Math-1920 · mathematics · Analysis

Math-1920

Equilibrium for infinitely many positive masses

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Fix a dimension $n\geq2$. Let $(x_k)_{k\geq1}$ be an infinite sequence of distinct points of $\mathbb R^n$ with no finite accumulation point, and let $a_k>0$ be their masses. Assume the convergence condition \[ \sum_{k=1}^{\infty} \frac{a_k}{(1+|x_k|)^{n-1}}<\infty. \] The use of $1+|x_k|$ allows a mass at the origin; outside a fixed ball this is equivalent to the denominator in the source formulation.

For $x$ outside the set of masses define the force \[ F(x)=\sum_{k=1}^{\infty} \frac{a_k(x_k-x)}{|x-x_k|^n}. \] The displayed summability condition makes this vector series absolutely and locally uniformly convergent away from the mass points: its tail is bounded by a constant multiple of $\sum_k a_k/|x_k|^{n-1}$ on each fixed compact set.

Must there exist a point $x\in\mathbb R^n\setminus\{x_k:k\geq1\}$ with $F(x)=0$? Such a point is an equilibrium of the logarithmic force when $n=2$ and of the Newtonian force when $n\geq3$. The question concerns an infinite discrete configuration and allows masses tending to zero, arbitrary geometry of their positions, and infinite total mass whenever the weighted sum converges. No uniform positive lower bound on the $a_k$ and no stronger convergence assumption for an unrenormalized potential is imposed.

Short English statement

Must the convergent force field of infinitely many positive, locally discrete masses have an equilibrium point?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 0

No attempts yet. Significant partial results and substantial reductions are welcome. Contribute an attempt.