Dans cet article, nous allons démontrer la chose suivante :

 

LE PROBLEME DIT DE LA SIGNATURE DE L’ESPACE-TEMPS EST UN FAUX PROBLEME, QUI DISSIMULE EN REALITE LA NATURE FONDAMENTALEMENT DOUBLE, AUSSI BIEN DE L’ESPACE QUE DU TEMPS ET DONC, DE L’ESPACE-TEMPS.

 

Et nous allons résoudre complètement le problème, avant de l’appliquer à notre propos.

Avant cela, un peu d’algèbre élémentaire.

Soient x₁ et x₂ des réels et x un nombre tel que :

 

1. x² = x₁x₂

 

Le couple (x₁,x₂) est dans R². Si x₁x₂ >= 0, alors x₁x₂ = |x₁x₂| et le nombre x est réel. Dans ce cas, (1) fournit deux possibilités :

 

2. x = +|x₁x₂|^1/2  et  x = -|x₁x₂|^1/2

 

Si x₁x₂ =< 0, alors x₁x₂ = -|x₁x₂| et le nombre x est imaginaire pur. Là encore, deux possibilités :

 

3. x = +i|x₁x₂|^1/2  et  x = -i|x₁x₂|^1/2

 

Ce nombre x en question n’est autre que la moyenne géométrique des réels x₁ et x₂. Selon le signe du produit x₁x₂, x sera dans R ou dans C, algébriquement isomorphe à R². Rappelons que c’est Cardan qui introduisit les nombres complexes pour résoudre l’équation du 3^ème degré x³ + 3px + 2q = 0. Ceci ouvre sur deux choix possibles : soit raisonner avec des nombres complexes z = x+iy et identifier les réels à des complexes de partie imaginaire nulle, soit continuer à ne raisonner qu’avec des réels et considérer alors des couples (x,y). Le passage de R² à C s’accompagne quand même d’une loi produit [z = x+iy , z’ = x’+iy’ => zz’ = xx’ – yy’ + i(xy’+x’y)], alors que le produit scalaire dans R² donnerait (x,y) x (x’,y’) = (xx’ + yy’) et le produit vectoriel, xy’-yx’. Aucun des deux ne coïncide avec le produit complexe.

Inversons maintenant la situation et soit x un nombre réel. Alors, x² est toujours >= 0. En conséquence, si nous écrivons de nouveau x² = x₁x₂ avec x₁ et x₂ réels, nous aurons x₁x₂ >= 0.

 

DEFINITION 1 :

LES REELS x, MOYENNES GEOMETRIQUES DES REELS x₁ ET x₂ TELS QUE x₁x₂ >= 0 SERONT DITS « DU GENRE TEMPS ».

 

Soit toujours x, x₁ et x₂ des réels et posons cette fois x² = -x₁x₂. Alors, x₁x₂ =< 0.

 

DEFINITION 2 :

LES REELS x, MOYENNES GEOMETRIQUES DES REELS x₁ ET x₂ TELS QUE -x₁x₂ >= 0 SERONT DITS « DU GENRE ESPACE ».

 

Ainsi, pour chaque réel x, je peux toujours remplacer x² par un produit x₁x₂ de réels. Si x₁x₂ >= 0, j’aurai x² = x₁x₂ et si x₁x₂ =< 0, j’aurai x² = -x₁x₂. Ceci me fournit deux hyperboles :

 

4. x₂ = x²/x₁  ,  x₂ = -x²/x₁

 

C'est-à-dire qu’en place d’un seul réel x, j’élargis à une double infinité de réels x₁ et x₂ tels que : x₁x₂ = x² si x₁x₂ >= 0 (1^ère infinité, solution 4a) ou bien x₁x₂ = -x² si x₁x₂ =< 0 (2^ème infinité, solution 4b).

 

MOYENNANT LE REMPLACEMENT DES CARRES DE REELS PAR DES PRODUITS DE DEUX REELS, JE N’AI PLUS BESOIN DES NOMBRES COMPLEXES.

 

Pour un x réel donné, je raisonne dans R² et j’ai une double infinité de valeurs possibles, en place d’un seul complexe z, avec une loi de multiplication spécifique.

Je n’ai plus non plus à me poser la lancinante question du signe dans les formes quadratiques. Considérons, en effet, la forme Q(x,y) = x² + y². Cette forme est définie positive, elle est dite euclidienne. Son tenseur métrique associé est d_ij tel que d_ii = 1 et d_ij = 0 pour i <> j, (i,j = 1,2). Je peux réécrire cette forme comme une somme de produits :

 

5. Q(x₁,x₂,y₁,y₂) = x₁x₂ + y₁y₂

 

Dans ce cas, j’aurai x₁x₂ = x² >= 0 et y₁y₂ = y² >= 0, soit x et y tous deux du genre temps. Mais toute somme de réels pouvant encore s’écrire comme un produit de réels, même si le résultat est un premier (p premier => p = 1xp = px1), j’ai aussi :

 

6. Q(x₁,x₂,y₁,y₂) = x₁x₂ + y₁y₂ = qq’

 

Et comme x₁x₂ et y₁y₂ sont tous deux >= 0, j’ai qq’ >= 0, soit Q elle-même du genre temps.

Considérons maintenant la forme Q(x,y) = x² - y². Cette forme n’est plus de signe définie, elle est dite pseudo-euclidienne. Son tenseur métrique associé est g_ij tel que g₁₁ = -g₂₂ = 1, g₁₂ = g₂₁ = 0. Je l’écris comme somme de produits (et surtout pas différence, sinon je tournerais en rond !)

 

7. Q(x₁,x₂,y₁,y₂) = x₁x₂ + y₁y₂

 

J’aurai cette fois x₁x₂ = x² >= 0 et y₁y₂ = -y² =< 0, soit x du genre temps et y du genre espace. De nouveau, j’écris Q comme produit :

 

8. Q(x₁,x₂,y₁,y₂) = qq’

 

Mais ce produit peut être, soit positif, soit négatif, soit nul.

