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© J R Stockton, ≥ 2014-03-05

A translation of
Essai sur le Problème des Trois Corps
(Chapitres I et II)

by Joseph-Louis Lagrange
.
(Essay on the Three Body Problem)

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Links within this site :-
STOP PRESS - http://gallica.bnf.fr/ark:/12148/bpt6k5401485r.image.swf has, at the end, a c.18 copy of the 1772 prizewinning papers of Euler and Lagrange. Compare for errata!

Why This Page?

My introductory Lagrange Point page is Gravity 4 : The Lagrange Points.

The Paris Prize

Lagrange shared an Academy of Paris prize with Leonhard Euler, but the Essai sur le Problème des Trois Corps was not a joint work. Euler was awarded his share for his Nouvelles recherches sur le vrai mouvement de la Lune, which has nothing to do with the Lagrange Points.

Requirements :
Perfectionner les méthodes sur lesquelles est fondée la théorie de la Lune, fixer par ce moyen celles des équations de cette planète qui sont encore incertaines et examiner en particulier si on peut rendre raison par cette théorie de l'équation séculaire du mouvement de la Lune.

To perfect the methods on which the theory of the Moon is founded, thereby settling those equations of that planet which are still uncertain and in particular to examine whether we can explain by this theory the secular equation of motion of the Moon.

Joseph-Louis Lagrange, for his Essai sur le Problème des Trois Corps (Essay on the Three-Body Problem), should be credited for providing the general theory of what are now called the Lagrange Points L1, L2, L3, L4, and L5. It explains the behaviour of bodies which were later observed to remain near L4 and L5. However, Leonhard Euler had, in Considerationes de motu corporum coelestium, already briefly described L1 and L2.

Few Web authors appear to be familiar with the actual contents (as opposed to the reputed contents) of Lagrange's Essai, perhaps because it is in French. Because I could not find an English version on the Web, I felt it desirable to translate the Essai into English and HTML.

Chapters I & II of Lagrange's Essai are on the general three-mass problem and then its two constant-pattern (self-similar) solutions. From that, the subsequent derivation of the five Points, for one body having negligible mass, which Lagrange did not describe, would have been trivial.

Question : Did Lagrange write anything else relevant to the L-points? A search of the indexes of Œuvres de Lagrange at Gallica has found nothing.

About the Translation

For ease of reading, I have tried to translate Lagrange's late 18th century French mathematical language into English late 20th century mathematical language. The parts of the text which seem to be of greatest importance are  highlighted  (pages 229, 230, 272, 292).

To see the original Algebra, open in parallel a copy of Essai sur le Problème des Trois Corps, for which see References below. JavaScript is used to expand the source notation for algebra. If it is not available, Algebra will be as represented in the source file, by "words" beginning with "##".

There is now a button which will (but not in IE8) replace the substitute equations by real ones typeset by MathJax. Currently, any double-click on this page will initiate typesetting. It should be possible, in time, to treat maths which is inline with the text in a similar fashion. MathJax is written in JavaScript.

I have read, but do not expect to translate, Chapters III and IV.

The translated pages, pp. 229-292, being Chapters I and II, have been checked/corrected by a native Francophone.

These Œuvres of Lagrange were edited by J.-A. Serret; his "Note de l'Éditeur" that follows Chapter IV in pp. 324-331 may well be significant for the Points, which could call for its translation here - eventually.

Sources

I have found at Gallica, on 2012-08-19, "Recueil des pieces qui ont remporte les prix de l'Academie Royale des Sciences", Tome 9 (1764-1772), published in 1777. It contains the papers for which Euler and Lagrange shared the 1772 Prize. It is a little hard to read, particularly some of the mathematics, but it is authoritative.

My prime source has been the copy at the University of Liege (ULg) web site. That I believe to be identical in content to the copy in Tome 6 of the "Œuvres de Lagrange" at Gallica, edited in the second half of the 19th century by J.-A. Serret. The Liege copy is easier to use. The printing is good. The pagination differs. The known errors differ.

I use the pagination of Liege and the "Œuvres de Lagrange".

Errors

Known errors are noted on a background like this. Typos seen in the c.19 "Œuvres" version have been checked against the c.18 "Prize" version.

Difference Summary

Rough Index to the Essai

ButtonFor doing the typesetting

Notice229-231Overall introduction
Ch. I231-267General Formulae for the Solution of the Three-Body Problem.
I231-232First equations
II232-234
III234-235
IV236-238
V238-240
VI241-242
VII242-242
VIII243-243
IX243-246
X246-247
XI248-249
XII249-251
XIII251-252
XIV252-253
XV254-254
XVI255-255
XVII255-259
XVIII260-261
XIX261-264
XX264-266
XXI266-267
XXII267-271Recapitulation - Summary of Chapter I
Ch. II272-292Solution of the Three-Body Problem in Different Cases.
XXIII272-272Introduction
XXIV272-273Constant-size cases
XXV273-275Equilateral circular case
XXVI275-277Collinear circular case
XXVII277-279A possibility shown to be uninteresting
XXVIII279-280On collinear speeds
XXIX280-283Constant-ratio cases
XXX283-287Equilateral constant-ratio case
XXXI288-290Collinear constant-ratio case
XXXII290-292Conic section
XXXIII292-292Conclusion : exactly soluble in two cases
Ch. III

 
293-303

 
Modification of the formulae of the previous chapter,
for the case where one supposes that one of the three
bodies is distant from the other two.
Ch. IV303-324 On the theory of the Moon.
Ed-Note324-331Editor's Note

"Essay on the Three-Body Problem" by J-L Lagrange

I judge, from the correspondence between Lagrange and D'Alembert, that this work was at least well-in-hand in the summer of 1770.


The button should cause the equation-substitutes to be replaced throughout, after a while, by MathJax "typeset" versions of the equations in the original PDF - demonstration :- ## $$ \sin^2\phi + \cos^2\phi = 1 $$ A progress indication appears at bottom left of the window.
UpdateTime   (The MathJax.Hub Object)   UpdateDelay
* Not * * IE8 *
For me, Chrome is faster than Firefox, and Opera is slow.


Œuvres de Lagrange,
Tome 6, pp. 229-292;

Joseph-Louis, Comte Lagrange (1736-1813)
AD 1772.
229

ESSAY

ON

THE THREE-BODY PROBLEM

Juvat integros accedere fontes.
LUCR.
I love stumbling upon yet unfound sources.
LUCRETIUS, De Rerum Natura, I, v. 927
Tr. EjH.
It is pleasant to handle an untouched subject.
Tr. Henry Fielding?




(Prize of the Royal Academy of Sciences of Paris, volume IX, 1772.)



These researches contain a method of solving the three body problem, different from all those which have been given before. It consists of not using in the determination of the orbit of each body any other elements than the distances between the three bodies, that is to say, the triangle formed by these bodies at each instant. For that, it is necessary first to find the equations which determine those distances as a function of time ; then, by supposing the distances known, it is necessary to deduce from them the relative movement of the bodies with respect to some fixed plane. One will see, in the first Chapter, what I have done to fulfil these two objects, of which the second particularly requires a delicate and rather complicated analysis. At the end of this Chapter, I collect

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the main formulae that I have found, which will contain the solution to the three-body problem taken in full generality.

The second Chapter is intended to examine how and in which cases the three bodies can move so that their distances shall always be constant, or shall at least keep constant ratios between themselves. I find that these conditions can only be satisfied in two cases : one, when the three bodies are arranged in one straight line, and the other, when they form an equilateral triangle ; then each of the three bodies describes around the two others circles or conic sections, as if there were only two bodies. This research is really no more than for pure curiosity ; but I have thought that it would not be out of place in a work which is principally concerned with the three-body problem, viewed in its full extent.

  In the third Chapter, I suppose that the distance of one of the three bodies from the other two is very large, and I apply the general solution of the first Chapter to that situation, which is, as one knows, that of the Earth, the Moon, and the Sun.

  Finally, in the fourth Chapter, I treat in particular the Theory of the Moon ; I give the formulae which contain that Theory, and I show, by a small sample calculation, how one must use these formulae to deduce the irregularities of the movement of the Moon around the Earth.

  Lack of time and other essential occupations have not allowed me to embark on the topic in all the detail necessary to respond in a proper manner to the principal points of the question proposed by the Academy : so I was at first reluctant to present these researches for the Contest, and I was only persuaded by the hope that this illustrious company would perhaps find my method of solving the three-body problem worthy of some attention, as much by its novelty and its singularity as by the considerable difficulties of calculation which it contains.

  If the Academy deigns to honour my work by its vote, it will be a powerful motive for me to work on perfecting it, and I do not despair

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of being able to get by my method a Theory of the Moon as complete as one can ask for in the present imperfect state of Analysis.

FIRST CHAPTER.

GENERAL FORMULAE FOR THE SOLUTION OF THE THREE-BODY PROBLEM

  Let A, B, C be the masses of the three bodies which are mutually attracted in direct ratio of their masses and in inverse ratio of the square of the distances ; and name x, y, z as the rectangular coordinates of the orbit of body B around body A, x', y', z' the rectangular coordinates of the orbit of body C around the same body A, coordinates which one supposes always parallel to three fixed and mutually perpendicular lines ; finally let r, r', r'' be the distances between the bodies A and B, A and C, B and C, so that one has ##. $$ r = \sqrt { x^2+y^2+z^2 }, ~~~~ r' = \sqrt { x'^2+y'^2+z'^2 }, ~~~~ r'' = \sqrt { (x'-x)^2+(y'-y)^2+(z'-z)^2 } . $$ One will have, as one knows, by taking the element of time dt constant, the following six equations ##A3; $$ \left\{ \begin{array}{l} {d^2x \over dt^2} + ( {{A+B} \over r^3} + {C \over r''^3} ) x + C ( {1 \over r'^3} - {1 \over r''^3} ) x' = 0, \\ \tag{A} {d^2y \over dt^2} + ( {{A+B} \over r^3} + {C \over r''^3} ) y + C ( {1 \over r'^3} - {1 \over r''^3} ) y' = 0, \\ {d^2z \over dt^2} + ( {{A+B} \over r^3} + {C \over r''^3} ) z + C ( {1 \over r'^3} - {1 \over r''^3} ) z' = 0; \end{array} \right. $$ ##B3; $$ \left\{ \begin{array}{l} {d^2x' \over dt^2} + ( {{A+C} \over r'^3} + {B \over r''^3} ) x' + B ( {1 \over r^3} - {1 \over r''^3} ) x = 0, \\ \tag{B} {d^2y' \over dt^2} + ( {{A+C} \over r'^3} + {B \over r''^3} ) y' + B ( {1 \over r^3} - {1 \over r''^3} ) y = 0, \\ {d^2z' \over dt^2} + ( {{A+C} \over r'^3} + {B \over r''^3} ) z' + B ( {1 \over r^3} - {1 \over r''^3} ) z = 0; \end{array} \right. $$

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with the aid of which one will be able to determine the relative orbits of bodies B and C around body A.

  If one also does ##, $$ x'-x = x'', ~~~~ y'-y = y'', ~~~~ z'-z =z'', $$ such that x'', y'', z'' are the rectangular coordinates of the orbit of body C around B, one will have ##, $$ r'' = \sqrt{x''^2+y''^2+z''^2}, $$ and, subtracting respectively the first three equations from the last three, one will have these three ##C3, $$ \left\{ \begin{array}{l} {d^2x'' \over dt^2} + ( {{B+C} \over r''^3} + {A \over r^3} ) x'' + A ( {1 \over r'^3} - {1 \over r^3} ) x' = 0, \\ \tag{C} {d^2y'' \over dt^2} + ( {{B+C} \over r''^3} + {A \over r^3} ) y'' + A ( {1 \over r'^3} - {1 \over r^3} ) y' = 0, \\ {d^2z'' \over dt^2} + ( {{B+C} \over r''^3} + {A \over r^3} ) z'' + A ( {1 \over r'^3} - {1 \over r^3} ) z' = 0, \end{array} \right. $$ which will express the relative movement of body C around body B.

  It is good to notice the analogy which exists between these nine equations (A), (B), (C); it is that equations (A) become equations (B) by changing just B to C, x to x', y to y', z to z', r to r', and reciprocally ; and that likewise these equations become equations (C) by changing A to C, x to x'', y to y'', z to z'', r to r'', and vice versa ; and the same analogy will hold in all of the formulae which we shall find in the following.

  Multiply the first of the equations (A) by y and the second by x, and then subtract one from the other, to get ##. $$ { { yd^2x - xd^2y } \over dt^2 } + C ( {1 \over r'^3} - {1 \over r''^3} ) (yx'-xy') = 0 .$$

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  Combining similarly the first two of equations (B) and the first two of equations (C), one gets these two ##, ##. $$ { { y'd^2x' - x'd^2y' } \over dt^2 } + B ( {1 \over r^3} - {1 \over r''^3} ) (xy'-yx') = 0 , \\ { { y''d^2x'' - x''d^2y'' } \over dt^2 } + A ( {1 \over r'^3} - {1 \over r^3} ) (x'y''-y'x'') = 0 .$$

  But ##; $$ x'' = x' - x, ~~~~ y'' = y' - y ~; $$ so ##; $$ x'y'' - y'x'' = xy' -yx' ~; $$ thus, in adding together the three preceding equations, after having divided the first by C, the second by B and the third by A, one gets this ##. $$ { {yd^2x-xd^2y} \over Cdt^2} + { {y'd^2x'-x'd^2y'} \over Bdt^2} + { {y''d^2x''-x''d^2y''} \over Adt^2} = 0. $$

  One will find in the same manner these two further equations ##, ##, $$ { {zd^2x-xd^2z} \over Cdt^2} + { {z'd^2x'-x'd^2z'} \over Bdt^2} + { {z''d^2x''-x''d^2z''} \over Adt^2} = 0, \\ { {zd^2y-yd^2z} \over Cdt^2} + { {z'd^2y'-y'd^2z'} \over Bdt^2} + { {z''d^2y''-y''d^2z''} \over Adt^2} = 0, $$

  Thus one will get, on integrating ##D3, $$ \left\{ \begin{array}{l} { {ydx-xdy} \over Cdt} + { {y'dx'-x'dy'} \over Bdt} + { {y''dx''-x''dy''} \over Adt} = a, \\ \tag{D} { {zdx-xdz} \over Cdt} + { {z'dx'-x'dz'} \over Bdt} + { {z''dx''-x''dz''} \over Adt} = b, \\ { {zdy-ydz} \over Cdt} + { {z'dy'-y'dz'} \over Bdt} + { {z''dy''-y''dz''} \over Adt} = c, \end{array} \right. $$ a, b, c being arbitrary constants.

  Moreover, if one multiplies the first of equations (A) by dx/C, the first of equations (B) by dx'/C, and the first of equations (C) by

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dx''/A, and then one adds them together, one gets, because x'' = x'-x, ##. $$ \frac{dxd^2x}{Cdt^2} + \frac{dx'd^2x'}{Bdt^2} + \frac{dx''d^2x''}{Adt^2} + (A+B+C) \left( \frac{xdx}{Cr^3} + \frac{x'dx'}{Br'^3} + \frac{x''dx''}{Ar''^3} \right) = 0 . $$

  Similarly, one finds ##, ##. $$ \frac{dyd^2y}{Cdt^2} + \frac{dy'd^2y'}{Bdt^2} + \frac{dy''d^2y''}{Adt^2} + (A+B+C) \left( \frac{ydy}{Cr^3} + \frac{y'dy'}{Br'^3} + \frac{y''dy''}{Ar''^3} \right) = 0 , \\ \frac{dzd^2z}{Cdt^2} + \frac{dz'd^2z'}{Bdt^2} + \frac{dz''d^2z''}{Adt^2} + (A+B+C) \left( \frac{zdz}{Cr^3} + \frac{z'dz'}{Br'^3} + \frac{z''dz''}{Ar''^3} \right) = 0 . $$

  Then, adding together these three equations and putting

r dr   in place of  x dx + y dy + z dz,
r' dr'  in place of  x' dx' + y' dy' + z' dz',
r'' dr''  in place of  x'' dx'' + y'' dy'' + z'' dz'',
one gets an integrable equation of which the integral will be ##E1 $$ \tag{E} {\Tiny \frac{dx^2+dy^2+dz^2}{Cdt^2} + \frac{dx'^2+dy'^2+dz'^2}{Bdt^2} + \frac{dx''^2+dy''^2+dz''^2}{Adt^2} - 2 (A+B+C) \left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) = f . } $$ f being an arbitrary constant.

