An Original Demonstration of the Bertrand's Theorem?

An original demonstration of the Bertrand’s theorem? Eric Guiot

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Eric Guiot. An original demonstration of the Bertrand’s theorem? . 2015. hal-01261983

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An original demonstration of the Bertrand’s theorem ?

Eric Guiot Independent researcher [email protected]

Abstract: In this paper we present an original demonstration of the Bertrand’s theorem. This demonstration is the consequence of a generalization of the Binet’s equation. Moreover we present a family of central and non conservative forces which allows obtaining closed trajectories.

Keywords: Bertrand’s theorem; Central force; conic; PACS Number:45.50.Dd; 45.00.Pk

1. Introduction

At the 19th century the mathematician Bertrand published an important theorem of the classical mechanics (Bertrand,[1]). This theorem states that it is only two conservative and central forces for which all orbits radially bounded are closed, the force of Newton and the Hooke’s. These forces leads to conic trajectories with the difference that the Newton’s is directed towards one of the foci and the Hooke’s towards the geometrical centre of the conic. The proof of this theorem has continued to draw attention and several different demonstrations has been recently published (Santos[2], Grandati [3]). In this paper we present another demonstration of this theorem. This demonstration seems original and is the consequence of a generalization of the Binet’s equation. Moreover we present a family of central and non conservative forces which allows obtaining closed trajectories. This family seems unknown. Consequently, in a first part of this paper we study the forces which lead to conics trajectories. In a second part we study the central forces and we determine if these forces are conservative.

2. Generalization of the Binet’s equation

To obtain these forces we use the equation of the acceleration in polar coordinate. Therefore as distinguished from the Binet’s equation we consider not only the radial acceleration but also the tangential acceleration. This original method (at our knowledge) allows obtaining all the forces which lead to conic trajectories and not only the central forces. We present here this method which has been published elsewhere (Guiot [4], [5]).

  As usual in celestial mechanics we will use the polar system of coordinate (F;eR ;e ) where F (foci of the conic) is the origin of the repair, r is the radial distance to the origin with the relation   FM  r.eR And the angle  is measured from the periapsis of the orbit. In this repair the acceleration is given by the classical relations   2     a  (r  r )eR  (r  2r)e 1 But the orbital shape is more concisely described by the reciprocal u  as a function of  . r And by using the relations

E.Guiot An original demonstration of the Bertrand’s theorem?

d d 1 u r  r    dt dt u u 2 d d du u  u  u.u' dt dt d d d d u' u' u .u" dt dt d 2 We obtain a generalization of the Binet’s equation.   u"u 2 2  u 3 2  2u'2 u2  u'u 2   2u'u 2 2  u 3  a  eR  e u 4 u 4 By noticing that this equation can be written   u"u 22  u 32 u' 2u'u 22  u 3  2u'u 22  u 3  a   e  e  4 4  R 4   u u u  u We introduce two functions given by  u"u 2 2  u 3 2 Y(u)  u 4 And  2u'u 22  u 3 Z(u)  u 4 And the acceleration becomes   u'   a  Y(u)  Z(u)eR  Z(u)e  u  We can now write the system of equation

 u' aR  Y(u)  Z(u)  u  a  Z(u) We introduce a new function f (u) definite by Y(u)  A. f (u) Where Ais constant. To obtain r() as a conic, we have to solve a differential equation as u"u  B Where B is a second constant. Consequently we have now to introduce a relation between Y(u) and  . This relation is   Cu f (u) Where C is a constant of the motion. Indeed with this relation we obtain u" 2  u 2  Au 2 f (u) And A u"u  B  C 2 This differential equation leads now to the classical solution p r()  1 eCos The parameter p of the conic is

E.Guiot An original demonstration of the Bertrand’s theorem?

C 2 p  a(1 e2 )  A Where e is the eccentricity and a the semi major axis. Thus we obtain C  A a(1 e2 ) We have now to determine the tangential component of the acceleration and by using  C.r 1     f 'r  f  r 2 f 2  Where d f ' f (r) dr We obtain   1 Cf 'r r a  r  2r   C f 2 f r Consequently our family of force (per unit mass) is with respect for the Newton’s law of dynamics   2 1 f ' 1 1 f ' 1  F  Af .eR  r   eR  Cr f   .e 2 f r  2 f r  Or, more simply   2 1 f ' 1 dr  2 1 f ' 1  F  Af .eR  r   eR  r    .e 2 f r  d 2 f r  Where A and C are two constants. Their physical dimensions depend on the choice of f (r) .

3. Determination of the central forces

A force is central if this force is always directed to the same point. Consequently to determine the family of central forces we introduce a point, called I which is located on the right (AB) . The distance FI is called  (Figure 1).

 eR

 M e  F O  A B M F O I F

  IF

Figure 1. Representation of the central force

E.Guiot An original demonstration of the Bertrand’s theorem?

