A COMPARISON OF TWO SYSTEMS DESCRIBING

STUDIA UNIV. “BABES ¸ –BOLYAI”, MATHEMATICA, Volume XLVIII, Number 3, September 2003

A COMPARISON OF TWO SYSTEMS DESCRIBING ELECTROMAGNETIC TWO-BODY PROBLEM V. G. ANGELOV, AND L. GEORGIEV

Dedicated to Professor Gheorghe Micula at his 60th anniversary Abstract. Two systems of equations of motion describing two charged particles in the framework of classical electrodynamics are compared.

1. Introduction The main purpose of the present paper is to compare two systems equations of motion describing the two-body problem of classical electrodynamics [1], [2] (cf. also [3], [4]). One can see at the end of the paper that even a small difference in the right-hand sides of the equations generates various solutions and completely different physical conclusions for the two-body system. In 1940 [5] J. L. Synge proposes equations of motion describing the behaviour of two charged particles. His derivations are based on the relativistic form of the pondermotive Lorentz force given by W. Pauli [6] by means of Lienard-Wiechert retarded potentials. J. L. Synge formulates the problem in the Minkowski space, that is, in the framework of the special theory of relativity. Consequently the finite velocity of the propagation of interaction generates delays which are, although implicitly, in the arguments of the unknown velocities of the moving particles in the equations of motion [5]. This does not come as a surprise because the theory of differential equations with retarded argument is formulated about twenty years later (cf. A. D. Myshkis [7]). In order to overcome this difficulty J. L. Synge [5] builds a sequence of successive approximations such that on every step one has to solve a system of ordinary differential equations. Although there is no a convergence theorem for the successive approximations he proposes some idea for solving of the system. In a recent paper [8], however, we have shown that not only a convergence theorem cannot be proved, but even a sequence of successive approximations could not be constructed in such a way. On the base of the same method [5] J. L. Synge calculates the energy on every step (of successive approximations) and makes a conclusion that the two-body system is not stable (cf. p.139, [5]). Received by the editors: 10.12.2002.

11

V. G. ANGELOV, AND L. GEORGIEV

Later in 1963 R. D. Driver [9] recognizes the system obtained in [5] as a functional differential system with delays depending on the unknown trajectories and obtains a correct formulation of the Synge problem even in 1-dimensional case. Since we have taken the same point of view for the type of the delays we are able to compare the systems for 3-dimensional case considered in [1] and [2]-[4]. Prior to begin the main exposition we want to discuss one more difficulty concerning Synge equations. They are 8 in number, while the unknown functions are 6 in number. The problem mentioned is not considered in [1] and related known papers. In [4] (cf. also [2]) we show that the system of equations of motion is equivalent to the one consisting of 6 equations. More precisely, the 4-th and 8-th equation are a consequence of the rest ones. In the present paper we recall some formulation from [3] and [2] in order to obtain 3-dimensional case of J. L. Synge equations. Here we succeed to simplify the right-hand sides of the equations in more extent than [2]-[4]. We present the equations of motion from [1] using our denotations which makes the comparison of both systems easier. Thus we see that equations from [1] can be turned into our ones (we pretend 1 they are Synge’s equations) if the constant k (from [1]) is chosen to be k = 3 ( c the c speed of light). Then it is not surprise that the right-hand sides of equations from [1] ( c3 times larger than ours) generates unstable solutions. At the same time it is shown in [2] that Kepler problem for two charged particles has a circle solution. 2. J. L. Synge’s equations of motion (p)

(p)

(p)

(p)

As in [2]-[5] we denote by x(p) = (x1 (t), x2 (t), x3 (t), x4 (t) = ict) (p = 1, 2) (i = −1) the space-time coordinates of the moving particles, by mp - their proper masses, by ep - their charges, c - the speed of light. The coordinates of the (p) (p) (p) velocity vectors are u(p) = (u1 (t), u2 (t), u3 (t))(p = 1, 2). The coordinates of the unit tangent vectors to the world-lines are (cf. [2], [3]): 2