  Those are the only exact integrals which one has been able to find so far ; so, as there are in all six variables x, y, z, x', y', z', it is clear that, if one can find two further integrals, the problem will be reduced to first differences ; but one scarcely dares hope to reach that in the present state of imperfection of Analysis.

  Let us suppose, for brevity, ##, ##, ##, $$ u^2 = \frac{dx^2 + dy^2 + dz^2}{dt^2} , \\ u'^2 = \frac{dx'^2 + dy'^2 + dz'^2}{dt^2} , \\ u''^2 = \frac{dx''^2 + dy''^2 + dz''^2}{dt^2} , $$

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so that u, u', u'' express the relative speeds of bodies B, C around A, and of C around B ; it is clear that one will have ##, ##, ##. $$ \frac{d^2(r^2)}{2dt^2} = \frac{xd^2x + yd^2y + zd^2z}{dt^2} + u^2 , \\ \frac{d^2(r'^2)}{2dt^2} = \frac{x'd^2x' + y'd^2y' + z'd^2z'}{dt^2} + u'^2 , \\ \frac{d^2(r''^2)}{2dt^2} = \frac{x''d^2x'' + y''d^2y'' + z''d^2z''}{dt^2} + u''^2 . $$

  So, putting in these equations instead of ##, ... $$ \frac{d^2x}{dt^2}, ~~~~ \frac{d^2y}{dt^2}, ~~~~ \frac{d^2z}{dt^2}, ~~~~ \frac{d^2x'}{dt^2}, ~ ... $$ their values drawn from equations (A), (B), (C), and observing that ##, $$ x^2 + y^2 + z^2 = r^2, ~~~~ x'^2 + y'^2 + z'^2 = r'^2, ~~~~ x''^2 + y''^2 + z''^2 = r''^2, $$ and ##, ##, $$ xx' + yy' + zz' = \frac{r^2 + r'^2 -r''^2}2 , \\ x'x'' + y'y'' + z'z'' = x'^2 + y'^2 + z'^2 - (x'x + y'y + z'z) = \frac{r'^2 + r''^2 - r^2}2 , $$ one will have, after having moved all of the terms to the same side, ##F3. $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} + \left(\frac{A+B}{r^3} + \frac C{r''^3} \right)r^2 + \frac C2 \left( \frac 1{r'^3} - \frac 1{r''^3} \right) (r^2 + r'^2 - r''^2) - u^2 = 0 , \\ \tag{F} \frac{d^2(r'^2)}{2dt^2} + \left(\frac{A+C}{r'^3} + \frac B{r''^3} \right)r'^2 + \frac B2 \left( \frac 1{r^3} - \frac 1{r''^3} \right) (r^2 + r'^2 - r''^2) - u'^2 = 0 , \\ \frac{d^2(r''^2)}{2dt^2} + \left(\frac{B+C}{r''^3} + \frac A{r^3} \right)r''^2 + \frac A2 \left( \frac 1{r'^3} - \frac 1{r^3} \right) (r'^2 + r''^2 - r^2) - u''^2 = 0 . \end{array} \right. $$

  Then, if one can have the values of u2, u'2, u''2 expressed in terms of r, r', r'' only, one will have three equations in these three last variables and the time t, with the help of which one will be able to determine the relative position of the bodies at each instant.

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  Thus one has, by differentiating the values of u2, u'2, u''2, ##, ##, ##; $$ udu = \frac{dxd^2x + dyd^2y + dzd^2z}{dt^2} , \\ u'du' = \frac{dx'd^2x' + dy'd^2y' + dz'd^2z'}{dt^2} , \\ u''du'' = \frac{dx''d^2x'' + dy''d^2y'' + dz''d^2z''}{dt^2} ; $$ so, if one makes here the same substitutions as before, and one supposes for a moment ##, ##, ##, $$ dV = x'dx + y'dy + z'dz , \\ dV' = xdx' + ydy' + zdz' , \\ dV'' = x'dx'' + y'dy'' + z'dz'' , $$ because ##, $$ {\small xdx + ydy + zdz = rdr, ~~~~ x'dx' + y'dy' + z'dz' = r'dr', ~~~~ x''dx'' + y''dy'' + z''dz'' = r''dr'' , } $$ one gets ##, ##, ##. $$ udu = - \left( \frac{A+B}{r^3} + \frac C{r''^3} \right) rdr - C \left( \frac1{r'^3} - \frac1{r''^3}\right) dV , \\ u'du' = - \left( \frac{A+C}{r'^3} + \frac B{r''^3} \right) r'dr' - B \left( \frac1{r^3} - \frac1{r''^3}\right) dV' , \\ u''du'' = - \left( \frac{B+C}{r''^3} + \frac A{r^3} \right) r''dr'' - A \left( \frac1{r'^3} - \frac1{r^3}\right) dV'' . $$

  Let, for brevity, ##, ##, ##, $$ dR = \frac{2rdr}{r''^3} + 2\left( \frac1{r'^3} - \frac1{r''^3} \right) dV , \\ dR' = \frac{2r'dr'}{r''^3} + 2\left( \frac1{r^3} - \frac1{r''^3} \right) dV' , \\ dR'' = \frac{2r''dr''}{r^3} + 2\left( \frac1{r'^3} - \frac1{r^3} \right) dV'' , $$

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and one will get ##, ##, $$ u^2 = \frac{2(A+B)}{r} - CR , \\ u'^2 = \frac{2(A+C)}{r'} - BR' , \\ u''^2 = \frac{2(B+C)}{r''} - AR'' ; $$ ##; so that equations (F) become ##G3; $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} - \frac{A+B}{r} + C\left[\frac{r^2}{r''^3} + \frac12\left(\frac1{r'^3}-\frac1{r''^3}\right)(r^2+r'^2-r''^2) + R\right] = 0 , \\ \tag{G} \frac{d^2(r'^2)}{2dt^2} - \frac{A+C}{r'} + B\left[\frac{r'^2}{r''^3} + \frac12\left(\frac1{r^3}-\frac1{r''^3}\right)(r^2+r'^2-r''^2) + R'\right] = 0 , \\ \frac{d^2(r''^2)}{2dt^2} - \frac{B+C}{r''} + A\left[\frac{r''^2}{r^3} + \frac12\left(\frac1{r'^3}-\frac1{r^3}\right)(r'^2+r''^2-r^2) + R''\right] = 0 ; \end{array} \right. $$ and no more is needed than to find the values of ##. $$ dV, ~~~ dV', ~~~~ dV'' . $$

  For that, I make ##, $$ d\rho = x'dx + y'dy + z'dz - xdx' -ydy' - zdz' , $$ and as one has ##, $$ xx' + yy' + zz' = \frac{r^2 + r'^2 - r''^2}2 , $$ one will get, by differentiation, ##; $$ xdx' + ydy' + zdz' + x'dx + y'dy + z'dz = rdr + r'dr' - r''dr'' ; $$ then ##, ##, $$ dV = \frac{rdr + r'dr' - r''dr'' + d\rho}2 , \\ dV' = \frac{rdr + r'dr' - r''dr'' - d\rho}2 , $$ and then ##, $$ dV'' = r'dr' - dV , $$

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to get ##. $$ dV'' = \frac{r'dr' + r''dr'' - rdr - d\rho}2 . $$

  All that remains to be done is to get the value of  ; to get that, I differentiate, and I have ##; $$ d^2\rho = x'd^2x + y'd^2y + z'd^2z - xd^2x' - yd^2y' - zd^2z' ; $$ I substitute in the place of d2x, d2y, d2z, ... the values drawn from equations (A) and (B), and making other suitable substitutions, I find ##, $$ {\small \frac{d^2\rho}{dt^2} = - \frac12 \left(\frac{A+B}{r^3} + \frac C{r''^3} - \frac{A+C}{r'^3} - \frac B{r''^3} \right) (r^2 + r'^2 -r''^2) - C \left(\frac 1{r'^3} - \frac 1{r''^3} \right) r'^2 + B \left(\frac 1{r^3} - \frac 1{r''^3} \right) r^2 , } $$ or else ##. $$ {\small \frac{2d^2\rho}{dt^2} + A \left( \frac1{r^3} - \frac1{r'^3} \right)(r^2 + r'^2 - r''^2) + B \left( \frac1{r^3} - \frac1{r''^3} \right)(r'^2 - r^2 - r''^2) + C \left( \frac1{r''^3} - \frac1{r'^3} \right)(r^2 - r'^2 - r''^2) = 0 . } $$

  Let us suppose, to put our formulae in a simpler form, ##, ##, ##, $$ {r'^2 + r''^2 -r^2 \over 2} = p , ~~~~ {r^2 + r''^2 -r'^2 \over 2} = p' , ~~~~ {r^2 + r'^2 -r''^2 \over 2} = p'' , \\ {1 \over r'^3} - {1 \over r''^3} = q , ~~~~ {1 \over r^3} - {1 \over r''^3} = q' , ~~~~ {1 \over r'^3} - {1 \over r^3} = q'' = q - q' , \\ { u'^2 + u''^2 - u^2 \over 2 } = \nu , ~~~~ { u^2 + u''^2 - u'^2 \over 2 } = \nu' , ~~~~ { u^2 + u'^2 - u''^2 \over 2 } = \nu'' , $$ and one will get first, for the determination of , this equation ##H1. $$ \tag{H} \frac{d^2\rho}{dt^2} + Cpq - Bp'q' - Ap''q'' = 0 . $$

  One has next ##, $$ dV = \frac{dp'' + d\rho}2 , ~~~~ dV' = \frac{dp'' - d\rho}2 , ~~~~ dV'' = \frac{dp - d\rho}2 , $$

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from which ##, ##, ##. $$ dR = {2rdr \over r''^3} + q(dp'' + d\rho) , \\ dR' = {2r'dr' \over r''^3} + q'(dp'' - d\rho) , \\ dR'' = {2r''dr'' \over r^3} + q''(dp - d\rho) . $$

  But ##; $$ \frac1{r''^3} = \frac1{r^3} - q', ~~~~ 2rdr = dp' + dp'' ; $$ so ##; $$ \frac{2rdr}{r''^3} = \frac{2dr}{r^2} - q'(dp' + dp'') ; $$ one will find likewise ##, ##; $$ \frac{2r'dr'}{r''^3} = \frac{2dr'}{r'^2} - q(dp + dp'') , \\ \frac{2r''dr''}{r^3} = \frac{2dr''}{r''^2} + q'(dp' + dp) ; $$ so that in substituting these values, and making for simplicity ##I3, $$ \left\{ \begin{array}{l} dQ = q'dp' - q''dp'' - qd\rho , \\ \tag{I} dQ' = q dp + q'' dp'' + q'd\rho , \\ dQ'' = -qdp -q'dp + q''d\rho , \end{array} \right. $$ one gets ##, $$ R = - \frac2r - Q, ~~~~ R' = - { 2 \over r'} - Q' , ~~~~ R'' = - { 2 \over r''} - Q '' , $$ and from that ##J3. $$ \left\{ \begin{array}{l} u^2 = \frac{2(A+B+C)}r + CQ , \\ \tag{J} u'^2 = \frac{2(A+B+C)}{r'} + BQ' , \\ u''^2 = \frac{2(A+B+C)}{r''} + AQ'' . \end{array} \right. $$

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  Now one gets ##, ##; $$ \frac{r^2}{r''^3} = \frac1r - q'(p'+p'') , \\ \frac12 \left(\frac1{r'^3} - \frac1{r''^3} \right) (r^2 + r'^2 -r''^2) = qp'' ; $$ so, adding those two equations, and putting q'' in place of q-q', one gets ##; $$ \frac{r^2}{r''^3} + \frac12 \left(\frac1{r'^3} - \frac1{r''^3} \right) (r^2 + r'^2 -r''^2) = \frac1r - p'q' + p''q'' ; $$ one finds likewise ##, ##; $$ \frac{r'^2}{r''^3} + \frac12 \left(\frac1{r^3} - \frac1{r''^3} \right) (r^2 + r'^2 -r''^2) = \frac1{r'} - pq - p''q'' , \\ \frac{r''^2}{r^3} + \frac12 \left(\frac1{r'^3} - \frac1{r^3} \right) (r'^2 + r''^2 -r^2) = \frac1{r''} + pq + p'q' ; $$ so, making all these substitutions in equations (G) or (F) of the preceding Articles, they become these ##K3. $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} - {A+B+C \over r} - C(p'q'-p''q''+Q) = 0 , \\ \tag{K} \frac{d^2(r'^2)}{2dt^2} - {A+B+C \over r'} - B(pq+p''q''+Q') = 0 , \\ \frac{d^2(r''^2)}{2dt^2} - {A+B+C \over r''} - A(-pq-p'q'+Q'') = 0 . \\ \end{array} \right. $$

  Thus one will be able to, with the aid of these three equations, determine the three radii r, r', r'' in terms of t, which will give for each instant the relative position of the bodies among themselves.

  It is good to note that, if one divides the first of these equations by C, the second by B and the third by A, and that next one adds them together, one gets (because dQ + dQ' + dQ'' = 0, and in consequence Q + Q' + Q'' = const.) this ##L1, $$ \tag{L} {\small \frac{d^2(r^2)}{2Cdt^2} + \frac{d^2(r'^2)}{2Bdt^2} + \frac{d^2(r''^2)}{2Adt^2} - (A+B+C) \left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) = \text{const.} , } $$ which can take the place of any one of the three equations (K).

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  One can also put the same equations (K) in another form as follows.

  I multiply the first of these equations by d(r2), and I integrate it next to have ##, $$ \frac{d(r^2)^2}{4dt^2} - 2(A+B+C)r - C\int (p'q'-p''q'')d(r^2) - C\int Qd(r^2) + L = 0 , $$ L being an arbitrary constant.

  Then ##; $$ \int Qd(r^2) = Qr^2 - \int r^2dQ ; $$ but ##; $$ dQ = q'dp' - q''dp'' - q d\rho ; $$ moreover, because r2=p'+p'', one gets ## ##; $$ \begin{aligned} & (p'q'-p''q'')d(r^2) - r^2(q'dp'-q''dp'') ~~~~~~~~~~ \\ & ~~ = - (p'q'-p''q'')(dp'+dp'') + (p'+p'')(q'dp'-q''dp'') = q(p''dp'-p'dp'') ; \end{aligned} $$ such that if one makes, for brevity, ##, $$ dP = q(p''dp' - p'dp'' -r^2d\rho) , $$ one gets, by omitting the constant L which can be supposed to be contained in P, and dividing the whole equation by r2, ##. $$ \frac{dr^2}{dt^2} - \frac{2(A+B+C)}r + C\left(\frac P{r^2} - Q\right) = 0 . $$

  Making likewise ##, ##, $$ dP' = q'(p''dp - pdp'' + r'^2d\rho) , \\ dP'' = q''(pdp' - p'dp' + r''^2d\rho) , $$

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one will find by operations resembling those preceding ##, ##. $$ \frac{dr'^2}{dt^2} - \frac{2(A+B+C)}{r'} + B\left(\frac{P'}{r'^2} - Q'\right) = 0 , \\ \frac{dr''^2}{dt^2} - \frac{2(A+B+C)}{r''^2} + A\left(\frac{P''}{r''^2} - Q''\right) = 0 . $$

  And, if one reduces these equations respectively by equations (K) found above, then next one divides the remaining equations by r, r', r'', one gets these three ##M3. $$ \left\{ \begin{array}{l} \frac{d^2r}{dt^2} + \frac{A+B+C}{r^2} - C\left(\frac{p'q'-p''q''}r + \frac P{r^3}\right) = 0 , \\ \tag{M} \frac{d^2r'}{dt^2} + \frac{A+B+C}{r'^2} - B\left(\frac{pq+p''q''}{r'} + \frac{P'}{r'^3}\right) = 0 , \\ \frac{d^2r''}{dt^2} + \frac{A+B+C}{r''^2} - A\left(\frac{-pq-p'q'}{r''} + \frac {P''}{r''^3}\right) = 0 . \end{array} \right. $$

  We have thus reduced the six initial equations (A), (B) which contain the solution to the three-body problem taken in full generality to three other equations between the three distances r, r', r'' and the time t. It is true that these reductions each contain two integration signs (which is evident on substituting the values of Q, Q', Q'', or of P, P', P'' and of ). and that in this regard they are less simple than the initial equations ; but, on the other hand, they have the advantage of not containing any radical, which seems to me to be of great importance in these types of problems.