Inversons les choses et considérons la forme bilinéaire sur R⁴ :

 

9. Q(x₁,x₂,y₁,y₂) = x₁x₂ + y₁y₂

 

Attention : ce n’est plus (7) ! Dans (7), nous nous sommes fixés x et y ; dans (9), nous nous fixons x₁, x₂, y₁ et y₂. Nous avons 4 possibilités (avec un léger abus d’écriture) :

 

10. x₁x₂ = x² >= 0 , y₁y₂ = y² >= 0 : Q(x₁,x₂,y₁,y₂) = Q(x,y) = x² + y²
11. x₁x₂ = x² >= 0 , y₁y₂ = -y² =< 0 : Q(x₁,x₂,y₁,y₂) = Q(x,y) = x² - y²
12. x₁x₂ = -x² =< 0 , y₁y₂ = y² >= 0 : Q(x₁,x₂,y₁,y₂) = Q(x,y) = -x² + y² = -(x² - y²)
13. x₁x₂ = x² =< 0 , y₁y₂ = y² =< 0 : Q(x₁,x₂,y₁,y₂) = Q(x,y) = -(x² + y²)

 

Avec une seule forme, (9), j’obtiens à présent 4 formes quadratiques, 2 euclidiennes et 2 pseudo-euclidiennes.

Quelle est ma métrique ?

 

POUR Q(x,y) = x² + y², LA METRIQUE EST EUCLIDIENNE [SIGNATURE (2,0)] ;

POUR Q(x,y) = x² - y², LA METRIQUE EST PSEUDO-EUCLIDIENNE [SIGNATURE (1,1)] ;

POUR Q(x₁,x₂,y₁,y₂) = x₁x₂ + y₁y₂, LA METRIQUE EST TOUJOURS EUCLIDIENNE. C’EST LE SIGNE DES PRODUITS ET NON PLUS DES COEFFICIENTS METRIQUES, QUI CHANGE.

 

Ça revient au même ? Pas du tout, même si les résultats sont les mêmes. Dans Q(x,y) = x² - y², on se situe dans R², muni d’une géométrie pseudo-euclidienne. Dans Q(x₁,x₂,y₁,y₂) = x₁x₂ + y₁y₂, on se situe dans R⁴ et on est ramené à une géométrie euclidienne, de tenseur métrique d_ijh_ab (i,j = 1,2 ; a,b = 1,2 ou +,- comme précédemment). On n’a plus à se préoccuper du signe des coefficients métriques, ils sont tous positifs. Je n’ai même plus à me préoccuper du signe de mes produits, je n’ai qu’à constater : si x₁x₂ >= 0 pour un couple (x₁,x₂) donné, alors il existe un réel x tel que x₁x₂ = x² ; si x₁x₂ =< 0, alors il existe un réel x tel que x₁x₂ = -x² ; et de même pour y₁y₂. Suivant sur quelles hyperboles x₁x₂ et y₁y₂ je me trouve, j’aurai une forme quadratique de signature (2,0) ou (1,1). Si je passe d’une hyperbole à l’autre, je change de signature.

Le problème de la signature de l’espace-temps est donc bel et bien un faux problème et ce qu’il masquait était en réalité la nature fondamentalement double de l’espace comme du temps. Algébriquement, ça s’interprète comme le fait que R peut toujours être considéré comme la moyenne quadratique de R² vis-à-vis du produit euclidien.

 

LES IMAGINAIRES N’APPARAISSENT QUE COMME CONSEQUENCE LOGIQUE DE LA CONTRACTION DE R² EN SA MOYENNE QUADRATIQUE.

 

C’est parce qu’au lieu de considérer des couples de réels (x₁,x₂), on ne considère qu’un seul nombre x. Alors, évidemment, x peut être réel ssi son carré est >= 0, mais si son carré est =< 0, on se voit dans l’obligation d’étendre le calcul algébrique et donc la structure d’algèbre et même de corps… ou la résolution de la cubique par radicaux est impossible, ce qui est absurde et a été confirmé par le théorème de Galois…

 

Je ne dis pas qu’il faut tirer un trait sur le calcul complexe, je dis simplement qu’il n’est plus nécessaire en physique et masque même les véritables structures réelles.

A quoi bon, pour ne citer que cet exemple, continuer à utiliser des solutions oscillantes en e^iw, par simple commodité de calcul, si c’est pour n’en retenir, de toute façon, que la partie réelle ?...

Quelle est l’unité imaginaire ? C’est i telle que :

 

14. i² = -1 = i₁i₂  ,  (i₁, i₂ réels)…

 

C’est un nombre auquel correspond en réalité une infinité de réels i₁ et i₂ tels que i₁i₂ = -1…

 

La généralisation à toute puissance entière n est immédiate. Soient (x₁,…,x_n) dans Rⁿ. Le nombre x tel que :

 

(15)     xⁿ = x₁…x_n

 

est soit réel, soit complexe. C’est, plus généralement, le résultat de la contraction de Rⁿ en R : on passe d’un n-uplet de réels à un nombre unique. Si on se donne x, on a alors x_n = xⁿ/(x₁…x_n-1). Pour n = 4, j’ai la relation x⁴ = (x₁x₂)(x₃x₄) <=> (x₃x₄) = y² = x⁴/(x₁x₂) = x⁴/z² = (x²/z)² <=> y = +/- x²/z, hyperboles soit réelles, soit complexes.

Sauf preuve contraire, pour nos applications, nous n’aurons besoin que du cas n = 2.

 

This work, in bidouille 27, shows that, when synthetically treated in dimension 4, the problem of the motion inside a Newtonian field is very different from its approach in dimension 3.

On this purpose, I would like to make a general comment about Jacobi’s construction. If you look twice on it, it is purely kinematical. Hamilton’s principal function, also named Jacobi’s function, derives from Jacobi’s principle, based itself on the action. If S = I_t1^t2 ½ m[dx(t)/dt]²dt is the Galilean action calculated from t₁ to t₂ > t₁ on a given trajectory x(t) in Euclidian 3-space E(t), then there is a one-to-one correspondence between S and Jacobi’s function. That is the base of Hamilton-Jacobi’s theory. Now, what is to be integrated is the KINEMATICAL part of the Lagrangian ONLY. It is clear in the above formula. So, if we extend it to more complicated situations involving gyroscopic terms and eventually a potential energy, such as in the relativistic lagrangian L = -mc²[1- v²(t)/c²]^1/2 + qv(t)A[x(t),t] – Phi[x(t),t] of the original version, then we normally would have to get back from the generalized 3-momentum P[x(t),t] = p(t) + qA[x(t),t], which is a FIELD and mixes kinetics and potential to the purely kinetic 3-momentum p(t) = P[x(t),t] - qA[x(t),t] to compute the action. This is actually what is done, however, the space-time derivatives of the relativistic action are usually define through ð_iS = P_i, not p_i. I don’t agree with such an extension, but I am ready to reconsider my point of view if seriously proved. If I refer to the basics of analytical mechanics, the correct relation, for me, seems to be ð_iS = p_i = P_i – qA_i. Besides, in the relativistic relation p_ipⁱ = m²c², one uses the p_i’s, not the P_i’s… The gyroscopic terms in the Lagrangian can always be absorbed in a rotating reference frame, as a transformation under canonical form shows. It remains that the rotation in question is precisely due, not to kinematical terms, but indeed to potential ones, as to know, qA[x(t),t], coupling to the velocity 3-vector, and this coupling is POTENTIAL… One has criticized Poincaré enough for suggesting to insert Newton’s gravitational potential into mass, but the very same has been done in general relativity… and is still fully accepted… In Einstein’s theory of gravity, POTENTIAL energy is incorporated into the space-time frame and taken responsible for its curvature… This is NOT what I understood of Jacobi’s principle: even in this principle, assumed to be the most general one (Poisson and Noether are consequences of it), we are set back to pure kinematics, the MOTION, the reference…

