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 ?

 

In the original version, the problem of the spontaneous decay of a point-like body (particle) of mass m (>=0) into two fragments of masses m₁ (>= 0) and m₂ (>= 0) can be completely solved making use of the conservation of both energy and 3-momentum. With :

 

E = mc²/(1-v²/c²)^1/2  ,   E₁ = m₁c²/(1-v₁²/c²)^1/2  , E₂ = m₂c²/(1-v₂²/c²)^1/2

 

the energies of the decaying particle and the fragments, respectively, v the velocity of the frame into which the observer places with respect to the decaying particle and p, p₁ and p₂ the associated 3-momenta, the conservation of energy E = E₁ + E₂ leads to :

 

m/(1-v²/c²)^1/2 = m₁/(1-v₁²/c²)^1/2 + m₂/(1-v₂²/c²)^1/2

 

while the conservation of 3-momentum leads to p = p₁ + p₂. As the Lorentz correction factors are always >= 1, the original theory forecasts m > m₁ + m₂ in the frame at rest (v = 0) for the particle to decay. If m =< m₁ + m₂, the particle is assumed to be stable. Relations between energies and 3-momenta completely solve for E₁ and E₂ in this particular case. Notice however two things:

1. the mass condition for spontaneous decay on m has nothing absolute at all, on the contrary, it is typically relative to the frame into which the observer places. m > m₁ + m₂ is only valid for v = 0. For 0 < v < c, (1 – v²/c²)^1/2 induces a contraction; for v = c, 0 = < v₁, v₂ < c, the mass condition would become m = 0, while still for v = c but v₁ = v₂ = c, it would become m = m₁ + m₂, meaning that the reaction is not at all seen the same way by different observers. Nothing universal.
2. In the original version, masses are always assumed to be non-negative, an additional restriction to physics, while m = 0 for v = c for the mass at rest is only an argument to avoid infinities. Actually, the mass at rest m being completely independent from the velocity v, there is no solid reason for having m = 0 when v = c.

 

Going from curves to surfaces, the problem enounces differently. From formulas (10) bidouille 23, the conservation of energy now writes:

 

1. H₊₊ = H₁₊₊ + H₂₊₊  and  H_-- = H_1-- + H_2—

 

leading to :

 

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

 

and :

 

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

 

In the 2-fragments case, the can completely solve for v₁ and v₂ only using these two equations. Inverting (2) and (3) gives :

 

4. (1–v₂/c)/(1+v₂/c) = {[m(1-v/c) – m₁(1+v/c)] + [m(1-v/c) + m₁(1+v/c]v₁/c}/m₂(1+v/c)(1+v₁/c)
5. (1+v₂/c)/(1-v₂/c) = {[m(1+v/c) – m₁(1-v/c)] - [m(1+v/c) + m₁(1-v/c]v₁/c}/m₂(1-v/c)(1-v₁/c)

 

Multiplying them ad organizing the terms gives the quadratic equation for v₁/c :

 

6. [(m²+m₁²-m₂²)(1-v²/c²) + 2mm₁(1+v²/c²)]v₁²/c² - 8mm₁vv₁/c² - (m²+m₁²-m₂²)(1-v²/c²) + 2mm₁(1+v²/c²) = 0

 

The discriminant of this equation is :

 

7. D₁ = 4(m+m₁+m₂)(m+m₁-m₂)(m-m₁+m₂)(m-m₁-m₂)(1-v²/c²)² = 4Q₄(m)(1-v²/c²)²

 

One sees it does NOT depend on any restriction on the velocity v, since (1-v²/c²)² is always positive or zero, it depends on the sign of the mass quartic Q₄(m). We need to find at least one real-valued root of (6), two at the best. So, the mass condition becomes:

 

8. Q₄(m) >= 0

 

For m >= m₁ + m₂ >= m₂ >= m₁ >= 0, it is automatically satisfied. But this is now just ONE possibility among others. Let us evaluate all possible situations, assuming m₂ >= m₁. Then we will discuss the solutions.

