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Challenge Physics Challenge by QuantumQuest #1

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  1. Mar 28, 2017 #1
    Submitted and Judged by @QuantumQuest
    Solution credited to: @TSny

    RULES:
    1) In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored.
    2) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
    3) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
    4) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

    CHALLENGE:
    Three point masses ##m_1, m_2## and ##m_3## which are located at the non-collinear points ##P_1, P_2## and ##P_3## respectively can interact only through gravitational attractions. The masses are isolated in space and they have no interaction with other objects. We suppose that an axis ##\sigma## is passing through the center of mass of the system of the three given masses and is perpendicular to the plane of the triangle ##P_1P_2P_3##. Which conditions must angular velocity of the system (regarding given axis) and distances ##P_1P_2 = d_{12}##,##P_2P_3 = d_{23}## and ##P_1P_3 = d_{13}## must satisfy in order the shape and the size of the triangle ##P_1P_2P_3## stay constant, as the system is rotating?
     
    Last edited: May 5, 2017
  2. jcsd
  3. Mar 28, 2017 #2
    Let ##O## stand for the center of mass to the system. Introduce a coordinate frame that rotates about the axis ##\sigma## with angular velocity ##\omega=const##.
    With respect to this moving frame the potential energy of the system is
    $$W(\overline {OP}_1,\overline {OP}_2,\overline {OP}_3)=-\gamma\sum_{1\le i<j\le 3}\frac{m_im_j}{|\overline {OP}_i-\overline {OP}_j|}+\frac{\omega^2}{2}\sum_{i=1}^3|\overline {OP}_i|^2$$
    This is a function of six variables -- coordinates of the vectors ##\overline {OP}_i##.
    We are given with a configuration of particles such that the particles rest with respect to the rotating system.
    To solve the problem one should find conditional extrema of the function ##W## under the following constrain
    $$\sum_{i=1}^3m_i \overline {OP}_i=0.\qquad (*)$$
    These are three scalar equations.
    Using the Lagrange multipliers
    we have
    $$\frac{\partial W}{\partial \overline {OP}_i}+\overline\lambda m_i=0,\quad i=1,2,3.\qquad (**)$$
    Therefore we have a system of 4 algebraic vector equations (*)-(**) with respect to unknowns $$\overline {OP}_i,\quad i=1,2,3,\quad \overline\lambda,\quad\omega.$$
    One also should take into account the additional three scalar equations ##|\overline {OP}_i-\overline {OP}_j|=d_{ij}##.


    I just wonder if why does the author think that such an overdetermined system must have a solution :)
     
    Last edited: Mar 28, 2017
  4. Mar 28, 2017 #3

    Charles Link

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    @zwierz Please compute the frequency along with any other additional necessary constraints. Before posting, we looked over this problem quite carefully and also solved it. I think you will find the solution quite interesting.
     
  5. Mar 28, 2017 #4

    mfb

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    It has solutions, and it has useful relations you did not find so far.
     
  6. Mar 28, 2017 #5
    Very well, it would be interesting to look at your solution once. For now I suspect that you just implicitly assume that the particles are connected by weightless rods to provide the conditions ##|\overline{OP}_i-\overline{OP}_j|=d_{ij}##. In other case the problem is hardly solvable for arbitrary ##d_{ij}##
     
  7. Mar 28, 2017 #6

    mfb

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    There are no rods.
    The problem is solvable.

    Did you see the condition given in the last sentence of the problem?
     
  8. Mar 28, 2017 #7
    ok. I await with great interest how will you solve this problem for arbitrary ##d_{ij}##
     
  9. Mar 28, 2017 #8

    Charles Link

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    You are allowed to assign constraints, if necessary, to the ## d_{ij} ##.
     
  10. Mar 28, 2017 #9
    that's quite another matter.
    thus it is a linear system relative to ##\overline {OP}_i##
     
  11. Mar 28, 2017 #10

    QuantumQuest

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    It does have a solution and I really fail to see how it is overdetermined. I think that it is rather a well determined system but you see it through an overwhelming way.

    Did you see anything that even remotely implies a such thing in the wording of the problem?

    Again, the wording of the problem is exact. There is absolutely nothing missing.
     
