An Approach to the Triangle

In this Section , we look at the { Three Body Problem } and created a special { mapping relation } :

• between the physical { Position-Vectors Triangle } & the { Mass * Position-Vectors Triangle } ,
  • with the latter taking responsibility for finding the { Center of Mass } ,

  via the matching of { Fermat Point } onto { Fermat Point } .

A possible advantage here is the super-imposition of the 3 { 120-degrees-apart Fermat-Point-Directions } for the 2 triangles on one-another .

In the { Three Body Problem } , we start-off with 3 mass's , { M1 } , { M2 } & { M3 } respectively , in a { 3-Dimensional Space } , with :

• the forces of attraction / replusion between each pair of mass's being equal & opposite .

• And the problem is to predict the positions of the 3 mass's at any time later ,
  • given an initial set of starting positions & velocities .

The key features of this { classic problem in physics } are :

• the { conservation of linear momentum } , so that :
  • the position of the { Center of Mass } moves at a constant velocity , along a straight-line ,

    and its position can be readily predicted at any time later .

• the { conservation of angular-momentum } about the { Center of Mass } , so that :
  • the { angular-momentum vector } for the 3-mass's is constant-&-invariant .

Our finidngs in { An Approach to the Three Body Problem } , back in 1996 , is that :

• the { angular momentum vector } may be split into two (2) orthogonal components :
  • a first component that is perpendicular to the { plane-of-the-triangle } formed by the 3 mass's , and

  • a second component that is perpendicular to the first component and therefore lies in the { plane-of-the-triangle } ,

  so that :

  • the distribution between the 2 components , at any subsequent time , is dependent on the characteristics of the { Moments-of-Inertia } , and in particular :
    • the ratio of [ { I-XX } vs. { I-YY } ] ,

      with { I-XX } & { I-YY } being the { Moments-of-Inertia } along the two (2) { Principal-Directions } lying in the { Plane-of-the-Triangle } .

Let us briefly review how we would go about finding the { Center of Mass } , as a lead-on / foundation for the next topic below .

First-of-all , we set up the quantity [ ] , such that :

We then select any point , { Point O } , in a { 3-Dimensional Space } as the { Reference Point } ,

• and we construct the { Position Vectors } , [ R1 ] , [ R2 ] , & [ R3 ] respectively , as per the diagram on-the-left , below .

We then construct the { Mass * Position-Vectors } , i.e. :

• [ M1 * R1 ] , [ M2 * R2 ] , & [ M3 * R3 ] respectively ,

  and add these together to arrive at { Point Q } and the vector { Vector O-Q } , as per the diagram on-the-right , above .

We then divide / down-size the vector { Vector O-Q } by the factor of [ 1 / MT ] to arrive at the position of the { Center of Mass } as per the diagram on-the-right , above .

And we present this diagram below , for slightly better clarity :

We notice here that if [ MT = 1 ] , down-sizing the vector { Vector O-Q } is not necessary .

We can therefore 'standardize' the 3 mass's by using the new quantities [ m1 ] , [ m2 ] & [ m3 ] respectively , such that :

• , , ,

  yielding :

  

And this is what we shall use in the next [ Topic ] immediately below .

Let us suppose that we were able to identify the positions of [ m1 ] , [ m2 ] , & [ m3 ] at a specific time { t } , so that :

• the 3 { point-mass's } form the triangle { Triangle D-E-F } , which then identifies the { plane-of-the-triangle } , as per the diagram on-the-left , below :

We then identify the { Fermat Point } / { pseudo Fermat Point } , { Point P } , via the traditional { Fermat Point Construction Method } ,

• and we also construct the [ { Reference Circle of Radius } with the { Equilateral Triangle R-S-T } ] ,

  based on the { Shape Classification System } , as discussed in [ Section V ] , as per the diagram on-the-right , above .

{ Point O } is then the { center of the circle } & the { centroid of Triangle R-S-T } , as well as the { centroid of Triangle D-E-F } .

The { Physical Triangle } , { Triangle D-E-F } , is then referenced by 4 quantities :

• The position of { Point O } & the orientation of the { Polar North Direction } ,
• the radius of the { Reference Circle } , denoted by { } ,
• the angle { angle } , and
• the ratio of [ { length of Line O-P } : { radius } ] .

