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Point Masses in Equilibrium

  1. Aug 12, 2007 #1
    This is a highly ideaized problem.

    Find a collection of stationary point masses on the unit line [0-1] such that:

    1) There is more than 1 mass
    2) The total mass = 1
    3) Each mass is >0
    4) There is no other mass present in the universe
    5) Each mass attracts each other mass with Newtonian inverse-square force
    6) The entire collection of masses is in equilibrium - that is, the total force
    on each mass is = 0
     
  2. jcsd
  3. Aug 13, 2007 #2

    mathman

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    Impossible! The two end masses will be attracted toward the middle, and eventually all will clump together.
     
  4. Aug 13, 2007 #3
    That's true for a finite number of masses, but for an infinity of masses there won't necessarily have to be end masses. So, use an infinity of masses.
    Bob
     
  5. Aug 13, 2007 #4

    olgranpappy

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    Otherwise one wouldn't really call it a "collection of... point masses" would one?

    Since you did not specify units, #2 doesn't make any sense... but you're among friends, so we won't hold it against you. In that case #2 is still a trivial normailzation issue.

    This also goes without saying, because we do *physics* here.

    Fair enough.

    Fair enough.

    But, of course, not a stable equilibrium.

    Anyways, this sounds like a nice riddle, but I'm not sure if it can be done. It seems like the mass distribution [tex]\lambda(x)[/tex] must satisfy

    [tex]
    \int_0^{x_0}dx\frac{\lambda(x)}{(x-x_0)^2}=\int_{x_0}^1dx\frac{\lambda(x)}{(x-x_0)^2}
    [/tex]
    for any [tex]x_0[/tex] within [0,1].
     
  6. Aug 13, 2007 #5
    Thanks for your corrections and comments to my statements.
    Are you saying that it's impossible or just that YOU don't
    know how to do it?
    Bob
     
  7. Aug 13, 2007 #6

    olgranpappy

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    Oh, I was just wondering whether or not you actually found a solution... my first impulse was that it would be impossible, but I haven't thought about it very hard. Cheers.
     
  8. Aug 13, 2007 #7

    olgranpappy

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    oops... no, it's impossible. my first impulse was right.

    cheers.
     
  9. Aug 13, 2007 #8
    How do you know it's impossible? Could you post a proof of that?
    Or references to one?
    Bob.
     
  10. Aug 14, 2007 #9

    AlephZero

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    I don't have a solution, but I don't see why it is impossible either. You can easily construct a system with the surprising behaviour that the masses move apart, not together!

    Clearly there is no solution with a finite number of masses. As mathman says the end masses must be attracted to the middle.

    My instinct says there is no solution for a sufficiently smooth continuous mass distribution either, for the same basic reason: there is no way you can get enough mass close to the end of the line to make the forces balance.

    That leaves the possibility of a discontinuous mass distribution - for example an infinite number of point masses.

    It seems plausible that if there are solutions, there will be solutions which are symmetrical about the center of the line. (There may also be unsymmetrical solutions of course)

    So, consider an sequence of points with a limit point at each end of the interval, where the masses decrease towards the end of the interval and the total mass is finite.

    As a specific example, take the symmetrical set of points at 1/2^k and 1-1/2^k for k = 2, 3, 4.... Another point at 1/2 might help, but it's irrelevant to what I have done so far.

    Suppose the masses at points 1/2^k and 1/2^k are ma^k, for some 0 < a < 1, which means the total mass is finite, and m is the scale factor to make the total mass = 1.

    A bit of numerical experimentation shows that when a is small, for example a = 1/3, the force F on points near the ends of the line are towards the centre, and F decreases as k increases. But when a is larger, say a = 2/3, in the limit the forces act AWAY from the centre and they INCREASE as k increases.

    It seems plausible to me that there is some configuration (not necessarily as simple as geometric ratios between the points!) where all the forces can be made to balance.

    But solving an infinite number of equations in an infinite number of variables is hard...