(p)

(p)

λ(p) α =

uα (t) γp uα (t) ic (p) = (α = 1, 2, 3), λ4 = iγp = c ∆p ∆p

(1)

3 3 X 1 1 X (p) 2 − 1 2 (p) 2 [u (t)] ) , ∆p = (c − [uα (t)]2 ) 2 . It follows γp = c/∆p . where γp = (1 − 2 c α=1 α α=1 By < ., . >4 we denote the scalar product in the Minkowski space, while by < ., . > - the scalar product in 3-dimensional Euclidean subspace. The equations of motion modeling the interaction of two moving charged particles are the following (cf. [5]): (p)

mp

dλr ep (p) (p) = 2 Frn λn (r = 1, 2, 3, 4) dsp c

(2)

c dt = ∆p dt(p = 1, 2). Recall γp (p) that in (2) there is a summation in n (n = 1, 2, 3, 4). The elements Frn of the

where the elements of proper time are dsp =

12

A COMPARISON OF TWO SYSTEMS DESCRIBING ELECTROMAGNETIC TWO-BODY PROBLEM

electromagnetic tensors are derived by the retarded Lienard-Wiechert potentials (p) (p) (p) ep λr ∂Ar ∂An (p) A(p) = − (p) (pq) (r = 1, 2, 3, 4), that is, Frn − . By ξ (pq) we = r (p) (p) hλ , ξ i4 ∂xr ∂xn denote the isotropic vectors (cf. [2], [5]) (p)

(q)

(p)

(q)

(p)

(q)

ξ (pq) = (x1 (t)−x1 (t−τpq (t)), x2 (t)−x2 (t−τpq (t)), x3 (t)−x3 (t−τpq (t)), icτpq (t)), where hξ (p,q) , ξ (p,q) i4 = 0 or   21 3 X 1 (p) (q) [xβ (t) − xβ (t − τpq (t))]2  , ((pq) = (12), (21)). τpq (t) =  c

(3pq )

β=1 (p) Frn

Calculating as in [5] we write equations from (2)in the form: ( (p) (pq) (q) (q) dλα Qp ξα hλ(p) , λ(q) i4 − λα hλ(p) , ξ (pq) i4 (pq) dλ 1+ ξ = 2 , + dsp c dsq 4 hλ(q) , ξ (pq) i34 " #) (q) 1 dλα dλ(q) ξ (pq) − λ(p) , (α = 1, 2, 3) + (q) (pq) 2 hλ(p) , ξ (pq) i4 dsq dsq 4 α hλ , ξ i4 ( (pq) (q) (p) (q) dλ4 Qp ξ4 hλ(p) , λ(q) i4 − λ4 hλ(p) , ξ (pq) i4 (pq) dλ = 2 + 1 + ξ , dsp c dsq 4 hλ(q) , ξ (pq) i34 " #) (q) (q) 1 (pq) (p) (pq) dλ4 (p) dλ + (q) (pq) 2 hλ , ξ i4 − λ , ξ dsq dsq 4 4 hλ , ξ i4

(4.α)

(4.4)

where Qp = e1 e2 /mp (p = 1, 2). Further on, we have u(q) ≡ u(q) (tpq ) (tpq = t − τpq ), (q)

(q)

(q)

(q)

(q)

(q)

λ(q) = (γpq u1 /c, γpq u2 /c, γpq u3 /c, iγpq ) = (u1 /∆pq , u2 /∆pq , u3 /∆pq , ic/∆pq )

where γpq =

3 1 X (q) [u (t − τpq (t)]2 1− 2 c α=1 α

!− 21 , ∆pq =

2

c −

3 X

! 21 [u(q) α (t

2

− τpq (t))]

α=1

u(p)

