Counting infinite sums with charges

Question

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.

                                       

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?

Answer 1

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.