    Thinking a bit more, possibly there is a solution for a continuous mass distribution, provided the mass density goes to infinity sufficiently fast at the ends. But solving integral equations with singularities at the end points of the range is also hard.
     
    Last edited: Aug 14, 2007
  11. Aug 14, 2007 #10

    olgranpappy

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    Yes, let me explain in words and then again in equations:

    Consider any "small piece" of the line mass distribution (e.g. some small piece right in the middle). This piece must have the forces on it which are pulling "to the left" (due to all the mass to the left of it) equal the forces on it which are pulling "to the right" (due to all the mass to the right of it). Otherwise the system wouldn't be in equilibrium because there would be a net force on the piece we are considering.

    Now, consider a small piece of the line-mass distribution that is located just a little further to the left of the piece we just considered. The force pulling this new piece to the right is necessarily *bigger* than the force which was pulling the previous piece we considered to the right, and the force pulling this new piece to the left is necessarily *smaller* than the the force which was pulling the previous piece we considered to the left. But since those two forces were equal, the sum of the forces on the new piece must *not* be equal.

    Thus the body is not in equilibrium.

    Here is another easy way to see this by exploiting the limiting behavior of the equation I wrote down, which is just an equation that says that the force pulling "to the right" equals the force pulling "to the left"
    [tex]
    \int_0^{x_0} \frac{\lambda(x)}{(x-x_0)^2}=\int_{x_0}^1\frac{\lambda(x)}{(x-x_0)^2}dx
    [/tex]

    Consider the equation for x_0 --> 0. The LHS vanishes, but the RHS goes to
    [tex]
    \int_0^1\frac{\lambda(x)}{(x-x_0)^2}dx\;.
    [/tex]

    But, the integrand is manifestly non-negative (\lambda is positive def), thus the above integral can only be zero if lambda is zero for all values of x.
     
    Last edited: Aug 14, 2007
  12. Aug 14, 2007 #11

    AlephZero

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    You are assuming there is a continuous mass distribution with mass density > 0 everywhere on the line.

    The proof is OK given those assumptions, but if there are an infinite number of point masses, The forces only need to balance at those discrete points. In between the points, there is no mass so the gravitational force is zero.

    Of course the gravitational potential is not constant in the gaps between the point masses, but the OP's question doesn't talk about the potential, it talks about the force on the mass. No mass at a point means no gravitational force on that point.
     
  13. Aug 14, 2007 #12

    olgranpappy

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    No I am not. I am assuming the the mass density is greater than *or equal to* zero everywhere, which is necessarily true for a mass density.
     
  14. Aug 14, 2007 #13

    olgranpappy

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    ... there is either zero mass "to the left" of some point considered, or there is nonzero mass "to the left". If the former, then keep going "to the left" further until you run into some mass, if the latter than that mass is not in equilibrium. Either way we find that the only way to satisfy the equilibrium condition is for the mass to be zero everywhere.
     
  15. Aug 14, 2007 #14

    olgranpappy

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    ...anyways, it should be very easy for you to prove me wrong by simply writing down a solution. But, I'm claiming that no solution exists. Prove me wrong.
     
  16. Aug 14, 2007 #15

    AlephZero

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    Well, here's one set of masses which are all in equilibrium. Granted it doesn't satisfy all the OP's requirements, but it's a counter-example to your proof using integrals:

    For every integer k (positive negative and zero), there is a mass of 2^k at position x = 2^k.

    For each mass k, the force between masses (k+j) and (k-j) are equal and opposite. Therefore the total force on each mass k = 0.

    There's no trickery with summing non-convergent infinite series here. It's easy to show that the total force to the right and left of each mass are finite.

    If you put an infinitesimally small "test mass" at any point where there isn't a mass already, then of course it would move. Your integral proof shows that. But it's irrelevant to the OP's question.
     