(p) γ (p) (p) d( ∆αp ) d( cp uα ) 1 (p) uα dλα = = = u ˙ + hu(p) , u˙ (p) i(α = 1, 2, 3) and c 2 α 4 dsp dt ∆ dt ∆ ∆ p p p γp (p) icd( ∆1p ) dλ4 d(iγp ) ic = c = = 4 hu(p) , u˙ (p) i, where the dot means a differentiation in dsp dt ∆ dt ∆ p p γp t. dλ(q) dt In order to calculate we need the derivative ≡ Dpq which should dsq dtpq be calculated from the relation ! 21 3 1 X (p) 2 t − tpq = [x (t) − x(q) (tpq < t by assumption). α (tpq )] c α=1 α

13


3 X

So we have

dt −1= dtpq

(p) (p) [xα (t) − x(q) α (tpq )][uα (t)

α=1

c

3 X

dt − u(q) α (tpq )] dtpq . ! 21

(q) 2 [x(p) α (t) − xα (tpq )]

α=1

Since (3pq ) has a unique solution (cf. [3]) we obtain 2 c τpq (Dpq − 1) = hξ (pq) , u(p) iDpq − hξ (pq) , u(q) i and we can solve the above equation c2 τpq − hξ (pq) , u(q) i d d with respect to Dpq : Dpq = 2 . We have also = . dsp ∆p dt c τpq − hξ (pq) , u(p) i d 1 d 1 d dt Dpq d Then = = = ; dsq ∆pq dtpq ∆pq dt dtpq ∆pq dt # " (q) (q) (q) (q) uα Dpq dλα Dpq d uα /∆pq dλα (q) 1 (q) (q) = = = Dpq u˙ α 2 + 4 hu , u˙ i dsq ∆pq dt ∆pq dt ∆pq ∆pq (α = 1, 2, 3); (q) dλ4 hu(p) , u(q) i − c2 icDpq (q) (q) hu , u˙ i; hλ(p) , λ(q) i4 = = ; 4 dsq ∆pq ∆p ∆pq hu(p) , ξ (pq) i − c2 τpq hu(q) , ξ (pq) i − c2 τpq hλ(p) , ξ (pq) i4 = ; hλ(q) , ξ (pq) i4 = ; ∆p ∆pq (q) hξ (pq) , u(q) i − c2 τpq (q) (q) 1 (pq) dλ (pq) (q) hξ , i4 = Dpq hξ , u˙ i + hu , u˙ i ; dsq ∆2pq ∆4pq dλ(q) Dpq hu(p) , u(q) i − c2 (q) (q) (p) (q) hλ(p) , hu , u ˙ i . i4 = hu , u ˙ i + dsq ∆p ∆2pq ∆2pq We note that in the last expressions ξ (pq) is 4-dimensional vector in the lefthand sides, while in the right-hand sides ξ (pq) is 3-dimensional part of the first three coordinates. Replacing the above expressions in (4.α) and (4.4) and performing some obvious transformations we obtain for (pq) = (12), (21), α = 1, 2, 3 : ( (q) (pq) (p) Qp [c2 − hu(p) , u(q) i]ξα − [c2 τpq − hu(p) , ξ (pq) i]uα 1 (p) uα (p) (p) . u˙ α + 3 hu , u˙ i = 2 ∆p ∆p c [c2 τpq − hu(q) , ξ (pq) i]3 ∆4pq + Dpq ∆2pq hξ (pq) , u˙ (q) i + (hξ (pq) , u(q) i − c2 τpq )hu(q) , u˙ (q) i . + (5pα ) ∆2pq (q)

+Dpq −Dpq

1 (p) (p) Qp hu , u˙ i = 2 ∆3p c 14

(q)

[hu(p) , ξ (pq) i − c2 τpq ][u˙ α + uα hu(q) , u˙ (q) i/∆2pq ] − [c2 τpq − hu(q) , ξ (pq) i]2 ) (pq) [hu(p) , u˙ (q) i + (hu(p) , u(q) i − c2 )hu(q) , u˙ (q) i/∆2pq ]ξα (pq)

