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I have tough problem I am not sure how to solve:

For this question, we are confined to a plane. Consider a gravitational field that is proportional to $\frac{1}{r^3}$ instead of $\frac{1}{r^2}$, and consider its potential which is then proportional to $\frac{1}{r^2}$. Suppose that I put $n$ identical point masses in this plane, all at different locations and let $U(x,y)$ be the potential function.

What can we say about the critical points of $U(x,y)$? Specifically, can we show that the number of critical points is always $\leq n$?

Thank you for your help!

Remark: By critical point, I mean the usual definition in vector calculus, that is a place where both partial derivatives are zero.

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So you are confined to a plane that has a gravitation field pointing to $(0,0)$ in that plane with a magnitude of $1/r^3$? I think people will be confused by the mention of a "plane", thinking that this is a planar source. Could you be more clear about the geometry? –  Alan Rominger Oct 31 '11 at 0:50
    
@Zassounotsukushi: I don't understand your question at all. Imagine you put $n$ point masses at distinct locations on a 2 dimensional plane. Think of the gravitational potential function. Now imagine that for a single mass, it decayed like $\frac{1}{r^2}$ instead of $\frac{1}{r}$. –  EPN Oct 31 '11 at 15:28
    
and $r$ is defined how? I could say $r$ is referenced to a point outside that plane, and that's not quite going to achieve your intention, but I don't think this is clear in your question. My comment is mostly establishing that if you're referencing some $U(x,y)$ then the $r$ you refer to is probably $\sqrt{x^2+y^2}$. Someone reading your question might see that this is probably the case. –  Alan Rominger Oct 31 '11 at 15:46
    
@Zassounotsukushi: Yes, I mean that if there was a single point mass at the point $(x_0,y_0)$ then $$U(x,y)=\frac{C}{(x-x_0)^2+(y-y_0)^2}.$$ ($C$ is some constant that doesn't matter. Might as well make it one) –  EPN Oct 31 '11 at 18:17

2 Answers 2

up vote 4 down vote accepted

Take 4 points arranged in a square. The middle of the square is a critical point by symmetry, and the midpoint of the four sides of the square would be a critical point just for the two vertices it joins together, ignoring the other two. But the other two are little more than twice as far away, so eight times weaker force, so if you bring the point closer to the center of the square by an amount about 1/8 of the way to the other side, you will cancel out the force from the far pair from the force on the near pair. So there are 5 critical points for four points on a square, and this holds for all sufficiently fast-falling-off forces.

Using squares-of-squares, I believe it is easy to establish that the number of critical points is generically $n^2$. I believe it is a difficult and interesting mathematical problem to establish any sort of nontrivial bound on the number of critical points. The trivial bound is from the order of the polynomial equation you get, and it's absurdly large---- it grows like $2^n$.

EDIT: The correct growth rate

The answer for a general configuration is almost certainly N+C critical points (this should be an upper bound and a lower bound for two different C's --- I didn't prove it, but I have a, perhaps crappy, heuristic). For the case of polygons, there are N+1 critical points. For squares where the corners are expanded to squares, and so on fractally, the number of critical points is N+C where C is a small explicit constant. The same for poygons of polygons.

I found a nifty way of analyzing the problem, and getting some good estimates, but I want to know how good the mathematicians are at this before telling the answer. Perhaps you can ask this question on MathOverflow?

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1  
I don't think that this is true. I like the reasoning, but can you provide a proof of what you claim? I somehow suspect that for a square there is only exactly one critical point, and that it is at the center of the square. Please be careful! –  EPN Oct 31 '11 at 15:25
    
@Eric: there are five critical points--- you can easily prove it by the fact that there are three critical points along the vertical bisector of the square. Critical points are topological, and it takes a big perturbation to get rid of them. For the critical point half-way between the two vertices of the square, the other two vertices are too far to get rid of the critical point, they just move it about 1/6 of the way to the center. –  Ron Maimon Oct 31 '11 at 19:24
    