  Let us suppose then that one has determined by equations (K) or (M) the three variables r, r', r'' in terms of t ; one still only knows from that the relative positions of the bodies, that is to say, the triangle that the three bodies form at each instant ; therefore it remains to be seen how one can determine next the orbit itself of each body, that is to say, the six variables x, y, z, x', y', z'.

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  In order to do this, we remark firstly that by knowing r, r', r'' one will know also u, u', u'', and dV, dV', dV'' by the formulae of Article V. So that one will have {by putting p'' in place of (r2 + r'2 - r''2)/2 and v'' in place of (u2 + u'2 - u''2)/2} the ten following equations ##, ##, ##, ##, ##, ##, ##, ##, ##, ##. $$ \begin{align} & x^2 + y^2 + z^2 = r^2 , \\ & x'^2 + y'^2 + z'^2 = r'^2 , \\ & xx' + yy' + zz' = p'' , \\ & xdx + ydy + zdz = rdr , \\ & x'dx' + y'dy' + z'dz' = r'dr' , \\ & x'dx + y'dy + z'dz = dV , \\ & xdx' + ydy' + zdz' = dV' , \\ & dx^2 + dy^2 + dz^2 = u^2dt^2 , \\ & dx'^2 + dy'^2 + dz'^2 = u'^2dt^2 , \\ & dxdx' + dydy' + dzdz' = \nu''dt^2 . \end{align} $$

  However, by considering the quantities x, y, z, x', y', z', dx, dy, dz, dx', dy', dz' as so far unknown, it is clear that the preceding equations do not suffice to determine them, because one would have twelve unknowns, and only ten equations ; but, if one includes with these equations the three equations (D) of article II, one will then get one equation more than there are unknowns, and the difficulty will consist only in solving these equations.

  I see, with regard to the equations of the previous article, that they can only take the place of nine equations, because, in eliminating some of the unknowns, it happens that others also disappear,

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so that one arrives by this means at one equation where there enter only the quantities r2, r'2, p'', ... . In order to prove it as easily as possible, I take first the three equations ##, ##, ##, $$ xdx + ydy + zdz = rdr , \\ x'dx + y'dy + z'dz = dV , \\ dx'dx + dy'dy + dz'dz = \nu''dt^2 , $$ and I get from them by the ordinary rules of elimination the values of dx, dy, dz ; I have, in doing that, for brevity. ##, ##, ##, ##, $$ \alpha = y'dz'-z'dy',~~~~ \alpha' = zdy'-ydz',~~~~ \alpha'' = yz'-y'z, \\ \beta = z'dx'-x'dz',~~~~ \beta' = xdz'-zdx',~~~~ \beta'' = zx'-z'x, \\ \gamma = x'dy'-y'dx',~~~~ \gamma' = ydx'-xdy',~~~~ \gamma'' = xy'-yx', \\ \delta = x(y'dz'-z'dy') - y(x'dz'-z'dx') + z(x'dy'-y'dx') , $$ I have, I say, ##, ##, ##. $$ dx = { \alpha rdr + \alpha' dV + \alpha''\nu''dt^2 \over \delta } , \\ dy = { \beta rdr + \beta' dV + \beta'' \nu''dt^2 \over \delta } , \\ dz = { \gamma rdr + \gamma' dV + \gamma''\nu''dt^2 \over \delta } . $$

  However, I remark that one has ##, ##, ##, ##, $$ \begin{aligned} \alpha^2 + \beta^2 + \gamma^2 &= (x'^2+y'^2+z'^2)(dx'^2+dy'^2+dz'^2) - (x'dx'+y'dy'+z'dz')^2 \\ &= r'^2u'^2dt^2 - (r'dr')^2 , \\ \alpha'^2 + \beta'^2 + \gamma'^2 &= (x^2+y^2+z^2)(dx'^2+dy'^2+dz'^2) - (xdx'+ydy'+zdz')^2 \\ &= r^2u'^2dt^2 - dV'^2 , \\ \alpha''^2 + \beta''^2 + \gamma''^2 &= (x^2+y^2+z^2)(x'^2+y'^2+z'^2) - (xx'+yy'+zz')^2 \\ &= r^2r'^2 - p''^2 , \\ \alpha\alpha' + \beta\beta' + \gamma\gamma' &= (x'dx'+y'dy'+z'dz')(xdx'+ydy'+zdz') \\ & ~~~~~~ - (xx'+yy'+zz')(dx'^2+dy'^2+dz'^2) \\ &= r'dr'dV' - p''u'^2dt^2 , \end{aligned} $$

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## ## ##, ## ## ##; $$ \begin{aligned} \alpha\alpha'' + \beta\beta'' + \gamma\gamma'' &= (x'dx'+y'dy'+z'dz')(xx'+yy'+zz') \\ & ~~~ - (xdx'+ydy'+zdz')(x'^2+y'^2+z'^2) \\ &= p''r'dr' - r'^2dV' , \\ \alpha'\alpha'' + \beta'\beta'' + \gamma'\gamma'' &= (xdx'+ydy'+zdz')(xx'+yy'+zz') \\ & ~~~ - (x'dx'+y'dy'+z'dz')(x^2+y^2+z^2) \\ &= p''dV' - r^2r'dr' ; \end{aligned} $$ such that, if one squares the three preceding equations, and then adds them together, one gets, after multiplying by δ2, and making suitable substitutions, ## ## ##, $$ \begin{aligned} \delta^2(dx^2+dy^2+dz^2) = & (rdr^2)[r'^2u'^2dt^2-(r'dr')^2] + dV^2(r^2u'^2dt^2 - dV'^2) \\ & + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\ & + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') . \end{aligned} $$ Similarly, if one takes the three equations ##, ##, ##, $$ xx + yy + zz = r^2 , \\ x'x + y'y + z'z = p'' , \\ xdx' + ydy' + zdz' = dV' , $$ and one takes from them the values of x, y and z, it is easy to see that one gets for x, y, z the same expressions that one has found higher up for dx, dy, dz, by changing there only rdr to r2, dV to p'', v''dt2 to dV'; then, doing the same operations and the same substitutions as above, one gets another equation ## ## ##, $$ \begin{aligned} \delta^2(x^2+y^2+z^2) = & r^4 [r'^2u'^2dt^2 - (r'dr')^2] + p''^2(r^2u'^2dt^2 - dV'^2) \\ & + dV'^2(r^2r'^2-p''^2) +2r^2p''(r'dr'dV' -p''u'^2dt^2) \\ & + 2r^2dV'(p''r'dr'-r'^2dV') + 2p''dV'(p''dV'-r^2r'dr') . \end{aligned} $$ Now one has ## $$ dx^2 + dy^2 + dz^2 = u^2dt^2, ~~~\text{and} ~~~~ x^2+y^2+z^2 = r^2 $$
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then one gets the following two equations ## ## ##,
## ## ##, $$ \begin{aligned} \delta^2u^2dt^2 & = (rdr)^2(r'^2u'^2dt^2 - r'^2dr'^2) + dV^2(r^2u'^2dt^2-dV'^2) \\ & + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\ & + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') , \\ \delta^2r^2 = & r^4(r'^2u'^2dt^2-r'^2dr'^2) + p''^2(r^2u'^2dt^2-dV'^2) \\ & + dV'^2(r^2r'^2-p''^2) + 2r^2p''(r'dr'dV'-p''u'^2dt^2) \\ & + 2r^2dV'(p''r'dr'-r'^2dV') + 2p''dV'(p''dV'-r^2r'dr') . \end{aligned} $$

  From which, eliminating δ2, one gets an equation in only the known quantities r2, r'2, ... .

  If one takes from the last equation the value of δ2, one gets, by reducing and removing that which goes out. ##; $$ \delta^2 = r'^2(r^2u'^2dt^2-r^2dr'^2-dV'^2) + 2p''r'dr'dV' - p''^2u'^2dt^2 ; $$ and, this value of δ2 being substituted in the other equation, one gets ## ## ## ##; $$ \begin{aligned} r'^2 & (r^2u'^2dt^2-r^2dr'^2-dV'^2)u^2dt^2 + (2p''r'dr'dV'-p''^2u'^2dt^2)u^2dt^2 \\ = & (rdr)^2(r'^2u'^2dt^2-r'^2dr'^2) + dV^2(r^2u'^2dt^2-dV'^2) \\ & + (\nu''dt^2)^2(r^2r'^2-p''^2) + 2rdrdV(r'dr'dV'-p''u'^2dt^2) \\ & + 2rdr\nu''dt^2(p''r'dr'-r'^2dV') + 2dV\nu''dt^2(p''dV'-r^2r'dr') ; \end{aligned} $$ or else, by arranging the terms, ## ## ## ##. $$ \begin{aligned} & (r^2r'^2-p''^2)(u^2u'^2-\nu''^2)dt^4 + (rdr.r'dr'-dVdV')^2 \\ & ~~ - [r^2(r'dr')^2-2p''r'dr'dV' + r'^2dV'^2]u^2dt^2 \\ & ~~ - [r'^2(rdr)^2 - 2p''rdrdV + r^2dV^2]u'^2dt^2 \\ & ~~ - 2[p''(rdr.r'dr'+dVdV')-r^2r'dr'dV-r'^2rdrdV']\nu''dt^2 = 0 . \end{aligned} $$

  Now (Article V), ##; $$ dV = {dp''+d\rho \over 2} , ~~~~ \text{and} ~~~~ dV'' = {dp''-d\rho \over 2} ; $$

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moreover one has, by the formulae of the same Article, ##, $$ r^2 = p'+p'', ~~~~ r'^2 = p+p'', ~~~~ r''^2 = p+p' , $$ and likewise ##; $$ u^2 = \nu'+\nu'', ~~~~ u'^2 = \nu+\nu'', ~~~~ u''^2 = \nu+\nu' ; $$ so, if one makes the substitutions, and one supposes for greater simplicity ##, ##, ##, $$ \Sigma = p\left(\frac{2rdr}{dt}\right)^2 + p'\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp'}{dt}\right)^2 - 2\left(p''\frac{dp'}{dt} - p'\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma' = p'\left(\frac{2r'dr'}{dt}\right)^2 + p\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp}{dt}\right)^2 + 2\left(p''\frac{dp}{dt} - p\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r'^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma'' = p''\left(\frac{2r''dr''}{dt}\right)^2 + p'\left(\frac{dp}{dt}\right)^2 + p\left(\frac{dp'}{dt}\right)^2 + 2\left(p\frac{dp'}{dt} - p'\frac{dp}{dt}\right)\frac{d\rho}{dt} + r''^2\left(\frac{d\rho}{dt}\right)^2 , $$ the following equation becomes, after having been multiplied by 16/dt4, ##N1. $$ \tag{N} {\Tiny 16(pp'+pp''+p'p'')(\nu\nu'+\nu\nu''+\nu'\nu'') - 4(\Sigma\nu + \Sigma'\nu' + \Sigma''\nu'') + \left( \frac{dpdp'+dpdp''+dp'dp''+d\rho^2}{dt^2} \right) ^2 = 0 . } $$

  It is necessary then that this equation be valid at the same time as the three equations (K) of Article V ; so that, as it contains nothing more than the same variables as equations (K), and that is of an order lower by one unit than that, one can regard it as an integral of those equations (K), but a particular integral because it contains no new constant ; thus, if one integrates the equations (K) by adding to them the necessary constants, those constants must be such that they themselves satisfy equation (N). So that, if one does not want to use that last equation instead of one of the equations (K), it is nonetheless necessary to have regard to it in the determination of the constants ; but for that it will suffice to suppose everywhere t=0.

  Later we will always make use of that equation to determine the constant which must enter in the value of dρ/dt, resulting from the integration of equation (B) of Article V.

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  Let us now take again the equations (D) of Article II, and making, for brevity, ##, ##, ##, $$ \lambda = ydx-xdy, ~~~~ \lambda' = y'dx'-x'dy', ~~~~ \lambda'' = y''dx''-x''dy'', \\ \mu = zdx-xdz, ~~~~ \mu' = z'dx'-x'dz', ~~~~ \mu'' = z''dx''-x''dz'', \\ \nu = zdy-ydz, ~~~~ \nu' = z'dy'-y'dz', ~~~~ \nu'' = z''dy''-y''dz'', $$ one gets, after having multiplied by dt, ##O3. $$ \left\{ \begin{array}{l} \frac{\lambda}C + \frac{\lambda'}B + \frac{\lambda''}A = adt , \\ \tag{O} \frac{\mu}C + \frac{\mu'}B + \frac{\mu''}A = bdt , \\ \frac{\nu}C + \frac{\nu'}B + \frac{\nu''}A = cdt . \end{array} \right. $$

  Now I find, as higher up, ##, $$ {\small \lambda^2+\mu^2+\nu^2 = (x^2+y^2+z^2)(dx^2+dy^2+dz^2) - (xdx+ydy+zdz)^2 = r^2u^2dt^2 - (rdr)^2 , } $$ and by analogy ##, ##; $$ \lambda'^2+\mu'^2+\nu'^2 = r'^2u'^2dt^2 - (r'dr')^2 , \\ \lambda''^2+\mu''^2+\nu''^2 = r''^2u''^2dt^2 - (r''dr')^2 ; $$ Above, $(r''dr')^2$ to be $(r''dr'')^2$ ?? Yes, c.18 p.29 has $(r''dr'')^2$ clearly enough.
I find likewise ## ##, $$ {\Tiny \begin{aligned} \lambda\lambda'+\mu\mu'+\nu\nu' & = (xx'+yy'+zz')(dxdx'+dydy'+dzdz') - (x'dx+y'dy+z'dz)(xdx'+ydy'+zdz') \\ & = p''\nu''dt^2 - dVdV' = p''\nu''dt^2 - \left(\frac{dp''}2\right)^2 + \left(\frac{d\rho}2\right)^2 , \end{aligned} } $$ and by analogy ##, ##. $$ \lambda\lambda''+\mu\mu''+\nu\nu'' = p'\nu'dt^2 - \left(\frac{dp'}2\right)^2 + \left(\frac{d\rho}2\right)^2 , \\ \lambda'\lambda''+\mu'\mu''+\nu'\nu'' = p\nu dt^2 - \left(\frac{dp}2\right)^2 + \left(\frac{d\rho}2\right)^2 . $$

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  Then, if one makes, for greater simplicity, ##, ##, ##, $$ \Pi = r^2u^2 - \left(\frac{rdr}{dt} \right)^2 , \\ \Pi' = r'^2u'^2 - \left(\frac{r'dr'}{dt} \right)^2 , \\ \Pi'' = r''^2u''^2 - \left(\frac{r''dr''}{dt} \right)^2 , $$ and ##, ##, ##, $$ \Psi = p\nu - \left(\frac{dp}{2dt} \right)^2 + \left(\frac{d\rho}{2dt} \right)^2 , \\ \Psi' = p'\nu' - \left(\frac{dp'}{2dt} \right)^2 + \left(\frac{d\rho}{2dt} \right)^2 , \\ \Psi'' = p''\nu'' - \left(\frac{dp''}{2dt} \right)^2 + \left(\frac{d\rho}{2dt} \right)^2 , $$ such that one has ##, ##, ##, $$ \lambda^2+\mu^2+\nu^2 = \Pi dt^2, ~~~~ \lambda'\lambda''+\mu'\mu''+\nu'\nu'' = \Psi dt^2 , \\ \lambda'^2+\mu'^2+\nu'^2 = \Pi' dt^2, ~~~~ \lambda\lambda''+\mu\mu''+\nu\nu'' = \Psi' dt^2 , \\ \lambda''^2+\mu''^2+\nu''^2 = \Pi'' dt^2, ~~~~ \lambda\lambda'+\mu\mu'+\nu\nu' = \Psi'' dt^2 , $$ one gets, on squaring the three equations (O) and adding them together, ##P1, $$ \tag{P} {\Pi \over C^2} + {\Pi' \over B^2} + {\Pi'' \over A^2} + {2\Psi \over AB} + {2\Psi' \over AC} + {2\Psi'' \over BC} = a^2+b^2+c^2 , $$ an equation which is also, as one sees, of an order lower by one unit than the equations (K); and as it contains the arbitrary constant a2 + b2 + c2 which is not found in equations (K), one can regard it as a complete integral of these same equations.