 

Case 1 : m₂ >= m₁ >= 0. Ordered roots of (8): -(m₁+m₂), -(m₂–m₁), (m₂–m₁), (m₁+m₂).

Q₄(m) >= 0 for m =< -(m₁+m₂), -(m₂–m₁) =< m =< (m₂–m₁) and m >= m₁+m₂.

m >= m₁+m₂: confirmed;

m =< -(m₁+m₂) : a negative mass (antimatter) would be likely to spontaneously decay into 2 fragments of matter (positive masses);

-(m₂–m₁) =< m =< (m₂–m₁) : passes through m = 0, so a massless matter (or antimatter) particle would be likely to decay into 2 fragments of matter.

 

 

Case 2 : m₂ >= 0 >= m₁. Ordered roots of (8): -(m₂-m₁), -(m₂+m₁), (m₂+m₁), (m₂-m₁).

Q₄(m) >= 0 for m =< -(m₂-m₁), -(m₂+m₁) =< m =< (m₂+m₁) and m >= m₂-m₁.

m =< -(m₂-m₁) : antiparticle likely to decay into antiparticle + particle;

-(m₂+m₁) =< m =< (m₂+m₁) : passes through m = 0, massless (anti)particle likely to decay into particle + antiparticle;

m >= m₂-m₁: matter particle likely to decay into antimatter and matter.

 

Case 3 : 0 >= m₂ >= m₁. Ordered roots of (8): (m₁+m₂), -(m₂–m₁), (m₂–m₁), -(m₁+m₂).

Q₄(m) >= 0 for m =< (m₁+m₂), -(m₂–m₁) =< m =< (m₂–m₁) and m >= -(m₁+m₂).

m =< (m₁+m₂): antimatter decaying into 2 antiparticles, with –m >= -(m₁+m₂) >= 0, confirmed;

-(m₂–m₁) =< m =< (m₂–m₁): passes through m = 0, massless (anti)particle likely to decay into 2 antiparticles;

m >= -(m₁+m₂) >= 0: matter to decay into 2 antiparticles.

 

This sounds pretty exotic. Should some processes not be observed, other conservation laws should be invoked. Here, we are only dealing with the conservation of energy. Notice that all these mass conditions are INDEPENDENT OF THE CHOICE OF FRAME. Condition (8) is now universal.

Consider from now on that (8) is satisfied and let us examine the solutions for velocities. Changing m₁ with m₂ leads to a similar equation for v₂, namely:

 

9. [(m²+m₂²-m₁²)(1-v²/c²) + 2mm₂(1+v²/c²)]v₂²/c² - 8mm₂vv₁/c² - (m²+m₂²-m₁²)(1-v²/c²) + 2mm₂(1+v²/c²) = 0

 

Solutions of (6) are:

 

10. v₁/c = [4mm₁v/c +/- Q₄^1/2(m)(1-v²/c²)]/[(m+m₁+m₂)(m+m₁-m₂) – (m-m₁+m₂)(m-m₁-m₂)v²/c²]

 

So, solutions of (9) are straightforward:

 

11. v₂/c = [4mm₂v/c +/- Q₄^1/2(m)(1-v²/c²)]/[(m+m₁+m₂)(m-m₁+m₂) – (m+m₁-m₂)(m-m₁-m₂)v²/c²]

 

The presence of both signs (+/-) is necessary, as the only requirements are v₁ >= 0 and v₂ >=0, since they are moduli. So, according to the choice on v, we will keep the + or the – sign.

Already notice this:

 

12. v = c  =>  v₁ = v₂ = c  WHATEVER THE MASSES.

 

One can see it better on the following equivalent expressions for v₁ and v₂ :

 

v₁/c = [4mm₁v/c +/- Q₄^1/2(m)(1-v²/c²)]/[(m²+m₁²-m₂²)(1-v²/c²) + 2mm₁(1+v²/c²)]

v₂/c = [4mm₂v/c +/- Q₄^1/2(m)(1-v²/c²)]/[(m²+m₂²-m₁²)(1-v²/c²) + 2mm₂(1+v²/c²)]

 

Imagine a hypothetical observer moving at the speed of light with respect to the decaying mass, than he will observe both fragments ejected at the speed of light. The whole process will look to him as moving at the speed of light. A confirmation of the universality of c.