[c2 τpq − hu(q) , ξα

i]2

hu(p) , ξ (pq) i − τpq hu(p) , u(q) i . [c2 τpq − hu(q) , ξ (pq) i]3

,


h

i . ∆2pq + Dpq (hξ (pq) , u˙ (q) i + (hξ (pq) , u(q) i − c2 τpq )hu(q) , u˙ (q) i/∆2pq ) + + Dpq

(5p4 ) )

hu(p) , ξ (pq) ihu(q) , u˙ (q) i/∆2pq − τpq hu(p) , u˙ (q) i − τpq hu(p) , u(q) ihu(q) , u˙ (q) i/∆2pq [c2 τpq − hu(q) , ξ (pq) i]2

.

One can prove (as in [4]) that (5p4 ) is a consequence of (5pα ). Indeed, multi(p) plying (5pα ) by uα , summing up in α and dividing into c2 we obtain (5p4 ). Therefore we can consider a system consisting of the 1st , 2nd , 3rd , 5th , 6th and 7th equations. The last equations form a nonlinear functional differential system of neutral type with respect to the unknown velocities (cf. [10]-[13]). The delays τpq depend on the unknown trajectories by the relations (3pq ). Now we are able to present (5pα ) in a suitable form in order to make further simplifications (Recall that we shall not consider (5p4 ) because it is a consequence of (5pα )): 2 Qp ∆p c − hu(p) , u(q) i (pq) hu(p) , u˙ (p) i (p) u = ξ − u˙ pα + α 2 2 2 (q) (pq) 2 2 ∆p c (c τpq − hu , ξ i) c τpq − hu(q) , ξ (pq) i α −

c2 τpq − hu(p) , ξ (pq) i (q) uα c2 τpq − hu(q) , ξ (pq) i

∆2pq + Dpq hξ (pq) , u˙ (q) i + Dpq

hu(q) , ξ (pq) i − c2 τpq (q) (q) hu , u ˙ i + ∆2pq

hu(q) , u˙ (q) i (q) uα − (6pα ) ∆2pq c2 − hu(p) , u(q) i (q) (q) (pq) (p) (q) (pq) −Dpq hu , u˙ iξα + Dpq hu , u˙ iξα . ∆2pq Let us recall that if (pq) = (12) then u(1) = u(1) (t) and u(2) = u(2) (t − τ12 ), while when (pq) = (21), then u(2) = u(2) (t) and u(1) = u(1) (t − τ21 ). Further on from (6pα ) we obtain hu(p) , u˙ (p) i (p) u˙ pα + uα = ∆2p ( c2 − hu(p) , u(q) i ∆2pq + Dpq hξ (pq) , u˙ (q) i (pq) Qp ∆p = 2 2 ξα − c (c τpq − hu(q) , ξ (pq) i)2 c2 τpq − hu(q) , ξ (pq) i (p) (pq) +Dpq (hu(p) , ξ (pq) i − c2 τpq )u˙ (q) ,ξ i − c2 τpq ) α + Dpq (hu

−Dpq −

(c2 − hu(p) , u(q) i)hu(q) , u˙ (q) i (pq) ξα − ∆2pq

c2 τpq − hu(p) , ξ (pq) i hu(q) , ξ (pq) i − c2 τpq (q) (q) 2 (pq) (q) (q) ∆ + D hξ , u ˙ i + D hu , u ˙ i uα + pq pq pq ∆2pq c2 τpq − hu(q) , ξ (pq) i

(hu(p) , ξ (pq) i − c2 τpq )hu(q) , u˙ (q) i (q) uα − ∆2pq (c2 − hu(p) , u(q) i)hu(q) , u˙ (q) i (pq) (p) (q) (pq) −Dpq hu , u˙ iξα + Dpq ξα . ∆2pq

