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A translation of
De motu rectilineo trium corporum
se mutuo attrahentium

E.327, by Leonhard Euler
(On the rectilinear motion of three bodies
mutually attracted to each other

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Latinists, mathematicians, and others are invited to offer improvements to the translation.
Material common to Euler and the Lagrange Points, including References, is now consolidated in About Euler on Lagrange Points.

Why This Page?

Judgung by its title, E.327,De motu rectilineo trium corporum se mutuo attrahentium (On the rectilinear motion of three bodies mutually attracted to each other) seemed the most likely of the papers of Leonhard Euler to contain material related to the Lagrange Points.

Since I could not find an English version of E.327 on the Web, I have translated the PDF images into English and HTML. The translation was initially by Google Translate, and subsequently improved. The original equations are not 100% legible, and their transcription will need to be checked by considering the mathematics. They are also not 100% correct.

In E.327, Euler considered only linear motion. Therefore, E.327 may suggest, but does not predict, L1, L2 and L3.

About This Page

There is a later edited transcription, with notes by Max Schürer, in Leonhardi Euleri Commentationes Astronomicae (1860).
Partial Preview - Google Books, images - but none of E.327 is shown.
Likewise for "Look inside" at Amazon.

I have  highlighted  the parts of the text of E.327 which seem to be of greatest importance. Euler's three  marginal notes  are indicated thus. Notes by me are on page body background colour.

For the original algebraic expressions, open in parallel a copy of De motu rectilineo trium corporum se mutuo attrahentium, for which see References elsewhere. Expressions in this text represent those of Euler, without necessarily preserving the layout.

In the original text, it can be hard to see any difference between italic 'a' and Greek 'α'; also, 'u' and 'v' are generally equivalent; 'i' and 'j' can be; and 'ſ' represents 's'. Greek beta occurs in two forms, ϐ/old-beta and β, assumed equivalent and both shown here as β. Where a represents a simple variable, Euler used aa rather than a2. Between equations, Euler's "et" is here transcribed as "&" to obviate translation.

There is now a button which shows the text as a single column in a fresh tab or window, a form which may be more convenient for suggesting changes. This is constructed, by code shown below, on your computer from the three-column version; so no manual copying or duplicate transmission is involved. Details may change.

A similar button omits the Latin.

My main Lagrange page is Gravity 4 : The Lagrange Points.

Using Windows XP pro sp3, Chrome 22.0 or Safari 5.1.7, within the main table, strange things are sometimes happening to the vertical positioning of italic characters within their line of text. But not in Firefox 3.6.22+, Opera 11.51+, or IE 8.

Leonhard Euler - E.327 -
"On the rectilinear motion of three bodies mutually attracted to each other"

Euler Archive : De motu rectilineo trium corporum se mutuo attrahentium

Eneström Number E.327 : pages 144-151 (of Tome IX?).
Written 1763, presented 1765, published 1767.

by JRS
by Google, JRS, et al
            A           B                 C
Sint A, B, C massae trium corporum eorumque distantiae a puncto fixo O ad datum tempus t ponantur Let A, B, C be the masses of three bodies, and their distances from a fixed point O at a given time t be
OA = x,   OB = y   &   OC = z
ubiquidem sumitur y>x et z>y. Hinc motus principia praebent has tres aequationes : with the condition that y>x and z>y. Then the laws of motion give these three equations :
(using units such that the gravitational constant G in Newton's F = GMm/r2 is unity)
  I.   ddx/dt2 =   B/(y−x)2 + C/(z−x)2
  II.   ddy/dt2 = −A/(y−x)2 + C/(z−y)2
  III.   ddz/dt2 = −A/(z−x)2 − B/(z−y)2
Character d is never a variable; dd would now be written d2.
unde facile deducuntur binae aequationes integrabiles : 1 from which are easily deduced two integrable equations :
In the following, dx2 means (dx)2, not d2x which is shown as ddx.
Prior : Multiply I, II, III respectively by A, B, C and add ; integrate ; integrate dt :
Ad2x/dt2 + Bd2y/dt2 + Cd2z/dt2   =   0
Adx/dt + Bdy/dt + Cdz/dt   =   E
Ax + By + Cz   =   Et + F