This alone already smells nice.

 

13. m = m₁+m₂  =>  v₁ = v₂ = v.

 

And this is logical, for we can always get back to the frame at rest, all velocities being constant. In the frame at rest, v₁ = v₂ = v = 0 and the whole process would appear as NON-EXISTING to the observer’s eye: everything steady => nothing happens. No decay.

 

14. m = -(m₁+m₂)  =>  v₁ = v₂ = c²/v.

 

Thus, for an observer in a subluminic frame (0 =< v =< c^-), both fragments will have superluminic speeds. Can we handle that ? Assume, for instance, m₂ >= m₁ >= 0. An antiparticle, seen as being subluminic, would then decay into two particles, seens as being superluminic. In the frame at rest (v = 0), the process is even seen to be instantaneous.

Can we handle that, honestly ? I don’t know. It’s allowed, that’s all I can say.

A result that now confirms my feelings :

 

15. m = m₁ – m₂  =>  v₁ = v  ,  v₂ = c²/v

 

Consequently, if v is subluminic, fragment 1 will be seen as subluminic, while fragment 2 as superluminic. There is no contradiction with logic: remember both H₊₊ and H_— become NEGATIVE at superluminic regime, making the energy balance of the process DECREASE and allowing, at least theoretically, the decay of a mass into a lighter fragment and a heavier one, as long as this last fragment is ejected at superluminic speed, compensating for in energy. Finally:

 

16. m = m₂ – m₁  =>  v₁ = c²/v  ,  v₂ = v

 

(simple permutation of m₁ and m₂).

The other possibilities are less interesting, as they lead to more complicated formulas. Thus, for v = 0, we get:

 

v₁/c = [(m-m₁+m₂)(m-m₁-m₂)/(m+m₁+m₂)(m+m₁-m₂)]^1/2

v₂/c = [(m+m₁-m₂)(m-m₁-m₂)/(m+m₁+m₂)(m-m₁+m₂)]^1/2

 

Here, it is clear the (+) sign must be retained in (10) and (11). More interesting are the following properties in the frame at rest :

 

17. v₁v₂/c² = |(m-m₁-m₂)/(m+m₁+m₂)|
18. v₁/v₂ = |(m-m₁+m₂)/(m+m₁-m₂)|

 

which immediately lead to v₁ =< c and v₂ =< c for m >= m₁+m₂ >= m₂ >= m₁ >= 0.

 

NOWHERE do we now see any restriction to be imposed on the velocities. The only restriction to be applied is (8) and it merely involves masses. There no longer is an “allowed area” and a “forbidden one”. No dichotomic cutting-off whatsoever. The process might look very different in very different frames.

 

To end with the subject, notice that in an instantaneous frame (v = +oo), there still remains finite solutions for v₁ and v₂, as (10) and (11) give:

 

19. (v₁/c)_v=+oo = +Q₄^1/2(m)/(m-m₁+m₂)(m-m₁-m₂) = [(m+m₁+m₂)(m+m₁-m₂)/(m-m₁+m₂)(m-m₁-m₂)]^1/2 = (c/v₁)_v=0
20. (v₂/c)_v=+oo = +Q₄^1/2(m)/(m+m₁-m₂)(m-m₁-m₂) = [(m+m₁+m₂)(m-m₁+m₂)/(m+m₁-m₂)(m-m₁-m₂)]^1/2 = (c/v₂)_v=0

 

THE VALUES OF v₁ AND v₂ IN THE FRAME AT INFINITY (v = +oo) ARE THE INVERSES (OR RECIPROCAL TO) OF THEIR VALUES IN THE FRAME AT REST (v = 0).

 

This again sounds logical.