If we agree with this, then any natural extension should take this into account and thus, get back to pure kinematics. Sure, it changes the sign before qA_i: instead of having ð_iS - qA_i, we find ð_iS + qA_i and p_ipⁱ = m²c² remains ð_iSðⁱS = m²c² instead of becoming (ð_iS – qA_i)( ðⁱS – qAⁱ) = m²c². Let S’ be such that ð_iS’ = P_i. Then ð_iS’ = ð_iS + qA_i and S’ = S + q I A_idxⁱ: it is always possible to go from S to S’ and back. The question is therefore academic, S only looks more natural to me than S’ and that is what I will use. Until formal proof of the opposite.

Let us now go back to our formulas (6) to (11), bidouille 23. We can always express the p_i’s in terms of the P_i’s and the A_i’s. That is precisely the point. If we use the definition of H_ab, we should take ðL/ðv^ib = P_ib, not p_ib and we would have to symmetrise. We would also loose all the properties (7) and (8) (separation of kinetic energy and potential energy), (10) (energy ratios) and (11). However, if we take into account the fact that external fields over 4D space-time M explicitly depend on the xⁱs only and not on (t⁺,t^-), which are parameters of MOTION, then formula (4), bidouille 27, extends to:

 

18. A_i^a(x) = (mu₀/4pi) I_M J_i^a(x’)d⁴x’/(xⁱ - x’ⁱ)(x_i – x’_i)

 

with:

 

19. J_i^a(x’) = rho(x’)v_i^a = rho(x’)p_i^a/m

 

That is the neverending dilemma: should J_i^a also depend on (t⁺,t^-) or not? In the original version, J_i do not depend on s/c. So, in the extended one, it should not depend on (t⁺,t^-), despite v_i^a can. But it is exactly the same in the original version: v_i can depend on s/c… If it does not, it is constant. We will come back on this in another work. For the time being, let us assume (19). Then,

 

20. A_i^a(x) = p_i^aPhi(x)/mc²

 

The Hamiltonian matrix is:

 

H_ab = ½ (vⁱ_aP_ib + vⁱ_bP_ia) – Lh_ab = (pⁱ_aP_ib + pⁱ_bP_ia)/2m – Lh_ab = (1/2m)[2pⁱ_ap_ib – pⁱ_cp_i^ch_ab + q(pⁱ_aA_ib + pⁱ_bA_ia – 2pⁱ_cA_i^ch_ab)]

 

That is,

 

21. H_ab = (1/2m)[2(1 + qPhi/mc²)pⁱ_ap_ib – (1 + 2qPhi/mc²)pⁱ_cp_i^ch_ab]

 

In components:

 

22. H₊₊ = (1 + qPhi/mc²)pⁱ₊p_i+/m
23. H_-- = (1 + qPhi/mc²)pⁱ_-p_i-/m = (1 + qPhi/mc²)²m²c⁴/H₊₊
24. H_+- = (1/2m)[2(1 + qPhi/mc²)pⁱ₊p_i- – 2(1 + 2qPhi/mc²)pⁱ₊p_i-] = -qPhi

 

since pⁱ₊p_i- = m²c². Following that,

 

25. H_a^a = -2qPhi
26. Det(H_ab) = (1 + qPhi/mc²)²m²c⁴ – q²Phi² = m²c⁴(1 + 2qPhi/mc²)

 

As

 

27. L = (pⁱ_cp_i^c + 2qPhipⁱ_cp_i^c/mc²)/2m = (1 + 2qPhi/mc²)mc²

 

we deduce:

 

28. H = mc² = pⁱ_ap_i^a/2m

 

Quite logical, since all gyroscopic terms qvⁱ_aA_i^a should vanish in the Hamiltonian functional.

Let us turn to the Jacobi functions. We now have TWO S_a such that:

 

29. H_ab = ð_aS_b + ð_bS_a
30. ð_iS_a = p_ia

 

and 2 Jacobi equations:

 

31. ð_aS_b + ð_bS_a = (1/2m)[2(1 + qPhi/mc²)ðⁱS_að_iS_b – (1 + 2qPhi/mc²)h_abðⁱS_cð_iS^c]

 

since we only have two independent coefficients H₊₊ and H_+-. From (22) to (26), we have:

 

32. ð₊S₊ = (1 + qPhi/mc²)ðⁱS₊ð_iS₊/2m
33. ð_-S_- = (1 + qPhi/mc²)ðⁱS_-ð_iS_-/2m
34. ð₊S₊ð_-S_- = ¼ (1 + qPhi/mc²)²m²c⁴
35. ð₊S_- + ð_-S₊ = -qPhi
36. (ðⁱS₊ð_iS₊)(ð^jS_-ð_jS_-) = m⁴c⁴

 

The last equation being a consequence of the first 3 ones. Anyway, it is obvious, since pⁱ₊p_i- = m²c² implies:

 

37. ðⁱS₊ð_iS_- = m²c²

 

I leave the problem of the elastic scattering of two (anti)material objects to the interested reader. Today, I’m rather going to focus on the attraction of a mass m by a field with source m’. My motivation, in all this work on space-time relativity, is mainly NDE’s. The question is : do we already have incontestable proves of accelerated motions reaching the speed of light first, then moving faster ? The answer is actually yes, but they don’t come from physics, they come from medicine. Before books were written on the subject of near-death experiment, numerous testimonies had been collected on patients having experienced an accelerated motion in a post-coma state (phase 4), moving through a “Tunnel”, then reaching “a large White Light”, then entering a “Garden of Colours”. At that time, there was still no means for all these people, to invent anything : first, they were dispatched nearly everywhere in the world, second they did not know each one another and third, there were constants that came back again and again in their reports, whatever their education, social level, culture and behaviour. Today, after a lot of publication has been written on the subject, one can unfortunately object the fact that “new experiencers” might only copy the “former” ones.

