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Given a configuration $C$ of $N$ distinct fixed points of equal mass in the plane (eventually in space), let $f_C(N)$ denote the number of points $P$ for which the gravitational field at $P$ vanishes. The gravitational force is Newtonian, i.e $1/r^2$.

For example $f_C(2)=1$ for all $C$ and for a configuration $C$ of $3$ unequally spaced collinear points, $f_C(3)=2$.

Conjecture: $f_C(N)$ is always finite and nonzero.

Is this problem known? Can we establish an upper bound on $f_C(N)$?

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    $\begingroup$ one question is: what is the power law that you have in mind? It's like gravity in plane can mean both $g\sim \frac{1}{r^2}$ or $g\sim \frac{1}{r}$. In any event, the definining equations for the zero gravity locus are $d$ polynomials and at least generically there are finitely many points (in fact, for transversal case you can compute the number of points from Bezout theorem). Also could you specify what is your base field: $\mathbb{R}$ or $\mathbb{C}$? $\endgroup$
    – user74900
    Commented Oct 5, 2017 at 16:42
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    $\begingroup$ There is a trivial counterexample with two points and the line/plane of points that are equidistant from them. More generally, one expects the zero points to lie on surfaces of codimension 1. @AknazarKazhymurat, the zero points do satisfy a polynomial, but it is a polynomial in two or three variables, so it does not necessarily have finitely many zeros. $\endgroup$
    – Noah Stephens-Davidowitz
    Commented Oct 5, 2017 at 17:29
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    $\begingroup$ I think, notation $f(N)$ is misleading: this function depends not on $N$ only, but on configuration. $\endgroup$
    – Fedor Petrov
    Commented Oct 5, 2017 at 17:42
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    $\begingroup$ @NoahStephens-Dawidowitz the field is vector, so we have three polynomial equations in three variables, in general this should define a discrete set. $\endgroup$
    – Fedor Petrov
    Commented Oct 5, 2017 at 17:45
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    $\begingroup$ @NoahStephens-Davidowitz I'm afraid I don't understand your counterexample. If the masses are located at $(\pm 1,0)$, then the gravitational field at $(0,1)$ is a nonzero vector directed towards the origin, isn't it? $\endgroup$
    – Wojowu
    Commented Oct 5, 2017 at 19:17

1 Answer 1

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Since you are talking about gravitation (rather than electrostatics) I assume that all charges are positive. (With charges of different sign it is easy to arrange a whole curve of equilibrium points).

It is certainly non-zero if the number of points is at least 2. (From very general consideration it must be at least $N-1$ if you count multiplicitis properly. Without counting multiplicities it can be $1$ for any $N$).

But an exact upper estimate is not known. J. C. Maxwell conjectured that is is at most $(N-1)^2$, but this is wide open. Even finiteness is not known. I know only computer-assisted proof for $N=3$ but I have not checked it. If one assumes finiteness, then there are upper estimates but they are much worse than expected.

Reference: http://www.math.purdue.edu/~eremenko/dvi/equil2.pdf

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  • $\begingroup$ "Since you are talking about gravitation (rather than electrostatics) I assume that all charges are positive": isn't this statement a little shady? Given that gravitational forces are always attractive? $\endgroup$
    – Konstantinos Kanakoglou
    Commented Oct 5, 2017 at 23:32
  • $\begingroup$ @KonstantinosKanakoglou Depends on your definition of positive and the scalar potential. $\endgroup$
    – zibadawa timmy
    Commented Oct 6, 2017 at 3:08
  • $\begingroup$ @KonstantinosKanakoglou At its root, the question is just about Newtonian $r^{-2}$ potential fields; aside from charge, there's no real difference between gravity and EM fields in that regard. $\endgroup$
    – Steven Stadnicki
    Commented Oct 6, 2017 at 3:09
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    $\begingroup$ @AlexandreEremenko: Why is it obvious that the number of points is nonzero? For example even with $N=3$, if the configuration does not consist of 3 collinear points or an equilateral triangle, the existence of a point $P$ where the field vanishes is not obvious. $\endgroup$
    – math_lover
    Commented Oct 6, 2017 at 8:51
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    $\begingroup$ @JoshuaBenabou How about the following? If you consider a sufficiently large ball $B$ which contains all the points in its interior, it is clear that the gravitational field $F$ is continuous on $B$ and pointing into $B$ everywhere on its surface. Then for a sufficiently small $\varepsilon$, $x+\varepsilon F(x)$ is in $B$ for all $x\in B$. As an endomorphism of a convex compact subset, it has a fixpoint (where $F$ has to be $0$) by the Schauder fixpoint theorem. $\endgroup$
    – Klaus Draeger
    Commented Oct 6, 2017 at 13:00

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