 

 

Gros cafouillage la semaine dernière : j’ai manifestement eu un gros coup de pompe… Tout n’est pas à rejeter dans la bidouille 23, mais les choses nécessitent approfondissement, d’une part, et amélioration d’autre part. Je refais donc cette bidouille 23 pour la deuxième fois, mais ce n’est pas grave, et j’y apporte les compléments nécessaires, notamment le passage à un cadre mieux adapté.

On repart de la formule (1), bidouille 22 et on fait apparaître explicitement la dépendance fonctionnelle (nous en aurons besoin par la suite) :

 

1. ds²(t) = c²dt² - dx²(t) = [1 – v²(t)/c²]c²dt² = 2c²dt⁺(t)dt^-(t)

 

Dans cette représentation-là, l’élément de surface ds² est fonction du temps t, de même que dt⁺ et dt^-, qu’on pourrait qualifier « d’instants avancé et retardé », respectivement :

 

2. dt⁺(t) = [1 + v(t)/c]dt  ,  dt^-(t) = [1 – v(t)/c]dt  ,  v(t) = |v(t)|

 

Il est alors évident que les dérivées partielles ð₊(t) = ð/ðt⁺(t) et ð_-(t) = ð/ðt^-(t) étant des dérivées fonctionnelles n’ont aucune raison de commuter, sauf dans le cas v = cte, i.e. le mouvement 3D libre dans l’espace. Cette non-commutativité n’est due qu’au fait que nous nous plaçons dans une représentation fonctionnelle t⁺(t) et t^-(t). Cette représentation nous a été utile pour retrouver les équations du mouvement dans l’espace-temps, pour autant ce n’est plus la représentation naturelle. Car ces équations de la dynamique, on le sait, s’unifient en la paramétrisation en courbe xⁱ(s/c). Et nous proposons d’étendre cette représentation en une représentation en surface xⁱ(t⁺,t^-). Dans cette dernière, les paramètres temporels t⁺ et t^- deviennent indépendants et la dépendance fonctionnelle porte désormais sur les xⁱ, c’est-à-dire sur x et t, qui deviennent des fonctions de t⁺ et de t^-. Conséquence : les dérivées partielles ð₊ = ð/ðt⁺ et ð_- = ð/ðt^- ne sont plus fonctionnelles et n’ont, de ce fait, plus aucune raison de ne pas commuter. Ainsi, ð₊ð_- = ð_-ð₊ dans cette représentation et on peut dire que c’est désormais le 4-gradient ð_i(t⁺,t^-) = ð/ðxⁱ(t⁺,t^-) qui y devient fonctionnel, de sorte que ce sont les opérateurs ð_i(t⁺,t^-) et ð_j(t⁺,t^-) qui ne commutent plus pour i différent de j. Mais cela n’a aucune importance, car nous sommes passés à une description en champ à 2 dimensions externes, temporelles, t⁺ et t^- et 4 dimensions internes, spatio-temporelles, xⁱ (i = 0,1,2,3).

Il n’y a plus qu’à appliquer les résultats établis sur la dynamique des champs. D’abord, le lagrangien (20) bidouille 22 :

 

3. L[xⁱ(t⁺,t^-),vⁱ_a(t⁺,t^-),t⁺,t^-] = ½ mvⁱ_a(t⁺,t^-)v_i^a(t⁺,t^-) – 2W[xⁱ(t⁺,t^-),t⁺,t^-]

 

s’exprime en Joules et non pas en J/s² : ce n’est pas une densité temporelle ! Introduisons les dérivées totales :

 

4. d/dt^a = d_a = ð_a + vⁱ_a(t⁺,t^-)ð_i(t⁺,t^-) + d_avⁱ_b(t⁺,t^-)ð/ðvⁱ_b(t⁺,t^-) 

 

avec ð_i(t⁺,t^-) = ð/ðxⁱ(t⁺,t^-) et calculons d_aL :

 

d_aL = ð_aL + vⁱ_að_iL + d_avⁱ_bðL/ðvⁱ_b = ð_aL + vⁱ_ad_bðL/ðvⁱ_b + d_avⁱ_bðL/ðvⁱ_b

