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I need to have a scalar function $\psi(\mathbf{r}, \sigma)$ of the coordinate $\mathbf{r}$ in euclidean 3-D space, which depends on a scalar adjustment parameter $\sigma$, such that in the limit $\sigma \to 0$ the $\psi$ function becomes the Coulomb potential of an electric charge on the origin,

$$\phi(\mathbf{r}) = \dfrac{C}{|\mathbf{r}|},$$

where $C$ is a constant that depends on the charge of the particle. So that

$$\lim_{|\mathbf{r}| \to \, 0} \psi{(\mathbf{r}, \sigma)} = \psi{(0, \sigma)}$$

for $\sigma > 0$, $\psi$ has derivatives of all orders everywhere, and

$$\lim_{\sigma \to \, 0} \psi{(\mathbf{r}, \sigma)} = \phi(\mathbf{r}).$$

That is, a function which shows Coulomb potential behaviour away from the origin, but is continuous at the origin, infinitely differentiable everywhere, and that can be made arbitrarily close to the Coulomb potential.

What are the smooth forms that have this property?

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  • $\begingroup$ Lots of options exist. A simple case is $\psi(r,\sigma)=\begin{cases} \phi(r) & r>\sigma \\ \phi(\sigma) & r \leq \sigma \end{cases}$. $\endgroup$
    – Ian
    Commented Jan 20, 2022 at 1:55
  • $\begingroup$ Yes, of course, I need to restate the problem. $\endgroup$
    – Arc
    Commented Jan 20, 2022 at 1:58
  • $\begingroup$ $\phi(r) = C/(\vert r\vert + a), a > 0$. $\endgroup$ Commented Jan 20, 2022 at 2:03
  • $\begingroup$ @Aruralreader, thanks for the comment, but sorry I phrased it incorrectly. I have restated the question. $\endgroup$
    – Arc
    Commented Jan 20, 2022 at 2:07
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    $\begingroup$ Still a lot of options, but harder to write them down explicitly. You may want to look up the notion of a mollifier. $\endgroup$
    – Ian
    Commented Jan 20, 2022 at 2:29

1 Answer 1

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While following the suggestions in the comments to my question, I found the soft-core Coulomb interaction:

$$\phi_{soft}(\mathbf{r},\sigma) = \dfrac{C}{(\mathbf{r}^2 + \sigma^2)^{1/2}}$$

which is obviously smooth everywhere and recovers the original Coulomb interaction in the limit $\sigma \to 0$.

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