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First of all, I am not sure of the terminology here, I am interested in the function $$F(x)=x|x|^{-d},x\in \mathbb{R}^d\setminus \{0\}$$ in dimension $d\geq 2$. I read somewhere that this is called the gravitational force (and I agree in dimension 3, but I am not sure for other dimensions), and elsewhere that this comes from the Coulomb potential.

Anyway, the divergence of this force is $-c_d\delta_0$ in the distributional sense (for some constant $c_d$), i.e. for any sufficiently smooth domain $D\subset \mathbb{R}^d$ not having $0$ in the boudnary and with normal vector $n_D$ on the boundary, $$ \int_{\partial D}n_D(x)\cdot F(x)S(dx)=-c_d1_{0\in D}. $$

First question: does anyone knows a good keyword to look upon this function and this kind of facts in the literature? Or a textbook?

Second question: consider now the force of mass spread uniformly over some smooth set $A\subset \mathbb{R}^d$: $$ F_A(x)=\int F(y-x)1_A(y)dy. $$ I'm pretty sure this is well defined, at least outside $\partial A$. Has the divergence of $F_A$ also a good interpretation in the distributional sense? Ultimately I would like to replace $1_A(y)dy$ by some finite measure $\mu(dy)$...

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  • $\begingroup$ $\text{div}\,F_A(x)=1_A(x)$, so well-defined for $x$ not on the boundary of $A$. $\endgroup$ Jan 28, 2020 at 15:31
  • $\begingroup$ That is surprising because if $A$ is a ball, then $F_A$ is the force of the corresponding "planet", and I think I remember from physics class that $F_A=0$ inside $A$, which would give $div F(x)=0$ for $x$ in the interior of $A$. I was more expecting some measure lying on $\partial A$... $\endgroup$ Jan 28, 2020 at 15:59
  • $\begingroup$ no, the force of gravity inside a planet is not zero, see, for example, Journey through the center of the Earth $\endgroup$ Jan 28, 2020 at 16:51
  • $\begingroup$ I see, thanks, I was mistaken between the mass on a sphere and the mass on a ball. When you say $div F_A=1_A$, this is true in any dimension? You think it extrapolates to any measure $\mu$? $\endgroup$ Jan 28, 2020 at 17:28
  • $\begingroup$ yest, it holds in any dimension, and for any measure $d\mu(y)=\rho(y)dy$, see answer box. $\endgroup$ Jan 28, 2020 at 18:23

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In $d$ dimensions, with a charge/mass density $\rho(\mathbf{x})$, we have \begin{align} &F(\mathbf{x})=\frac{1}{2-d}\,\text{grad}\,\frac{1}{|\mathbf{x}|^{d-2}},\;\;\text{div}\,F(\mathbf{x})=-S_{d}\,\delta(\mathbf{x}),\\ &F_\rho(\mathbf{x})=\int F(\mathbf{y}-\mathbf{x}) \rho(\mathbf{y})\,d\mathbf{y},\\ &\Rightarrow\text{div}\,F_\rho(\mathbf{x})=S_d\int\delta(\mathbf{y}-\mathbf{x}) \rho(\mathbf{y})\,d\mathbf{y}=S_d\,\rho(\mathbf{x}). \end{align} Here $S_d$ is the surface area of the $d$-dimensional unit sphere.

In the electrostatic context, $F_\rho$ is the electric field from an electrical charge density $\rho$, and the statement that the divergence of the electric field equals the charge density (times a constant) is the first Maxwell equation (the differential form of Gauss's law).

To more specifically answer the question in the OP: the divergence of the force field is discontinuous but finite at the boundary $\partial A$ of a uniform charge density in $A$.

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  • $\begingroup$ Ok thanks, but how do you justify switching $\int$ and $div$? $\endgroup$ Jan 28, 2020 at 20:27
  • $\begingroup$ Well, there is an issue here beause $F$ is not absolutely integrable $\endgroup$ Jan 29, 2020 at 8:53
  • $\begingroup$ I am probably just missing your point, but you are basically asking for Gauss' law of electrostatics, that the divergence of the electric field equals the charge density; I thought that relation was true even for singular charge densities, at least I have never seen it conditioned on smooth densities. $\endgroup$ Jan 29, 2020 at 9:06
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    $\begingroup$ Basically I agree with all you said, but I am looking maybe for a textbook doing things in a mathematically rigourous fashion in any dimension. I guess gravitation and electrostatics are more something that happen in dimension 3. $\endgroup$ Jan 29, 2020 at 9:10
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This is an addendum to Carlo's answer, about how to make it rigorous, as per the OP's comments.

Although one can probably do it by hand or brute force, the commutation of ${\rm div}$ and $\int$ is best shown using Fubini's Theorem for distributions as explained in my answer to Can distribution theory be developed Riemann-free?

Careful also about what ${\rm div}$ means. Inside the integral it is made of derivatives in the sense of distributions, while outside these are classical derivatives. In the notations used for my answer to the other MO question, $F_{\rho}$ is an element of $\mathscr{O}_{\rm M}$, the space of temperate smooth functions which embeds into the space of temperate distributions $\mathscr{S}'$. Moreover, one has a commutative diagram, with this embedding turning classical derivatives in $\mathscr{O}_{\rm M}$ into derivatives in the sense of distributions in $\mathscr{S}'$.

Summary:

In the first line of Carlo's answer, everything in sight (and in particular $F$ and ${\rm div}$) is to be understood in the sense of distributions. However, in the second line $F_{\rho}$ is to be understood as a classical function while $\int$ is in the sense of distributions (I prefer the notation $\langle\ ,\ \rangle_{\mathbf{y}}$). Finally in the third line ${\rm div}$ is first understood as a classical derivative in $\mathscr{O}_{\rm M}$, then in the sense of distributions in $\mathscr{S}'$, before commutation with the formal integral sign.

Remark: In the above I took $\rho$ in $\mathscr{S}$.

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