 

en vertu des équations de Lagrange (27) bidouille 22. Pour toutes les quantités ne dépendant que des t^a, comme les 4-vitesses vⁱ_a et les 4-impulsions pⁱ_a, d_a coïncide avec ð_a. A présent que [ð_a,ð_b] = 0, d_avⁱ_b = d_ad_bxⁱ = d_bd_axⁱ = d_bvⁱ_a (c’est la condition d’intégrabilité retrouvée). Il en résulte que :

 

d_aL = ð_aL + vⁱ_ad_bðL/ðvⁱ_b + (d_bvⁱ_a)ðL/ðvⁱ_b = ð_aL + d_b(vⁱ_aðL/ðvⁱ_b)

 

Finalement :

 

5. d_b(vⁱ_aðL/ðvⁱ_b – Ld_a^b) = d_bH_a^b = -ð_aL

 

comme on s’y attendait. La matrice hamiltonienne est :

 

6. H_ab = vⁱ_aðL/ðv^ib – Lh_ab = vⁱ_ap_ib – Lh_ab = pⁱ_ap_ib/m – (pⁱ_cp_i^c/2m – 2W)h_ab

 

En composantes :

 

7. H₊₊ = pⁱ₊p_i+/m  ,  H_-- = pⁱ_-p_i-/m  ,  H_+- = H_-+ = 2W

 

Cette matrice possède l’invariant :

 

8.  H = H_a^a = 4W

 

H_ab s’exprimant en Joules, l’intégrale de surface II d_bH_a^bdt⁺dt^- = I H_a^bdt_b s’exprime en Js. Il s’agit donc d’une action. Pour autant, les équations de conservation locales :

 

9.  d_bH_a^b = 0  <=>  ð_aL = 0

 

continuent bien d’exprimer la conservation de l’énergie, comme on aurait été en droit de s’attendre. Quant aux intégrales de surface II Ldt⁺dt^- et II H_abdt^adt^b, elles s’expriment en Js² et représentent donc des inerties. Noter les résultats intéressants suivants :

 

10. H₊₊ = mc²(1-v/c)/(1+v/c) = mc²E₊/E_-  ,  H_-- = m²c⁴/H₊₊ = mc²(1+v/c)/(1-v/c)

 

Avec E₊ = mc²/(1+v/c) et E_- = mc²/(1-v/c), de sorte que seules 2 des 3 composantes de H_ab sont indépendantes : H₊₊ et H_+- = H_-+. Encore plus intéressant :

 

11. det(H_ab) = H₊₊H_-- - (H_+-)² = m²c⁴ – 4W² = (mc² + 2W)(mc² - 2W) = HL

 

où H = mc² + 2W est l’hamiltonien scalaire associé habituellement à L :

 

LE DETERMINANT DE LA MATRICE HAMILTONIENNE EST EGAL AU PRODUIT DU LAGRANGIEN PAR L’HAMILTONIEN SCALAIRE ASSOCIé.

 

On a aussi le moment cinétique :

 

12.  M_abc = t_aH_bc – t_bH_ac = -M_bac  en Js

 

Je vais maintenant dire un mot sur la notion de référentiel de repos.

 

Pour un même corps, ce référentiel n’est plus unique, puisqu’il y a maintenant deux dimensions de temps et, par suite, deux vitesses, vⁱ₊ et vⁱ_-. Si l’on a vⁱ₊ = 0, c’est que xⁱ ne dépend pas de t⁺ : le corps est immobile dans la direction t⁺ (mais il peut se mouvoir dans la direction t^-). Inversement, si vⁱ_- = 0, xⁱ ne dépend pas de t^- : le corps est immobile dans la direction t^- (mais peut se mouvoir dans la direction t⁺). Et c’est seulement si vⁱ₊ = vⁱ_- = 0 qu’on peut parler d’immobilité totale, car xⁱ est alors constant.

