Consider a potential for charge distribution:

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

where $\rho(\mathbf{r'})$ is the charge density.

This potential is known to be finite at any $\textbf{r}$.

How to prove it analytically just based on its definition?

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 any $\mathbf{r}$?


Kohn-Sham DFT is surely way too advanced, quantum chemistry, terminology for the simple question you asked. This potential for a charge distribution is about the basic electrostatics.

When you just calculate $\Delta v_H(\vec r)$ for your $v_H$, you will get $4\pi \rho(\vec r')$ which is right because it's the right Poisson equation for the potential (a combination of ${\rm div} \vec D = \rho$ and $\vec D = \epsilon \nabla v_H$). This boils down to the fact that $$ \Delta \frac{1}{|\vec r|} = 4\pi\delta^{(3)}(\vec r) $$ in three dimensions. In particular, if you compute the value of the Laplacian of $1/r$ for any point $\vec r\neq 0$, you get zero. The right delta-function, the distribution-related difficult part of the calculation, may be calculated via Gauss' law.

The formula above couldn't be true for any other power of $r$, of course, so any modification (more precisely, messing up with) the formula would obviously produce a wrong one.

In $D$ spatial dimensions, $1/r$ would be replaced by $1/r^{D-2}$, up to an overall coefficient that has to be substituted correctly either to the Poisson equation or to the definition of $v_H$. The Laplacian of this right power of $r$ is against vanishing everywhere except for the origin. Functions like this potential are called "harmonic functions".

The potentials are finite simply because far away from any charge distribution, the charge distribution may be expanded into multipole expansions. In the leading approximation, the charge distribution may be replaced by the overall charge $Q=\int \rho\,d^3 x$ located at $\vec r=0$ and the potential from this simple charge distribution clearly goes like $Q/r$ or, in $D$ spatial dimensions, $Q/r^{D-2}$ which is convergent. Even with different exponents, you could get convergent integrals for localized charge distribution but random formulae like that would obviously be invalid equations for the electrostatic potential.

  • $\begingroup$ yep, the question is in fact much simpler. edited the post. $\endgroup$
    – molkee
    Nov 21 '12 at 17:02

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