In mathematics there are certain infinite sums that converge (are conditionally convergent) but the number they converge to depends on the ordering of the sum (not absolutely convergent). I reckon this goes under the name Riemann rearrangement theorem — that a conditionally convergent sum can be rearranged to sum to any real number $M$.

Now consider the following setup where filled circles denote positive unit charges and hollow circles denote negative unit charges. The line of charges is embedded in a three-dimensional space.

                                       enter image description here

In the above diagram I want to calculate the potential due to the surrounding charges. It comes naturally that the contributions to the potential at $\times$ due to the left and right side are equal. One can thus write the potential as a sum:

$$V = \frac{2}{4\pi\epsilon_0}\left(1-\frac{1}{2}+\frac{1}{3} - \frac{1}{4} +\cdots \right).$$

The sum in the bracket is conditionally convergent, but not absolutely convergent. This means that one can change the order in which the charges are summed to arrive at any real number $M$ that is the potential due to this geometric arrangement at $\times$.

I understand that potential is not a physical quantity — it is the potential difference between two points that matters. Now consider a point at infinity along a coordinate axis that is perpendicular to the axis formed by the line of charges. Is the potential there fixed by choosing the counting rule? I.e. is the boundary condition at infinity equivalent to choosing a counting rule for $V$?

Additionally, if this is not a well-behaved problem as originally posed — what are the axioms/requirements that need to be satisfied for an EM problem to be well-behaved?

  • $\begingroup$ Isn't there usually an issue for thinking about the potential at infinity for charge distributions that exist at infinity? $\endgroup$ – Aaron Stevens Sep 9 '19 at 15:18
  • $\begingroup$ There's no spherical symmetry in this problem, so "at infinity" is ambiguous. Are you standing infinitely far away from X, but 1 meter from the closest part of the wire? If so, your potential "at infinity" will be different than if you stood infinitely far away from X and 2 meters from the closest part of the wire (which is also "at infinity"). $\endgroup$ – probably_someone Sep 9 '19 at 15:53
  • $\begingroup$ @probablysomeone The OP already explains this: Now consider a point at infinity along a coordinate axis that is perpendicular to the axis formed by the line of charges. Start on the wire and move infinitely far away perpendicular to the wire. $\endgroup$ – Aaron Stevens Sep 9 '19 at 21:28
  • $\begingroup$ As Aaron already mentioned I put my reference point (zero potential) at infinity perpendicular to the infinite line of charge. I also specifically mention that this is done in three dimensions to assure that the electric field at that point drops to zero - therefore the potential has zero gradient at the reference point. $\endgroup$ – Akerai Sep 9 '19 at 21:46
  • $\begingroup$ @AaronStevens I am not certain if there is such issue - to be completely honest I've never done axiomatic classical electromagnetism, but I would be very curious to know. I am not shocked that the potential would be ill-defined at infinity along the line of charge, but I attempted to choose a reference point for the potential that would be well-behaved as commented upon above. $\endgroup$ – Akerai Sep 9 '19 at 21:47

The issue is that your infinite sum assumes that the potential of each point charge goes to $0$ at infinity. This is a problem for your series to absolutely converge, because the charge distribution itself extends to infinity. This is similar to (but not exactly the same thing) what happens when dealing with the infinite line charge.

Instead, it will be sufficient to first determine the electric field along a line through point $X$ perpendicular to the axis of charges. Due to symmetry, the field must point along this line, so we only need to add up the field components along the line. Therefore, starting with charge $n=1$ and moving along the line (not the same numbers as labeled in your figure. I suppose my $n=1$ would be your charge $5$) $$E_n(x)=\frac{k\,q_n}{r_n^2}\cos\theta_n=\frac{(-1)^{n-1}kq}{(an)^2+x^2}\cdot\frac{x}{\sqrt{(an)^2+x^2}}$$ Where $k=1/4\pi\epsilon_0$, $x$ is the distance along the line from point $X$, $q$ is the magnitude of a single charge, and $a$ is the distance between successive charges. We can make some simple cosmetic changes by saying $q=1$ and $a=1$: $$E_n(x) =\frac{(-1)^{n-1}k}{n^2+x^2}\cdot\frac{x}{\sqrt{n^2+x^2}}$$

The total field at point $x$ is then just double the sum over the field contributed to the charges on one half of the axis $$E(x)=\sum_{n=1}^\infty E_n(x)=2k\sum_{n=1}^\infty\frac{(-1)^{n-1}x}{(n^2+x^2)^{3/2}}$$

For large $n$ the terms go like $1/n^3$, so this series absolutely converges.