no external forces,
constant momentum,
steady mean motion.
prior | first Adx + Bdy + Cdz = Edt   &   Ax + By + Cz = Et + F
Posterior : as in E.626 §6 :   Multiply I, II, III in turn by Adx, Bdy, Cdz & add, evaluating LHS & RHS independently :
Adxd2x/dt2 + Bdyd2y/dt2 + Cdzd2z/dt2 = + AB(dxdy)/(yx)2 + AC(dxdz)/(zx)2 + BC(dydz)/(zy)2
= − AB(d(yz))/(yx)2 − AC(d(zx))/(zx)2 − BC(d(zy))/(zy)2
Now integrate both sides dx, dy, dz, etc.
(note : d (dx/dt)2 / dt = 2 (dx/dt) d2x/dt2) :
A(dx)2/dt2 + B(dy)2/dt2 + C(dz)2/dt2 = ( K + AB/(yx) + AC/(zx) + BC/(zy) ) × 2
posterior | after  (Adx2 + Bdy2 + Cdz2) / dt2 = G + 2AB/(y−x) + 2AC/(z−x) + 2BC/(z−y) .
Hinc autem ob defectum tertiae aequationis integralis parum ad motus cognitionem concludere licet. But from this, due to the lack of a third integral equation, little knowledge of the motion can be inferred.
  2. Statuamus x=y−p et z=y+q, ut p et q sint quantitates positiuae : et prima integralis praebet : 1  2. Let us set x=y−p and z=y+q, such that p and q are positive quantities : the first integral gives :
O           A           B                 C 
From O :    x           y                 z 
            |     p     |         q       | 