  17. Aug 14, 2007 #16
    Very clever counter-example to olgranpappy's proof, AlephZero!

    And I'm thankful to all members who responded (and who will respond) to
    my original post. There is a bit of trickery here, for which I'm sorry! I saw a statement by Stanislaw Ulam that it was possible, but he didn't show how to do it.
    I'm trying to find out how it is done.
    Bob
     
  18. Aug 15, 2007 #17

    olgranpappy

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    But, this is *not* a counter-example. Granted it is a clever example. But, again, it is not a counter-example because there is no way to set up this arrangement of point masses along any finite line (or a line of "length 1")
     
  19. Aug 15, 2007 #18

    AlephZero

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    Of course it's not an existence proof of a solution to the OP's question, but it does demostrate the error in your "integral" proof.

    You proposed
    [tex]\int_0^{x_0}dx\frac{\lambda(x)}{(x-x_0)^2}=\int_{x_0}^1dx\frac{\lambda(x)}{(x-x_0)^2}[/tex].

    Why does it make any difference to your argument to take
    [tex]\int_0^{x_0}dx\frac{\lambda(x)}{(x-x_0)^2}=\int_{x_0}^\infty dx\frac{\lambda(x)}{(x-x_0)^2}[/tex]
    instead?

    What is wrong with your proof is that you should have said
    [tex]\int_0^{x_0}dx\frac{\lambda(x_0)\lambda(x)}{(x-x_0)^2}=\int_{x_0}^1dx\frac{\lambda(x_0)\lambda(x)}{(x-x_0)^2}[/tex].

    When [tex]\lambda(x_0) = 0[/tex] that equality always holds since both integrals are zero.

    For any arrangement of point masses, [tex]\lambda(x_0) = 0[/tex] everywhere, except at the mass points. And at the mass points, [tex]\lambda(x_0)[/tex] is a constant times a dirac delta function, which makes the integrals a discontinuous function of [tex]x_0[/tex].

    The situations of a sequence of point masses tending to a limit point, and the "smoothed out" equivalent of a constant mass density along a line, are completely different.

    Anyway, apart from that, if Ulam says it's possible, I'm not going to pick a fight with him about maths... :smile:
     
  20. Aug 15, 2007 #19

    olgranpappy

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    Oh, you are correct. There is an error in both of my "proofs". Sorry.

    After further consideration I think that it might be possible to find a stable arrangement, but I still can't think of one.

    cheers.
     
    Last edited: Aug 16, 2007
  21. Aug 16, 2007 #20

    AlephZero

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    Well, I think I see how to construct a solution, but my topology is too rusty to be sure I'm right, and a proper proof would belong somewhere in the maths section not "classical physics". The magic word "Ulam" was a big clue to inventing this!

    The idea is to construct a sequence of partial solutions with a finite number of masses, which tend to a limit of a complete solution with an infinite number of masses.

    Sketch of the method:

    Start with two points P1 and P2 with m = 1/4, x = 1/4 and 3/4.

    Now add two more points P3 and P4, with m = 1/8, x = 1/8 and 7/8. Adjust the positions of P1 and P2, preserving symmetry about the mid point, so P1 and P2 are in equilibrium (but P3 and P4 are not). In other words move P1 by some distance dx, and move P2 by -dx.

    Now add two more points P5 and P6, with m = 1/16, x = 1/16 and 15/16. Adjust the postions of P1 thru P4 so they are in equlibrium, maintaining symmetry. (Move the P1 thru P4 distances dx1, dx2, -dx2, -dx1).

    Repeat an infinite number of times...

    When you add each pair of points, you can restore equilibrium by moving all the other points slightly towards the middle of the line. (This assertion needs a rigorous proof, but thinking about which direction the unbalanced forces act makes it very plausible, and we are only considering a finite set of masses at each stage so there's nothing too tricksy going on).

    Therefore the sequence of positions of each point is monotonic and bounded, so it converges to a limit. The limit is an infinite set of points in equlibrium (that assertion also needs proof!!)
     
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