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So, taking the quite well known identity: \begin{equation} \nabla\cdot\left(\frac{\mathfrak{\hat r }}{\mathfrak{r}^2}\right)=4\pi\delta^3(\mathfrak{r}) \end{equation} $\mathfrak{r}=||r-r'||$ being the separation vector, which is widely used in physical contexts. The thing is, I see, sometimes, people using it as: \begin{equation} \nabla\cdot\left(\frac{\mathfrak{\hat r }}{\mathfrak{r}^n}\right)=4\pi\delta^3(\mathfrak{r}) \end{equation} Letting $n\in\mathbb{Z}_{\geq2}$. One example is this proof of Gauss's law for Gravity, with $n=3.$ I never saw a detailed demonstration of this relation, though. Could it be simply stated that $n$ is, informally, a "scaling factor" of the vector field, and therefore there's no reason to believe that the divergence will be somewhat different? I'm not quite sure.

That's pretty much it, any response will be appreciated. Tks in advance!

Johann Wagner
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I don't see why the identity should hold for all $n\geq 2$. It is only for $n=2$. The Wikipedia page you linked to is equivalent to the first identity. Perhaps you're mixing up the normalizations. Let $\mathbf{r},\mathbf{r}'\in\Bbb{R}^3$, and define $\mathbf{R}=\mathbf{r}-\mathbf{r}'$ (I don't want to use that weird $\mathfrak{r}$ because it's not easy to bold, so the vector, the normalized vector, and the norm visually aren't distinguished). Let $R=\|\mathbf{R}\|$ be the Euclidean norm of this vector, and for non-zero $\mathbf{R}$, we define $\widehat{\mathbf{R}}:=\frac{\mathbf{R}}{R}$ the normalized vector.

The first identity in this notation says $\nabla\cdot\left(\frac{\widehat{\mathbf{R}}}{R^2}\right)=4\pi \delta^{(3)}(\mathbf{R})$. Or equivalently, $\nabla\cdot\left(\frac{\mathbf{R}}{R^3}\right)=4\pi \delta^{(3)}(\mathbf{R})$ (notice there's no hat's in this equation). This is the way the Wikipedia page is using the identity.

To repeat again: your first equation and the Wikipedia page are using exactly the same fact, expressed in different notation (you with the unit vector, Wikipedia doesn't use unit vectors).

As for detailed demonstrations of this relation, see this answer (you obviously first need a rigorous definition before you can provide a proof).

peek-a-boo
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The well-known identity you start with looks wrong.

With $$ \mathfrak{\hat r}=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}\quad\text{ and }\quad\quad \mathfrak r^2=x_1^2+x_2^2+x_3^2 \quad\text{ and }\quad\quad \mathfrak r^3=(x_1^2+x_2^2+x_3^2)^{3/2} $$ I get \begin{align} \frac{\partial}{\partial x_i}\left(\frac{\mathfrak{\hat r }}{\mathfrak{r}^2}\right)=\frac{1}{\mathfrak r^2}-\frac{2\,x_i^2}{(x_1^2+x_2^2+x_3^2)^2}=\frac{1}{\mathfrak r^2}-\frac{2\,x_i^2}{\mathfrak r^4} \end{align} which does not lead to a divergence of zero away from the origin. In fact, it gives $$ \nabla\cdot\left(\frac{\mathfrak{\hat r }}{\mathfrak{r}^2}\right)=\frac{3}{\mathfrak r^2}-\frac{2}{\mathfrak r^2}=\frac{1}{\mathfrak r^2}\,. $$

In contrast, \begin{align} \frac{\partial}{\partial x_i}\left(\frac{\mathfrak{\hat r }}{\mathfrak{r}^3}\right)=\frac{1}{\mathfrak r^3}-\frac{3\,x_i^2}{(x_1^2+x_2^2+x_3^2)^{5/2}}=\frac{1}{\mathfrak r^3}-\frac{3\,x_i^2}{\mathfrak r^{5}} \end{align} which leads to $$ \nabla\cdot\left(\frac{\mathfrak{\hat r }}{\mathfrak{r}^3}\right)=\frac{3}{\mathfrak r^3}-\frac{3\,\mathfrak r^2}{\mathfrak r^5}=0\quad\text{ for }\mathfrak r\not=0\,. $$

Kurt G.
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  • The starting identity is correct; you seem to have misunderstood the notation (the hat means the normalized vector) and hence obtained a non-zero divergence. – peek-a-boo Mar 28 '22 at 18:03
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    @peek-a-boo . Correct. I assume (as I wrote) that $\hat{\mathfrak{r}}$ is not normalized. Thanks for pointing that out. Since almost two weeks I consider it as the duty of OP to make some comments on that. – Kurt G. Mar 28 '22 at 18:40