3
$\begingroup$

For instance, for a two-body system, there will always be a point in-between the two in which the forces of gravity completely cancel, and if I were there then I would experience no net force. I believe there is also a point, or a region, for three body systems as well. Assuming we treat the stars or planets as simply point-masses, will there always be, for any number of objects, a point, or a region, of equilibrium? Or no?

$\endgroup$
  • $\begingroup$ Intuitively yes, because the gravitational field is continuous (isn't it?) and therefore there must be local maxima between the minima at the point masses. Or vice versa if you are not thinking about rubber sheets? $\endgroup$ – RedGrittyBrick Nov 26 '14 at 11:15
  • 1
    $\begingroup$ @RedGrittyBrick Yes, but in the two planet system this is a quasi one dimensional problem. The 2 other dimensions are trivial due to symmetry. With More masses in 3D space one must assure that the potential shows a saddle point (a maximum or minimum is forbidden as this must be on the boundary) $\endgroup$ – mikuszefski Nov 26 '14 at 11:32
  • $\begingroup$ The question can also be: does the harmonic function, constructed as the superposition of n Newton potentials always have at least one critical point (apart from infinity). One can map this on the problem: Is there always one $\vec s$ such that $\sum (\vec{r}_i-\vec s) f_i(\vec s)=0$, where the $f_i$ are continuous positive definite functions. Maybe this is a question for a mathematician. $\endgroup$ – mikuszefski Nov 27 '14 at 16:12
  • $\begingroup$ Start at any location and always travel in the direction opposite that of the net gravitational force vector. You'll either hit infinity or an equilibrium point. That's a rough way to do it. If you can, travel in the direction of decreasing magnitude of the force. That will make you less likely to miss the equilibrium point that, intuitively, must exist for a multiple body system $\endgroup$ – Jim Nov 28 '14 at 19:51
  • $\begingroup$ Thanks for accepting my answer as correct. However, I am not satisfied with it, yet, and might put some updates in the future. $\endgroup$ – mikuszefski Dec 15 '14 at 13:17
4
$\begingroup$

Yes, and I think it is even better. This might not be a full proof, but I think it gives a good idea.

I will consider $n$ point sources. We know that the potential is harmonic, so there is no maximum and no minimum in the space we are interested in. All critical points, if they exist are saddle points. We are looking for those saddle points, or for points where the force is zero. Note that due to the fact that these are saddle points I would not call it equilibrium; it is meta-stable.

The force at point $\vec s$ is

$\vec F(\vec s)= \sum_i^n m_i \frac{\vec r_i-\vec s}{\vert\vec r_i-\vec s\vert^3}$

To make it easier for now I will assume that all masses $m_i=1$, but the idea can be generalized without problem. We want that all three components of the force are zero at the same time, i.e.

$\vec F(\vec s)= \sum_i^n \frac{\vec r_i-\vec s}{\vert\vec r_i-\vec s\vert^3}=0$

Let us rewrite this by multiplying with all denominators

$\sum_i^n (\vec r_i-\vec s) \prod_{j\backslash i} \vert\vec r_j-\vec s\vert^3=0 $

Now what are the properties of this function? First we notice that the product is a positive semi-definite function. Then let's say $\vec s$ approaches $\vec r_k$. Here the produkt of the $k$th term is non-zero and sort of slowly varying as it does not contain $\vec r_k - \vec s$, while the first term approaches linearly zero. All other terms in the sum contain the factor $\vert \vec r_k - \vec s \vert^3$ so they do not play a role, as they approach zero much faster. Hence at mass point k the solution to the equation for component $\alpha$ is a plane perpendicular to $\vec e_\alpha$. So near $ \vec r_k $ the solution for $s_x$ is $s_x = r_{i,x}$ and the same for $y,z$. To illustrate this I generated some images for four equal masses randomly distributed. The first image shows the solution for $s_x$ and the position of the masses as spheres. The second image combines all solutions. Here one can see the solution for $s_x$ for four randomly distributed (equal) masses

Combination of all sulutions

Next, let us look at the solution far away. In this case we have $\vert \vec s \vert \gg \vert \vec r_i\vert \forall_i$. So the equation simplifies to

$\sum_i^n (\vec r_i-\vec s) \prod_{j\backslash i} \vert\vec s\vert^3=\sum_i^n (\vec r_i-\vec s) \vert\vec s\vert^{3(n-1)}=0 $

and we get, hence,

$\sum_i^n (\vec r_i-\vec s)=0 $

or

$\vec s = \frac {\sum_i^n \vec r_i}{n} $

If we would have taken into account the $m_i$, we would have the center of mass here. The solutions are planes through the center of mass. This is what one would expect, as in the far field the force is approximately the force of the total mass concentrated in the center of mass. One can see this behaviour in the next figure.

Solution for $sx$ far away

So at the end, we know that the solutions to the components of $\vec s$ to each component are infinite surfaces that approximate a plane at infinity but are distorted at the positions of the masses in such a way that locally they pass the mass parallel to the infinite plane. This is because everything is continuous and smooth.

In the final step we must understand that there is always a point where these three surfaces cross. It is clear that all three solutions always cross at the $\vec r_i$. In case of an even number of masses it is easy to see that there must be an crossing apart from the $\vec r_i$ (due to the fact that the surfaces come from $-\infty$ and go to $+\infty$) So two surfaces intersect in such a way that they produce an infinite line. The third surface crosses this infinite line as it is infinite and more or less perpendicular to the first two. Hence, there must be at least one saddle point. To visualize this, one might look at the following sketch. Here, for simplicity it is assumed that one solution is the drawing plane and the two other solutions are perpendicular to this, following the red and green line,respectively. Even numbers definitively produce a crossing

But what about an odd number of masses. In this case one can go from $-\infty$ and go to $+\infty$ only by passing through the masses. This can give something like this:

Odd number might not have a result?

However, this problem is avoided by the fact that there is always an odd number of zeros between to masses. This can easily be seen from the above equations. Let us consider the line between two masses. Without restriction we can put the origin in the position of the first mass and the second mass we put on the $x$-axis. So we must solve

$\ldots+ (-s_x) \prod ()+\ldots+ (x_0- s_x) \prod ()+\ldots=0$

We know from above that near zero only the $-s_x$ term plays a role, so the function becomes negative when going direction $x_0$. When approaching $x_0$ only the second term is of importance, but as $s_x<x_0$ this is positive. As everything is continuous there must be an odd number of zeros in between. It is this bending behaviour that results in the intersection of each pair of surfaces and as each intersection is an infinite line, there is a crossing point. So it is more like this (and the sketch for the even number is actually wrong and should have at least four crossings)

No, odd numbers also have a solution

Eventually, it is quiet likely that you might find more than one critical point as every pair should give rise to one (actually I am still thinking about this, whether it should be $n-1$ or ${ n \choose 2}$ ). There might be less if some coincide, either accidentally or due to symmetry.

Edit The statement on plane intersections generate infinite lines should be different. I think it can be one of the following three possibilities: an infinite line, a line going from on mass to the next or a closed loop. It is easy to see that the closed loop must intersect the third plane in a second point additionally to the position of the mass. A closed loop through a second mass in considered as case two. Case two then should show an equilibrium point due to the odd numbers of zeros in the third plane, i.e. the fact that the plane sort of bends back. (if one looks carefully you find case 2 and 3 in the second figure)

What I am also wandering is: can I deduce from the smooth behaviour that none of the surfaces is crossing itself?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.