+Dpq (hu(p) , ξ (pq) i − c2 τpq )u˙ (q) α + Dpq

15


Obviously the second summand and the last one cancel each other and after a re-arrangement of the rest summands we have: hu(p) , u˙ (p) i (p) uα = u˙ pα + ∆2p =

c2 (c2 τpq

   c2 − hu(p) , u(q) i ∆2pq + Dpq hξ (pq) , u˙ (q) Qp ∆p (p) (q)  − Dpq hu , u˙ i ξα(pq) + − hu(q) , ξ (pq) i)2  c2 τpq − hu(q) , ξ (pq) i

2 c τpq − hu(p) , ξ (pq) i 2 + − 2 ∆pq + Dpq hξ (pq) , u˙ (q) i − (q) (pq) c τpq − hu , ξ i hu(q) , ξ (pq) i − c2 τpq (q) (q) c2 τpq − hu(p) , ξ (pq) i Dpq hu , u˙ i+ 2 (q) (pq) ∆2pq c τpq − hu , ξ i (hu(p) , ξ (pq) i − c2 τpq )hu(q) , u˙ (q) i (q) (p) (pq) 2 (q) u + D (hu , ξ i − c τ ) u ˙ . + Dpq pq pq α α ∆2pq

−

(q)

It is easy to see that the second and third summands before uα cancel each other so that we obtain the following simplified form of (6pα ): Qp ∆p hu(p) , u˙ (p) i (p) uα = 2 2 u˙ pα + · ∆2p c (c τpq − hu(q) , ξ (pq) i)2 (" # c2 − hu(p) , u(q) i ∆2pq + Dpq hξ (pq) , u˙ (q) i (p) (q) · − Dpq hu , u˙ i ξα(pq) − (7pα ) c2 τpq − hu(q) , ξ (pq) i ) c2 τpq − hu(p) , ξ (pq) i ∆2pq + Dpq hξ (pq) , u˙ (q) i (q) (p) (pq) 2 (q) uα + Dpq hu , ξ i − c τpq u˙ α . − c2 τpq − hu(q) , ξ (pq) i In the same way we can obtain more suitable (in view of next section) form of the equations (5p4 ) (although we proved that (5p4 ) is a consequence of (5pα )): ( (p) (pq) 1 (p) (p) Qp ∆p hu , ξ i − τpq hu(p) , u(q) i 2 (pq) (q) hu , u ˙ i = ∆ + D hξ , u ˙ i − pq pq 3 ∆2p c2 c2 τpq − hu(q) , ξ (pq) i −

Dpq hu(q) , u˙ (q) i hu(p) , ξ (pq) i − τpq hu(p) , u(q) i · 2 + ∆2pq c2 τpq − hu(q) , ξ (pq) i

Dpq hu(q) , u˙ (q) i hu(p) , ξ (pq) i − τpq hu(p) , u(q) i Dpq τpq hu(p) , u˙ (q) i + · − 2 2 2 ∆pq c2 τpq − hu(q) , ξ (pq) i c2 τpq − hu(q) , ξ (pq) i

) , i.e.

1 (p) (p) Qp ∆p hu , u˙ i = (7p4 ) 2 · 2 2 ∆2p c c τpq − hu(q) , ξ (pq) i " # hu(p) , ξ (pq) i − τpq hu(p) , u(q) i ∆2pq + Dpq hξ (pq) , u˙ (q) i · − Dpq τpq hu(p) , u˙ (q) i c2 τpq − hu(q) , ξ (pq) i (p = 1, 2).