I believe you now since I checked via computer algebra. It gave $(\sqrt{2(\sqrt{2}-1)},0)\approx (0.91,0)$ as another location. (And of course the 3 other points given by symmetry) I see why it follows if there are 3 critical points on the vertical bisector, but why must this be true in the first place? –  EPN Oct 31 '11 at 20:25
2  
Seeing a plot of the potential at an axis equidistant from the sources did the trick for me. –  mmc Oct 31 '11 at 22:11
    
@Eric: the 3 points are easy because the force is projected to 1d by symmetry, and you can count the zeros by the number of sign changes. The force is clearly down on the edge of the square, and clearly transitions to up, then to zero, then to down, then to up again. The general problem of giving a polynomial bound for the number of critical points (which I am sure exists) seems really interesting. There are topological methods, which go by the winding number of the vector field, and there are variety methods because the problem is algebraic, but mathoverflow might be best here. –  Ron Maimon Nov 1 '11 at 1:42

In this answer we analyze a generalization of OP's question. In the last section 6 we will argue heuristically that one should expect an upper bound of critical points of the form

$$c(M)~\leq~~5n-11 \qquad {\rm for} \qquad n\geq 3. $$

1) Let us identify the plane $\mathbb{R}^2\cong \mathbb{C}$ with the complex plane $z=x+\mathrm{i} y$. Let

$$Z:=\{z_1, \ldots, z_n\}\subseteq \mathbb{C}$$

be a set of $n$ different punctures in the complex plane, where $n\in\mathbb{N}$. Consider the punctured complex plane $M:=\mathbb{C} \backslash Z$ and the Riemann sphere $S^2:=\mathbb{C} \cup \{\infty\}$ with Betti numbers

$$b_0(M)~=~1, \qquad b_1(M)~=~n, \qquad b_2(M)~=~0,$$ $$b_0(S^2)~=~1, \qquad b_1(S^2)~=~0, \qquad b_2(S^2)~=~1,$$

and Euler characteristics

$$\chi(M)~=~b_0(M)-b_1(M)+b_2(M)~=~ 1-n, $$ $$\chi(S^2)~=~b_0(S^2)-b_1(S^2)+b_2(S^2)~=~ 2, $$

respectively.

2) Let $p>0$ and $k_1, \ldots, k_n >0$ be $n+1$ positive constants. Let the potential $V:M \to]0,\infty[$ and its extension $\tilde{V}: S^2 \to [0,\infty]$ be

$$V(z)~:=~\sum_{i=1}^n \frac{k_i}{p|z-z_i|^p},$$

$$ \tilde{V}(z)~:=~\left\{ \begin{array}{rcl} V(z) &{\rm for}& z \in M, \cr +\infty &{\rm for}& z \in Z, \cr 0 &{\rm for}& z \in\{\infty\}. \end{array} \right. $$

Let $$ c_0~:=~ \#{\rm minimum~pts}, \qquad c_1~:=~ \#{\rm saddle~pts}, \qquad c_2~:=~ \#{\rm maximum~pts}, $$

So

$$c_0(S^2)~=~c_0(M)+1, \qquad c_1(S^2)~=~c_1(M), \qquad c_2(S^2)~=~c_2(M)+n, $$

because $z=\infty$ is a minimum point and $Z$ are maximum points for $\tilde{V}$.

3) Define two positive functions $E, F:M \to\mathbb{R}_{+}$ as

$$E(z)~:=~\sum_{i=1}^n\frac{ k_i(x-x_i)^2}{|z-z_i|^{p+4}}~>~0,\qquad F(z)~:=~\sum_{i=1}^n\frac{ k_i(y-y_i)^2}{|z-z_i|^{p+4}}~>~0, \qquad z \in M. $$

The $2\times 2$ Hessian matrix $H$ for the potential $V$ is

$$H_{xx}~=~\sum_{i=1}^n k_i\frac{(p+1)(x-x_i)^2-(y-y_i)^2}{|z-z_i|^{p+4}} ~=~(p+1)E-F,\qquad z \in M, $$ $$H_{yy}~=~\sum_{i=1}^n k_i\frac{(p+1)(y-y_i)^2-(x-x_i)^2}{|z-z_i|^{p+4}} ~=~(p+1)F-E,\qquad z \in M, $$ $$H_{xy}~=~(p+2)\sum_{i=1}^n \frac{k_i(x-x_i)(y-y_i)}{|z-z_i|^{p+4}}, \qquad z \in M,$$

with positive trace

$$ {\rm tr}H ~=~H_{xx}+H_{yy}~=~p(E+F) ~=~p\sum_{i=1}^n \frac{k_i}{|z-z_i|^{p+2}}~>~0, \qquad z \in M.$$

and determinant

$$ {\rm det}H ~=~H_{xx}H_{yy}-H^2_{xy} ~=~\underbrace{(p^2+2p+2)EF}_{>0} - \underbrace{\left((p+1)(E^2+F^2)+H^2_{xy}\right)}_{>0}.$$