In my opinion, this “White Light” in question is nothing else than the physical light. And the “Garden of Colours”, whatever it is, is located OUTSIDE the light cone. As I’m more and more deeply convinced that NDEs are definitely at the very centre of all parapsychological phenomena, because it makes use of all (or nearly all) new physical properties (that’s what I think at present), investigating NDEs amounts to investigate EXTENDED properties, not of some more “new physics”, but first of already-existing physics.

Some of our great thinkers of late 19^th century asserted us that “classical physics had nothing more to learn us”.

A century after, some of our great thinkers of late 20^th century asserted us that “classical physics had nothing more to learn us FOR LONG”.

We know now that is not true. Even at the classical level, physics still DOES have something new to learn us. At large scales, the quantum level can only CONFIRM (or not) what classical physics predicts. So, may it seems “elementary” to a lot of specialists, there is no need in quantizing everything systematically, we should first examine classical phenomena and ONLY THEN quantize.

Let us now get to today’s subject. To begin with, a remark on Maxwell’s linear field theory. The field equations with sources:

 

1. ðⁱð_iA_j(x,t) = mu₀j_j(x,t)

 

have general solutions :

 

2. A_i(x,t) = (mu₀/4pi) I_B3 j_i(x’,t’)d³x’/|x-x’| + A_i^(wav)(x,t)  ,  t’ = t - |x-x’|/c

 

where B₃ is the Kirchhoff 3-ball |x-x’|² = (x-x’)² =< c²t² and A_i^(wav) is the solution to the wave (i.e. sources-free) equations ðⁱð_iA_j(x,t) = 0. The integral kernel is in 1/|x-x’|. This is essentially due to the fact that, despite not saying it, the space parameters x are kept distinct from the time parameter t. As a result, 1/|x-x’| is the kernel of the laplacian operator ð^að_a. Consequently, integration must be done over the causal domain B₃ and physics only keep the delayed solution j_i(x’,t’), corresponding to an electromagnetic field emitted by a charge source rho(x,t) at time t and propagating from this source into space. The advanced solution j_i(x’,t+|x-x’|/c) is rather interpreted as an external EM field falling down onto the emitting source and interacting with it. However, it clearly appears in (1) that, even if we are not talking about space-time relativity yet, time has been inserted into the frame under a “time-like” distance x⁰ = ct, extending E³ to Minkowski pseudo-euclidian space-time M. The characteristics of (1) are:

 

3. s² = c²t² - (x²+y²+z²) = c²t² - r² = 0

 

that is, the light cone. Besides, the Kirchhoff 3-sphere x² = c²t², with variable radius R = ct, identifies with the light cone in dimension 4. It remains that the REAL integration domain of (1) is NO LONGER E³, but M. The distance from point to point is no longer x², but s², which has no definite sign anymore. And the treatment of (1) MUST BE hold in FOUR dimensions and not 3, if we want to keep space and time on an equal footing, as required by the very form of the second-order differential operator in (1). This means that the integral kernel of the d’Alembertian is 1/s² and the general solution takes the 4D form:

 

4. A_i(x) = (mu₀/4pi) I_M j_i(x’)d⁴x’/(xⁱ-x’ⁱ)(x_i-x’_i) + A_i^(wav)(x)

 

with a potential now in 1/s² and not 1/r. This is absolutely normal: in dimension n, the kernel of the laplacian is in 1/r^n-2, with r² = (x₁)²+…+(x_n)². In (4), integration now includes ALL M: there is no restrictions on the physical domain anymore, since the Casimir square (xⁱ-x’ⁱ)(x_i-x’_i) has no definite sign. There are also neither “advanced” nor “delayed” solution anymore, there are only points x’ assumed to be distinct from the observation point x.

However, it should remain clear that (4) and (2) must be formally equivalent, since they are both complete solutions to the very same equations, (1). So?

So, in (2), we deal with dynamical sources and fields showing a space behaviour. Their dynamics is expressed in their explicit time-dependence. Whereas in (4), as time has been included into the frame, the solutions of (1) become static, with a space-time behaviour. There is no more time-dependence. This is a theorem:

 

THEOREM :

THE 4D STATIC SOLUTIONS OF MAXWELL EQUATIONS, WITH POTENTIALS IN 1/s², ARE FORMALLY EQUIVALENT TO THEIR 3D DYNAMICAL SOLUTIONS, WITH POTENTIALS IN 1/r.

 

The reader will easily verify that ðⁱð_i(1/s²) is indeed zero. More generally, in any pseudo-euclidian space of dimension n endowed with a metric of signature (p,q), p+q = n, the kernel of ðⁱð_i = ð²/(ðx¹)² +…+ ð²/(ðx^p)² - ð²/(ðx^p+1)² -…- ð²/(ðxⁿ)² is 1/s^n-2, with s² = (x¹)² +…+ (x^p)² - (x^p+1)² -…- (xⁿ)².

To conclude with this remark, the 4D static potential 1/s² is much more natural to the field equations (1) than the 3D (restricted) dynamical potential 1/r.

 

CAUTION : the static 4D potentials A_i(x) are now in TESLAS and the field intensities in T/m. We could introduce 4D units, such as 4D Tesla-meters (4Tm) and 4D Teslas (4T) so as to keep the familiar units, with 1 4Tm = 1 3T and 1 4T = 1 3T/m.

 

Let us now turn to the Lagrange functional of a massive point-like body under the influence of a Maxwell-like external field.