If the force is directed to I then the vector product    IM * F  0    By noticing that, in our repair FeR e the vector IM is given by  Cos  r IM  Sin We obtain

Sin.FR  (Cos  r).F   0 With the relations specific to the conics a(1 e2 ) eCSin r  and r  r f 1 eCos a(1  e 2 ) We obtain ae(2 f  rf ')   f '(r  a)  2 f

To obtain a central force we are looking for the family of functions f (r) which leads to  as a constant. Consequently we write the equation d   0 dr This relation leads to ea 2 (2 ff "3 f '2 )  0 ( f '(r  a)  2 f ) 2 And the solving is 1 f (r)  2 (C1r  C2 )

Where C1 and C2 are constant. The position of I is given by aC e   2 C2  aC1 And the force becomes  ar(C  rC )  C (a 2 (1  e 2 )  r 2 )  eC  F  A 2 1 2 e  A 2 Sin.e 3 R 3  ar(C1r  C2 ) (C1r  C2 )

Consequently we have determined a family of central forces which lead to conic trajectories. These forces are depending on two constants C1 and C2 are directed toward the point I . The distance IF is given by aC e IF  2 C2  aC1 And the distance OI by aC e a 2C e OI  OF  IF  ae  2  1 C2  aC1 C2  aC1 Note we obtain a simple ratio 4

E.Guiot An original demonstration of the Bertrand’s theorem?

OI aC  1 IF C2

To simplify the expression we call R the distance IM . This one is given by R  IM  (  r cos)2  (r sin)2  2  r 2  2r.cos And by using a(1 e2 ) r  1 eCos We obtain 2aC C (ar  r 2  a 2 (1 e2 ))  C 2 (2a 2  e2a 2  r 2  2ar)  C 2 (a 2r 2 ) R  1 2 2 1 (1) (C1a  C2 ) We determine now the magnitude of our force. This one is given by

2 2 ar(C  rC )  C (a 2 (1 e2 )  r 2 )  eC  F  A 2 1 2  2 Sin  3   3   ar(C1r  C2 )  (C1r  C2 )  And by using

2 e2r 2  a(1 e2 )  r Sin  er We obtain 2 2 A 2aC C (ar  r 2  a 2 (1 e2 ))  C (2a 2  e2a 2  r 2  2ar)  C (a 2r 2 ) F  1 2 2 1 3 a (C1r  C2 ) Consequently we can write A C a  C F  1 2 R 3 a (C1r  C2 ) Note that we give the expressions of the kinetic momentum and of the equation of time of this force in another paper (Guiot [6]).

4 – Our demonstration of the Bertrand’s theorem.

A force is conservative if its magnitude only depends on the distance of the object from the origin. With this definition we can now study our force. By introducing the unity vector   IM e   IM This force is given by  A C a  C  F   1 2 R.e 3  a (C1r  C2 )

Thus if the force is conservative r has to disappear from this expression. A first solution is to choose

C1  0 And the total force becomes  A  F   R.e 2 R aC 2 5

E.Guiot An original demonstration of the Bertrand’s theorem?

The center of force is given by aC e IF  2  ae C2  aC1 This indicates that I and O are combined. Consequently this force is the Hooke’s. A second solution is to inverse the relation between R and r given by equation (1). We obtain

aC (C 2  2C C a)(R 2  a 2 (1 e2 ))  R 2a 2C 2 r  2  2 2 1 1 (2)  (C1a  C2 ) (C1a  C2 )

It appears that we can’t choose the sign of the second part of this relation. Indeed, by noticing that the equation 2 2 dR d 2aC C (ar  r 2  a 2 (1 e2 ))  C (2a 2  e2a 2  r 2  2ar)  C (a 2r 2 )  1 2 2 1  0 dr dr (C1a  C2 ) Admits a solution for aC r  2 (C1a  C2 ) We see that for this value of r the distance IM is minimal. Therefore is defined by a(1 e)  r  a(1 e) And consequently aC a(1 e)  2  a(1 e) (C1a  C2 ) Thus the sign of the second part of the equation (2) has to be alternatively positive and negative. Thus the expression of the force is depending on the position of M and consequently on the angle of revolution. This fact indicates that this force isn’t conservative because we can’t obtain a potential which is only depending on R . Consequently the only solution is given by the condition r  R The solving of equation (1) leads to

C2  0 Or 2 aC1 (ae  a  R) C2  2 (2a  ae 2  2R)

We have to choose the first solution because C2 has to be constant. Consequently the force is given by  A  F   .e 2 R (C1R) And the position of center of force is given by aC e IF  2  0 C2  aC1 Consequently I and F are combined. This force is the Newton’s.

We have demonstrated that only two central and conservative forces lead to conics trajectories, the Newton’s and the Hooke’s. This result is naturally in agreement with the theorem of Bertrand but is obtained with an original way. 6

E.Guiot An original demonstration of the Bertrand’s theorem?

5. Conclusion

In this paper we showed that the force given by  ar(C  rC )  C (a 2 (1  e 2 )  r 2 )  eC  F  A 2 1 2 e  A 2 Sin.e 3 R 3  ar(C1r  C2 ) (C1r  C2 ) Is a central force which leads to conic and closed trajectories. This force isn’t conservative except in the two limiting cases:

if C1  0the force becomes the Newton’s.

if C2  0 the force becomes the Hooke’s. This result is an original demonstration of the Bertrand’s theorem. Moreover the force we present here seems unknown until today. Consequently it could be an improvement of our knowledge on the relation between the conic and the central forces.

REFERENCES

[1] Bertrand J C.R.Acad. Sci. Paris (1873) 77 849 [2] Santos F.C., V.Soares, A.C.Tort, Phys. Rev. E 78 (2009) 036605 [3] Grandati Y., Berard A. Ménas. F. Am. J. Phys. 76, 782 (2008) [4] Guiot E., Int.J.Mod.Phys.D Vol. 24, N°. 5 (2015) 1550036 [5] Guiot.E. Int.J.Mod.Phys.D Vol.24, N°.13 (2015) 1550088 [6] Guiot.E. Nov 2015 https://hal.archives-ouvertes.fr/hal-01245759