A sufficient condition for negative determinant is:

$$F> (p+1)E \qquad \vee \qquad E> (p+1)F\qquad\Rightarrow \qquad {\rm det}H<0. $$

The off-diagonal element $H_{xy}$ is bounded by

$$ \frac{|H_{xy}|}{p+2} ~\leq~\sum_{i=1}^n \frac{k_i|x-x_i||y-y_i|}{|z-z_i|^{p+4}} ~\leq~\frac{E+F}{2}. $$

4) The positive trace ${\rm tr}H>0$ implies that there cannot be any maximum points in $M$,

$$c_2(M)~=~0 \qquad \Leftrightarrow \qquad c_2(S^2)~=~n .$$

A minimum point $z$ must have positive determinant ${\rm det}H(z)>0$, while a saddle point $z$ must have non-positive determinant ${\rm det}H(z)\leq 0$. Generically, one may assume that all critical points $z$ in $M$ are non-degenerate ${\rm det}H(z)\neq 0$, so that $V$ is a Morse function. This implies that all critical points in $M$ are isolated points. Morse theory yields that$^{1)}$

$$ c_2(M)-c_1(M)+c_0(M)~=~\chi(M)\qquad \Leftrightarrow \qquad c_2(S^2)-c_1(S^2)+c_0(S^2)~=~\chi(S^2),$$

or equivalently,

$$ c_1(M)-c_0(M)~=~n-1\qquad \Leftrightarrow \qquad c_1(S^2)-c_0(S^2)~=~n-2.$$

5) Lemma: The determinant ${\rm det}H(z)<0$ is negative sufficiently close to the set $Z$.

The open set

$$ U := \{z\in M \mid {\rm det}H(z)>0\} ~=~\cup_a U_a $$

consists of a number of connected components $U_a$. Since the restriction $V|_{U}$ is concave, each connected component $U_a$ can contain at most one minimum point.

6) Up until now our analysis is rigorous. In the rest of the answer we speculate. Heuristically, there should exists a triangularization of the Riemann sphere $S^2$ with the set $Z$ as vertices, such that each face contains at most one minimum point. Assume that this is true. Let the number of faces, edges, and vertices be $f$, $e$, and $n$, respectively. Assume $n\geq 3$. Then

$$ f-e+n ~=~\chi(S^2)~=~ 2, \qquad 2e~=~\sum_{j\geq 3} j f_j~\geq~3f, \qquad f~=~\sum_{j\geq 3} f_j, $$

implies that $\frac{f}{2}~=~\frac{3f}{2}-f~ \leq~ e-f~=~n-2$, and hence,

$$c_0(M)+1~=~c_0(S^2)~\leq~f~\leq~ 2(n-2)\qquad \Rightarrow \qquad c_0(M)~\leq~ 2n-5.$$

Then an upper bound for the number of critical points in $M$ becomes

$$c(M)~=~ c_{0}(M) + c_{1}(M) + c_{2}(M) ~=~ 2c_{0}(M)+n-1~\leq~ 5n - 11. $$

--

$^{1)}$ Naively, the Morse inequality implies that $c_{0}(M) \geq b_{0}(M)=1$, which is obviously wrong for $n=1$. This can be cured, if one takes boundary contributions in $M$ properly into account. The formulation in terms of the Riemann sphere $S^2$ doesn't have this problem, because it has no boundary.

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Very cool +1. I will try to thoroughly read this tomorrow. –  EPN Nov 9 '11 at 20:51
    
Correction to the answer(v2): The word concave should be concave upwards according to Wikipedia. –  Qmechanic Jan 17 at 19:40

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