Substitute in   Ax + By + Cz   =   Et + F   to get :  
(A+B+C)y−Ap+Cq = Et+F       ideoque | and therefore
y = (Ap−Cq+Et+F) / (A+B+C) ;   dy = (Adp−Cdq+Edt) / (A+B+C)
x = (−(B+C)p−Cq+Et+F)/ (A+B+C) ;   dx = (−(B+C)dp−Cdq+Edt) / (A+B+C)
z = (Ap+(A+B)q+Et+F) / (A+B+C) ;   dz = (Adp+(A+B)dq+Edt) / (A+B+C)
Hinc integralis secunda hanc induit formam : ? From this the second integral takes this form :
(A(B+C)dp2 + C(A+B)dq2 + 2ACdpdq + EEdt2 / (A+B+C)dt2
=   G + 2AB/p + 2AC/(p+q) + 2BC/q
  3. Faciamus vero easdem substitutiones in primis aequationibus differentio-differentialibus, quae iam ad duas revocabantur : ?  3. But let us make the same substitutions in the first double-differential equations, which were already reduced to two :
(−(B+C)ddp − Cddq) / (A+B+C) dt2 = B/pp + C/(p+q)2
(Addp + (A+B)ddq) / (A+B+C) dt2 = −A/(p+q)2 − B/qq
unde colligitur : (ddp+ddq)/dt2 = (−A−C)/(p+q)2 −B/pp −B/qq . ? whence is deduced : (ddp+ddq)/dt2 = (−A−C)/(p+q)2 −B/pp −B/qq .
Deinde utrumque elementum ddp et ddq seorsim ita exprimitur : ? Then each element ddp and ddq is expressed separately
  1°.   ddp/dt2 = (−A−B)/pp − C/(p+q)2 + C/qq
  2°.   ddq/dt2 = A/pp − A/(p+q)2 − (B−C)/qq
unde oritur una aequatio integralis ad hanc formam perducta ? whence arises one integral equation reduced to this form*
(B(Adp2+Cdq2) + AC(dp+dq)2) / (A+B+C)dt2 = G + 2AB/p + 2AC/(p+q) + 2BC/q
quandoquidem in G postremus ille terminus EE comprehenditur. ? since that previous term EE is included in G.
  4. Cum igitur solutio sit perducta ad duas aequationes differentio-differentiales inter p, q et t ; insigne lucrum obtineri est censendum, si has aequationes ad duas alias primi tantum gradus differentiales revocare liceret. Hoc autem singulari artificio sequentem in modum praestari posse comperi. Statuo q=pu, et bonae aequationes differentio- differentiales ita repraesententur : ?  4. Since with the solution is reduced to two double-differential equations in p, q and t; we can consider ourselves to have obtained a substantial gain if these equations may be reduced to two others of only first order differentials. But by this singular artifice I have found out that it can be shown in the following way. I set q=pu, and good double-differential equations may be >represented thus :
d dp/dt = dt/pp (− A − B − C/(u+1)2 + C/uu)
d (udp+pdu)/dt = dt/pp (A − A/(u+1)2 − (B−C)/uu) .
Iam artificium in hac substitutione consistit, ut ponam dp/dt = r/√p et dq/dt = (udp+pdu)/dt = s/√p ; mox enim patebit his substitutionibus binas variabiles p et t ex calculo elidi posse ita ut tantum hae tres r, s et u reliquantur, per prima differentialia determinandae. Statim vero aequatio illa integralis supra inuenta adeo ad formam finitam redit hanc (B(Arr+Css)+AC(r+s)2) / (A+B+C) = Gp + 2AB + 2AC/(u+1) + 2BC/u quae insignem usum afferre poterit. ? Now the artifice in this substitution consists in making dp/dt = r/√p and dq/dt = (udp+pdu)/dt = s/√p ; for soon it will be clear that, these substitutions having been made, the two variables p and t can be eliminated so that there are only these three r, s and u left to be determined by the first differentials. But immediately that integral equation found above can even revert to this finite form (B(Arr+Css)+AC(r+s)2) / (A+B+C) = Gp + 2AB + 2AC/(u+1) + 2BC/u which will be able to prove itself extraordinarily useful.
  5. Cum sit dp/dt=r/√p erit dt=dp√p/r, unde nostrae aequationes differentio-diffentiales praebent: ?  5. Since dp/dt=r/√p, then dt=dp√p/r, and so our double-differential equations give:
dr/√p − rdp/2p√p = dp/pr√p (− A − B − C/(u+1)2 + C/uu)
ds/√p − sdp/2p√p = dp/pr√p (A − A/(u+1)2 − (B−C)/uu)
unde sit : ? thence :
dr = rdp/2p + dp/pr (− A − B − C/(u+1)2 + C/uu)
ds = sdp/2p = dp/pr (A − A/(u+1)2 − (B−C)/uu) .
Praeterea vero habebitur : ? Further, however, we will have :
udp + pdu = sdt/√p = sdp/r
sicque sit dp/p = rdu/(s−ru), quo valore ibi substituto sit ? and so it is udp + pdu = sdt/√p = sdp/r, which value having been substituted
dr(s−ru) = ½rrdu + du(−A − B − C/(u+1)2 + C/uu)
ds(s−ru) = ½rsdu + du(A − A/(u+1)2 − (B−C)/uu)  
ex quarum combinatione nascitur : ? from the combination of which arises :
½r(rds−sdr) + ds(−A − B − C/(u+1)2 + C/uu)
dr(A − A/(u+1)2 − (B−C)/uu) = 0 .