In the original version, this functional writes:

 

5. L[xⁱ(s/c),vⁱ(s/c),s/c] = ½ mvⁱ(s/c)v_i(s/c) + qvⁱ(s/c)A_i[x(s/c)]  ,  vⁱ(s/c) = cuⁱ(s/c) = cdxⁱ(s/c)/ds

 

The Lagrange equations of motion are:

 

cdpⁱ/ds + qcdA_i/ds = cdpⁱ/ds + qcv^jð_jA_i = qv^jð_iA_j

 

that is:

 

cdpⁱ(s/c)/ds = q(ð_iA_j – ð_jA_i)[x(s/c)]v^j(s/c) = qF_ij[x(s/c)]v^j(s/c)

 

The electromagnetic field equations are:

 

6. ð_[iF_jk](x) = (ð_iF_jk + ð_jF_ki + ð_kF_ij)(x) = 0
7. ðⁱF_ij(x) = mu₀J_j(x)  ,  J_j(x) = rho(x)v_j(s/c)

 

they derive from the Bianchi identities and the lagrangian density:

 

8. L_EM[A_i(x),F_ij(x),xⁱ] = F_ij(x)F^ij(x)/2mu₀ + J_i(x)Aⁱ(x)

 

In (5), the field variables are the xⁱ and the dynamical parameter, s/c; in (8), the field variables are the A_i and the dynamical parameters, the xⁱ. Consequently, the field equations (6) and (7) depend on the xⁱ only and so does the 4-current density J_i(x).

The natural extension to time surfaces is:

 

9. L[xⁱ(t⁺,t^-),vⁱ_a(t⁺,t^-),t⁺,t^-] = ½ mvⁱ_a(t⁺,t^-)v_i^a(t⁺,t^-) + qvⁱ_a(t⁺,t^-)A_i^a[x(t⁺,t^-)]

 

Except for the 2-components (a = + and -), there is nothing more to add. The equations of motion are thus similar to those of the original version:

 

10. d_ap_i^a(t⁺,t^-) = md_ad^ax_i(t⁺,t^-) = q(ð_iA_j^a – ð_jA_i^a)[x(t⁺,t^-)]v^j_a(t⁺,t^-) = qF_ij^a[x(t⁺,t^-)]v^j_a(t⁺,t)

 

The electromagnetic field equations become:

 

11. ð_[iF_jk^a_](x) = (ð_iF_jk^a + ð_jF_ki^a + ð_kF_ij^a)(x) = 0
12. ðⁱF_ij^a(x) = mu₀J_j^a(x)  ,  J_j^a(x) = rho(x)v_j^a(t⁺,t^-)

 

In the gauges:

 

13. ðⁱA_i^a = 0

 

the complete solutions of (12) are therefore:

 

14. A_i^a(x) = (mu₀/4pi) I_M J_j^a(x’)d⁴x’/(xⁱ-x’ⁱ)(x_i-x’_i) + A_i^a(wav)(x)

 

just as in (4) above, but for 2 components (+) and (-). Let us try to apply these results to a point-like source in motion with charge q’. It shows easier to first examine the problem in the original version, then replacing s² with c²t⁺t^-. For rho(x) = q’d(x), J_j(x) = q’d(x)v_j(s/c) and:

 

15. A_i(x) = (mu₀q’c/s²)dx_i(s/c)/ds
16. F_ij(x) = -(2mu₀q’c/s⁴)[x_i(s/c)dx_j(s/c)/ds – x_j(s/c)dx_i(s/c)/ds]
17. d²x_i(s/c)/ds = du_i(s/c)/ds = -(2mu₀qq’/ms⁴)[x_i(s/c) - x_j(s/c)u_i(s/c)u^j(s/c)]

 

Equation (17) is quadratic in the u's: Riccati. Non integrable.

I won’t pursue any further in this direction, as I finally understood how it all works. I’m not surprised any longer I spent my time turning in circles : the whole stuff looked inconsistent and contradictory. It took me no less than two weeks of hard work to understand why. The explanation is simple : when you assume the energy matrix H_ab is conserved, it sounds logical to treat both H₊₊ conserved and H_— conserved at the same time, doesn’t it ? Well, it appears to be wrong… you actually have to treat both conditions separately. For H₊₊ and H_— are reciprocal to each other and treating them simultaneously leads you to contradictions…

Today, we will see how it actually works, but I leave bidouilles 24 and 25 so as to show how difficult basic problems can become when expanding our vision of things and getting out of habits.

We first go back to H_ab, bidouille 23, formula (7) and (10) and go a bit deeper. H₊₊ is the projection of H_ab on the AXIS t⁺ (advanced time, this is important for what follows), H_—is the projection of H_ab on the AXIS t^- (delayed time), while H_+- = H_-+ is the projection of H_ab on the PLANE (t⁺,t^-). Now, we see from (7) or (10) that:

 

THE KINETIC ENERGY LAYS ON THE ADVANCED AND DELAYED TIME AXIS,

WHILE THE POTENTIAL ENERGY LAYS ON THE (t⁺,t^-) PLANE.

 

This is a rather interesting property of how energy is dispatched in the (t⁺,t^-) representation.

Following that, we have:

 

1. H₊₊ >= 0  for  (m >= 0 and 0 =< v =< c)  and for  (m =< 0 and v >= c)
2. H₊₊ <= 0  for  (m >= 0 and v >= c)  and for  (m =< 0 and 0 =< v =< c)
3. While H₊₊ = 0 holds whether for m = 0 or v = c or even both.
4. If m >= 0, then 0 =< H₊₊ =< mc² for 0 =< v =< c and –mc² =< H₊₊ =< 0 for v >= c
5. If m =< 0, then -|m|c² =< H₊₊ =< 0 for 0 =< v =< c and 0 =< H₊₊ =< |m|c² for v >= c

 

Thus, H₊₊ is always bounded. Properties a, b and c obviously hold for H_--, but H_— is NOT bounded. Instead of d and e, we have:

 

6. If m >= 0, then H_-- > mc² for 0 =< v < c and H_-- < -mc² for v > c
7. If m =< 0, then H_-- < -|m[c² for 0 =< v < c and H_-- > |m|c² for v > c
8. While H_-- = +oo for (m >= 0 , v = c^-) and for (m =< 0 , v = c⁺) and H_-- = -oo for (m >= 0 , v = c⁺) and for (m =< 0, v = c^-)

 

Let us now consider a 3-bodies problem. The conservation of H₊₊ implies:

 

1. H₊₊ = H₁₊₊ + H₂₊₊

 

The only requirement, for the time being, is that the sum on the right-hand side have same sign as H₊₊. In terms of velocities:

 

2. m(1-v/c)/(1+v/c) = m₁(1-v₁/c)/(1+v₁/c) + m₂(1-v₂/c)/(1+v₂c)

 

This equality is SYMMETRIC, right? It can be read in both directions. Well, this is WRONG… because we are located on the t⁺ axis. And this means that the SIGNAL can only move from the PRODUCT of a reaction BACK TO its origin. So, if m is the original mass and m₁, m₂ the products, the process along the t⁺ axis can ONLY be m₁ + m₂ -> m, which corresponds to a COLLISION PROBLEM between two masses m₁ and m₂ giving a resulting mass m and we will see in a moment that it satisfies the mass condition.