16


3. Equations of motion from [1] Now we consider the equations of motion, considered in [1], p.79-80, using his denotations: xk (s) - the position of the particle k in R3 at the instant s (k = 1, 2); ek - the charge of the particle k (k = 1, 2); mk - the mass of the particle k (k = 1, 2); ri = ri (t) - the delays, which satisfy the equations: (∗) cri (t) = |xi (t) − xj (t − ri )| (j 6= i); vi - the normalized velocities, where x0i = cvi for i = 1, 2; xi (t) − xj (t − ri ) , γi = 1 − vj (t − ri ).ui (j 6= i), where ”.” indicates the ui = cri 3 dot of scalar product in R . REMARK 1: xk (s) is the restriction of the space-time vector (k) (k) (k) (k) (x1 (s), x2 (s), x3 (s), x4 (s) = ics), while ui corresponds to the restriction of the ξ (pq) in 3-dimensional Euclidean subspace of the Minkowski space. isotropic vector cτpq The delays ri = ri (t) and the equations (∗) correspond to τpq = τpq (t) and to the equations (3pq ) respectively. Finally, we use the notation u(k) = u(k) (s) = (k) dxα (s) (k) (k) (k) (α = (u1 (s), u2 (s), u3 (s)) for the velocity vectors, where u(k) α (s) = ds (k) u (s) 1, 2, 3), k = 1, 2 so the normalized vector vk (s) would be equaled to , if we c should use the notation for it (k = 1, 2). The equations of motion, given in [1], are the following: vi0 =

ei (1 − vi2 )1/2 [Ej + (vi .Ej )(ui − vi ) − (vi .ui )Ej ], mi c

(∗∗)

where vi2 = |vi |2 (= vi .vi ) and kcej kcej Ej = 2 3 [ui − vj (t − ri )][1 − vj2 (t − ri )] + ui × ([ui − vj (t − ri )] × vj0 (t − ri )), ri γi ri γi3 where ” × ” stands for the cross product in R3 , ”k > 0 is a constant depending on the units used”, and the denotation vj0 (t − ri ) most probably means a derivative with respect to the argument of vj (t − ri ), (in [1] there is no explanation). Rewrite the right-hand side of (∗∗) in the form: ei (1 − vi2 )1/2 {[1 − (vi .ui )]Ej + (vi .Ej )(ui − vi )} mi c and calculate the vector cross product from Ej : kcej 0 kcej kcej (ui .vj0 (t−ri )[ui −vj (t−ri )]− v (t−ri ). Ej = 2 3 [1−vj2 (t−ri )][ui −vj (t−ri )]+ ri γi ri γi3 ri γi2 j (since ui .[ui − vj (t − ri )] = |ui |2 − vj (t − ri ).ui = 1 − vj (t − ri ).ui = γi !). Consequently the equations (∗∗) are equivalent to the following ones: kcej ei (1 − vi2 )1/2 [1 − (ui .vi )] 2 3 [1 − vj2 (t − ri )][ui − vj (t − ri )]+ vi0 = mi c ri γi 17


kcej 0 kcej 0 (ui .vj (t − ri ))[ui − vj (t − ri )] − v (t − ri ) + + ri γi3 ri γi2 j kcej + 2 3 [1 − vj2 (t − ri )][(vi .ui ) − (vi .vj (t − ri ))]+ ri γi +

(8)

kcej (ui .vj0 (t − ri ))[(vi .ui ) − (vi .vj (t − ri ))]− ri γi3 kcej 0 − (vi .vj (t − ri )) (ui − vi ) . ri γi2

Then we can arrange the symbols, including the vectors ui , vi , vj (t − ri ) and vj0 (t − ri ) respectively and we obtain the equivalent equations: ( " [1 − (ui .vi )][1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))] kei ej (1 − vi2 )1/2 0 vi = u + i mi ri γi2 ri γi # [1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))][(vi .ui ) − (vi .vj (t − ri ))] 0 + − (vi .vj (t − ri )) − ri γi # " [1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))][(vi .ui ) − (vi .vj (t − ri ))] 0 − (vi .vj (t − ri )) − −vi ri γi ) [1 − (ui .vi )][1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))] 0 − vj (t − ri ) − vj (t − ri )[1 − (ui .vi )] ri γi and finally one has (