  12. Mar 30, 2017 #11
    there are misprints in my first post, it should be written as follows
    $$W(\overline{OP}_1,\overline{OP}_2,\overline{OP}_3)=-\gamma\sum_{1\le i<j\le 3}\frac{m_im_j}{|\overline{OP}_i-\overline{OP}_j|}-\omega^2\sum_{i=1}^3\frac{m_i}{2}|\overline{OP}_i|^2$$

    If the particles do not lie on the same line then from equations (*)-(**) we get

    $$\omega^2=\frac{\gamma(m_1+m_2+m_3)}{d^3_{13}}$$
    and the
    equilateral
    triangle is solely possible ##d_{12}=d_{13}=d_{23}##

    The case when all the particles lie on the same line is boring
     
    Last edited: Mar 30, 2017
  13. Mar 30, 2017 #12

    QuantumQuest

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    @zwierz

    I really don't want to be any discouraging and I see that you put good efforts in an (unnecessarily in my opinion) cumbersome way. Now, while this is perfectly acceptable, you have to also abide by the rules in order to get credit. We all abide by these rules when we participate in a challenge here at PF. So, I want you to give a full proof including the necessary math, as of rule #1. I cannot really see how you get to ##\omega## and that the triangle is equilateral.

     
  14. Mar 30, 2017 #13
    I don't care about credits. I think about problem if find it interesting.
    If you wish the details they are as follows. Equations (**) give
    $$(a_{12}+a_{13}-\omega^2 m_1)\overline{OP}_1-a_{12}\overline{OP}_2-a_{13}\overline{OP}_3=-\overline{\lambda}m_1,\qquad (***)$$
    $$(a_{21}+a_{23}-\omega^2 m_2)\overline{OP}_2-a_{21}\overline{OP}_1-a_{23}\overline{OP}_3=-\overline{\lambda}m_2,\qquad (!)$$
    $$(a_{31}+a_{32}-\omega^2 m_3)\overline{OP}_3-a_{31}\overline{OP}_1-a_{32}\overline{OP}_2=-\overline{\lambda}m_3,\qquad (!!)$$
    here
    $$a_{ij}=\frac{\gamma m_im_j}{d^3_{ij}}$$
    If you sum these three equations and use (*) you obtain ##\overline{\lambda}=0##.
    Now express ##\overline{OP}_3## from (*) and put it into (***) this gives
    $$(a_{12}+a_{13}-\omega^2 m_1)\overline{OP}_1-a_{12}\overline{OP}_2+a_{13}(\frac{m_1}{m_3}\overline{OP}_1+\frac{m_2}{m_3}\overline{OP}_2)=0$$
    Since ##\overline{OP}_1## and ##\overline{OP}_2## are linearly independent you get two scalar equations to express ##\omega##.
    Then do the same for (!) (!!)
     
    Last edited: Mar 30, 2017
  15. Mar 30, 2017 #14

    Charles Link

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    @zwierz I have a couple of questions about your solution=and yes, you did get the correct answer. ## \\ ## 1) In your first posting, you describe the energy in a rotating coordinate system, so that apparently there is no kinetic energy term. Is that why the term that looks like the kinetic energy has a "-" sign in your second posting? I am familiar with ## [\frac{d \vec{r}}{dt}]_o=\frac{d \vec{r}}{dt}+\omega \times \vec{r} ##, but have never seen a potential energy term transformed or introduced in a rotating system due to what is kinetic energy in the stationary frame. ## \\ ## 2) Your energy term ## W ## is a scalar=I found it difficult to take a vector derivative (gradient) on the vector-type Lagrange multiplier constraint term. Instead, I squared each vector component and set the sum of the squares equal to zero as the constraint to try and check your solution. ## \\ ## 3) The resulting set of equations does not appear to be simple. If you assume ## d_{12}=d_{13}=d_{23} ##, the result follows for ## \omega^2 ##, but reaching that result appears non-trivial. Did you see an easy shortcut to this result? ## \\ ## Additional item you may find of interest, as was pointed out by @mfb in discussions regarding the equilateral triangle solution, is that for ## m_3 ## negligible, the system describes the Lagrangian points L4 and L5 for the sun, earth, and satellite system. ## \\ ## Anyway, I'm glad you got the correct result.
     
  16. Mar 30, 2017 #15
    this term is a potential energy of the inertial force.
    I have proved that but not assumed
    Yes, I also recollected that but unfortunately after I had solved the problem
    Actually I do not understand which nontriviality you mean. A term expressed from one equation is substituted to another equation that are absolutely standard things.
    I did not check that but I almost sure that equations (***),(!),(!!) with ##\lambda=0## are the second Newton laws for each particle relative to an inertial frame
     
    Last edited: Mar 30, 2017
  17. Mar 30, 2017 #16

    Charles Link

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    @zwierz Thank you. I'm going to need to take a second look at it. The result may follow directly upon setting equal the results for ## \omega^2 ##.
     