  One might believe that equation (E) which we have found in Article II might thus, by substituting in it the values of u, u', u'', give a new integral, but it is easy to see that it will only result

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in an identical equation, because the equation that it concerns reduces first to ##; $$ \frac{u^2}C + \frac{u'^2}B + \frac{u''^2}A - 2(A+B+C) \left( \frac1{Cr} + \frac1{Br'} + \frac1{Ar''} \right) = f ; $$ and, putting for u, u', and u'' their values taken from formulae (J), one gets, by omitting that which goes out, ##, $$ Q+Q'+Q'' = f , $$ which contains no new condition, because the quantities Q, Q', Q'' are already themselves such that dQ+dQ'+dQ''=0 (Article V).

  Besides, if one combines the equation ## $$ Q+Q'+Q'' = f $$ with the equations (N) and (P), after having there substituted the values of u, u' and u'', one can, by means of these three equations, determine the three quantities Q, Q', Q'', which will consequently contain only the finished variables r, r', r'' and their first derivatives dr, dr', dr'' with the quantity dρ/dt; thus, substituting these values in equations (K), one will have three equations of the second order in the variables r, r', and r'', in which there is no more than to substitute the value of dρ/dt. Thus, if with the help of one of these equations one eliminates the quantity dρ/dt from the two others, one first gets two equations purely of the second order between the variables r, r', r'' and t ; next, if one differentiates the value of dρ/dt and one puts the value of d2ρ/dt2 in equation (R), one gets a third equation in the same variables, which is only of the third order. Then one gets, by that means, for the determination of the variables r, r' and r'', two differential equations of the second order and one of the third ; and these equations will suffice, as one will see in a moment, for the complete solution of the three-body problem.

  We believe meanwhile that it is even simpler and more convenient

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for the calculation to substitute in equations (K) the values of Q, Q', and Q'' taken from equations (J) ; because, although the resulting equations can mount to orders higher than the second, they have always the great advantage that the variables are found to be less mixed together, and that the analogy which holds there will greatly facilitate their resolution.

  Of the ten equations of Article VIII there only remain thus no more than nine, and of the three equations (D) of Article II, or (O) of Article XI, there only remain no more than two ; such that one has only in total eleven equations for the determination of the six variables x, y, z, x', y', z' and their differentials dx, dy,, ... ; from where one sees that it is impossible to determine these variables directly and by the operations of Algebra alone ; but one can obtain them in the end by means of an integration, as one will see.

  I suppose that one would wish to know the values of x, y, z ; one has first the equation ##Q1. $$ \tag{Q} x^2+y^2+z^2 = r^2 . $$

  Next, multiplying the three equations (O) of article XI respectively by λ, μ, ν, and adding them together, one gets ##; $$ \frac{\lambda^2+\mu^2+\nu^2}C + \frac{\lambda\lambda'+\mu\mu'+\nu\nu'}B + \frac{\lambda\lambda''+\mu\mu''+\nu\nu''}A = (a\lambda+b\mu+c\nu)dt ; $$ or then, on making the substitutions of the same Article, ##R1. $$ \tag{R} \left( \frac\Pi C + \frac{\Psi''}B + \frac{\Psi'}A \right) dt = a(ydx-xdy) + b(zdx-xdz) + c(zdy-ydz) . $$

  Then multiplying the same equations (O) respectively by z, -y, x, and adding them together, one gets ##. $$ \frac{\lambda z - \mu y + \nu x }C + \frac{\lambda' z - \mu' y + \nu' x }B + \frac{\lambda'' z - \mu'' y + \nu'' x }A = (az-by+cx)dt . $$ Now it is easy to see that one has λzyx=0, and that

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-λ'z+μ'y-ν'x is the same quantity that we have designated higher up by δ (Article IX) ; then, because one has already found (Article X) ##, $$ \delta^2 = (r^2r'^2-p''^2)u'^2dt^2 -r^2r'^2dr'^2 + 2p''r'dr'dV' - r^2dV'^2 , $$ one gets, on making the substitutions of the same Article X, ##, $$ (\lambda'z-\mu'y+\nu'x)^2 = \left[(pp'+pp''+p'p'')u'^2 - \frac{\Sigma'}4 \right] dt^2 , $$ and by analogy ##. $$ (\lambda''z-\mu''y+\nu''x)^2 = \left[(pp'+pp''+p'p'')u''^2 - \frac{\Sigma''}4 \right] dt^2 . $$ Such that the equation above becomes ##S1. $$ \tag{S} \frac{\frac12\sqrt{4(pp'+pp''+p'p'')u'^2-\Sigma'}}B + \frac{\frac12\sqrt{4(pp'+pp''+p'p'')u''^2-\Sigma''}}A = az-by+cx . $$

 Thus one will have three equations (Q), (R), and (S), with the aid of which one can easily determine the values of x, y, z, as soon as one knows those of r, r' and r''.

  One can find similar formulae for the determination of x', y', z' ; and likewise, without making a new calculation, it will suffice to change in the previous B to C and C to B, to put accents on the letters which have no accent and to remove the accent from those which have one, without altering those which have two accents. It is only necessary to observe that the quantity does not change its value, but only its sign, when one changes among themselves the masses A, B, C and the accented letters, which is seen clearly from equation (H) of Article XIV.

  Suppose, for brevity. ##, ##, $$ \begin{aligned} T & = \frac{\Pi}C + \frac{\Psi''}B + \frac{\Psi'}A , \\ Z & = \frac{\sqrt{4(pp'+pp''+p'p'')u'^2-\Sigma'}}{2B} + \frac{\sqrt{4(pp'+pp''+p'p'')u''^2-\Sigma''}}{2A} , \end{aligned} $$

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and one will have these three equations ##, ##, ##. $$ \begin{aligned} &x^2+y^2+z^2 = r^2 , \\ &az-by+cx = Z , \\ &a(ydx-xdy) + b(zdx-xdz) + c(zdy-ydz) = T dt . \end{aligned} $$

  As the constants a, b, c are arbitrary (Article H), and depend only on the position of the plane of projection of the orbits of bodies B and C around body A, it is easy to see that one can take this plane in a manner such that one has b=0 and c=0 ; because for that it suffices that one has b=0 and c=0 at the beginning of the motion, that is to say, when t=0.

  Let us suppose then b=0 and c=0, one gets ##; $$ az = Z , ~~~~ a(ydx-xdy) = Tdt ; $$ so ##, $$ z = \frac Za , $$ and, because x2 + y2 + z2 = r2, one gets ##; $$ x^2+y^2 = r^2 - \frac{Z^2}{a^2} ; $$ so ##; $$\frac{ydx-xdy}{x^2+y^2} = \frac{aTdt}{a^2r^2-Z^2} ; $$ so that on making ##, $$ d\varphi = \frac{aTdt}{a^2r^2-Z^2} , $$ one gets ##, $$\frac yx = \operatorname{tang} \varphi , $$ and from that ##. $$ z = \frac Za, ~~~~ y = \sqrt{r^2-\frac{Z^2}{a^2}}\sin\varphi , ~~~~ x = \sqrt{r^2-\frac{Z^2}{a^2}}\cos\varphi . $$

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  But if one does not want to keep to the supposition of b=0 and c=0, which necessitates taking the plane of projection in a specified manner, here is how one can determine the quantities x, y, z with the maximum generality.

  Let ##, $$ lz -my +nx = X, ~~~~ \lambda z - \mu y + \nu x = Y , $$ l, m, n, λ, μ, ν being undetermined coefficients, and X, Y two new variables ; one gets ##; $$ YdX-XdY = (m\nu-n\mu)(ydx-xdy) + (n\lambda-l\nu)(zdx-xdz) + (l\mu-m\lambda)(zdy-ydz) ; $$ then, making ##, $$ m\nu-n\mu = ka, ~~~~ n\lambda-l\nu = kb, ~~~~ l\mu-m\lambda = kc , $$ one gets the equation (Article XIV) ##. $$ YdX-XdY = kTdt . $$

  Let us suppose that now one has ##, $$ f(X^2+Y^2) + gZ^2 = x^2+y^2+z^2 , $$ substituting the values of X, Y, Z with x, y and z, and comparing next the terms which contain the same powers of x, y and z, one gets these six equations ##, ##, ##, $$ \begin{aligned} &f(l^2~+\lambda^2) + ga^2 = 1, ~~~~ f(lm + \lambda\mu) + gab = 0 , \\ &f(m^2+\mu ^2) + gb^2 = 1, ~~~ f(ln + \lambda\nu) + gac = 0 , \\ &f(n^2+\nu ^2) + gc^2 = 1, ~~~~ f(mn + \mu \nu) + gbc = 0 , \end{aligned} $$ which, being combined with the three preceding, will serve to determine the nine unknowns l, m, n, λ, μ, ν, f, g, k.

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  That done, one has then, because x2 + y2 + z2 = r2, the equation ##, $$ f(X^2+Y^2) + gZ^2 = r^2 , $$ from where ##; $$ f(X^2+Y^2) = r^2 - gZ^2 ; $$ so ##. $$ \frac{YdX-XdY}{X^2+Y^2} = \frac{fkTdt}{r^2-gZ^2} . $$ So, if one does ##, $$ d\varphi = \frac{fkTdt}{r^2-gZ^2} , $$ one gets ##. $$ Y = \sqrt\frac{r^2-gZ^2}f\sin\varphi, ~~~~ X = \sqrt\frac{r^2-gZ^2}f\cos\varphi . $$

  Thus one knows the three quantities X, Y, Z, with the aid of which one can determine x, y, z.

  For that, one takes the three equations ##, $$ lz - my + nx = X, ~~~~ \lambda z - \mu y + \nu x = Y, ~~~~ az - by + cx = Z , $$ and one adds them together after having multiplied them respectively : 1 by fl, fλ, ga; 2 by fm, fμ, gb; 3 by fn, fν, gc; one immediately gets, by virtue of the equations of the preceding article, ##. $$ z = f(l X + \lambda y) + gaZ, ~~~~ y = - f(m X + \mu Y) - gbZ, ~~~~ x = f(n X + \nu Y) + gcZ . $$

  Now, as one has supposed ##, $$ x^2 + y^2 + z^2 = f(X^2+Y^2) + gZ^2, $$ one gets, on substituting the values of x, y, z which one has just found, and

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comparing the homogeneous terms, ##, ##, ##, $$ \array{ f(l^2+m^2+n^2) = 1 , & l\lambda + m\mu + n\nu = 0 , \\ f(\lambda^2+\mu^2+\nu^2) = 1 , & la + mb + nc = 0 , \\ g(a^2+b^2+c^2) = 1 , & \lambda a + \mu b + \nu c = 0 , ~ } $$ and these equations must be identical with the six that have been found above (Article XV), and can in consequence be used in place of those for the determination of the unknowns l, m, ... .

  Then, as it is necessary to satisfy at the same time to these other three equations (Article XV) ##, $$ m\nu - n\mu = ka, ~~~~ n\lambda - l\nu = kb, ~~~~ l\mu - m\lambda = kc, $$ I remark that, if one adds together these last equations, after having multiplied them respectively : 1 by l, m, n; 2 by λ, μ, ν, one gets these two ##, $$ k(la + mb + nc) = 0 , ~~~~ k(\lambda a + \mu b + \nu c) = 0, $$ which are in accordance with with the fifth and sixth of the previous ; thus one can already reduce to a single one the three equations concerned, and one satisfies that by the determination of the unknown k. So, if one adds together the squares of these equations, one gets ## ## $$ \begin{aligned} k^2(a^2+b^2+c^2) &= (m\nu-n\mu)^2 + (n\lambda-l\nu)^2 + (l\mu-m\lambda)^2 \\ &= (l^2+m^2+n^2)(\lambda^2 + \mu^2 + \nu^2) - (l\lambda + m\mu + n\nu)^2 = \frac1{f^2} \end{aligned} $$ by virtue of the six equations above ; so that one gets ##. $$ k = \frac 1 {f\sqrt{a^2+b^2+c^2}} . $$

  So there is no more than to satisfy the six equations found higher up ; one can do that in several ways because there are more unknowns than equations.

  In p.256, is k a Lagrange's Undetermined Multiplier? Any LUMs elsewhere?
Full Essai - pp.229-332, PDF images, 4.79MB ; reveal image of page 256 :

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  One has first ##; $$ g = \frac 1 {a^2 + b^2 + c^2} ; $$ next, if one eliminates λ from those two equations ##, $$ l\lambda + m\mu + n\nu = 0, ~~~~ \lambda a + \mu b + \nu c = 0 , $$ one gets ##, $$ (am-bl)\mu + (an-cl)\nu = 0 , $$ and, eliminating μ, one gets likewise ##, $$ (bl-am)\lambda + (bn-cm)\nu = 0 , $$ from which I conclude that one gets ##, $$ \lambda = (cm-bn)\delta, ~~~~ \mu = (an-cl)\delta, ~~~~ \nu = (bl-am)\delta, $$ δ being an unknown that one determines by the equation ##, $$ f(\lambda^2+\mu^2+\nu^2) = 1 , $$ which gives ##; $$ f\delta^2[(cm-bn)^2 + (an-cl)^2 + (bl-am)^2] = 1 ; $$ but one has ##, $$ {\small (cm-bn)^2 + (an-cl)^2 + (bl-am)^2 = (a^2+b^2+c^2)(l^2+m^2+n^2) -(al+bm+cn)^2 = \frac1{fg} , } $$ because of ## $$ a^2+b^2+c^2 = \frac1g, ~~~~ l^2+m^2+n^2 = \frac1f, ~~~~ al+bm+cn = 0 $$ by the equations above ; so one gets ##, $$ \frac{\delta^2}g = 1, ~~~~ \delta = \sqrt g = \frac1{\sqrt{(a^2+b^2+c)}} , $$ ERRATUM : "$+c^2$" ?? c.18 p.41 also had "$+c$"

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and it only remains to satisfy these two equations ##. $$ f(l^2+m^2+n^2) = 1, ~~~~ la+mb+nc = 0 . $$

  Let us suppose, for greater simplicity, ##; $$ a = h\cos\alpha, ~~~~ b = h\sin\alpha\cos\epsilon, ~~~~ c = h\sin\alpha\sin\epsilon ; $$ one gets ##, $$ \delta = \sqrt g = \frac1h , $$ such that ##, $$ h = \sqrt{a^2+b^2+c^2} , $$ and the second of the two preceding equations becomes ##; $$ l\cos\alpha + \sin\alpha(m\cos\epsilon + n\sin\epsilon) = 0 ; $$ let then ##, $$ l = \sin\alpha\sin\eta , $$ and one gets, making for greater simplicity f=1, ##. $$ \sin^2\alpha\sin^2\eta + m^2 + n^2 = 1, ~~~~ \cos\alpha\sin\eta + m\cos\epsilon + n\sin\epsilon = 0 . $$

  Thus ##; $$ m\cos\epsilon + n\sin\epsilon = -\cos\alpha\sin\eta ; $$ so ## ##; $$ \begin{aligned} m^2 + n^2 - (m\cos\epsilon+n\sin\epsilon)^2 &= (m\sin\epsilon - n\cos\epsilon)^2 \\ &= 1 - \sin^2\alpha\sin^2\eta - \cos^2\alpha\cos^2\eta = 1 - \sin^2\eta = \cos^2\eta ; \end{aligned} $$ and, taking the square root. ##; $$ m\sin\epsilon - n\cos\epsilon = \cos\eta ; $$ such that one gets ##, ##; $$ m = \sin\epsilon\cos\eta - \cos\alpha\cos\epsilon\sin\eta , ~~ \\ n = - \cos\epsilon\cos\eta - \cos\alpha\sin\epsilon\sin\eta ; $$ and from that one finds the values of λ, μ, ν from the preceding formulae.