"

[1 − (vi .vj (t − ri ))][1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))] − ri γi − (vi .vj0 (t − ri )) − (9) # " [1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))][(vi .ui ) − (vi .vj (t − ri ))] − (vi .vj0 (t − ri )) − −vi ri γi ) [1 − (ui .vi )][1 − vj2 (t − ri ) + ri (ui .vj0 (t − ri ))] 0 − vj (t − ri ) − vj (t − ri )[1 − (ui .vi )] . ri γi To compare both systems we present the equations (9), using the denotations from our previous section II. We have for i ≡ p, j ≡ q, ri ≡ τpq , and in view of Remark 1: (p) (p) 1/2 u u ∆p u(p) ; (1 − vi2 )1/2 = 1 − , = ; vi = vi (t) ≡ c c c c vi0

kei ej (1 − vi2 )1/2 = mi ri γi2

vi0 = 18

ui

d(u(p) /c) = u˙ (p) /c (u(p) = u(p) (t)); dt


ui ≡

x(p) (t) − x(q) (t − τpq ) 1 (pq) (pq) (pq) ξ , ξ2 , ξ3 ; = cτpq cτpq 1 1 du(q) dt Dpq u˙ (q) d(u(q) /c) = · · = dtpq c dt dtpq c

vj0 (t − ri ) =

vj (t − ri ) ≡ u(q) /c;

(u(q) = u(q) (t − τpq ) = u(q) (tpq )); (q) (pq) u ξ c2 τpq − hu(q) , ξ (pq) i γi ≡ 1 − , = ; c cτpq c2 τpq (p) (q) u u c2 − hu(p) , u(q) i 1 − (vi .vj (t − ri )) = 1 − ; , = c c c2 D (pq) E D (q) (q) E Dpq u˙ (q) ξ u u 2 0 , + τ , 1 − pq 1 − vj (t − ri ) + ri (ui .vj (t − ri )) c c cτpq c E D = ≡ u(q) ξ (pq) ri γi τ 1− , pq

= 1 − (ui .vi ) ≡ 1 −

ξ (pq) u(p) , cτpq c

(vi .ui ) − (vi .vj (t − ri )) ≡

=

c

cτpq

∆2pq + Dpq hξ (pq) , u˙ (q) i ; c2 τpq − hu(q) , ξ (pq) i

c2 τpq − hu(p) , ξ (pq) i ; c2 τpq

hu(p) , ξ (pq) i − τpq hu(p) , u(q) i c2 τpq

and replacing in (9) we obtain the following 6 scalar equations: kep eq ∆p (c2 τpq )2 1 p u˙ α = · c cmp τpq (c2 τpq − hu(q) , ξ (pq) i)2 " # ( (pq) c2 − hu(p) , u(q) i ∆2pq + Dpq hξ (pq) , u˙ (q) i Dpq (p) (q) ξα · 2 · − 2 hu , u˙ i − cτpq c2 c c τpq − hu(q) , ξ (pq) i " # (p) ∆2pq + Dpq hξ (pq) , u˙ (q) i hu(p) , ξ (pq) i − τpq hu(p) , u(q) i Dpq (p) (q) uα − · − 2 hu , u˙ i − c c2 τpq c c2 τpq − hu(q) , ξ (pq) i (q) uα c2 τpq − hu(p) , ξ (pq) i ∆2pq + Dpq hξ (pq) , u˙ (q) i · · 2 + c c2 τpq c τpq − hu(q) , ξ (pq) i ) (q) Dpq u˙ α hu(p) , ξ (pq) i − c2 τpq + · (α = 1, 2, 3; (pq) = (12), (21)), c c2 τpq The above system is obviously equivalent to the following one (with Qp = ep eq ): mp