Now, let’s determine the potential at points along our line by setting $V(x_0)=0$ $$V(x)=-\int_{x_0}^xE(x’)\,\text dx’=-2k\int_{x_0}^x\sum_{n=1}^\infty\frac{(-1)^{n-1}x'}{(n^2+x'^2)^{3/2}}\,\text dx’$$

This is a simple integral to perform $$V(x)=2k\sum_{n=1}^\infty(-1)^{n-1}\cdot\left(\frac{1}{\sqrt{n^2+x^2}}-\frac{1}{\sqrt{ n^2+x_0^2}}\right)$$

This sum is actually absolutely convergent because for large $n$ the terms in the series $1/\sqrt{n^2+x^2}$ and $1/\sqrt{ n^2+x_0^2}$ end up canceling. Notice how this is only true when $x_0$ is finite. As soon as $x_0\to\infty$ this canceling no longer occurs, which is what we expected: we cannot say $V=0$ at infinity if we want an absolutely convergent series.

We can also determine the potential at point $X$ where $x=0$.

$$V(0)=2k\sum_{n=1}^\infty(-1)^{n-1}\left(\frac{1}{n}-\frac{1}{\sqrt{ n^2+x_0^2}}\right)$$

So, I think your issue with the order of adding terms is somewhat of a red herring. The sum is conditionally convergent due to the assumption of an infinite distribution of charge and potential $0$ at infinity. When you set $V=0$ at a finite distance from the line of charges then everything works out fine, and we don’t have the issue of conditional convergence.

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    $\begingroup$ This is wrong from the first paragraph. If you add a hollow circle at the point X in figure, you would have invariance under translations by $2 n a$. So this (modified) problem is equivalent to Poisson equation on $\mathbb{R}^2 \times S_1$, but unlike infinite line the total charge here is zero. So the potential at infinity of $\mathbb{R}^2$ is well defined, finite and could be taken as zero. So your problem is in assuming that the infinity must always be an infinitely distant sphere. $\endgroup$ – A.V.S. Sep 28 '19 at 4:47
  • $\begingroup$ @A.V.S. If you let $x_0\to\infty$ then you get a series that doesn't absolutely converge. This is the issue I was trying to fix. I agree, you do still get a convergent series when $x_0\to\infty$. I'll try to soften my language though. If you still think there is an issue, then I would love to see you make an answer showing the right way to think about this problem. $\endgroup$ – Aaron Stevens Sep 28 '19 at 12:17
  • $\begingroup$ I'm not sure if this answer is correct, but there are some ways to count dipoles in this question, and different ways will leave out different "net charges" (no net charge, or net charge 4, or net charge 5, or both net charges 4 & 5). Hence the answer does not seem to be unique to me. $\endgroup$ – Shing Sep 28 '19 at 14:35
  • $\begingroup$ from infinity, the potential to point X for one positive net charge surely different to zero net charge. $\endgroup$ – Shing Sep 28 '19 at 14:40
  • $\begingroup$ @Shing There isn't anything here that is incorrect. I calculated the field along the line I specify in the answer, then determine the potential along that line from where $V(x_0)=0$. And it produces a series that is not conditionally convergent. Just because there are other ways to view the problem doesn't mean this way is wrong. If I have made an error please point it out. All I was attempting to do was determine the potential in terms of an absolutely convergent series since the OP had an issue with a conditionally convergent series $\endgroup$ – Aaron Stevens Sep 28 '19 at 15:38

Real crystals like NaCl have surfaces that are neutral. The crystals are not composed of spherical shells. Instead they are cubic (or at least cuboid).

So to compute the Madelung sum in a way that converges, one needs to take shells that are physical. Then the only net charge is from the edges and corners. There number does not increase with the size of the shell, the distance does, so the series has decreasing terms with alternating sign. This will converge.

  • $\begingroup$ The text of your answer is irrelevant to the question. We are discussing here one dimensional charge distribution. So for either sphere or cube the results would be converging in the same way. $\endgroup$ – A.V.S. Sep 28 '19 at 5:02
  • $\begingroup$ @A.V.S. My point is that the question is not physical. This looks like a question about mathematics. $\endgroup$ – Pieter Sep 28 '19 at 18:05

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