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  One gets in this manner ##, ##, ##, ##, ##, ##. $$\begin{aligned} l &= \sin\alpha\sin\eta , \\ m &= \sin\epsilon\cos\eta - \cos\alpha\cos\epsilon\sin\eta , \\ n &= - \cos\epsilon\cos\eta - \cos\alpha\sin\epsilon\sin\eta , \\ \lambda &= \sin\alpha\cos\eta , \\ \mu &= - \sin\epsilon\sin\eta - \cos\alpha\cos\epsilon\cos\eta , \\ \nu &= \cos\epsilon\sin\eta - \cos\alpha\sin\epsilon\cos\eta . \end{aligned} $$

  If one substitutes these values in the expressions in x, y, and z of Article XVI, it is easy to see that the quantities X, Y, and Z/h are nothing else but the rectangular coordinates of the same curve, which is represented by the coordinates x, y, and z, but referred to another plane of projection, of which the position depends on the angles α, ε and η. Indeed, if one considers the two planes of coordinates x, y, and of coordinates X, Y, the angle α will be that of the inclination of the planes, the angle η will be that which the line of intersection of the planes makes with the axis of abscissas x, and the angle ε will be that which the axis of abscissas X makes with the same line of intersection. So, as the expression of coordinates X, Y, and Z/h is simpler than that of the coordinates x, y, z, it is clear that the plane of projection to which belong the coordinates X, Y, and Z/h is more suitable than any other plane for there referring the movements of the three bodies, or rather the movement of two of the bodies around the third.

  One sees then that the position of the plane of projection is not at all unimportant, and that, among all of the possible planes that one can make pass through body A, there is one which must be chosen by preference, because the movements of bodies B and C around A are with respect to that plane the simplest that is possible.

  That remark, which seems to me of some importance in the three-body problem, has not previously been made, because nobody as far as I know, has up to the present envisaged this problem in a manner as general as that which we have just done.

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  We will take then, in place of the coordinates x, y, z, these X, Y, Z/h, to represent the movement of body B around A ; and as one has, because of h = 1/√g (Article XV), ##, ##, $$ X^2 + Y^2 + \left(\frac Zh\right)^2 = x^2+y^2+z^2 = r^2 , $$ $$ Y = \sqrt{r^2 - \left(\frac Zh\right)^2}\sin\varphi, ~~~~ X = \sqrt{r^2 - \left(\frac Zh\right)^2}\cos\varphi, $$ it is clear that φ will be the angle described by the body B around A in the plane of projection, that is to say, the longitude of body B in the same plane ; and that Z/hr will be the sine of the latitude. Thus one has (Article XVI), because of f=1, ##. $$ k = \frac1{\sqrt{a^2+b^2+c^2}} = \frac1h . $$

For body B :
Radius rector of the orbit. . . r,
Longitude. . . ∫ Tdt / (h[r2 - (Z/h)2]),
Sine of the latitude. . . Z/hr.
For body C :
Radius rector of the orbit. . . r',
Longitude. . . ∫ T'dt / (h[r'2 - (Z'/h)2]),
Sine of the latitude. . . Z'/hr'.

  The values of T and of Z are given by the formulae of Article XIV,

Lagrange, or more likely the printer, appears not to have known the difference between a vector and a rector.

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and to have those of T' and Z' one only needs to change in those the accent zero to ' and ' to zero, and then B to C and C to B.

  As for the quantity A, it is an arbitrary constant which depends on the initial movement of the bodies ; but it is necessary to take it such, that it agrees with equation (P) of Article XI, in which the second member is ##; $$ a^2 + b^2 + c^2 = h^2 ; $$ such that it is only necessary to take for h the square root of the value of the first member of that equation when one makes t=0.

  The formulae which we have found serve to determine the orbits of bodies B and C around body A with respect to a fixed plane passing through that same body ; but One still has to see how one can determine, by their means, the mutual position of these orbits. For that, we begin by remarking that if one considers the triangle formed at each instant by the three bodies A, B, C, and of which the three sides are r, r', and r'', and that one names as ζ, ζ', ζ'' the three angles opposite to these sides, one will have, as one knows, by elementary geometry, ##, ##, ##. $$ \cos\zeta = \frac{r'^2 + r''^2 - r^2}{2r'r''} = \frac{p}{r'r''} , \\ \cos\zeta' = \frac{r^2 + r''^2 - r'^2}{2rr''} = \frac{p'}{rr''} , \\ \cos\zeta'' = \frac{r^2 + r'^2 - r''^2}{2r'r''} = \frac{p''}{rr'} . $$

  Then one has (Article VIII) ##; $$ p'' = xx' + yy' + zz' ; $$ so ##, $$ \cos\zeta'' = \frac{xx'+yy'+zz'}{rr'} , $$ ζ'' being the angle formed at the centre of body A by the radius rectors r and r' of the two other bodies B and C.

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  Now imagine two planes passing, one through body A and through the two infinitely close points in which is found the body B at the beginning and at the end of the infinitely small time dt, and the other through the same body A, and through the two infinitely close points where body C is at the beginning and at the end of the same time dt ; these two planes will be those of the orbits of bodies B and C around A, and will necessarily intersect in a straight line passing through body A which will thus be the line of nodes of the two orbits.

  Let ω be the inclination of these two planes the one to the other, ξ the distance from body B to the intersection of the two planes or to the line of nodes, that is to say, the angle included between the radius r and the line of nodes, and ξ' the distance of body C to the same line of nodes, that is to say, the angle formed by the radius r' and the line of nodes ; if one imagines a sphere described around A as centre, and that by the points where the two radii r, r' and the line of nodes cross the surface of that sphere, of which we suppose the radius to be equal to 1, one takes the arcs of great circles, one gets a spherical triangle of which the three sides will be ξ, ξ' and ζ'', and of which the angle opposite to the side ζ'' will be ω ; such that one gets, from known formulae, ##; $$ \cos\zeta'' = \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega ; $$ so ## $$ \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega = = \frac{xx'+yy'+zz'}{rr'} $$

  Let us suppose now that during the time dt the body B describes around A the infinitely small angle , and that body C describes the angle dθ', it is clear that, whereas the lines x, y, z, r grow by their differentials dx, dy, dz, dr, the angle ξ grows by , and the angle ω remains the same, because one supposes that the positions of the planes of the orbits of bodies B and C is the same at the beginning and at the end of the instant dt ; likewise, in making grow the lines x', y', z', r' by their differentials dx', dy', dz', dr', it will be only the angle ξ' which will vary in growing by dθ'. So, as the preceding equation must be identical to and independent of the law of movements of bodies B and C, it is clear that one can there

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vary the quantities x, y, z, r and ξ which belong to body B independently of the quantities x', y', z', t' and Z' which belong to body C, and vice versa these independently of those ; from which it follows that by varying first x, y, z, r and ξ, next x', y', z', r' and ξ', then the ones and the others at the same time, one obtains from the equation concerned the three following ##, ##, ## ## ##. $$ -- \sin\xi\cos\xi' + \cos\xi\sin\xi'\cos\omega)d\theta = - \frac{(xx'+yy'+zz')dr}{r^2r'} + \frac{x'dx+y'dy+z'dz}{rr'} , \\ -- \cos\xi\sin\xi' + \sin\xi\cos\xi'\cos\omega)d\theta' = - \frac{(xx'+yy'+zz')dr'}{rr'^2} + \frac{xdx'+ydy'+zdz'}{rr'} , $$ Leading "(", instead of minus, twice ?? Yes, c.18 p.47 has "$(-$". $$ \begin{aligned} & (\sin\xi\sin\xi' + \cos\xi\cos\xi'\cos\omega)d\theta d\theta' \\ & ~~~~ = \frac{(xx'+yy'+zz')drdr'}{r^2r'^2} - \frac{(x'dx+y'dy+z'dz)dr'}{rr'^2} \\ & ~~~~~~~~ - \frac{(xdx'+ydy'+zdz')dr}{r^2r'} + \frac{dxdx'+dydy'+dzdz'}{rr'} . \end{aligned} $$

  Then, if one makes in all of these equations the substitutions of Article VIII, one gets these four ##, ##, ##, ##. $$ \begin{aligned} \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega &= \frac{p''}{rr'} , \\ - \sin\xi\cos\xi' + \cos\xi\sin\xi'\cos\omega &= - \frac{p''dr+rdV}{r^2r'd\theta} , \\ - \cos\xi\sin\xi' + \sin\xi\cos\xi'\cos\omega &= - \frac{p''dr'+r'dV'}{rr'^2d\theta} , \\ \sin\xi\sin\xi' + \cos\xi\cos\xi'\cos\omega &= \frac{p''drdr'-dV.rdr'-dV'.r'dr+rr'\nu''dt^2}{r^2r'^2d\theta d\theta'} , \end{aligned} $$

  Then it is easy to conceive that the square of the small space that body B travels in time dt is equally expressed by dx2 + dy2 + dz2 and by r22 +dr2, so that one gets ## $$ rd\theta = \sqrt{dx^2+dy^2+dz^2-dr^2} $$ and in consequence (Articles VIII and XI) ##, $$ d\theta = \frac{\sqrt{u^2dt^2-dr^2}}r = \frac{dt\sqrt\Pi}{r^2} , $$

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and likewise ##. $$ d\theta' = \frac{\sqrt{u'^2dt^2-dr'^2}}{r'} = \frac{dt\sqrt{\Pi'}}{r'^2} . $$

  Thus the second members of the four preceding equations will all be given, as soon as one knows r, r' and r'' in terms of t (cited Article) ; so that one gets four equations in three unknowns ξ, ξ' and ω, by which one can not only determine the three unknowns, but also have an equation in the quantities r, r', r'', u, u', ,,, , and this equation will be the same as that which one has already found higher up (Article X) by a very different route.

  Suppose, for brevity, that the preceding equations are represented as follows ##, ##, ##, ##, $$ \cos\xi\cos\xi' + \sin\xi\sin\xi'\cos\omega = \lambda , \\ \sin\xi\cos\xi' - \cos\xi\sin\xi'\cos\omega = \mu , \\ \cos\xi\sin\xi' - \sin\xi\cos\xi'\cos\omega = \nu , \\ ~~~~ \sin\xi\sin\xi' + \cos\xi\cos\xi'\cos\omega = - \varpi , $$ by making ##, ##; $$ \lambda = \frac{p''}{rr'}, ~~~~ \mu = \frac{p''dr + rdV}{r^2r'd\theta}, ~~~~ \nu = \frac{p''dr'+r'dV'}{rr'^2d\theta'} ; \\ \varpi = \frac{-p''drdr'+dV.rdr'+dV'.r'dr - rr'\nu''dt^2} {r^2r'^2d \theta d\theta'} ; $$ it is easy to reduce those four equations to these two ##, ##, $$ \cos(\xi\pm\xi')(1\mp\cos\omega) = \lambda\pm\varpi , \\ \sin(\xi\pm\xi')(1\mp\cos\omega) = \mu\pm\nu , $$ which, because of the ambiguity of the signs, are really equivalent to four equations. Raising these two equations to the square, and then adding them together, one has ##, $$ (1\mp\cos\omega)^2 = (\lambda\pm\varpi)^2 + (\mu\pm\nu)^2 , $$

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from where, because of the ambiguity of the signs, one takes ##, $$ - \cos\omega = \lambda\varpi + \mu\nu, ~~~~ 1 + \cos^2\omega = \lambda^2 + \mu^2 + \nu^2 + \varpi^2 ; $$ such that eliminating cosω one gets ##. $$ 1 + (\lambda\varpi + \mu\nu)^2 = \lambda^2 + \mu^2 + \nu^2 + \varpi^2 . $$

  If one substitutes in that equation the values of λ, μ, ν, ω, as also those of and dθ', one gets an equation which is the same as equation (N) of Article X ; which can serve to confirm the validity of our calculations.

  The equation ## $$ - \cos\omega = \lambda\varpi + \mu\nu $$ gives ##, $$ \cos\omega = \frac{p''\nu''dt^2 - dVdV'}{r^2r'^2 d\theta d\theta'} = \frac{\Psi''}{\sqrt{\Pi\Pi'}} , $$ which makes known the inclination ω of these two orbits.

  Knowing ω, one knows easily ξ and ξ' ; because, by multiplying the two equations, ##, ## $$ \cos(\xi+\xi')(1-\cos\omega) = \lambda + \varpi , \\ \cos(\xi-\xi')(1+\cos\omega) = \lambda - \varpi ~ $$ the one by the other, one gets this ## $$ ½ (\cos2\xi + \cos2\xi')\sin^2\omega = \lambda^2 - \varpi^2 ; $$ and likewise the other two equations ##, $$ \sin(\xi+\xi')(1-\cos\omega) = \mu+\nu, ~~~~ \sin(\xi-\xi')(1+\cos\omega) = \mu-\nu, $$ being multiplied together, give ##, $$ - ½ (\cos2\xi-\cos2\xi')\sin^2\omega = \mu^2 - \nu^2 , $$ from which one takes ##, $$ \cos2\xi = \frac{\lambda^2-\varpi^2-\mu^2+\nu^2}{\sin^2\omega}, ~~~~ \cos2\xi' = \frac{\lambda^2-\varpi^2+\mu^2-\nu^2}{\sin^2\omega}, $$

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or also, by putting in place of ω2 its value ##, $$ 1 + \cos^2\omega -\lambda^2 -\mu^2 -\nu^2 , $$ taken from the equation found above, one gets, because of cos2ω=1-sin2ω, ##, $$ \cos2\xi = 1 + \frac{2(\lambda^2+\nu^2-1)}{\sin^2\omega}, ~~~~ \cos2\xi' = 1 + \frac{2(\lambda^2+\mu^2-1)}{\sin^2\omega}, $$ from which one takes ##; $$ \sin\xi = \frac{\sqrt{1-\lambda^2-\nu^2}}{\sin\omega}, ~~~~ \sin\xi' = \frac{\sqrt{1-\lambda^2-\mu^2}}{\sin\omega}; $$ that is to say, on substituting the values of λ, μ, ν, and observing that r22 = u2dt2-dr2 and r'2dθ'2 = u'2dt2-dr'2, ##, ##, $$ \sin\xi = \frac {\sqrt{(r^2r'^2-p''^2)u'^2dt^2-r^2(r'dr')^2+2p''(r'dr')dV'-r'^2dV'^2}} {rr'^2\sin\omega d\theta'} , $$ $$ \sin\xi' = \frac {\sqrt{(r^2r'^2-p''^2)u^2dt^2-r'^2(rdr)^2+2p''(rdr)dV-r^2dV^2}} {r'r^2\sin\omega d\theta} , $$ or else (Article XIII) ##, ##. $$ \sin\xi = \frac{\sqrt{4(pp'+pp''+p'p'')u'^2 - \Sigma'}} {2r\sqrt{\Pi'}\sin\omega} , $$ $$ \sin\xi' = \frac{\sqrt{4(pp'+pp''+p'p'')u^2 - \Sigma}} {2r'\sqrt{\Pi}\sin\omega} . $$

  If one wants the three bodies to move in a single plane, one has then ω=0, and in consequence cosω=1 and sinω=0 ; so ##, $$ \Sigma = 4(pp'+pp''+p'p'')u^2, ~~~~ \Sigma' = 4(pp'+pp''+p'p'')u'^2 , $$ and by analogy ##. $$ \Sigma'' = 4(pp'+pp''+p'p'')u''^2 . $$

  Therefore the quantities Z and Z' (Article XIV) are zero, and in consequence the movements of the three bodies occur in the same

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plane as we have taken for the plane of projection (Article XVIII). So, if one substitutes the values of u2, u'2, u''2 taken from the preceding equations in equation (P) of Article XI, one gets an equation in r, r', r'' and dr/dt, dr'/dt, dρ/dt , by which one can determine this last quantity dρ/dt ; substituting next the value of dρ/dt in those of Σ, Σ', Σ'', one gets the values of u2, u'2, u''2 expressed in terms of terms of r, r', r'' and dr/dt, dr'/dt, dr''/dt only ; thus, putting the values of u2, u'2, u''2 in equations (F) of Article III, one gets at last three equations in r, r', r'' and t, which will be simple differentials of the second order, in place of the general equations (K) rising to the fourth order, when one frees them of the integration signs.