−

u˙ pα +

h hu(p) , ξ (pq) i − τpq hu(p) , u(q) i ∆2 + Dpq hξ (pq) , u˙ (q) i pq − 2 · 2τ (q) , ξ (pq) i 2 (q) (pq) c − hu pq c τpq − hu , ξ i kcQp ∆p

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i (p) −Dpq τpq hu(p) , u˙ (q) i uα = =

(c2 τpq

  n c2 − hu(p) , u(q) i ∆2pq + Dpq hξ (pq) , u˙ (q) i kcQp ∆p  − Dpq hu(p) , u˙ (q) i ξα(pq) − 2· c2 τpq − hu(q) , ξ (pq) i − hu(q) , ξ (pq) i)

c2 τpq − hu(p) , ξ (pq) i ∆2pq + Dpq hξ (pq) , u˙ (q) i (q) uα + (10pα ) c2 τpq − hu(q) , ξ (pq) i o +Dpq hu(p) , ξ (pq) i − c2 τpq u˙ (q) (α = 1, 2, 3; (pq) = (12), (21)). α Conclusion remarks Our goal is to point out the difference between the system of equations of motion (9) (or equivalently (10pα )) from [1] and Synge’s equations (7pα ), (7p4 ). 1) equations (9) are obtained under assumption that (7p4 ) should be identities which is not discussed in [1]. On the other hand in [4] we have already proved that (7p4 ) is a consequence of (7pα ), α = 1, 2, 3. It is not obvious that (7p4 ) is an identity. 2) the right-hand sides of (10pα ) and Synge’s equations (7pα ) differ each 1 other by the multiplier c3 which is a consequence from kc = 2 . But this means that c the right-hand sides of (10pα ) are c3 -times larger than the right-hand sides of Synge equations (7pα ), that is, they have another dimension. Therefore it is not surprised that they possess only unstable solutions, while (7pα ) have a circle solution [2]. References [1] Driver, R. D., A neutral system with state-dependent delay, J. Diff. Equations, v.54, N1 (1984), 73-86. [2] Angelov, V. G., Plane orbits for Synge’s electromagnetic two-body problem (I). Seminar on Fixed Point Theory, Cluj-Napoca, v.3 (2002), 3-12. [3] Angelov, V. G., On the Synge equations in three-dimensional two-body problem of classical electrodynamics, J. Math. Anal. Appl. v.151, N2 (1990), 488-511. [4] Angelov, V. G., Escape trajectories of J. L. Synge equations, J. Non Anal. RWA, ser. B, v.1 (2000), 189-204. [5] Synge, J. L., On the electromagnetic two-body problem, Proc. Roy. Soc. (London), ser. A, 177 (1940), 118-139. [6] Pauli, W., Relativitatstheorie, Encycl. der Math. Wissenschaften. B.5, H.S. Art. 19, 1921. [7] Myshkis, A. D., On some probrlem of the theory of differential equations with deviating argument, Uspekhi Mat. Nauk, v.4, N5 (1949), 99-141 (in Russian). [8] Angelov, V. G., On the method of sucsesssive approximations for J. L. Synge electromagnetic two-body problem (to appear). [9] Driver, R. D., A two-body problem of classical electrodynamics: the one-dimensional case, Ann. Physics, v.21, N1 (1963), 122-142. [10] El’sgoltz, L. E., Norkin, S. B., Introduction in the theory of differential equations with deviating argument, Moscow, Nauka, 1971 (in Russian). [11] Myshkis, A. D., Linear differential equations with retarded argument, Moscow, Nauka, 1972 (in Russian). [12] Kamenskii, G. A., Skubachevskii, A. L., Linear boundary value problems for differentialdifference equations, Moscow, 1992 (in Russian). 20


[13] Hale, J.K., Verduyn-Lunel, S.M., Functional Differential Equations, Springer Verlag, 1993. University of Mining and Geology ”St. I. Rilski”, Department of Mathematics, 1700 Sofia, Bulgaria E-mail address: [email protected]

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