On the contrary, the conservation of H_— implies:

 

3. m(1+v/c)/(1-v/c) = m₁(1+v₁/c)/(1-v₁/c) + m₂(1+v₂/c)/(1-v₂c)

 

Again, this equality LOOKS symmetric. In fact, as we are now located on the t^- axis, the signal can ONLY move from the ORIGIN of the reaction to its PRODUCTS: m -> m₁ + m₂, which corresponds to a DECAY PROBLEM. Again, this satisfies the mass condition.

So, we have something very powerful: in the same energy matrix, we now find TWO CLASSES OF PROBLEMS. And H₊₊ and H_—being RECIPROCAL, these classes are RECIPROCAL to each other, which is indeed the case. That is why the conservation of H₊₊ and that of H_—MUST BE treated separately. Or, we turn in circles…

Equations (2) and (3) only LOOK symmetric. Actually, (2) must be read from RIGHT TO LEFT, while (3) must be read from LEFT TO RIGHT.

They are 16 possibilities, according to the signs of m₁, m₂ and the regimes v₁, v₂. Table 1 below resumes all possibilities and give the sign of corresponding H₊₊.

 

 

v1 v2 ->	sub sub	sub super	super sub	super super
m1 m2
+ +	+	+/-	+/-	-
+ -	+/-	+	-	+/-
- +	+/-	-	+	+/-
- -	-	+/-	+/-	+

 

Table 1: sign of H₊₊ or possibilities for couples (m₁,v₁), (m₂,v₂) for a given sign of H₊₊. Sub = subluminic (0 =< v₁ , v₂ =< c), super = superluminic (v₁, v₂ >= c).

 

We have exactly the same table for H_--. The differences lay in the mass conditions we get. Take for instance (m₁,m₂ ) = (+,+), (v₁,v₂) = (sub, sub). Then, H₊₊ >= 0, meaning that either (m >= 0 , v sub) or (m =< 0, v super). If m >= 0, then its regime is automatically subluminic and from d) above, we get the mass condition 0 =< m =< m₁+m₂, while for H_-- >= 0 and m >= 0, f) gives m >= m₁+m₂. It should now be clear that the t⁺ process can only be a COLLISION between two subluminic material objects giving a subluminic material object of mass 0 =< m =< m₁+m₂. A certain quantity of matter is lost in the collision and emitted in the outside under the form of (positive) energy. Opposite to this, the t^- process can only be a DECAY of a subluminic material object of mass m >= m₁+m₂ into two subluminic material objects with, similarly, a certain quantity of matter lost under the form of energy. And everything is right again.

What about the other situations? First, we have a symmetry between quadruplets (m₁,m₂,v₁,v₂) and their counterparts (-m₁,-m₂,c²/v₁,c²/v₂), which reduces the possibilities to 8 instead of 16. Then, we invoke two PRINCIPLES:

 

RESPECT OF POLARITIES:

TWO MASSES HAVING SAME SIGN CAN ONLY GIVE BIRTH TO A MASS WITH THIS SIGN.

 

A principle equivalent to the conservation of charge, but not so, as we will see in a minute. The reason is, mass can AGREGATE, whereas other (known) charges cannot.

 

RESPECT OF REGIMES:

TWO PROCESSES HAVING THE SAME VELOCITY REGIME CAN ONLY GIVE A PROCESS HAVING THIS REGIME.

 

These two principles sound quite logical, don’t they? If, say m₁ and m₂ are both subluminic and collide, the resulting mass m should rather be subluminic as well. One wouldn’t see very well how m could be given a superluminic speed, as velocities no longer simply add in space-time relativity. Same, if m₁ and m₂ are superluminic, it sounds logical to have m superluminic as well, so that the whole process is not observable.

On the other side, matter + matter can hardly give antimatter (polarities are not respected) and antimatter + antimatter can hardly give matter (for the same reason).

We then get the (not definitive) following table:

 

 

v1 v2 ->	sub sub	sub super	super sub	super super
m1 m2
mat mat	sub mat	+/-	+/-	super mat
mat antimat	+/-	+	-	+/-
antimat mat	+/-	-	+	+/-
antimat antimat	sub antimat	+/-	+/-	super antimat

 

Table 2: allowed processes respecting both mass polarities and velocity regimes.

 

What of the rest? The result depends on the values of H₁₊₊ and H₂₊₊. For H₊₊ to be positive or zero, we need H₁₊₊ >= -H₂₊₊, which allows processes like matter + antimatter -> matter or antimatter or massless (and conversely for H_-- >= 0). This is why we are not formally equivalent to charge conservation: because of aggregation. For instance, (sub mat) + (super antimat) gives H₊₊ >= 0, so (sub mat) if H₁₊₊ dominates or (super antimat) if H₂₊₊ dominates. We can even get m = 0 (massless – pair annihilation) or, a bit more exotic, v = c with m non zero. In my opinion, a massive object moving at exactly the speed of light is unlikely to be found. But my opinion does not account for in the real world.

Are polarities respected in matter – antimatter processes? Yes, they are: if the quantity of matter is greater than that of antimatter, the collision will give matter (a material residue); if the quantity of antimatter is greater than that of matter, the residue will be antimaterial; and if we have equal quantities of matter and antimatter, the residue will be massless.

Again, everything stands right up again.