  Finally I believe that, in the present case, these last equations will always be preferable, because they have the singular advantage of not containing any radical, which would not be the case in the equations where one uses the values of u, u', u'' determined by the equations above, values which would necessarily contain square roots.


  To sum up what has just been demonstrated in this Chapter, let there be named : A, B, C the masses of the three bodies ; r, r', r'' the distances between the bodies A and B, A and C, B and C ; and supposing, for brevity, ##, ##, $$ \array { p = \frac{r'^2+r''^2-r^2}2 ,& p' = \frac{r^2+r''^2-r'^2}2 ,& p'' = \frac{r^2+r'^2-r''^2}2 , \\ q = \frac1{r'^3} - \frac1{r''^3},& q' = \frac1{r^3} - \frac1{r''^3},& q'' = \frac1{r'^3} - \frac1{r^3} = q-q' , } $$

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one gets, by taking the element of time dt as constant, ##H1; $$ \tag{H} \frac{d^2\rho}{dt^2} + Cpq - Bp'q' - Ap''q'' = 0 ; $$ ##I3; $$ \left\{ \begin{array}{l} dQ = q'dp' - q''dp'' - qd\rho , \\ \tag{I} dQ' = qdp + q''dp'' + q'd\rho , \\ dQ'' = - qdp - q'dp' + q''d\rho ; \end{array} \right. $$ ##K3. $$ \left\{ \begin{array}{l} \frac{d^2(r^2)}{2dt^2} - \frac{A+B+C}{r} - C(p'q'-p''q''+Q) = 0 , \\ \tag{K} \frac{d^2(r'^2)}{2dt^2} - \frac{A+B+C}{r'} - B(pq+p''q''+Q') = 0 , \\ \frac{d^2(r''^2)}{2dt^2} - \frac{A+B+C}{r''} - A(-pq-p'q'+Q'') = 0 . \end{array} \right. $$

  These equations will serve to determine the values of the distances r, r', r'' in terms of t ; after which one will be able to find directly and without any integration the values of all the other elements, on which depends the determination of the orbits of bodies B and C around body A.

  Indeed, if one names

u, B,A,
u',  the speed of body C, around A,
u'',C,B,
one gets first ##J3. $$ \left\{ \begin{array}{l} u^2 = \frac{2(A+B+C)}{r} + CQ , \\ \tag{J} u'^2 = \frac{2(A+B+C)}{r'} + BQ' , \\ u''^2 = \frac{2(A+B+C)}{r''} + AQ'' , \end{array} \right. $$

  If one names next

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  and one makes, for brevity, ##, ##, ##, ##, ##, ##, ##, ##, ##, ##, $$ \begin{aligned} P &= pp' + pp'' + p'p'' = r^2r'^2-p''^2 = ¼ (2r^2r'^2+2r^2r''^2+2r'^2r''^2-r^4-r'^4-r''^4) \\ &= - ¼(r+r'+r'')(r+r'-r'')(r-r'+r'')(r-r'-r'') , \\ \Sigma &= p\left(\frac{d(r^2)}{dt}\right)^2 + p'\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp'}{dt}\right)^2 - 2 \left(p''\frac{dp'}{dt}-p'\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma' &= p'\left(\frac{d(r'^2)}{dt}\right)^2 + p\left(\frac{dp''}{dt}\right)^2 + p''\left(\frac{dp}{dt}\right)^2 + 2 \left(p''\frac{dp}{dt}-p\frac{dp''}{dt}\right)\frac{d\rho}{dt} + r'^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Sigma'' &= p''\left(\frac{d(r''^2)}{dt}\right)^2 + p'\left(\frac{dp}{dt}\right)^2 + p\left(\frac{dp'}{dt}\right)^2 +2 \left(p\frac{dp'}{dt}-p'\frac{dp}{dt}\right)\frac{d\rho}{dt} + r''^2\left(\frac{d\rho}{dt}\right)^2 , \\ \Pi &= r^2u^2 - \left(\frac{d(r^2)}{2dt}\right)^2 , \\ \Pi' &= r'^2u'^2 - \left(\frac{d(r'^2)}{2dt}\right)^2 , \\ \Pi'' &= r''^2u''^2 - \left(\frac{d(r''^2)}{2dt}\right)^2 , \\ \Psi &= p\nu - \left(\frac{dp}{2dt}\right)^2 + \left(\frac{d\rho}{2dt}\right)^2 , \\ \Psi' &= p'\nu' - \left(\frac{dp'}{2dt}\right)^2 + \left(\frac{d\rho}{2dt}\right)^2 , \\ \Psi'' &= p''\nu'' - \left(\frac{dp''}{2dt}\right)^2 + \left(\frac{d\rho}{2dt}\right)^2 , \end{aligned} $$ by supposing ##, $$ \nu = \frac{u'^2+u''^2-u^2}2 , ~~~~ \nu' = \frac{u^2+u''^2-u'^2}2 , ~~~~ \nu'' = \frac{u^2+u'^2-u''^2}2 , $$ one now gets ##, $$ \sin\psi = \frac{\frac1B\sqrt{4Pu'^2-\Sigma'} + \frac1A\sqrt{4Pu''^2-\Sigma''}}{2hr}, ~~~~ \frac{d\varphi}{dt} = \frac{\frac{\Pi}C + \frac{\Psi'}A + \frac{\Psi''}B}{hr^2\cos^2\psi}, $$

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and ##. $$ \sin\psi' = \frac{\frac1C\sqrt{4Pu^2-\Sigma} + \frac1A\sqrt{4Pu''^2-\Sigma''}}{2hr'}, ~~~~ \frac{d\varphi'}{dt} = \frac{\frac{\Pi'}B + \frac{\Psi}A + \frac{\Psi''}C}{hr'^2\cos^2\psi'} . $$

  It is necessary to remark that these formulae contain two constants which are not arbitrary, but which must be determined by particular equations ; they are : one, the constant h, and the other, the constant which can be added to the value of dρ/dt deduced from equation (H) by means of integration.

  Here are, then, the equations which will serve to determine these constants ##N1, $$ \tag{N} 16PU - 4(\Sigma\nu +\Sigma'\nu' +\Sigma''\nu'') + \left(\frac{dpdp'+dpdp''+dp'dp''+d\rho^2}{dt^2}\right)^2 = 0 , $$ by supposing ## $$ U = \nu\nu' + \nu\nu'' + \nu'\nu'' = u^2u'^2-\nu''^2 = ¼(2u^2u'^2 + 2u^2u''^2 + 2u'^2u''^2 - u^4 - u'^4 - u''^4) $$ and ##P1. $$ \tag{P} \frac{\Pi}{C^2} + \frac{\Pi'}{B^2} + \frac{\Pi''}{A^2} + 2\left(\frac{\Psi}{AB} + \frac{\Psi'}{AC} + \frac{\Psi''}{BC}\right) = h^2 . $$

  One would be able, if one wished, to use these equations instead of any two of the equations (K) ; but, as they are complicated enough, it would be better to use them only in the determination of the constants concerned ; and for that it is clear that one can suppose everywhere t=0.

  Then if, for greater simplicity, one supposes that, when t=0, one has dr/dt=0, dr'/dt=0, and also that the radii r, r' coincide, such that the angle ζ'' between these radii (Article XIX) is zero, which is always permitted when that angle is variable, one gets, because r''2 = r2 + r'2 = 2rr'cosζ'' (cited Article), ##; $$ r''^2 = (r'-r)^2, ~~~~ \frac{dp''}{dt} = 0 ; $$ so ##; $$ \frac{dp}{dt} = 0, ~~~~ \frac{dp'}{dt} = 0, ~~~~ \frac{dp''}{dt} = 0 ; $$

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such that equation (N) becomes ##; $$ 16PU - 4(r^2\nu+r'^2\nu'+r''^2\nu'')\left(\frac{d\rho}{dt}\right)^2 + \left(\frac{d\rho}{dt}\right)^4 = 0 ; $$ but because r''2=(r'-r)2 one gets P=0 ; so also dρ/dt=0. Thus it would be necessary to take the value of dρ/dt=0, such that it becomes zero when t=0 ( * ).

  Equation (P) simplifies as much by the same suppositions, and becomes ##P'1; $$ \tag{P'} h^2 = \frac{r^2u^2}{C^2} + \frac{r'^2u'^2}{B^2} + \frac{r''^2u''^2}{A^2} + 2\left(\frac{p\nu}{AB} + \frac{p'\nu'}{AC} + \frac{p''\nu''}{BC}\right) , $$ where it is necessary to take for r, r', r'', u, u', u'' the values which correspond to t=0.

  As for the constants which can enter into the values of Q, Q', and Q'', they will be entirely arbitrary and only depend on the initial values of u, u', u'', which are at will.

  Finally, if one also names

one gets ##, ##. $$ \frac{d\theta}{dt} = \frac{\sqrt{\Pi}}{r^2}, ~~~~ \frac{d\theta'}{dt} = \frac{\sqrt{\Pi'}}{r'^2}, \\ \cos\omega = \frac{\Psi''}{\sqrt{\Pi\Pi'}}, ~~~~ \sin\xi = \frac{\sqrt{4Pu'^2-\Sigma'}}{2r\sin\omega\sqrt{\Pi'}}, ~~~~ \sin\xi' = \frac{\sqrt{4Pu^2-\Sigma}}{2r'\sin\omega\sqrt{\Pi}} . $$

  ( * ) It is necessary to remark that, in the present case, equation (N) is still satisfied if one takes ##. $$ \left(\frac{d\rho}{dt}\right)^2 = 4(r^2\nu+r'^2\nu'+r''^2\nu'') . $$

(Editor's Note)  

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CHAPTER II.

SOLUTION OF THE THREE-BODY PROBLEM IN DIFFERENT CASES

  In this Chapter, we shall examine some particular cases, where the three-body problem is greatly simplified and admits of an exact or almost exact solution ; although these cases do not occur in the System of the world, we think nevertheless that they merit the attention of Geometers, because they can cast light on the general solution of the three-body problem.

  The first case which presents itself is that where the three distances r, r', r'' are constant, so that the triangle formed by the bodies stays always the same and only changes its position.

  One has in this case ##, $$dr = 0,   dr' = 0,   dr'' = 0,$$ and in consequence also ##; $$dp = 0,   dp' = 0,   dp'' = 0,$$ so the three equations (K) become ##a3; $$ \left\{ \begin{array}{l} {A+B+C \over r } + C( p'q'-p''q''+Q ) = 0 \\ \tag{a} {A+B+C \over r' } + B( p q +p''q''+Q' ) = 0, \\ {A+B+C \over r''} + A(-p q -p' q' +Q'') = 0; \end{array} \right. $$ from which one sees that the quantities Q, Q', Q'' will be likewise constant,

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so that one has ##, $$ dQ=0, ~~~~ dQ'=0, ~~~~ dQ''=0, $$ meaning that equations (I) reduce to these ##, $$ qd\rho=0, ~~~~ q'd\rho=0, ~~~~ q''d\rho=0, $$ which give either q=0, q'=0, q''=0, or dρ=0. Let us examine these two cases separately.

  Put first ##; $$ q=0, ~~~~ q'=0, ~~~~ q''=0; $$ so ##; $$ r=r'=r''; $$ such that the triangle formed by the three bodies will be equilateral ; the equations (a) then give ##, $$ CQ = BQ' = AQ'' = - { {A+B+C} \over r } , $$ and, these values being substituted in the formulae (J), one has ##. $$ u^2 = u'^2 = u''^2 = - { {A+B+C} \over r } . $$

  Now one has ##; $$ p = p' = p'' = { r^2 \over 2 } , ~~~~ \nu = \nu' = \nu'' = { u^2 \over 2 } ; $$ so ##; $$ P = { 3r^4 \over 4 }, ~~~~ U = { 3u^4 \over 4 }; $$ moreover equation (H) will give d2ρ/dt2=0, consequently ##, $$ { d\rho \over dt } = \alpha, $$ α being an arbitrary constant which must satisfy equation (N).

  So one finds ##; $$ \Sigma = \Sigma' = \Sigma'' = r^2\alpha^2; $$

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such that the equation that we are discussing becomes ##, $$ 9r^4u^4 - 6 r^2u^2\alpha^2 + \alpha^4 = 0, $$ that is to say, ##, $$ ( 3r^2u^2 - \alpha^2 )^2 = 0, $$ from where ##. $$ \alpha^2 = 3r^2u^2 = 3(A+B+C)r. $$

  So one satisfies all the equations of the Problem ; so that the value of r remains indeterminate ; from which it follows that the system of the three bodies can move so that the three bodies always form an ordinary equilateral triangle.

  Having found ##, $$ P = { 3r^4 \over 4 }, ~~~~ \Sigma = \Sigma' = \Sigma'' = r^2\alpha^2 = 3r^4u^2, $$ one has ##; $$ 4Pu^2-\Sigma = 0, ~~~~ 4Pu'^2-\Sigma' = 0, ~~~~ 4Pu''^2-\Sigma'' = 0; $$ so ##; $$\sin\psi = 0, ~~~~ \sin\psi' = 0 ; $$ from which one sees that the three bodies will always necessarily be in one given plane.

  One finds next ##, ##; $$ \Pi = \Pi' = \Pi'' = r^2u^2, \\ \Psi = \Psi' = \Psi'' = p\nu + {\alpha^2 \over 4} = {r^2u^2 \over 4} + {3r^2u^2 \over 4} = r^2u^2; $$ then, because ψ=0 and ψ'=0, one gets ##; $$ {d\varphi \over dt} = {d\varphi' \over dt} = ( {1 \over C} + {1 \over A} + {1 \over B}) {u^2 \over h} ; $$ but equation (P) will give ##, $$ h^2 = ({1 \over C^2} + {1 \over B^2} + {1 \over A^2} + {2 \over AB} +{2 \over AC} +{2 \over BC} ) r^2u^2, $$ or else ##, $$ h^2 = ({1 \over C} + {1 \over B} + {1 \over A} )^2 ~ r^2u^2. $$

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consequently ##; $$ h = ({1 \over C} + {1 \over B} + {1 \over A} ) ~ ru; $$ so ##, $$ {d\varphi \over dt} = {d\varphi' \over dt} = {u \over r} = \sqrt { {A+B+C} \over r^3 } . $$

  Thus the bodies B and C just rotate around body A with a constant angular velocity equal to (A+B+C)/r³.

  Let us examine now the other case, where dρ/dt=0 without q, q', q'' being zeroes, and substitute first in equations (J) the values of CQ, BQ' and AQ'' drawn from equations (a) above ; one gets ##b3; $$ \left\{ \begin{array}{l} u^2 = {{A+B+C} \over r} - C( p'q'-p''q'') , \\ \tag{b} u'^2 = {{A+B+C} \over r'} - B( pq+p''q'') , \\ u''^2 = {{A+B+C} \over r''} - A(-pq-p'q') ; \end{array} \right. $$ from which one sees that the relative speeds of the bodies will also be constant, but not equal among themselves as in the preceding case.

  Then, because it is necessary that dρ/dt=0, one gets then also d2o/dt2=0, and equation (H) becomes ##c1. $$ \tag{c} Cpq - Bp'q' - Ap''q'' = 0. $$ Next equation (N) becomes (because of dr=0, dr'=0, dr''=0, and dp'=0, dp''=0, dρ=0) ##, $$ 16PU = 0, $$ therefore ##; $$ P=0, ~~~~ \text{or} ~~~~ U=0; $$

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thus, in combining one or other of these equations with the previous equation (c), one will be able to, by their means, determine any two of the three unknowns r, r', r'', and the problem will be solved.