  18. Mar 30, 2017 #17
    Exactly. It is time to go to bed for me 00.20 :)
     
  19. Mar 30, 2017 #18

    mfb

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    Here is an alternative approach that might be easier to follow (I saw the problem before, don't count that as new solution):

    Let the center of mass be the origin of the coordinate frame: ##m_1 r_1 + m_2 r_2 + m_3 r_3 = 0##
    To keep shape and size, the masses have to rotate at constant ##\omega## and constant radius. This leads to three equations for the three masses:

    $$\omega^2 \vec r_1 = \frac {Gm_2 \vec d_{12}}{|d_{12}|^3} + \frac {Gm_3 \vec d_{13}}{|d_{13}|^3}$$
    $$\omega^2 \vec r_2 = \frac {Gm_1 \vec d_{21}}{|d_{12}|^3} + \frac {Gm_3 \vec d_{23}}{|d_{23}|^3}$$
    $$\omega^2 \vec r_3 = \frac {Gm_1 \vec d_{31}}{|d_{13}|^3} + \frac {Gm_2 \vec d_{32}}{|d_{23}|^3}$$
    Here ##d_{12} = r_1 - r_2 = -d_{21}## is pointing from mass 2 to mass 1. We can see ##\omega \propto \sqrt{G}## already.

    By taking differences, we can get rid of the bare vectors r. Equation 1 minus equation 2 and dividing by G:
    $$\frac{\omega^2}{G} \vec d_{12} = \frac{(m_1+m_2) \vec d_{12}}{|d_{12}|^3} + \frac {m_3 \vec d_{13}}{|d_{13}|^3} + \frac {-m_3 \vec d_{23}}{|d_{23}|^3}$$
    We can investigate this along the d12 vector and perpendicular to it. Take the cross product with d12 to study the perpendicular direction. The first two terms vanish.
    $$0 = \frac {m_3 \vec d_{13} \times \vec d_{12}}{|d_{13}|^3} + \frac {-m_3 \vec d_{23} \times \vec d_{12}}{|d_{23}|^3}$$
    As ##\vec d_{13} = \vec d_{12}+\vec d_{23}##, we have ##\vec d_{13} \times \vec d_{12} = (\vec d_{12}+\vec d_{23}) \times \vec d_{12} = \vec d_{23} \times \vec d_{12}##. Use this:
    $$ \frac {m_3 \vec d_{23} \times \vec d_{12}}{|d_{13}|^3} = \frac {m_3 \vec d_{23} \times \vec d_{12}}{|d_{23}|^3}$$
    Simplify:
    $$|d_{23}|= |d_{13}|$$
    We can repeat this with other differences, but we can also use that the problem is invariant under permutations of the masses:
    $$ |d_{23}| = |d_{13}| = |d_{12}| = d$$
    We get an equilateral triangle of arbitrary size. To determine ##\omega##, take the scalar product with d12 where we took the cross product before, and use ##\vec d_{12}\vec d_{12}=d^2## and ##\vec d_{12}\vec d_{13}=\frac{d^2}{2}## and ##\vec d_{12}\vec d_{23}=\frac{-d^2}{2}##:
    $$\frac{\omega^2}{G} d^2 = \frac{(m_1+m_2) d^2}{d^3} + \frac {m_3}{2d^3} + \frac {m_3 d^2}{2d^3}$$
    Simplify: $$\omega = \sqrt{ \frac{G(m_1+m_2+m_3)}{d^3}}$$

    If one mass is negligible, this is a well-known result from the Kepler problem, and the small mass is in one Lagrangian point.
     
  20. Mar 30, 2017 #19

    QuantumQuest

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    @zwierz

    I want to be fair regarding challenge so I want to ask you some things about your solution overall, in order to have a complete picture

    What kind of extrema you look for and why? What does this constraint represent? (I take ##W## in the corrected form you provided in #11)

    How did you arrive there?

    Also, your math in #13 are severely shortcut but I see your point. What I don't see presented in a clear way is how do you conclude that distances are equal.
     
    Last edited: Mar 30, 2017
  21. Mar 30, 2017 #20

    Charles Link

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    @zwierz I successfully worked through the equations of your solution: Using the first gradient equation, and a substitution (as you had mentioned) for ## \vec{r}_3 ## into that equation from the center of mass equation resulted in an equation of the form ## A \vec{r}_1+B \vec{r}_2=0 ##. This gives the result that ## A=0 ## and ## B=0 ## if ## \vec{r}_1 ## and ## \vec{r}_2 ## are not colinear. Setting ## B=0 ## gives ## m_2( \frac{1}{d_{12}^3}-\frac{1}{d_{13}^3})=0 ## so that ## d_{12}=d_{13} ## Setting ## A=0 ## gives the equation for ## \omega^2 ##. By symmetry of the equations, the result ## d_{12}=d_{13}=d_{23} ## follows without further processing. ## \\ ## An additional comment is I think the minus sign that you used for the kinetic energy term is basically because in the Lagrangian formalism of mechanics, ## T ## and ## V ## have opposite sign=the Lagrangian ## L=T-V ##. ## \\ ## In any case, a very interesting method that you used.
     
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