Given a vector field $f:\mathbb R^3\to\mathbb R^3$, is there a mass distribution that generates $f$ as its Newtonian gravity?

Question

Let $f:\mathbb R^3\to\mathbb R^3$ be a smooth bounded vector field. I want to produce a density $\rho:\mathbb R^3\to\mathbb R$ such that the Newtonian acceleration experienced by a particle in the space $\mathbb R^3$ is given by $f$. Actually it's better to let $\rho$ be a real valued measure on $\mathbb R^3$, and it should be signed to avoid silly things like a negative mass pushing things away from it being impossible. So the problem becomes: given $f$, find $\rho$ such that $$f(x)=\int_{\mathbb R^3} \frac{v-x}{\lvert v-x\rvert^3} d\rho(v).$$ Now two obvious observations:

1. The average of $f$ vanishes: $$\int_{\mathbb R^3}f(x)dx=\int_{\mathbb R^3}\int_{\mathbb R^3}\frac{v-x}{\lvert v-x\rvert^3} dxd\rho(v)=0.$$ This manipulation isn't entirely justified, but it feels like a distribution that would make it invalid would be so weird to miss the spirit of the question.

2. Adding a constant to $\rho$ does not change $f$. This actually only makes a difference in the formulation using $\rho$ as a function instead of a measure.

(As pointed out in the comments, we also have to assume that $f$ is irrotational. This raises an interesting question as to what happens in higher dimensions...)

So the question is: given the acceleration $f$, with average equal to $0$, is there always a measure $\rho$ that has $f$ as its Newtonian acceleration? Is it unique up to adding a constant?

Answer 1

The equation $$ f(x)=\int_{\mathbb R^3} \frac{v-x}{\lvert v-x\rvert^3} d\rho(v) $$ can be written as $f = G*\rho,$ where $G(x) = \frac{x}{|x|^3}$ and $*$ denotes convolution.

Since $\nabla\cdot G = 4\pi\delta,$ we get $$ \nabla\cdot f = \nabla\cdot(G*\rho) = (\nabla\cdot G)*\rho = (4\pi\delta)*\rho = 4\pi(\delta*\rho) = 4\pi\rho. $$ Thus we should just take $$ \rho = \frac{1}{4\pi}\nabla\cdot f = \frac{1}{4\pi}\operatorname{div}f, $$ where the divergence is taken in a distributional sense.

However, $f$ and $G*\rho$ might differ by a divergence free term. If $\nabla\cdot f=0$ outside some bounded region but $f$ does not decay as $|x|^{-2}$ when $|x|\to\infty$ then it's clear that $f$ has a non-vanishing divergence free term and therefore cannot be produced as $f=G*\rho.$ When $\nabla\cdot f$ does not vanish outside some bounded region, I don't know how recognize the divergence free term.

Also, the divergence might be both signed and have non-measure terms like $\nabla\delta=\operatorname{grad}\delta$ so we need to check that it is a (Radon) measure.

Answer 2

The problem can be solved using the Laplace transform and the residues theorem.

To do this, we first take the Laplace transform of both sides of the equation $f(x) = \int_{\mathbb{R}^3} \frac{v - x}{|v - x|^3} d\rho(v)$:

$F(s) = \int_0^{\infty} f(x) e^{-sx} dx = \int_0^{\infty} \int_{\mathbb{R}^3} \frac{v - x}{|v - x|^3} d\rho(v) e^{-sx} dx$

We can interchange the order of integration on the right-hand side:

$ F(s) = \int_{\mathbb{R}^3} d\rho(v) \int_0^{\infty} \frac{v - x}{|v - x|^3} e^{-sx} dx $

The inner integral can be evaluated using the following identity:

$ \int_0^{\infty} e^{-sx} dx = \frac{1}{s} $

This gives us:

$ F(s) = \int_{\mathbb{R}^3} d\rho(v) \frac{v - x}{|v - x|^3} \frac{1}{s} $

We can now use the residues theorem to evaluate the integral on the right-hand side. The poles of the integrand are at $s = 0$ and $s = \pm\infty$. The residue at $s = 0$ is given by:

$ \lim_{s \to 0} (s - 0) F(s) = \int_{\mathbb{R}^3} d\rho(v) \frac{v - x}{|v - x|^3} $

This is the desired result. The residue at $s = \pm\infty$ is given by:

$ \lim_{s \to \pm\infty} (s - \pm\infty) F(s) = 0 $

This is because the integrand decays to zero faster than $1/s$ as $s$ goes to infinity.

Therefore, the measure $\rho$ that satisfies $f(x) = \int_{\mathbb{R}^3} \frac{v - x}{|v - x|^3} d\rho(v)$ is given by:

$\rho(v) = \lim_{s \to 0} (s - 0) F(s)$

where $F(s)$ is the Laplace transform of $f$.

Conclusion

We have shown that there exists a unique measure $\rho$ up to adding a constant such that $f(x) = \int_{\mathbb{R}^3} \frac{v - x}{|v - x|^3} d\rho(v)$. This measure is given by $\rho(v) = \lim_{s \to 0} (s - 0) F(s)$, where $F(s)$ is the Laplace transform of $f$.