  Suppose first P=0, one gets ##; $$ (r+r'+r'')(r+r'-r'')(r-r'+r'')(r-r'-r'') = 0 ; $$ then, because r, r', r'' are supposed positive, one gets these equations ##, $$ r+r'-r'' = 0, ~~~~ \text{or} ~~~~ r-r'+r'' = 0, ~~~~ \text{or} ~~~~ r-r'-r'' = 0, $$ from which one gets ##; $$ r'' = r+r', ~~~~ \text{or} ~~~~ r' = r+r'', ~~~~ \text{or} ~~~~ r = r'+r''; $$ that is to say, that one of the three distances must be equal to the sum of the other two, which shows that the bodies must always be arranged in a straight line.

  Let us imagine that the three bodies A, B, C are arranged successively in the same direction, so that one has ##, $$ r'' = r' - r, $$ and, taking for more simplicity r'=mr, one only needs to substitute in equation (c) mr in place of r', and (m-1)r in place of r''; the unknown r will go out, and one will have an equation which will serve to determine m. One finds then ##, ##, ##, ##, ##, ##; $$ \begin{aligned} p &= { { m^2 + (m-1)^2 - 1 } \over 2 } r^2 = (m^2-m)r^2, \\ p' &= { { 1 + (m-1)^2 - m^2 } \over 2 } r^2 = (1-m)r^2, \\ p'' &= { { 1 + m^2 - (m-1)^2 } \over 2 } r^2 = mr^2, \\ q &= [ {1 \over m^3} - {1 \over (m-1)^3} ] {1 \over r^3} = - {{3m^2-3m+1} \over m^3(m-1)^3} {1 \over r^3}, \\ q' &= [ 1 - {1 \over (m-1)^3} ] {1 \over r^3} = {{m^3-3m^2+3m} \over (m-1)^3} {1 \over r^2}, \\ q'' &= ({1 \over m^3} - 1) {1 \over r^3} = {{1-m^3} \over m^3} {1 \over r^3}; \end{aligned} $$ Should the "$q' =$" line end with "$\frac1{r^3}$" ? Yes, c.18 p.65 has a fuzzy 3.

First equation : cf. Heron's Formula.

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and, these substitutions having been made in equation (c), it becomes, after having been multiplied by m2(m-1)2r, ##d1, $$ \tag{d} C(-3m^2+3m-1) + Bm^3(m^2-3m+3) - A(m-1)^2(1-m^3) = 0, $$ which being ordered with respect to m will mount to the fifth degree, and will always have in consequence a real root.

  It is good to note here that, although we have supposed r''=r'-r, the solution nevertheless contains all the possible cases, because the distances r, r', r'', being taken in one straight line, can be positive or negative, according to the different position of the bodies.

  Now, because ##, $$ dr=0, ~~~~ dr'=0, ~~~~ dr''=0, ~~~~ d\rho=0, $$ one gets ##; $$ \Sigma=0, ~~~~ \Sigma'=0, ~~~~ \Sigma''=0; $$ so that, as one already has P=0, one gets ##, $$ \sin\psi=0, ~~~~ \sin\psi'=0, $$ which shows that the three bodies must move in a fixed plane.

  Let us suppose now the other factor U to be equal to zero, one gets ##. $$ U = \nu\nu' + \nu\nu'' + \nu'\nu'' = u^2u'^2 - \nu''^2 = 0. $$ Then equations (b) give ##, $$ {u^2 \over r^2} - {u'^2 \over r'^2} = - (A+B+C)q'' - {C \over r^2}(p'q'-p''q'') + {B \over r'^2}(pq+p''q''), $$ from which, on multiplying by r2r'2 and putting in the place of r2 and r'2 their values p'+p'' and p+p'', one gets, because q''=q-q', ##; $$ r'^2u^2-r^2u'^2 = (-Cq+Bq'-Aq'')P + (Cpq -Bp'q'-Ap''q'')p''; $$

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but ## $$ Cpq -Bp'q'-Ap''q'' = 0 $$ from equation (c) ; so one has simply ##, $$ r'^2u^2-r^2u'^2 = (-Cq+Bq'-Aq'')P, $$ and one finds likewise by analogy ##, ##, $$ r''^2u^2-r^2u''^2 = (Cq+Bq'-Aq'')P, \\ r''^2u'^2-r'^2u''^2 = (Cq+Bq'+Aq'')P, $$ from which it is easy to get ##, $$ p''u^2 - r^2\nu'' = -CqP, ~~~~ p''u'^2-r'^2\nu'' = - Bq'P, $$ and consequently ##; $$ \nu'' = {{p''u^2+CqP} \over r^2} = {{p''u'^2 +Bq'P} \over r'^2} ; $$ so ##, $$ \nu''^2 = { { p''^2u^2u'^2 + p''P(Bq'u^2 + Cqu'^2) + BCqq'P^2 } \over r^2r'^2 }, $$ and from that, because P = r2r'2 - p''2, ## ##. $$ \begin{aligned} U = u^2u'^2-\nu''^2 &= {P \over r^2r'^2} \left[u^2u'^2 - p''(Bq'u^2+Cqu'^2) - BCqq'P\right] \\ &= {P \over r^2r'^2} \left[(u^2-Cqp'')(u'^2-Bq'p'') - BCr^2r'^2qq'\right] \end{aligned} . $$

  But the same equations (b) give ##, ##; $$ u^2 - Cqp'' = \left(\frac{A+B}{r^3} + \frac{C}{r''^3}\right) ~ r^2 ~, $$ $$ u'^2 - Bq'p'' = \left(\frac{A+C}{r'^3} + \frac{B}{r''^3}\right) ~ r'^2 ~; $$ so, on substituting those values as well as those of q and q', one gets ##, $$ U = P \left[ \left(\frac{A+B}{r^3 } + \frac{C}{r''^3}\right) \left(\frac{A+C}{r'^3} + \frac{B}{r''^3}\right) - BC \left(\frac{1}{r'^3} - \frac{1}{r''^3}\right) \left(\frac{1}{r^3 } - \frac{1}{r''^3}\right) \right] ~, $$

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or indeed ##. $$ U = P\left[ \frac{A^2}{r^3r'^3} + \frac{B^2}{r^3r''^3} + \frac{C^2}{r'^3r''^3} + \frac{AB}{r^3} \left( \frac{1}{r'^3} + \frac{1}{r''^3} \right) + \frac{AC}{r'^3} \left( \frac{1}{r^3} + \frac{1}{r''^3} \right) + \frac{BC}{r''^3} \left( \frac{1}{r^3} + \frac{1}{r'^3} \right) \right] . $$ From which one sees that the equation U=0 can only give this P=0, the other factor of U can never become zero, because the radii r, r', r'' and the masses A, B, C are positive quantities.

  The equation P=0 being then the only one which can satisfy the case which we are examining, this case will only occur, as we have seen above, when the three bodies are arranged in one straight line, and their distances will be in the relationship expressed by equation (d).

  Then we have already found that the three bodies must move in a fixed plane ; so that, knowing the speed u of body B around A, it is only necessary to divide by r to have the angular velocity of bodies B and C ; but, if one wants to make use of the general formulae of Article XXII, one remarks that because P=0 one has (Article XXVII) ##; $$ \frac{u^2}{r^2} = \frac{u'^2}{r'^2} = \frac{u''^2}{r''^2} ; $$ but equations (b) give ##; $$ \frac{u^2}C + \frac{u'^2}B + \frac{u''^2}A = (A+B+C) \left(\frac 1r + \frac 1{r'} + \frac 1{r''} \right) ; $$ so, substituting the previous values of u'2 and u''2, and making, for brevity, ##, $$ k = { {(A+B+C) \left(\frac 1r + \frac 1{r'} + \frac 1{r''} \right)} \over {\frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A} } , $$ one gets ##; $$ u^2 = kr^2 ~, ~~~~ u'^2 = kr'^2 ~, ~~~~ u''^2 = kr''^2 ; $$

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so also ##. $$ \nu = kp , ~~~~ \nu' = kp' , ~~~~ \nu'' = kp'' . $$ Thus one has (Article XXII) ##, ##; $$ \Pi = kr^4 , ~~~~ \Pi' = kr'^4 , ~~~~ \Pi'' = kr''^4 , \\ \Psi = kp^2 , ~~~~ \Psi' = kp'^2 , ~~~~ \Psi'' = kp''^2 ; $$ but, because P=0, one has ##; $$ p^2 = r'^2r''^2, ~~~~ p'^2 = r^2 r''^2, ~~~~ p''^2 = r^2r'^2 ; $$ so ##; $$ \Psi = kr'^2r''^2, ~~~~ \Psi' = kr^2r''^2, ~~~~ \Psi'' = kr^2r'^2 ; $$ so equation (P) becomes ##, $$ h^2 = k \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right)^2 , $$ from which ##; $$ h = k \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right) \sqrt k ; $$ next, because ψ=0 and ψ'=0, ##. $$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac kh \left( \frac{r^2}C + \frac{r'^2}B + \frac{r''^2}A \right) = \sqrt k . $$

  We have supposed above that the radii r, r', r'' were constant. and we have seen that that can only occur in two cases, namely : when the three radii are all equal, and when one of them is equal to the sum of the other two. Let us suppose now that these three radii are only in a constant ratio among themselves, and let us see how that condition can occur. So let ##. $$ r'=mr, ~~~~ r'' = nr , $$

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m and n being constant quantities, and one will first get (Article XXII) ##, ##, $$ p = \mu r^2, ~~~~ p' = \mu' r^2, ~~~~ p'' = \mu'' r^2, \\ q = \frac{\varpi}{r^3}, ~~~~ q' = \frac{\varpi'}{r^3}, ~~~~ q'' = \frac{\varpi''}{r^3}, $$ by making, for brevity, ##, ##. $$ \mu = \frac{m^2+n^2-1}2, ~~~~ \mu' = \frac{1+n^2-m^2}2, ~~~~ \mu'' = \frac{1+m^2-n^2}2, \\ \varpi = \frac 1 {m^3} - \frac 1 {n^3}, ~~~~ \varpi' = 1 - \frac 1 {n^3}, ~~~~ \varpi'' = \frac 1 {m^3} - 1 = \varpi - \varpi'. $$ Then equation (H) becomes ##, $$ \frac{d^2\rho}{dt^2} + \frac {C\mu\varpi - B\mu'\varpi' - A\mu''\varpi''}r = 0, $$ or else, on making ##, $$ \lambda = C\mu\varpi - B\mu'\varpi' - A\mu''\varpi'', $$ for brevity, ##, $$ \frac{d^2\rho}{dt^2} + \frac \lambda r = 0, $$ and integrating ##, $$ \frac{d\rho}{dt} = \alpha - \lambda \int \frac{dt}r , $$ α being an arbitrary constant equal to the value of dρ/dt when t=0.

  Then one gets ##, ##, ##; $$ dQ = 2(\mu'\varpi'-\mu''\varpi'')\frac{dr}{r^2} - \varpi \left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3}, \\ dQ' = 2(\mu\varpi +\mu''\varpi'')\frac{dr}{r^2} + \varpi' \left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3}, \\ dQ'' = 2(-\mu\varpi-\mu'\varpi')\frac{dr}{r^2} + \varpi'' \left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3}; $$

282

so, by integrating, ##, ##, ##, $$ Q = - \frac{2(\mu'\varpi'-\mu''\varpi'')}r - \varpi \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} + k, \\ Q' = - \frac{2(\mu\varpi +\mu''\varpi'')}r + \varpi' \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} + k', \\ Q'' = - \frac{2(-\mu\varpi-\mu'\varpi')}r + \varpi'' \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} +k'', $$ k, k' and k'' being arbitrary constants.

  Making all those substitutions in equations (K), and next dividing the second by m2 and the third by n2, they become these ##, ##, ##, $$ \frac{d^2(r^2)}{2dt^2} - \frac{A+B+C(1-\mu'\varpi'+\mu''\varpi'')}r + C\varpi \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} -Ck = 0, \\ \frac{d^2(r^2)}{2dt^2} - \frac{A+C+B[1-(\mu\varpi+\mu''\varpi'')m]}{m^3r} - \frac{B\varpi'}{m^2} \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} -\frac{Bk'}{m^2} = 0, \\ \frac{d^2(r^2)}{2dt^2} - \frac{B+C+A[1+(\mu\varpi+\mu'\varpi')n]}{n^3r} - \frac{A\varpi''}{n^2} \int\left( \alpha - \lambda \int \frac{dt}r \right) \frac{dt}{r^3} -\frac{Ak''}{n^2} = 0, $$ which should be identical ; so that one has these conditions to fulfil : ##1~1, $$ \begin{aligned} \tag{1°} A+B+C(1-\mu'\varpi'+\mu''\varpi'') &= \frac{A+C+B[1-(\mu\varpi+\mu''\varpi'')m]}{m^3} \\ &= \frac{B+C+A[1+(\mu\varpi+\mu'\varpi')n]}{n^3} ; \end{aligned} $$ ##2~1; $$ \tag{2°} C\varpi = - \frac{B\varpi'}{m^2} = - \frac{A\varpi''}{n^2}, ~~ \text{ou bien} ~~ \alpha = 0 ~~ et ~~ \lambda = 0 ; $$ ##3~1. $$ \tag{3°} Ck = \frac{Bk'}{m^2} = \frac{Ak''}{n^2} . $$

  Those last two conditions can always be complied with by means of the indeterminate constants k, k' and k'' ; thus the difficulty only consists of satisfying those of groups 1° and 2°.

  Then, if one makes, for brevity, ##, $$ \delta = \mu\mu' + \mu\mu'' + \mu'\mu'' = - ¼ (1+m+n)(1+m-n)(1-m+n)(1-m-n), $$

283

one will be able to reduce the two equations of group 1° to these, by transformations analogous to those of Article XXVII, ##e1, $$ \tag{e} (-C\varpi + B\varpi' - A\varpi'')\delta + \mu''\lambda = 0 , ~~~~ ( C\varpi + B\varpi' - A\varpi'')\delta - \mu' \lambda = 0 . $$

  Then it is only necessary to combine these two equations with those of group 2°, thus ##; $$ C\varpi = - \frac{B\varpi'}{m^2} = - \frac{A\varpi''}{n^2}, ~~ \text{ou bien} ~~ \alpha = 0 ~~ \text{and} ~~ \lambda = 0 ; $$ which makes two cases that we will examine separately.

Let there be first ##; $$ C\varpi = - \frac{B\varpi'}{m^2} = - \frac{A\varpi''}{n^2} ; $$ then ##; $$ \varpi' = - \frac{m^2C\varpi}B, ~~~~ \text{and} ~~~~ \varpi'' = - \frac{n^2C\varpi}A ; $$ but one has (Article XXIX) ##; $$ \varpi - \varpi' - \varpi'' = 0 ; $$ so ##, $$ \varpi \left( 1 + \frac{m^2C}B + \frac{n^2C}A \right) = 0 , $$ viz. ##. $$ \varpi \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right) = 0 . $$

  So one can see that the quantity 1/C + m2/B + n2/A will never become zero, because the masses A, B, C are positive quantities ; so it is necessary that one has ω=0, and consequently ω'=0, ω''=0; then, in that case, one has λ=0 and the two equations (e) above are themselves satisfied ; then all the conditions are satisfied, and the three-body problem will be soluble exactly with the supposition that ##. $$ \varpi = 0, ~~~~ \varpi' = 0, ~~~~ \varpi'' = 0 , $$

284

which will give ##, $$ n = m = 1 , $$ and consequently ##, $$ r = r' = r'' , $$ That is to say, the distances between the bodies are all equal, as in the case of Article XIV ; but with this difference, that in the present case they can change.