Processes are reversed in the t^- direction, but we straightforwardly see the results are essentially the same. Let us take for instance H_-- >= 0. Then H_1-- + H_2—must be >= 0. H_-- >= 0 can be realized two ways: subluminic matter or superluminic antimatter. Going back to table 1 and respecting both polarities and regimes, we get:

 

Subluminic matter -> subluminic matter + subluminic matter + positive energy exceed

Superluminic matter -> superluminic matter + superluminic matter + negative energy exceed

 

Whereas for:

 

Subluminic matter -> subluminic matter + superluminic antimatter

 

the velocity regimes are not satisfied. So, we should expect this decay to be forbidden or to have a very small cross-section (which roughly amounts to the same). Again, conceptually, it sounds more logical subluminic matter can only decay into subluminic parts. The question is not really matter being able to eject antimatter: in the spontaneous neutron beta-decay, an ANTIneutrino is indeed emitted… This alone shows that matter CAN decay into matter AND antimatter (and antimatter into antimatter AND matter). The point is, there should be a tremendous acceleration for the antimatter part to be emitted FROM THE DEPARTURE faster than light… It seems much more consistent to respect the velocity regime. As in:

 

Superluminic matter -> superluminic matter + superluminic matter + negative energy exceed

 

The whole process is then superluminic and remains non observable. A process like:

 

Superluminic matter -> superluminic matter + subluminic antimatter

 

despite seducing, should not be allowed, for the emitted antimatter should be SLOWED DOWN FROM THE DEPARTURE. One hardly sees how a material object moving faster than light could decay into a material part moving faster than light (this, ok) and an antimaterial part ejected SLOWER THAN LIGHT…

 

Subluminic antimatter -> subluminic antimatter + subluminic antimatter + negative energy exceed

 

Allowed, obviously the conjugate of sub mat -> sub mat + sub mat + pos ener exceed above. Same for:

 

Superluminic antimatter -> superluminic antimatter + superluminic antimatter + positive energy exceed

 

In the end, we should keep only 4 possibilities on the 16, to sum up:

 

a)  Subluminic matter -> subluminic matter + subluminic matter + positive energy exceed

 

b)  Superluminic matter -> superluminic matter + superluminic matter + negative energy exceed

 

c)  Subluminic antimatter -> subluminic antimatter + subluminic antimatter + negative energy exceed

 

d)  Superluminic antimatter -> superluminic antimatter + superluminic antimatter + positive energy exceed

 

 

We now know the masses m₁ and m₂ of the two bodies, as well as their velocities v₁ and v₂ and we search for the condition on the mass m and the velocity v of the product of their collision. The former problem is thus reversed and we keep the notations.

From the conservation of energy, we get :

 

1. (1-v/c)/(1+v/c) = (H₁₊₊+H₂₊₊)/mc²
2. (1+v/c)/(1-v/c) = (H_1--+H_2--)/mc² = (m₁²c⁴/H₁₊₊+m₂²c⁴/H₂₊₊)/mc²

= c²(m₁²H₂₊₊+m₂²H₁₊₊)/mH₁₊₊H₂₊₊

 

This gives :

 

3. m₁²(H₂₊₊)² + m₂²(H₁₊₊)² + (m₁²+m₂²-m²)H₁₊₊H₂₊₊ = 0

 

We know that rotating the energy axes an angle alpha such that :

 

4. tan alpha = [m²-(m₁²+m₂²)]/(m₂²-m₁²)

 

Turns (3) into canonical form :

 

5. M₁(H₁₊₊)² + M₂(H₂₊₊)² = 0

 

with :

 

6. 2M₁ = m₁²+m₂² + [(m₁²+m₂²-m²)² + (m₂² - m₁)²]^1/2
7. 2M₂ = m₁²+m₂² - [(m₁²+m₂²-m²)² + (m₂² - m₁)²]^1/2
8. -4M₁M₂ = Q₄(m)  [bidouille 24, formula (7)]

 

We are thus set back to condition (8) bidouille 24 : Q₄(m) >= 0, independent of the velocities.

Having a condition on m, we can deduce one on the ratio v/c. Reversing (1) gives :

 

9. v/c = (1-H₊₊/mc²)/(1+H₊₊/mc²)  ,  H₊₊ = H₁₊₊ + H₂₊₊

 

We prefer to express this in terms of the ratios v₁/c and v₂/c. This will exhibit a rather interesting structure. We have :

 

H₊₊/c² = m₁(1-v₁/c)/(1+v₁/c) + m₂(1-v₂/c)/(1+v₂/c)

            = [m₁(1-v₁/c)(1+v₂/c) + m₂(1+v₁/c)(1-v₂/c)]/(1+v₁/c)(1+v₂/c)

 

Inserting this into (9) leads us to the following expression :

 

10. v/c = [m_-- + m_+-v₁/c + m_-+v₂/c + m₊₊v₁v₂/c²]/[m₊₊ + m_-+v₁/c + m_+-v₂/c + m_--v₁v₂/c²]

 

where m_ab is the 2x2 mass matrix, with components :

 

11. m₊₊ = m+m₁+m₂  ,  m_+- = m+m₁-m₂  ,  m_-+ = m-m₁+m₂  ,  m_-- = m-m₁-m₂

 

or, in condensed form :

 

12. m_ab = m + am₁ + bm₂   (a,b = +,-)

 

Factorizing with the help of this matrix, we finally obtain :

 

13. v/c = (m₊₊/m_--)[(v₁/c + m_-+/m₊₊)(v₂/c + m_+-/m₊₊) + det(m_ab)/m₊₊²] /

/[(v₁/c + m_+-/m_--)(v₂/c + m_-+/m_--) + det(m_ab)/m_--²]

14. det(m_ab) = m₊₊m_-- - m_+-m_-+

 

The mass condition Q₄(m) >= 0 together with (13-14) completely solve our problem. As masses can now be positive, negative or zero and velocities can be =< c or >= c, there are many possibilities. Again, the original version only retained positive masses (and even strictly for matter) and subluminic velocities. Unfortunately, neither det(m_ab) nor the mass ratios in (13) are always sign-definite. So, to obtain a condition on v/c, it is still simpler to use (9) and investigate all possible combinations.

Let us start with the familiar case m₂ >= m₁ >= 0. This corresponds to case 1, bidouille 24. We now see that we have 3 possibilities on m: m =< -(m₁+m₂), -(m₂-m₁) =< m =< (m₂-m₁) and m >= (m₁+m₂). For each one of them, there are 4 situations : (v₁ =< c, v₂ =< c), (v₁ >= c, v₂ =< c), (v₁ =< c, v₂ >= c) and (v₁ >= c, v₂ >= c). This already makes 12 possibilities only for case 1. There are two other cases, which makes a total of 36 possibilities. Instead of a single one…

We will do it for case 1 and let the reader deduce the two other cases. We will discuss the results later on. In case 1 alone, the most familiar one, there are already surprises…

 

a) m =< -(m₁+m₂).  This is acceptable, at least on the paper.

i) v₁ =< c, v₂ =< c : 0 =< H₁₊₊ =< m₁c² , 0 =< H₂₊₊ =< m₂c², so 0 =< H₊₊ =< (m₁+m₂)c² and since m =< 0, v >= c.

