{\varepsilon}~=~-\vec{\nabla}^2\Phi_{\varepsilon}~\in~ C^{\infty}(\mathbb{R}^3),\qquad \varepsilon>0,\tag{10}$$


$$ \rho_{\varepsilon}~=~\frac{3\varepsilon}{(r^2+\varepsilon)^{\frac{5}{2}}} \qquad \mathrm{and} \qquad \rho_{\varepsilon}~=~\frac{3\varepsilon-r^2}{(r^2+\varepsilon)^{3}} ~=~\frac{4\varepsilon}{(r^2+\varepsilon)^{3}}-\frac{1}{(r^2+\varepsilon)^{2}},\tag{11}$$

respectively. We can now smear with a test function $f$. One may check that the first charge density $\rho_{\varepsilon}$ in eq. (11) satisfies

$$\lim_{\varepsilon\to 0^+}\rho_{\varepsilon} ~=~4\pi\delta^3(\vec{r}) , \tag{12} $$

while the second charge density $\rho_{\varepsilon}$ in eq. (11) does not make sense as a distribution when $\varepsilon\to 0^+$, i.e. after smearing with a test function $f$, the limit $\varepsilon\to 0^+$ is not finite.

$^{1}$ In the second case, we could in principle define the smeared potential

$$ \Phi[f] ~:=~ \int_{\mathbb{R}^3} \!d^3r \ \Phi(\vec{r})f(\vec{r}),\tag{13}$$

because $\Phi=\frac{1}{2r^2}$, $r\neq 0$, is locally Lebesgue integrable in $\mathbb{R}^3$, and then define the electric field as a distribution

$$ \vec{E}[f] ~:=~ \Phi[\vec{\nabla} f] ~\stackrel{(13)}{=}~ \int_{\mathbb{R}^3} \!d^3r \ \Phi(\vec{r})\vec{\nabla}f(\vec{r}),\tag{14}$$

so that

$$ -\vec{\nabla} \cdot \vec{E}[f]~=~ \int_{\mathbb{R}^3} \!d^3r \ \Phi(\vec{r})\vec{\nabla}^2 f(\vec{r})~=~ \lim_{\varepsilon\to 0} \int_{\{r\geq\varepsilon\}} \!d^3r \ \Phi(\vec{r})\vec{\nabla}^2 f(\vec{r}).\tag{15}$$

However, this mathematical construct is not as useful in practice as one might naively have hoped for. For instance, if we try to integrate by parts, we essentially get back to the problem that the electric field $\vec{E}= \frac{\vec{r}}{r^4}$, $r\neq 0$, from eq. (2) is not locally Lebesgue integrable in $\mathbb{R}^3$.

  • $\begingroup$ This answer is too formal! $\endgroup$
    – Ron Maimon
    Sep 1, 2011 at 3:06
  • $\begingroup$ A distribution is a very useful tool (when it exists), although in practical physical situations, a mere cut-off/regularization is clearly more realistic. Note however, that when we introduce a cut-off the potential does strictly speaking not behave as $1/r^2$ at the origin as OP requests (v3). It still serves a point to investigate if the OP's idealized problem has a description in terms of a distribution. $\endgroup$
    – Qmechanic
    Sep 1, 2011 at 18:32

The electric field from your potential is:

$$E(r) = {2\over r^3}$$

Using Gauss's law, the total charge in a sphere of radius R is:

$$Q(r) = \oint E \cdot dS = 4\pi r^2 {2\over r^3} = {8\pi\over r}$$

The total charge is decreasing with r, so there is a negative charge cloud of density

$$ \rho(r) = {1\over 4\pi r^2} {dQ\over dr} = - {4\over r^4}$$

But the total charge at infinity is zero, so there is a positive charge at the origin, cancelling the negative charge cloud, of a divergent magnitude. If you assume this charge is a sphere of infinitesimal radius $\epsilon$, the positive charge at the origin is

$$Q_0 = \int_\epsilon^\infty 4\pi r^2 {4\over r^4} = {16\pi \over \epsilon}$$

This is not a distribution in the mathematical sense, but it is certainly ok to work with, so long as you keep the $\epsilon$ around and take the limit $\epsilon$ goes to zero at the end of the day. Mathematicians have not had the last word on the class of appropriate generalized solutions yet.


Vladimir's answer is off by factor of 2. The laplacian is $\nabla^2(\frac{1}{r^2}) = \frac{4}{r^4}$ A potential that falls of as $\frac{1}{r^2}$ is a dipole (In general, if it falls off as $r^{-n}$ its an ($2^{n-1}$)-pole, e.g $\frac{1}{r^3}$ bheaviour is quadrupole, etc).