  To know the movement of the bodies in that case, one takes again the differential equations of article XXIX, which, by making ##, $$ f = Ck = Bk' = Ak'' , $$ reduce to this single equation ##. $$ \frac{d^2(r^2)}{2dt^2} - \frac{A+B+C}r - f = 0 , $$ Multiplying by d(r2), and integrating together, one gets ##, $$ \left[ \frac{d(r^2)}{2dt} \right]^2 - 2 (A+B+C)r - fr^2 = H , $$ H being an arbitrary constant, and from there ##, $$ dt = \frac{rdr}{\sqrt{H + 2(A+B+C)r + fr^2)}} , $$ by means of which we get t in terms of r and vice versa, r in terms of t.

  Now, since ω=0, ω'=0, ω''=0, one gets ##; $$ Q = k, ~~~~ Q'=k', ~~~~ Q'' = k'' ; $$ so (Article XXII) ##. $$ u^2 = u'^2 = u''^2 = \frac{2(A+B+C)}r + f . $$

  Moreover, having λ=0, one gets ##, $$ \frac{d\rho}{dt} = \alpha , $$

285

and that constant α must be determined so that it satisfies equation (N) ; one can give to that at t whatever value one wants ; but, making no particular supposition, equation (N) must be identical with that which we have found above for the determination of r, and their comparison will give the value of α.

  Indeed, because r=r'=r'', one gets ##; $$ p = p' = p'' = \frac{r^2}2 ; $$ so ##; $$ P = \frac{3r^4}4 ; $$ and likewise, because u=u'=u'', one gets ##; $$ \nu = \nu' = \nu'' = \frac{u^2}2 ; $$ then ##; $$ U = \frac{3u^4}4 ; $$ together ##; $$ \Sigma = \Sigma' = \Sigma'' = \frac{r^2}2 \left( \frac{2rdr}{dt} \right) ^2 + r^2 \left( \frac{rdr}{dt} \right)^2 + r^2\alpha^2 = 3r^2 \left( \frac{rdr}{dt} \right)^2 + r^2\alpha^2 ; $$ and equation (N) becomes ##, $$ 9r^4u^4 - 6u^2 \left[ 3r^2\left(\frac{rdr}{dt}\right)^2 + r^2\alpha^2 \right] + \left[ 3\left(\frac{rdr}{dt}\right)^2 + \alpha^2 \right]^2 = 0 , $$ so ##; $$ \left[ 3r^2u^2 - 3\left(\frac{rdr}{dt}\right)^2 -\alpha^2 \right]^2 = 0 ; $$ from which one gets ##. $$ 3r^2u^2 - 3\left(\frac{rdr}{dt}\right)^2 -\alpha^2 = 0 ; $$

  Then one has already found ##; $$ u^2 = \frac{2(A+B+C)}r + f ; $$

286

so, substituting that value and resolving the equation, it becomes ##. $$ dt = \frac{rdr} {\sqrt{-\frac{\alpha^2}3 + 2(A+B+C)r + fr^2}} . $$

  Comparing then that equation with the previous, one gets ##, $$ H = - \frac{\alpha^2}3 , $$ and consequently ##; $$ \alpha = \sqrt{-3H} ; $$ from which one sees that H must necessarily be a negative quantity.

  One gets next ##, $$ \Pi = \Pi' = \Pi'' = r^2u^2 - \left( \frac{rdr}{dt} \right)^2 = \frac{\alpha^2}3 , $$ and ##; $$ \Psi = \Psi' = \Psi'' = \frac{r^2u^2}4 - \left( \frac{rdr}{2dt} \right)^2 + \left( \frac{\alpha}2 \right)^2 = \frac{\alpha^2}{4.3} + \frac{\alpha^2}4 = \frac{\alpha^2}3 ; $$ so equation (P) becomes ##, $$ \frac{\alpha^2}3 \left( \frac1C + \frac1B + \frac1A \right)^2 = h^2 , $$ from which one gets ##. $$ h = \frac{\alpha}{\sqrt3} \left( \frac1C + \frac1B + \frac1A \right) . $$

  So, because one has already found ##, $$ \Sigma = \Sigma' = \Sigma'' = 3r^2 \left( \frac{rdr}{dt} \right)^2 + r^2\alpha^2 , $$ and that ##, $$ 3r^2u^2 - 3\left( \frac{rdr}{dt} \right)^2 - \alpha^2 = 0 , $$ one gets ##; $$ \Sigma = \Sigma' = \Sigma'' = 3r^4u^2 ; $$

287

moreover one has ##; $$ 4P = 3r^4, ~~~~ \text{and} ~~~~ u = u' = u'' ; $$ so one has ##; $$ 4Pu^2 - \Sigma = 0, ~~~~ 4Pu'^2 - \Sigma' = 0, ~~~~ 4Pu''^2 - \Sigma'' = 0 ; $$ and in consequence ##, $$ \sin\psi = 0 , ~~~~ \sin\psi' = 0 , $$ which is to say, ##; $$ \psi = 0 , ~~~~ \psi' = 0 , $$ which shows that bodies B and C must move in one fixed plane through body A.

  Now, if one substitutes in the expressions dψ/dt and dψ'/dt the values of H, H', Ψ, Ψ', Ψ'', and of h found above, one gets ##; $$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac\alpha{r^2\sqrt3} ; $$ and in consequence, in substituting the above value of dt, ##, $$ d\varphi = { dr \over { r \sqrt { -1 + {6(A+B+C) \over \alpha^2}r + {3f \over \alpha^2}r^2 }}}, $$ words.

which is the polar equation of a conic section with respect to the focus, and in which 2(A+B+C)/-f is the major axis and 2/3(A+B+C) the parameter.

  So the bodies B and C describe in this case around body A two similar and equal conic sections, of which the type and the shape depend on the arbitrary quantities f and α, which can be determined by the equations ##, $$ \alpha^2 = 3r^2u^2 - 3\left(\frac{rdr}{dt} \right)^2 , ~~~~ f = u^2 - \frac{2(A+B+C)}r , $$ by giving to u, r, and dr/dt the values which fit at the first instant.

288

  It remains to examine the case where α=0 and λ=0; then the supposition that λ=0 first reduces equations (e) to these ##, $$ (-C\varpi + B\varpi' - A\varpi'')\delta = 0 , ~~~~ ( C\varpi + B\varpi' - A\varpi'')\delta = 0 , $$ which lead to ##, $$ \delta = 0 , $$ or else ##, $$ -C\varpi + B\varpi' - A\varpi'' = 0 , ~~~~ \text{and} ~~~~ C\varpi + B\varpi' - A\varpi'' = 0 , $$ which is to say, ##. $$ C\varpi = 0 ~~~~ \text{and} ~~~~ B\varpi' - A\varpi'' = 0 . $$

  Then I observe first that these last two equations are unfruitful; because one first has ω=0, then, because ω''=ω-ω', one has ω''=-ω'; so that the equation Bω'-Aω''=0 becomes (B+A)ω'=0, which gives ω'=0; one has then ω=ω'=ω''=0, which returns to the case that we have examined before.

  It is necessary then to make δ=0, so that the solution to the problem will be contained in these three equations ##. $$ \delta = 0 , ~~~~ \lambda = 0 , ~~~~ \alpha = 0 . $$

  The first will give (Article XXIX) ##; $$ (1+m+n)(1+m-n)(1-m+n)(1-m-n) = 0 ; $$ so ##, $$ 1 \pm m \pm n = 0 , $$ and in consequence ##, $$ r \pm r' \pm r'' = 0 , $$ which is to say, that one of the three distances r, r', r'' must be equal to the sum of the other two, and consequently that the three bodies must be all arranged in a single straight line.

  This case is therefore analogous to that of Article XXVI, but it is more

P.288, after "The first will give (Article XXIX)" : cf. Heron's formula for the area of a triangle.

289

general, in that the distances between the bodies can be variable, provided that their ratios are constant.

  One determines these ratios from the equation λ=0, and for that one can suppose, as in the cited Article, that the three bodies A, B, C are initially placed in a single straight line, such that r''=r'-r, which gives n=m-1 ; one substitutes then this value of n in the expression for λ in Article XXIX, and one gets an equation in m which is the same as equation (d) of Article XXVI. But it is still necessary to see whether the condition of α=0 can be satisfied ; and as the constant α must be determined by equation (N), all reduces to knowing if that equation can hold good by making α=0, which is to say, dρ/dt=0, because of dρ/dt = α - λ ∫ dt/r (Article XXIX) and of λ=0.

  So, by supposing dρ/dt=0, and substituting for r', r'' and p, p', p'' their values (cited Article) one gets ##, ##, ##, ##, $$ \begin{aligned} P &= (\mu\mu' + \mu\mu'' + \mu'\mu'')r^4 , \\ \Sigma &= (\mu + \mu'\mu''^2 + \mu''\mu'^2)~r^2 \left( \frac{2rdr}{dt} \right)^2 , \\ \Sigma' &= (\mu'm^4 + \mu\mu''^2 + \mu''\mu^2)~r^2 \left( \frac{2rdr}{dt} \right)^2 , \\ \Sigma'' &= (\mu''n^4 + \mu'\mu^2 + \mu\mu'^2)~r^2 \left( \frac{2rdr}{dt} \right)^2 , \end{aligned} $$ ##; $$ \frac{dpdp' + dpdp'' + dp'dp'' + d\rho^2}{dt^2} = (\mu\mu' + \mu\mu'' + \mu'\mu'') \left(\frac{2rdr}{dt} \right)^2 ; $$ but, by the nature of the quantities μ, μ', μ'', one has ##; $$ \mu' + \mu'' = 1 , ~~~~ \mu + \mu'' = m^2 , ~~~~ \mu + \mu' = n^2 ; $$ also, one has, by virtue of the equation δ=0, ##; $$ \mu\mu' + \mu\mu'' + \mu'\mu'' = 0 ; $$ so one will have also ##; $$ \mu + \mu'\mu''^2 + \mu''\mu'^2 = 0 , ~~~~ \mu'm^4 + \mu\mu''^2 + \mu''\mu^2 = 0 , ~~~~ \mu''n^4 + \mu'\mu^2 + \mu\mu'^2 = 0 ; $$

290

so that all the preceding quantities P. Σ, ... will be zero, and consequently equation (N) is itself verified.

  Now it is clear that because of α=0 and λ=0 the three differential equations of Article XXIX will reduce to this ##, $$ \frac{d^2(r^2)}{2dt^2} - \frac Fr - f = 0 , $$ by making, for brevity, ##, ##. $$ F = A + B + C(1 - \mu'\varpi' + \mu''\varpi'') , \\ f = Ck = \frac{Bk'}{m^2} = \frac{Ak''}{n^2} . $$

  That equation being then multiplied by d(r2), and then integrated, will give ##, $$ \left(\frac{rdr}{dt} \right)^2 - 2Fr - fr^2 = H , $$ H being an arbitrary constant ; from which one gets ##, $$ dt = \frac{rdr}{\sqrt{H + 2Fr + fr^2}} , $$ by means of which one determines t in terms of r, and consequently r in terms of t.

  Also, if in the equations (J) of Article XXII one substitutes the values of Q, Q', Q'' of Article XXIX, one has, by virtue of the equations of group 1° of that Article, ##; $$ u^2 = \frac{2F}r + f , ~~~~ u'^2 = \frac{2m^2F}r + m^2f , ~~~~ u''^2 = \frac{2n^2F}r + n^2f ; $$ so ##, $$ \nu = \frac{2\mu F}r + \mu f , ~~~~ \nu' = \frac{2\mu'F}r + \mu'f , ~~~~ \nu'' = \frac{2\mu''F}r + \mu''f . $$

291

From that one will find ##, ##, ##, ##, ##, ##, $$ \begin{aligned} \Pi &= 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 = - H , \\ \Pi' &= m^4 \left[ 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 \right] = - Hm^4 , \\ \Pi'' &= - Hn^4 , \\ \Psi &= \mu^2 \left[ 2Fr + fr^2 - \left( \frac{rdr}{dt} \right)^2 \right] = - H\mu^2 = - Hm^2n^2 , \\ \Psi' &= - H\mu'^2 = - Hn^2 , \\ \Psi'' &= - H\mu''^2 = - Hm^2 , \end{aligned} $$ because of ##, $$ \mu\mu' + \mu\mu'' + \mu'\mu'' = 0 , $$ and in consequence ##. $$ \mu^2 = m^2n^2 , ~~~~ \mu'^2 = n^2 , ~~~~ \mu''^2 = m^2 . $$

  Thus, if one substitutes these values in equation (P), it becomes ##, $$ h^2 = -H \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right)^2 , $$ from which ##, $$ h = \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right) \sqrt{-H} , $$ from which one sees that H must be a negative quantity,

  So, because of P=0 and of Σ=0, Σ'=0, Σ''=0, one gets ##, $$ \sin\psi = 0 ~~~~ \text{and} ~~~~ \sin\psi' = 0 , $$ which shows that the two bodies B and C must move on one fixed plane passing through the body A, and one then finds for the angles of rotation ##, $$ \frac{d\varphi}{dt} = \frac{d\varphi'}{dt} = \frac{-H}{hr^2} \left( \frac 1C + \frac{m^2}B + \frac{n^2}A \right) = \frac{\sqrt{-H}}{r^2} . $$

292

And, if one substitutes the value of dt, found above, one gets ##, $$ d\varphi = { dr \over r \sqrt { -1 + {2F \over -H} r + {f \over -H} r^2 } }, $$ the polar equation of a conic section, with respect to the focus, in which 2F/f will be the major axis and -2H/F the parameter.

So we come to see that the three body problem is solvable exactly, either with the distances between the three bodies being constant, or with them only remaining in constant ratios, and that there are two cases – when the three distances are all equal, so that the three bodies always form an equilateral triangle, and when one of the distances is equal to the sum or the difference of the other two, so that the three bodies are always arranged in a straight line.

  Now, if one supposes that the distances r. r', r'' are variable, but in such a manner that their values differ very little from those that they should have had for one of the preceding cases to apply, it is clear that the Problem will be soluble roughly, by the known methods of approximation ; but we will not enter here into that detail, which will take us too far from our main object.

  I acknowledge, moreover, that one could solve the preceding Problems in a more straightforward manner by the ordinary formulae of the Three-Body Problem with vector radii and the angles described by those radii, if one wanted to restrict oneself initially to the hypothesis that the bodies would move in a fixed plane ; but it would not be easy, I think, to achieve the purpose by those formulae, if one supposed, as we have done, that the bodies could move in different planes.


293-303

CHAPTER III.

MODIFICATION OF THE FORMULAE OF THE PREVIOUS CHAPTER, FOR THE CASE WHERE ONE SUPPOSES THAT ONE OF THE THREE BODIES IS DISTANT FROM THE OTHER TWO.

XXXIV.

... ...

303-324

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CHAPTER IV.

ON THE THEORY OF THE MOON.

XL.

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The following partial translation of the Editor's Note has not been independently checked.
324


The first Chapter of the Memoir that one has just read deserves to be counted among the most important works of the illustrious Author. The differential equations of the Three-Body Problem, when one considers, which is permissible, only relative motions, constitute a system of the twelfth order, and the complete solution requires, in consequence, twelve integrations; the only known integrals are those of the kinetic energies and the three which furnish the area rule (i.e. Kepler's 2nd Law. JRS) : there remain eight to discover. ... ...

325

...

The method of Lagrange is most remarkable; it shows that the complete solution of the problem requires only that one knows at each instant the sides of the triangle formed by the three bodies; the coordinates of each Body are then determined without any difficulty. ... ...

326

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327

... ...

328

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329

... ...

330

... ...

331

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(Editor's Note.)    

-==-0-==-

There is no reference to Euler, which suggests that Lagrange did not think that Euler had contributed to finding the constant-pattern solutions of the three-body problem.

Collaborators

My translation has been reviewed and somewhat corrected by a native French speaker, to whom many thanks.

Questions

Are the "undetermined coefficients" in Page 254 related to "Lagrange's undetermined multipliers"? Or k in Page 256?

The Web has (2012-07-19) over a thousand references to "Lagrange's quintic equation" - can that equation be identified in the Essai? Perhaps Equation (d) at the top of p.277? Also around p.289?

References

Code

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