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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

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

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?

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@Lagerbaer: thx for putting the nice equations... btw, what do I have to do to learn to do that by myself? –  becko Apr 28 '11 at 21:59
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Hi. The markup language used is LaTeX. You enclose inline formulae with single dollar signs and formulas that go on their own line with double dollar signs. I thought there was a FAQ on this somewhere on this site but currently I can't seem to find the link... I'll keep on searching. In the meantime, you can click edit on your post and look at what I've done. –  Lagerbaer Apr 28 '11 at 22:05
    
Here's a start: meta.physics.stackexchange.com/questions/529. There is probably better information on this posted on meta.stackoverflow.com. –  David Z Apr 29 '11 at 1:43

6 Answers 6

up vote 4 down vote accepted

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.

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Thanks. This is the sort of answer I usually expect when I ask a question at physics.stackexchange. Sometimes an intuitive and short argument can be very enlightening. –  becko Sep 1 '11 at 4:25

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 $r$!

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Can you also comment on whether such a charge distribution exists? –  Lagerbaer Apr 28 '11 at 22:16
    
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. –  becko Apr 28 '11 at 22:24
    
Yes, it is valid at the origin. It is also infinity but achieved "continuously". –  Vladimir Kalitvianski Apr 28 '11 at 22:27
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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. –  wsc Apr 28 '11 at 22:35
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@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. –  wsc Apr 28 '11 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

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I think you're assuming that $\Delta$ commutes with $\frac{\partial }{\partial r}$, but they don't commute. –  becko Apr 28 '11 at 22:59
    
@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. –  Hans de Vries Apr 28 '11 at 23:09
    
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 $\partial_z(1/r)=-z/r^3$ –  Hans de Vries Apr 29 '11 at 1:16

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.

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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. –  becko Apr 30 '11 at 13:15
    
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. –  becko Apr 30 '11 at 13:35

We put the permittivity $\varepsilon=1$ to one from now on. Let us first rephrase the question a bit. Instead of starting from the potential

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

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

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

respectively, and we want to know the charge density $\rho=\vec{\nabla} \cdot \vec{E}$, in particular, at the origin $r=0$. For $r\neq 0$, the charge density is

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

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 $1$.

A minor issue is that the electric field $\vec{E}$ has not been specified at $r=0$. 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

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

A Lebesgue majorant of the integrand is

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

respectively. Only the first case 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.

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, we can carry out the next step to define the derivative of $E^i$,

$$-\vec{\nabla} \cdot \vec{E}[f] = \int_{\mathbb{R}^3} d^3r \ \vec{E}(\vec{r})\cdot\vec{\nabla} f(\vec{r}) =\lim_{\varepsilon\to 0} \int_{\{r\geq\varepsilon\}} d^3r \ \vec{E}(\vec{r})\cdot\vec{\nabla} f(\vec{r}) $$ $$ =\lim_{\varepsilon\to 0} \int_{\{r\geq\varepsilon\}} d^3r \left[ \vec{\nabla} \cdot\left( \vec{E}(\vec{r})f(\vec{r})\right)-f(\vec{r}) \underbrace{\vec{\nabla} \cdot \vec{E}(\vec{r})}_{=0}\right] $$ $$ =-\lim_{\varepsilon\to 0} \int_{\{r=\varepsilon\}} d^2\vec{A} \cdot \vec{E}(\vec{r})f(\vec{r})=-\lim_{\varepsilon\to 0} \int_{\{r=\varepsilon\}} \frac{d^2A}{\varepsilon^2}f(\vec{r}) =-4\pi f(0), $$

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

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

Well, so much for distribution theory and mathematical idealization. In reality, the $\frac{-1}{r^4}$ charge density, $r\neq 0$, 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

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

respectively. The regularized charge density $\rho_{\varepsilon}=-\vec{\nabla}^2\Phi_{\varepsilon}\in C^{\infty}(\mathbb{R}^3)$ is

$$\rho_{\varepsilon}=\frac{3\varepsilon}{(r^2+\varepsilon)^{\frac{5}{2}}} \qquad \mathrm{and} \qquad \rho_{\varepsilon}=\frac{3\varepsilon-r^2}{(r^2+\varepsilon)^{3}}, \qquad \varepsilon>0,$$

respectively. We can now smear with a test function $f$. It is straightforward to check that the first $\rho_{\varepsilon}$ becomes $4\pi\delta^3(\vec{r})$ as $\varepsilon\to0$, while the second $\rho_{\varepsilon}$ does not make sense as a distribution when $\varepsilon\to0$.


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

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

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

$$ \vec{E}[f] = \Phi[\vec{\nabla} f] = \int_{\mathbb{R}^3} d^3r \ \Phi(\vec{r})\vec{\nabla}f(\vec{r}),$$

so that

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

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 part, we essentially get back to the problem that the electric field $\vec{E}= \frac{\vec{r}}{r^4}$, $r\neq 0$, is not locally Lebesgue integrable in $\mathbb{R}^3$.

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This answer is too formal! –  Ron Maimon Sep 1 '11 at 3:06
    
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. –  Qmechanic Sep 1 '11 at 18:32
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+1: Some readers find posts like this useful! –  joshphysics Jan 29 '13 at 16:38

I like the following approach to these kind of questions. Since we are dealing with the functions defined on $r \in [0,\infty>$, we must be aware that there might be a problem with differentiation at $r=0$. Therefore, instead of calculating the derivative of $f(r)$ and worrying about these kind of problems, why not extend the function on the domain $r \in <-\infty, \infty>$ and differentiate $f(r) \theta(r)$ with $\theta(r)$ being the Heaviside step function? In this way one can have a explicit overview of all the singularities. It can be easily shown that $$\nabla^2 \left((f(r) \theta(r)\right)) = \left(\frac{\partial^2 f(r)}{\partial r^2} + \frac{2}{r} \frac{\partial f(r)}{\partial r}\right)\theta(r) + 2\left(\frac{\partial f(r)}{\partial r} + \frac{1}{r} f(r)\right)\delta(r) + f(r) \delta'(r)$$ Therefore, $$\nabla^2 \frac{\theta(r)}{r^2} = \frac{4}{r^2} \theta(r) - \frac{2}{r^3} \delta(r) + \frac{1}{r^4} \delta'(r)$$ Indeed, the singularity is not integrable.

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That's a nice approach! –  Emilio Pisanty Feb 21 '13 at 17:36
    
+1 It's always good to see new ideas! –  becko Feb 21 '13 at 20:20
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The approach described in this answer (v1) seems to also wrongly predict that the Laplace operator $\nabla^{2} $ applied to the constant function $f(r)\equiv 1$ would be singular at $r=0$, whereas the correct result is, of course, that $\nabla^{2}1\equiv0$ is as smooth and regular as it can be, namely zero. –  Qmechanic Mar 8 '13 at 22:50

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