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An electrostatic potential associated with some delocalized charge $\int \rho(\mathbf{r}) d{\mathbf{r}}$ is given by:

$$v_H(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$$

This potential is finite at $\mathbf{r}=0$. Since $\frac{1}{|\mathbf{r}-\mathbf{r'}|}$ is a solution to a singular Poisson's equation, we can show that:

$$\nabla^2v_H(\mathbf{r})=-4\pi\rho(\mathbf{r})$$ where $\rho$ is a smooth function being finite everywhere.

Would a function

$$v(\mathbf{r}) = \int \frac{\rho(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|^n}d\mathbf{r'}$$

(where $n$ is some nonnegative integer)

be also finite at $\mathbf{r}=0$?

In this case, $\frac{1}{|\mathbf{r}-\mathbf{r'}|^n}$ does not represent a solution to Poisson's equation ($\mathbf{r} \in \mathbb{R}^3$) and the above analysis is not valid.

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Since you said yourself, it does not represent a solution to Poisson's equation, why would be interested it these functions? – namehere Nov 30 '12 at 0:43
I am working in the field of quantum chemistry. I would like to use a model potential like the one I am asking about in electronic structure calculations. – molkee Nov 30 '12 at 0:49
I don't see why this would NOT be finite at $r=0$... – Dylan Sabulsky Nov 30 '12 at 2:52
Let me see if I can't make some plots for this. – Dylan Sabulsky Nov 30 '12 at 2:56
nope, the question was inspired by some problem in DFT – molkee Nov 30 '12 at 3:32
up vote 1 down vote accepted

Maybe too simple, and certainly not a general solution, but anyway here goes.

The simplest case is a spherical charge distribution:

$$\rho= \left\{ \begin{array}{cc} \rho_0 & r \le a \\ 0 & r>a \end{array} \right. $$ Then

$$ V(0) = \rho_0 \int_0^a \frac{4 \pi r^2}{r^n} dr = 4 \pi \rho_0 \int_0^a r^{2-n} dr $$


$$ 0<n<3 \qquad V(0) = \frac{4 \pi \rho_0}{3-n} a^{3-n} \text{ , finite}$$

$$ n=3 \qquad V(0) = 4 \pi \rho_0 \lim_{\epsilon\rightarrow 0} \ln \left(\frac{r}{\epsilon} \right) \rightarrow \infty $$

$$ n>3 \qquad V(0) = \frac{4 \pi \rho_0}{n-3} \lim_{\epsilon\rightarrow 0} \left(\frac{1}{\epsilon^{n-3}} - \frac{1}{a^{n-3}} \right) \rightarrow \infty $$

In all these cases, the denominator of the original integral, $|r-r'|^n$, grows without bound for large $r$, and the charge is localized around the origin, so

$$ \lim_{r \rightarrow \infty} V(r) =0 $$

So, $n=3$ divides the finite from the infinite cases. (Technically, I suppose this calc constitutes a counter-example for $n \ge 3$ by exhibiting an unbounded result for those cases).

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thank you very much for your help – molkee Nov 30 '12 at 17:22
@molkee: You're welcome. – Art Brown Nov 30 '12 at 17:30

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