 

TWO SUBLUMINIC MATERIAL BODIES COLLIDING WOULD BE LIKELY TO PRODUCE TO A SUPERLUMINIC ANTIMATERIAL BODY.

 

ii) v₁ >= c, v₂ =< c : -m₁c² =< H₁₊₊ =< 0 , 0 =< H₂₊₊ =< m₂c², -m₁c² =< H₊₊ =< m₂c² and we have to distinguish two more cases : -m₁c² =< H₊₊ =< 0, then v =< c; 0 =< H₊₊ =< m₂c², then v >= c.

 

A SUBLUMINIC MATERIAL BODY COLLIDING WITH A SUPERLUMINIC ONE WOULD BE LIKELY TO PRODUCE, WHETHER A SUBLUMINIC OR A SUPERLUMINIC ANTIMATERIAL BODY, DEPENDING ON THE SIGN OF THE RESULTING ENERGY.

 

iii) v₁ =< c, v₂ >= c : same results, permuting m₁ and m₂.

iv) v₁ >= c, v₂ >= c : -m₁c² =< H₁₊₊ =< 0, -m₂c² =< H₂₊₊ =< 0, -(m₁+m₂)c² =< H₊₊ =< 0. Consequently, v =< c.

 

TWO SUPERLUMINIC MATERIAL BODIES COLLIDING WOULD BE LIKELY TO PRODUCE A SUBLUMINIC ANTIMATERIAL BODY.

 

This last result is quite astonishing… it undermeans observable antimatter could also be produced from collisions of non-observable matter and not only from the separation of pairs.

 

This could ask to some the question of the conservation of mass. Again, the original version asserts that mass is no longer conserved. The conserved quantities are energy and momentum. The argument given is : in non-interacting matter (a mere set of free particles), mass is indeed conserved, as in Galilean relativity; but, as soon as matter particles interact, mass is no longer conserved, for a part of it is converted into energy or conversely.

This is just fallacious.

For, if it was the case, then it should be the same for all types of charges, in particular for the electric charge. Now, the conservation of electric charge is, in turn, well accepted in the original version, should electric charges be free or interact. The only requirement is : the quantity of electric charge going out of a given 3-volume must be compensated for by an equal quantity of incoming charge. That’s all. So why, again, shouldn’t it be the same for mass ?... Mass IS conserved in space-time relativity, just the same as charge is. Again, one mistakes MASS with MASS EQUIVALENT: as has already been said, the relation E = mc² is a EQUIVALENCE between mass and energy, it says that a certain quantity of mass (m) IS CONVERTED INTO AN EQUIVALENT QUANTITY OF ENERGY, E/c². That’s all. It never asserts, and cannot do it, that mass IS energy and energy IS mass: these are two different things !

Restrictions, misinterpretations… the whole thing is not serious…

When a body spontaneous decay, it looses a certain amount of MASS, because this amount is CONVERTED INTO ENERGY. But if we add the masses of the products WITH the lost amount of mass, we retrieve the original mass… obviously…: m = m₁ + m₂ + dm and dm = dE/c², dE = link energy…

 

Why then two superluminic MATERIAL bodies could give birth to an ANTIMATERIAL one? Apparently, mass is not conserved. Of course, it is. Have a deeper look at what precedes and you will convince yourself that mass IS conserved. The explanation of this apparent paradox is simple:

 

SUPERLUMINIC MATTER (RESP. ANTIMATTER) IS FORMALLY EQUIVALENT TO SUBLUMINIC ANTIMATTER (RESP. MATTER).

 

I have a mass m >= 0, moving at a speed v such that v/c =< 1, this is formally equivalent to a mass –m =< 0, moving at a speed c²/v >= c. This can be deduced from H₊₊. As a result, my subluminic antimaterial collision product is formally equivalent to a superluminic material body. This may sound more “harmonic”, yet the result IS antimatter, not matter.

 

b) -(m₂-m₁) =< m =< (m₂-m₁). Here, m can either be positive, zero or negative.

i) v₁ =< c, v₂ =< c : 0 =< H₊₊ =< (m₁+m₂)c². If –(m₂-m₁) =< m < 0, v >= c; if 0 < m =< (m₂-m₁), v =< c and if m = 0, (9) gives v = -c, which is to be rejected, since v >= 0.

 

TWO SUBLUMINIC MATERIAL BODIES OF MASSES m₁ AND m₂ >= m₁ WOULD BE LIKELY TO PRODUCE A SUBLUMINIC MATERIAL BODY OF MASS 0 < m =< (m₂-m₁).

 

Precisely : here, we retrieve the LOSS OF MASS after a collision, m₂ looses a quantity of mass equal to m₁.

ii) v₁ >= c, v₂ =< c : -m₁c² =< H₊₊ =< m₂c² as before, we have to distinguish two more cases : -m₁c² =< H₊₊ =< 0 and 0 =< H₊₊ =< m₂c². However, –(m₂-m₁) =< m < 0 OR 0 < m =< (m₂-m₁), that is, 4 more cases…

[-m₁c² =< H₊₊ =< 0 , –(m₂-m₁) =< m < 0] and [0 =< H₊₊ =< m₂c² , 0 < m =< (m₂-m₁)] , v =< c; [-m₁c² =< H₊₊ =< 0 , 0 < m =< (m₂-m₁)] and [0 =< H₊₊ =< m₂c², –(m₂-m₁) =< m < 0], v >= c;

Same for matter, then :

 

A SUBLUMINIC MATERIAL BODY COLLIDING WITH A SUPERLUMINIC ONE WOULD BE LIKELY TO PRODUCE, WHETHER A SUBLUMINIC OR A SUPERLUMINIC MATERIAL BODY, DEPENDING ON THE SIGN OF THE RESULTING ENERGY.

 

iii) v₁ =< c, v₂ >= c : same conclusions, permuting m₁ and m₂.

iv) v₁ >= c, v₂ >= c : -(m₁+m₂)c² =< H₊₊ =< 0. –(m₂-m₁) =< m < 0 => v =< c; 0 < m =< (m₂-m₁) => v >= c.

 

TWO SUPERLUMINIC MATERIAL BODIES COLLIDING WOULD BE LIKELY TO PRODUCE A SUBLUMINIC MATERIAL BODY.

 

I will pursue later on.

Tedious, isn’t it ?