  6. En ergo duas aequationes simpliciter differentiales inter ternas variabiles r, s et u, unde si liceret r et s per u determinare, haberetur solutio problematis completa. Inde enim primo innotesceret quantitas p ex formula dp/p=rdu/(s−ru) hincque porro q=pu. Deinde vero tempus t daretur ex aequatione dt=dp√p/r=pdu√p/(s−ru); Ex quibus tandem pro dato tempore t colligerentur distantiae x, y, z ex formulis §. 2 datis. ?  6. See, then, two simpler differential equations in the three variables r, s and u from which if it were possible to determine r and s from u, there would be a complete solution of the problem. For from that, the quantity p would first become known from the formula dp/p=rdu/(s−ru) and hence moreover q=pu. But then the time t would be given from dt=dp√p/r=pdu√p/(s−ru); From which at last for a given time t, the distances x, y, z could be deduced from the formulas given in §. 2.
  7. Cum binae aequationes differentiales inventae sint : ?  7. Since two differential equations are found :
dr(s−ru) = ½rrdu + du(−A − BC/(u+1)2 + C/uu)
ds(s−ru) = ½rsdu + du(A − A/(u+1)2 − (B−C)/uu) .
In the first equation just above, as in the original, a sign is missing between B and C.
statim patet ambabus satisfieri sumendo quantitatem u constantem et s−ru=0, unde solutio obtinetur ? it is immediately clear that both are satisfied by assuming u constant and s−ru=0, from which the particular solution is obtained.
The above seems to be where constant-pattern comes in.
Solutio particularis Particular solution
particularis. Sit ergo u=α et s=αr, et aequatio ex combinatione nata praebet : ? Therefore, let u=α and s=αr, and the equation resulting from the combination gives :
− (A+B)α − Cα/(α+1)2 + C/α = A − A/(α+1)2 − (B−C)/αα   vel | or
0 = A(α + 1 − 1/(α+1)2) + B(α−1/αα) + C(α/(α+1)2 − 1/α −1/αα)   (?
"(" after C not in PDF? check.
seu | or   0 = A((α+1)3−1)/(α+1)2 + B(α3−1)/αα + C(α3−(α+1)3)/αα(α+1)2
ideoque ? and therefore
(1+3α+3αα) = Aα3(αα+3α+3) + B(α+1)23−1) .
Quaere quantitatem α ex hac aequatione quinti gradus definiri oportet : ? Look for quantity α from which an equation of the fifth degree should be defined :
(A+B)α5 + (3A+2B)α4 + (3A+B)α3 − (B+3C)α2 − (2B+3)α − B − C = 0 .
Deinde vero relatio, inter r et p ex hac aequatione est definienda : ? But then from this equation a relationship between r and p is to be defined :
dr = rdp/2p + dp/dr(−A − B − C/(α+1)2 + C/αα)
seu posito | or if we suppose   A + B + C/(α+1)2 − C/αα = ½D   ex hac | from this
2dr = dp/p (r−D/r)   seu | or   dp/p = 2rdr/(rr−D)   ita ut sit | so that it may be  
p = β(rr−D),   tum vero | then indeed   q = αβ(rr−D)   &   dt = dp√p/r
seu | or   dt = 2βdr√β(rr−D)   hinc | hence
t = βr√β(rr−D) − β2D∫dr/√β(rr−D) .
  8. Casus hic particularis, quo solutio succedit evolutionem diligentiorem meretur. Primum ergo observo ex aequatione illa quinti gradus pro α semper valorem realem positiuum, eumque unicum elici, cum unica signorum variatio occurrat, neque ?  8. This particular case, for which the solution follows, deserves a more exact working out. First, therefore, I observe that from that fifth degree equation in α a single real positive value is always obtained, wherever a single change of sign occurs,
igitur hic ulla ambiguitas locum habet, sed valor ipsius α tanquam determinatus spectari potest, pendens a massis trium corporum A, B, C. Inuento autem numero α colligitur quantitas D = 2(A+B) − 2C(2α+1)/αα(α+1)2, ubi animaduerto quantitatem D numquam evanescere posse. Si enim esset D=0 foret B = C(2α+1)/αα(α+1)2 − A  quo valore substituto prodiret : ? and so here there is no room for ambiguity, but the value of α itself can be considered as if determined, depending on the masses of the three bodies A, B, C. The number α having been found, the quantity D = 2(A+B) − 2C(2α+1)/αα(α+1)2 may be deduced, where I notice that the quantity D can never vanish. For if D=0, then   B = C(2α+1)/αα(α+1)2 − A   which value being substituted,
C(1+3α+3αα) = Aα3(αα+3α+3) + C(2α+1)(α3−1)/αα − A(α+1)23−1)
seu | or   C/αα (α4 + 2α3 + αα + 2α + 1) = A(α4 + 2α3 + αα + 2α + 1)
Ideoque | And so   C = Aαα   &   B = A(2α+1)/(α+1)2 = A = −Aαα/(α+1)2
foretque adeo massa B negativa, quod est absurdum. Multo minus quantitas D unquam fieri potest negativa. Posito enim : ? and mass B would even be negative, which is absurd. Much less can the quantity D ever be negative. For, given that
B = C(2α+1)/αα(α+1)2 − A − Δ ;   proveniret | would result
C/αα = A − Δ(α+1)23−1)/(α4 + 2α3 + αα + 2α + 1)   hincque | and hence
B = C(2α+1)/αα(α+1)2 − C/αα − Δ(α5 + 3α4 + 3α3)/(α4 + 2α3 + αα + 2α + 1)