Ce sont bien les quadri-vitesses qu’il faut prendre en compte. Retour, pour le comprendre, sur la paramétrisation t⁺(t), t^-(t). On avait vⁱ₊ = vⁱ/(1+v/c) et vⁱ_- = vⁱ/(1-v/c). Avec v = 0, je trouve bien v₊ = v_- = 0 : immobilité dans l’espace 3D. Pour les composantes i = 0 : v⁰₊ = c/(1+v/c) et v⁰_- = c/(1-v/c). Bon. Peut-on encore avoir zéro ? oui : pour v -> oo !!!

Résultat : on trouve maintenant deux référentiels de repos,

 

v = 0, immobilité spatiale : vⁱ₊ = vⁱ_- = vⁱ = (c,0), déplacement luminique le long de la direction temporelle ;

v = +oo, déplacement instantané dans l’espace : vⁱ₊ = -vⁱ_- = (0⁺,c) : immobilité temporelle.

 

Ça tombe un peu sous le sens, qu’on ne bouge plus dans le temps si l’on se déplace instantanément d’un endroit à un autre, puisqu’il n’y a alors plus de délai, ni retard, ni avance… On constate d’ailleurs qu’alors que v peut maintenant couvrir toutes les vitesses de 0 à +oo, les 4-vitesses vⁱ₊ et vⁱ_-, elles, restent comprises entre (c,0) et (0,c)… :)

 

Une autre possibilité consiste à passer des courbes aux surfaces. Non pas parce qu’on passe de la dimension 3 à la dimension 4, mais parce qu’on passe de la géométrie euclidienne à la géométrie pseudo-euclidienne. En géométrie euclidienne, une théorie basée sur des surfaces n’est qu’une « théorie du second ordre » : il n’y a pas de différences qualitatives entre « théorie du premier ordre », basée sur des courbes, et « théorie du second ordre », basée sur des surfaces. Ainsi, l’utilisation de surfaces en dynamique euclidienne est tout à fait superflue, si l’on peut se contenter de courbes. Par contre, elle devient indispensable en géométrie pseudo-euclidienne, parce que l’élément de surface ds² n’y est plus de signe défini. Seulement, ce n’est plus aussi simple que dans le cas euclidien. Comme toujours, les similitudes peuvent s’avérer trompeuses et masquer des difficultés plus profondes.

On commence donc par écrire toutes les décompositions canoniques possibles du ds² :

 

1. ds² = dxⁱdx_i = (1 – v²/c²)c²dt² = ds⁺ds^- = c²dt⁺dt^-

 

Seuls les éléments ds⁺, ds^- ou dt⁺, dt^- sont partout réels. Si les produits hyperboliques sont de même signe, le ds² est > 0 et 0 =< v < c ; si ces produits sont de signes opposés, le ds² est < 0 et v > c ; si dt^- = 0 avec dt non nul, le ds² est nul et v = c.

S’il s’agit de surfaces, on s’attend à une reparamétrisation xⁱ(s/c) -> xⁱ(t⁺,t^-) à deux paramètres réels. Ecrivons l’action correspondante, en différentielle :

 

2. dS² = m²c²ds² = dSⁱdS_i = dS⁺dS^- = L²dt⁺dt^- = L⁺L^-dt²

 

L’unique fonctionnelle de Lagrange L² est partout >= 0 : c’est le produit hyperbolique dt⁺dt^- qui porte le signe. De (1), on tire (à un signe global près) :

 

3. dSⁱ = mcdxⁱ  ,  dS⁺ = mc²dt⁺ = mc²(1 + v/c)dt  ,  dS^- = mc²dt^- = mc²(1 – v/c)dt
4. L² = m²c⁴  ,  L⁺ = mc²(1 + v/c)  ,  L^- = mc²(1 – v/c)

 

Dans toutes ces formules, v(t) = dx(t)/dt et, par suite, son module v(t) se réfèrent à des courbes. Ce n’est plus ce que l’on cherche. D’ailleurs, les fonctionnelles de Lagrange (4) y sont toutes linéaires en v, ce qui est inacceptable. Repartons de xⁱ(t⁺,t^-). Cette paramétrisation est aussi admissible que xⁱ(s/c) ou x(t), à la différence qu’elle se laisse étendre à toutes les valeurs de s². Par dérivation, on obtient une 4-vitesse à deux composantes :

