"long-range" and "singular" electrostatic potential?

[Glxblt76]

EDIT: Problem is FIXED.

Hello,

I'm trying to understand Ewald Summation and finally found a great link (http://micro.stanford.edu/mediawiki/images/4/46/Ewald_notes.pdf) that I could follow in the five first pages. But then I'm blocked by a rather odd formulation p. 5, after eq. (25):

"where erfc(z) ≡ 1 − erf(z). Because limz→∞ erf(z) = 1, we know that φ L i (r) is a long-range nonsingular potential and φ S i (r) is a short-range singular potential. (In comparison, the Coulomb potential of a point charge is both long-ranged and singular.) Given this result, we also have [...]"

Since the demonstration is based on this sentence, it's impossible for me to actually represent the next argument, and therefore I'm stuck at this point. I can't figure out what the following idioms actually mean:
- long-range nonsingular potential
- short-range singular potential
- long-ranged and singular potential

Is there a mathematical criterion for making the difference between short-range and long-range?
What does exactly mean "singular"? The wikipedia article on singular functions (https://en.wikipedia.org/wiki/Singular_function) does not help me at all to figure out the physical representation and meaning of this property, neither its useful mathematical implications for the demonstration at hand.

Could someone help me understand this and how it leads to eq. 26?

Thanks in advance.

 

[Delta2]

Cant help you with the singular term, but about long ranged and short ranged refers to how the potential survives at long distances (large r).

it is $erfc(z)\approx 0$ for large z so the potential $\phi_i^S(r)\approx 0$ is approximately zero for large r, while the long range potential is approximately $\phi_i^L(r)\approx \frac{1}{4\pi\epsilon_0}\frac{q_i}{|r-r_i|}$. Ofcourse for very large r, $r\to\infty$ both potentials become zero.

Another way to state this is that the sort ranged potential "vanishes faster" than then long ranged potential as $r\to\infty$

 

[vela]

$\phi_i^S(r)$ has a singularity at $r=0$. $\phi_i^L(r)$, on the other hand, doesn't because $\operatorname{erf}(z)$ is equal to $2z/\sqrt{\pi}$ to first order.

 

[Glxblt76]

Thanks for your responses. This helped me.