seu B multo magis esset quantitas negativa, cum valor ipsius α necessario sit positiuus. ? or B would be by much more a negative quantity, where the value of α itself would be necessarily positive.
  9. Cum ergo D necessario sit quantitas positiva, ponatur D=aa, si etiam numerus α spectetur ut datus, massae trium corporum ita se habebunt, ut sit : ?  9. Therefore, since D is necessarily a positive quantity, D=aa should be set, if also the number α should be regarded as given, the masses of the three bodies are such that :
B = α3(αα+3α+2)aa/2(α4+2α3+αα+2α+1) − C/[α+1]2   &
A = C/αα − (α+1)23−1)aa/2(α4+2α3+αα+2α+1)    
ex quo necesse est ut quantitas ? from which it is necessary that the quantity
2C(α4+2α3+αα+2α+1) / αα(α+1)2aa
intra hos limites (α+1)5−1 et α5−1 contineatur. Introducta ergo quantitate aa cum numero α in calculum, duo casus sunt perpendi, prout β fuerit quantitas positiva vel negativa, quos seorsim evolvamus. ? is contained within the limits (α+1)5−1 and α5−1. The quantity aa therefore having been introduced with the number α in the calculation, two cases are to be considered, for β a positive or negative quantity, which must be treated separately.
Casus I. Case 1.
  10. Sit primo β=+nn, erit p=nn(rr−aa) et q=αnn(rr−aa), unde cum constantes, E et F nihilo aequales statuere liceat, loca trium corporum A, B, C, quorum iam centrum gravitatis in O existit, ita per r definiuntur, ut sit : ?  10. First let β=+nn, so p=nn(rr−aa) and q=αnn(rr−aa), and so since the constants E and F can be set to zero, and the positions of the three bodies A, B, C, whose centre of gravity is O, and thus are defined by r, so that :
x = OA = −nn(rr−aa)/(A+B+C)   (B+C+Cα)   
y = OB =    nn(rr−aa)/(A+B+C)   (A−Cα)    
z = OC =    nn(rr−aa)/(A+B+C)   (A+(A+B)α).
At relatio inter r et tempus t ita se habet : ? But the relationship between r and the time t is this:
t = n3r√(rr−aa) − n3aa∫dr/√rr−aa)   seu | or
t = n3r√(rr−aa) − n3aa ? (r+√(rr−aa))/Δ. 
sumta constante Δ=a, tempore t=0, erat r=a, tumque omnia corpora in centro gravitatis erant conjuncta, unde quasi celeritatibus infinitis erant explosa, ut eae fuerint inter se ut quantitates −B−C−Cα, A−cα, A+(A+B)α tum vero labente tempore t quantitas r continuo magis increscit; ? Δ=a assumed constant, at time t=0 , we had r= a, and when all the bodies were together at the centre of gravity whence they were driven away as if at infinite speeds, as if they were to each other as the quantities −B−C−Cα, A−Cα, A+(A+B)α but then with the passage of time t the quantity r increases, more and more;
quouis autem tempore celeritas cuiusque corporis ex formula dt/dr=2n3√(rr−aa) innotescit. Notandum autem distantias corporum perpetuo inter se proportionem conservare. ? the speed of each body at any time is known from the formula dt/dr=2n3√(rr−aa). It should be noted, however, that the distances between the bodies perpetually preserve their ratio.
Casus II. Case II.
  11. Sit nunc β=−nn erit p=nn(aa−rr) et Q=αnn(aa−rr) et per r loca corporum ut ante ita determinantur, ut sit : ?  11. Now let β=−nn, so p=nn(aa−rr) and Q=αnn(aa−rr) and with r being the locations of the bodies as determined before, then :
x = OA = −nn(aa−rr)/(A+B+C)   (B+(1+α)C)   
y = OB =    nn(aa−rr)/(A+B+C)   (A−Cα)    
z = OC =    nn(aa−rr)/(A+B+C)   (A(α+1)+Bα).
Pro tempore autem t obtinetur, dt = 2n3dr√(aa−rr)
seu   t = n3r√(aa−rr) + n3aa∫dr/√(aa−rr)
hinc   t = n3r√(aa−rr) + n3aa Ang.sin.r/a.
? For the time t, on the other hand, is obtained dt = 2n3dr√(aa−rr)
or   t = n3r√(aa−rr) + n3aa∫dr/√(aa−rr)
hence   t = n3r√(aa−rr) + n3aa arcsin(r/a).
Quodsi ergo ponatur Ang.sin.r/a=Φ, ut sit r=asin.Φ erit t=n3(Φ+sin.Φcos.Φ), et distantiae inter se proportionales ad quodvis tempus sunt ut cos.Φ2. Unde si initio quo t=0 fuerit Φ=0, sicque r=0, et dt/dr=2n3a, erant tum distantiae : ? But if we let arcsin(r/a)=Φ, so that r=a sin(Φ), we have t=n3(Φ+sin(Φ)cos(Φ)), and the distances in proportion between themselves vary at any time as cos2(Φ). Hence, if at the beginning when t=0 we had Φ=0, and so r=0, and dt/dr=2n3a, then the distances would be :
x = −nnaa / (A+B+C)   (B+(1+α)C)   
y =    nnaa / (A+B+C)   (A−Cα)   
z =    nnaa / (A+B+C)   (A(α+1)+Bα)
ibique corpora in quiete. Sumto autem Φ=90°, seu elapso tempore t=n3aa. 90°, corpora in centro gravitatis conveniunt celeritate infinita. ? and there the bodies are at rest. But take Φ=90°, or elapsed time t=n3aa. 90°, the bodies will meet at the centre of gravity with infinite speed.