 

5. vⁱ₊(t⁺,t^-) = ðxⁱ(t⁺,t^-)/ðt⁺ = [1 + v(t)/c]^-1ðxⁱ(t⁺,t^-)/ðt
6. vⁱ_-(t⁺,t^-) = ðxⁱ(t⁺,t^-)/ðt^- = [1 - v(t)/c]^-1ðxⁱ(t⁺,t^-)/ðt

 

v(t) est (le module de) la vitesse linéique du mobile dans l’espace, alors que vⁱ₊ et vⁱ_- sont des 4-vitesses surfaciques. Dans les membres de droite, t⁺ et t^- sont des fonctions de t, puisque par intégration :

 

7. t⁺(t) = I₀^t [1 + v(t’)/c]dt’  ,  t^-(t) = I₀^t [1 - v(t’)/c]dt’

 

A la dualité de Minkowski vient s’ajouter une dualité de métrique h_ab de composantes :

 

8. h₊₊ = h_-- = 0  ,  h_+- = h_-+ = 1

 

On vérifie aussitôt que :

 

9. ds² = g_ijdxⁱdx^j = ½ h_abds^ads^b = ½ c²h_abdt^adt^b

 

de sorte que h_abdt^adt^b est un invariant scalaire au même titre que le ds²/c². En dérivant une nouvelle fois, on trouve une matrice accélération :

 

10. aⁱ₊₊(t⁺,t^-) = ðvⁱ₊(t⁺,t^-)/ðt⁺  ,  aⁱ_+-(t⁺,t^-) = ðvⁱ₊(t⁺,t^-)/ðt^-
11. aⁱ_-+(t⁺,t^-) = ðvⁱ_-(t⁺,t^-)/ðt⁺  ,  aⁱ_--(t⁺,t^-) = ðvⁱ_-(t⁺,t^-)/ðt^-

 

En développant :

 

12. aⁱ₊₊(t⁺,t^-) = [1 + v(t)/c]^-2{ð²xⁱ(t⁺,t^-)/ðt² - [1 + v(t)/c]^-1[dv(t)/cdt]ðxⁱ(t⁺,t^-)/ðt}
13. aⁱ_+-(t⁺,t^-) = [1 – v²(t)/c²]^-1{ð²xⁱ(t⁺,t^-)/ðt² - [1 + v(t)/c]^-1[dv(t)/cdt]ðxⁱ(t⁺,t^-)/ðt}
14. aⁱ_-+(t⁺,t^-) = [1 – v²(t)/c²]^-1{ð²xⁱ(t⁺,t^-)/ðt² + [1 - v(t)/c]^-1[dv(t)/cdt]ðxⁱ(t⁺,t^-)/ðt}
15. aⁱ_--(t⁺,t^-) = [1 - v(t)/c]^-2{ð²xⁱ(t⁺,t^-)/ðt² + [1 - v(t)/c]^-1[dv(t)/cdt]ðxⁱ(t⁺,t^-)/ðt}

 

On vérifie sans difficulté que :

 

16. det(aⁱ_ab) = 0  pour chaque i = 0,1,2,3

 

qui indique que seules 3 des 4 composantes de cette matrice sont indépendantes ou encore que aⁱ₊₊aⁱ_-- = aⁱ_+-aⁱ_-+ (i = 0,1,2,3). Ce résultat montre que les matrices accélération aⁱ_ab ne sont pas inversibles. D’autre part, si le produit différentiel dt⁺dt^- est commutatif, il n’en est plus de même du produit des opérateurs dérivés ð₊ = ð/ðt⁺ et ð_- = ð/ðt^-. On a un commutateur :

 

[ð₊,ð_-] = ð²/ðt⁺ðt^- - ð²/ðt^-ðt⁺ = [1 + v(t)/c]^-1(d/dt){[1 – v(t)/c]^-1d/dt} - [1 - v(t)/c]^-1(d/dt){[1 + v(t)/c]^-1 d/dt}