Is this a dirac delta? To find out, check : $$\int_{\mbox{All space}}\nabla^2(\frac{1}{r^2})d^3r $$ $$=16\pi\int_0^\infty \frac{1}{r^2}dr\neq 1$$ Yup, integral diverges, so it is NOT a delta function.

I think your confusion is regarding the nature of a delta function. If something blows up at the origin it does not mean it is necessarily a delta function.

  • 1
    $\begingroup$ My question was not whether $2/r^{4}$ was a dirac function. My question is how do you know that you don't have to add a dirac delta term at the origin, as you do with the laplacian of $1/r$. That is, you know that $\nabla^2(\frac{1}{r}) = 0 - 4\pi \cdot \delta^3(\vec{r})$, but how do you know that $\nabla^2(\frac{1}{r^2}) = 2/r^{4} -4\pi \cdot \delta^3(\vec{r})$ isn't true? That's what confounds me. $\endgroup$
    – becko
    Apr 30, 2011 at 13:15
  • $\begingroup$ The multipole expansion is applicable only to charge distributions that are confined to a finite region of space. I don't think we can automatically state that if the potential is $1/r^2$ then we have a dipole. Before that we would need to establish that the charge distribution is in a finite region of space, which isn't the case. $\endgroup$
    – becko
    Apr 30, 2011 at 13:35

Let us see:

$\Delta \frac{q}{r^2} = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial} {\partial r}\frac{q}{r^2}\right)=\frac{2q}{r^4}$.

Thus the charge density $\rho(r)$ is proportional to $r^{-4}$. Such a charge density is only possible to create at a macroscopic scale where the charge may be considered continuous (charged dielectric spherical layers).

EDIT: If the potential is spherically symmetric, it is not a dipole field but a monopole one with the charge continuously distributed along $r$!

  • $\begingroup$ Can you also comment on whether such a charge distribution exists? $\endgroup$
    – Lagerbaer
    Apr 28, 2011 at 22:16
  • $\begingroup$ How can I be sure that no DiracDelta functions show up in the laplacian of $q/r^2$? I mean, the laplacian of $q/r$ is zero, except at the origin. This calculation you've done, is it valid at the origin too? Thanks. $\endgroup$
    – becko
    Apr 28, 2011 at 22:24
  • $\begingroup$ Yes, it is valid at the origin. It is also infinity but achieved "continuously". $\endgroup$ Apr 28, 2011 at 22:27
  • 2
    $\begingroup$ You can see that it's valid at the origin by taking your potential to be $q/(r+\epsilon)^2$ -- take the derivatives and see if anything bad happens when you take $\epsilon$ to 0. $\endgroup$
    – wsc
    Apr 28, 2011 at 22:35
  • 1
    $\begingroup$ @becko, what you're supposed to get at is that the laplacian of 1/r in spherical coordinates is sort of pathological. If you correctly took the derivatives, you should see that as $r\rightarrow 0$ the charge distribution becomes infinite (as it must!) -- if you next take $\epsilon$ to 0 then you get the delta function. For any nonzero $r$ you recover 0 when you kill $\epsilon$. This pathology simply is not there for the $1/r^2$ potential. $\endgroup$
    – wsc
    Apr 28, 2011 at 23:33

The operator to go from one potential to the other is.

$-\frac{\partial }{\partial r}\left\{\frac{q}{4\pi r}\right\} ~~=~~ \frac{q}{4\pi r^2}$

and therefor the source in the center which is given by the Laplacian of the new potential is obtained by using the same operator on the source of the original potential.

$-\frac{\partial }{\partial r}\Big\{\nabla^2\frac{q}{4\pi r} \Big\} ~~=~~ -\frac{\partial }{\partial r}\Big\{q\,\delta(r)\Big\}$

So the source charge would be the radial derivative of the delta function.

Regards, Hans

  • 5
    $\begingroup$ I think you're assuming that $\Delta$ commutes with $\frac{\partial }{\partial r}$, but they don't commute. $\endgroup$
    – becko
    Apr 28, 2011 at 22:59
  • $\begingroup$ @becko: The question was: Is there a charge distribution at the center which has a $q/r^2$ potential field. The solution I gave here has this potential field. $\endgroup$ Apr 28, 2011 at 23:09
  • $\begingroup$ The radial differentiated point charge doesn't for fill the requirements for a static solution but only as an initial condition. Other then for example the case of a dipole charge $\partial_z(\delta(r))$ which has a potential dipole field $\partial_z(1/r)=-z/r^3$ $\endgroup$ Apr 29, 2011 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.