Let us now pick the { Fermat Point } , { Point P } , as our { Reference Point } for finding the { Center of Mass } :

• We then construct the { Position Vectors } , [ R1 ] , [ R2 ] & [ R3 ] respectively , as per the diagram on-the-right , above .

• We then construct the { Mass * Position-Vectors } , [ m1*R1 ] , [ m2*R2 ] & [ m3*R3 ] respectively , as per the diagram on-the-left , below .

And the { centroid of Triangle A-B-C } , { Point } , is then the { Center of Mass } for { m1 } , { m2 } & { m3 } .

Let us also construct the [ { Reference Circle of Radius } with the { Equilateral Triangle R'-S'-T' } ] , as per the diagram on-the-left , above ,

• and we also present this diagram on-the-right , above , same as the diagram on-the-left but magnified 3-times .

{ Point } is then the { center of the circle } & the { centroid of Triangle R'-S'-T' } , in addition to being the { centroid of Triangle A-B-C } .

The { M-R Triangle } , { Triangle A-B-C } , is then referenced by 4 quantities :

• The position of { Point } & the orientation of the { Polar North Direction } ,
• the radius of the { Reference Circle } , denoted by { } ,
• the angle { angle } , &
• the ratio of [ { length of Line -P } : { radius } ] .

Let us now bring back this key observation from [ Topic #2 ] above :

• the position of the { Center of Mass } can be readily predicted at any time ,

  because of the { conservation of linear momentum } .

We shall therefore pick the { Center of Mass } , { Point } , as our reference point ,

• and we can place a { translating Reference Frame } on { Center of Mass } , moving at a constant velocity , as our { base Reference Frame } .

The four (4) variables , in relation to this { base Reference Frame } anchored at { Point } , that fully describe the { M-R Triangle } , { Triangle A-B-C } , are then :

• the orientation of the { Polar North Direction } ,
• { } , the radius of the { Reference Circle } ,
• the angle { angle } ,
• the ratio of [ { length of Line -P } : { radius } ] .

And once we have established the { M-R Triangle } , the { Physical Triangle } can also be established via a { reverse mapping } procedure .

• And subsequently all { moments-of-inertia } characteristics / quantities associated with the { Physical Triangle } may also be established .

Let us now take note of two (2) key features here :

• The { Polar North Directions } for both the { M-R Triangle } & the { Physical Triangle } are always the same direction ,

  This is because we have used the same { Fermat Point } , { Point P } , and the same 3 x { 120-degrees-apart Fermat-Point-Directions } for our mapping , both forward & reverse ,

  • so that the 3-side of { Equilateral Triangle R'-S'-T' } & { Equilateral Triangle R-S-T } are always parallel .

  And we also make this important recognition , that :

  • if we rotate the { Physical Triangle } about its { Center of Mass } while keeping its { Size & Shape } constant ,
    • then both { Polar North Directions } move / rotate in an uniform fashion .

• If we double the size of the { Physical Triangle } , the size of the { M-R Triangle } also doubles , and visa-versa .
  • This is because the mapping relation is linear , both forward and reverse , i.e. :
    • [ R1 ] maps onto [ m1*R1 ] , and visa-vera ,
    • [ R2 ] maps onto [ m2*R2 ] , and visa-vera ,
    • [ R3 ] maps onto [ m3*R3 ] , and visa-vera ,

      so that doubling [ R1 ] , [ R2 ] & [ R3 ] will also double [ m1*R1 ] , [ m2*R2 ] & [ m3* R3 ] respectively , and visa-versa .

Our proposal here , for the analysis of the { Three Body Problem } , is therefore to use :

• the { angle } and the ratio of [ { length of Line -P } : { radius } ]

  as the two (2) { independent variables } .

And we can either construct the { differential equations } or do the { numerical methods analysis } therefrom ,

• depending on what kinds of { Force Fucntions } the reader may wish to investigate , for the moment at-hand .

In the { Three Body Problem } , we used the { Fermat Point } / { pseudo Fermat Point } .

Question :

• Is there a similar point we can use for the { Four Body } ?

  May be , and may be not ,

  • but the { Fermat Point } we've used in the { Congruent-Triangles Tetrahedron } can provide a good starting point .