Paper E.327 deals with three masses, affected only by their mutual gravity, moving along a straight line of fixed direction. There is no rotation. The bodies start from and/or finish at their barycentre.

Therefore, it cannot reasonably be considered as providing a discovery of L1, L2, and L3, since the Lagrange Points are generally considered to be a property of a system where the masses orbit around the barycentre.

But it does cover the special and not normally recognised case of L1, L2, and L3 when there is no angular momentum.

E.304 - Considerationes de motu corporum coelestium (Considerations on the motion of celestial bodies), predicts L1 and L2 only, by re-positioning our Moon. Other Euler papers appear to add nothing of significance concerning the Lagrange Points.

Open Questions

In Case I (section 10, β>0), it appears that the bodies are initially at the centre, and are blown apart (an early sort of Big Bang); Case II (section 11, β<0) appears to follow a Bang with a Big Crunch). Compare E.400 § 10 ??

Cases I & II imply a certain relationship between the speeds of the bodies. Where is this called for and/or supplied? At the beginning of § 7?


My thanks for assistance in the translation to Eduardus and other participants in news:alt.language.latin.


Compare this section with
The Lagrange Points Reconsidered
or Gravity 6 : Constant-Pattern Motion.

            A           B                 C

Evidently Euler wanted to solve the linear three-body problem and could only progress after § 6 by dropping generality and thereby finding constant-pattern solutions. But what if one seeks, ab initio, only constant-pattern solutions?

First, note that in a time-reversible system, "initial conditions" can be at any part of the motion. If the initial pattern is to be preserved, the initial velocities of the bodies must be proportional to their signed distances, at that instant, from some point on the line; and, unless the barycentre is to move, that point must be the barycentre. To preserve the pattern initially, the initial accelerations must likewise be in proportion — and, the pattern having been preserved, the preservation will continue, and will do so perpetually, unless something else happens.

If initially B is very close to C, its initial acceleration will be disproportionately to the right; likewise, if very close to A, disproportionately to the left. The variation between these conditions is manifestly monotonic. Therefore, there must be an intermediate position in which the acceleration of B is proportionate. So a constant-pattern solution exists and is calculable.

Since the pattern does not change, the accelerations of each body depend only on the scale, and must vary as the inverse square of the distance of the body from the barycentre. The motion of each body must take the same form as the one-dimensional motion of a particle attracted to a fixed mass. That will correspond to the limiting case of an orbiting particle with ever-increasing eccentricity.

See my Gravity 6 : Constant-Pattern Motion, which argues similarly for motion in a plane.

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