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This is a question triggered by this post

Madelung's constant is defined to the coefficient of electrostatic potential energy in a ionic crystal. In the example of $NaCl$, \begin{equation} M = \sum_{ijk}{}^{'}\frac{(-1)^{i+j+k}}{\sqrt{i^2+j^2+k^2}} \end{equation} is conditionally convergent.

Since this sum is conditionally convergent it is not suitable as definition of Madelung's constant unless the order of summation is also specified. There are two "obvious" methods of summing this series, by expanding cubes or expanding spheres. The latter, though devoid of a meaningful physical interpretation (there are no spherical crystals) is rather popular because of its simplicity. Thus, the following expansion is often found in the literature:[2] $$ M = -6 +12/ \sqrt{2} -8/ \sqrt{3} +6/2 - 24/ \sqrt{5} + \dotsb = -1.74756\dots. $$ However, this is wrong as this series diverges as was shown by Emersleben in 1951.[3][4] The summation over expanding cubes converges to the correct value. An unambiguous mathematical definition is given by Borwein, Borwein and Taylor by means of analytic continuation of an absolutely convergent series.

I have the following questions.

1) Expanding sphere leads to a divergent series. OK, what if I make a perfect spherical sample of $NaCl$, will the experimentally measured Madelung's constant to be infinity?

(EDIT: I maybe have some misunderstanding of divergence: it could be the case that the series is bounded but don't have a definite limit. So is the series for expanding sphere bounded? and if there is no definite limit, what's the experimentally measured value for a perfect spherical crystal?)

2) What physical principle dictates the order of summation? or why does finite number obtained by analytic continuation should be consistent with the observed value?

3) What is the role of charge neutrality here? I ever programed to compute Madelung's constant using the fractional charge idea((assign $\frac{1}{8}$ charge to the corner, $\frac{1}{4}$ to the edge, $\frac{1}{2}$ to the face ), which makes the expanding cube charge neutral. Does that mean charge neutrality is one of conditions that must be enforced? or it is just for the sake of computational efficiency?

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    $\begingroup$ I find the Wikipedia remark very strange: The latter, though devoid of a meaningful physical interpretation (there are no spherical crystals) is rather popular because of its simplicity. I think the physical interpretation is very clear: you order the contribution to the electrostatic potential of ions in the lattice by the distance to a given ion. On the other hand, I don't think this way of evaluating the sum is simple at all: it involves counting the number of ways a number can be written as a sum of three squares. $\endgroup$ – doetoe Jan 11 '14 at 2:57
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    $\begingroup$ Another comment: that the series diverges, doesn't necessarily mean that it becomes arbitrarily small or large, it simply means that it doesn't stay arbitrarily close to any number, like the sequence 0,1,0,1,0,1,... $\endgroup$ – doetoe Jan 11 '14 at 3:00
  • $\begingroup$ @doetoe, thanks for the clarification. So expanding sphere leads to a series with partial sum to be bounded, but not convergent? $\endgroup$ – anecdote Jan 11 '14 at 3:35
  • $\begingroup$ I think it could be quite difficult to show that of if the partial sums remain bounded (I'm not even sure they do). To bound the terms, you would need a tight enough asymptotic bound on the number of ways a number can be written as a sum of three cubes or (if this is not possible) a way to bound the difference between the number of ways to do this with $i+j+k$ even and those with $i+j+k$ odd. $\endgroup$ – doetoe Jan 11 '14 at 10:46
  • $\begingroup$ 1) Experimentally, NaCl sample has finite number of ions. Therefore the summation of series must be finite.. $\endgroup$ – user26143 Jan 11 '14 at 15:50
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This is not a rigorous answer to OP's questions, as we have not done any calculations or estimates.

The Madelung constant for the $NaCl$ crystal (of infinite size) is formally the sum

$$\tag{1} \sum_{\vec{r}\in \mathbb{Z}^3\backslash\{\vec{0}\}}\frac{(-1)^{x+y+z}}{||\vec{r}||_2} .$$

Here the $p$-norm is defined as

$$\tag{2} || \vec{r} ||_p ~:=~ \sqrt[p]{|x|^p+|y|^p+|z|^p}, \qquad || \vec{r} ||_{\infty}~:=~\max(|x|,|y|,|z|). $$

The procedure of arbitrarily truncating the summation (1) after completing a cube or ball of a certain size $a$ (possibly with an extra condition of only allowing electrically neutral sizes $a$), and then letting the size $a\to \infty$, is from a physical perspective rather crude.

Such truncation method seems ripe for systematic errors. If such truncation method happens to produce the correct answer, it seems more like an accident than a reliable scientific method.

Intuitively, it seems better to introduce a sufficiently smooth (or at least continuous) regulator function $\eta:[0,\infty[\to [0,\infty[$ with

$$\tag{3} \eta(0)~=~1\qquad \text{and}\qquad \lim_{r\to\infty}\eta(r)~=~0,$$

next consider the sum

$$\tag{4} \sum_{\vec{r}\in \mathbb{Z}^3\backslash\{\vec{0}\}}\frac{(-1)^{x+y+z}}{||\vec{r}||_2}\eta( ||\vec{r}||_p ~\varepsilon), $$

and finally removing the regulator $\varepsilon\searrow 0^{+}$.

In the spirit of Terence Tao blog entry, one would expect that the regularized sum (1) would not depend on the regulator function $\eta$ (or the $p\geq 1$ in the $p$-norm) once the regulator function $\eta$ belongs to a sufficiently nice class, and that such definition would agree with the (somewhat un-intuitive) definition by Borwein, Borwein and Taylor via analytic continuation.

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  • $\begingroup$ it's nice to see that the result is regulator independent. So as a rule of thumb, I should always apply a regulator whenever I have conditionally convergent/divergent series. Yet, I still believe regularization implies some physical process(or some measurement process) that's missing in the definition of the series. $\endgroup$ – anecdote Jan 12 '14 at 14:50

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