           = {[1 + v(t)/c]^-1[1 – v(t)/c]^-2dv(t)/dt + [1 - v(t)/c]^-1[1 + v(t)/c]^-2dv(t)/dt}d/cdt

           = [1 – v²(t)/c²]^-1{1/[1 – v(t)/c] + 1/[1 + v(t)/c]}dv(t)/dt d/cdt

 

soit :

 

17. [ð₊,ð_-] = 2[1 – v²(t)/c²]^-2[dv(t)/dt](d/cdt)

 

La contravariante de h_ab est la matrice h^ab de même composantes. On a donc :

 

18. h⁺⁺ = h^-- = 0  ,  h^+- = h^-+ = 1  ,  h_abh^bc = d_a^c (delta de Kronecker)  ,  h_abh^ab = 2

 

Ces relations nous conduisent à :

 

19. vⁱ⁺ = vⁱ_-  ,  v^i- = vⁱ₊  ,  v_i⁺ = v_i-  ,  v_i^- = v_i+

 

Ces propriétés vont nous aider à écrire la fonctionnelle de Lagrange. La masse 4D est susceptible de varier suivant t⁺ et/ou t^- : m₄(t⁺,t^-), ce qui ne modifie en rien sa répartition spatio-temporelle. Considérons pour le moment une masse constante et prenons pour lagrangien :

 

20.  L[xⁱ(t⁺,t^-), vⁱ₊(t⁺,t^-), vⁱ_-(t⁺,t^-), t⁺,t^-] = ½ m₄vⁱ_a(t⁺,t^-)v_i^a(t⁺,t^-) – 2W[xⁱ(t⁺,t^-),t⁺,t^-]

 

Les 4-impulsions sont :

 

21.  pⁱ_a = ðL/ðv_i^a = h_abðL/ðv_ib

 

On vérifie aussitôt que :

 

22.  pⁱ₊p_i- = m₄c²

 

Dérivons maintenant ces 4-impulsions par rapport aux temps t⁺ et t^- :

 

23.  ðpⁱ₊/ðt₊ = ð⁺pⁱ₊ = ðpⁱ₊/ðt^- = ð_-pⁱ₊ = m₄aⁱ_+-
24.  ðpⁱ₊/ðt_- = ð^-pⁱ₊ = ðpⁱ₊/ðt⁺ = ð₊pⁱ₊ = m₄aⁱ₊₊
25.  ðpⁱ_-/ðt₊ = ð⁺pⁱ_- = ðpⁱ_-/ðt^- = ð_-pⁱ_- = m₄aⁱ_--
26.  ðpⁱ_-/ðt_- = ð^-pⁱ_- = ðpⁱ_-/ðt⁺ = ð₊pⁱ_- = m₄aⁱ_-+

 

Les seules équations de Lagrange qui soient invariantes à la fois sous g_ij et sous h_ab sont de la forme :

 

27.  ð^apⁱ_a = ð^aðL/ðv_i^a = ðL/ðxⁱ

 

Des relations (22) à (25) ci-dessus, on tire les équations de mouvement :

 

28.  m₄(aⁱ_+- + aⁱ_-+) = m₄[1 – v²(t)/c²]^-1{dvⁱ(t)/dt + [1 – v²(t)/c²]^-1vⁱ(t)v(t)dv(t)/c²dt]} = -2ðW/ðxⁱ = 2Fⁱ[x(t⁺,t^-),t⁺,t^-]

 

En vertu de (13) et (14) et puisque ðxⁱ(t⁺,t^-)/ðt = dvⁱ(t)/dt.

 

Comparons avec la relativité restreinte. On reprend le 4-vecteur vitesse Vⁱ(s/c) = vⁱ/(1 – v²/c²)^1/2 et le lagrangien L = ½ m₄VⁱV_i – W. Le 4-vecteur accélération Aⁱ = cdVⁱ/ds a pour expression :

 

Aⁱ(s/c) = [1 – v²(t)/c²]^-1{dvⁱ(t)/dt + [1 – v²(t)/c²]^-1vⁱ(t)v(t)dv(t)/c²dt]}

 

qui ressemble étrangement ½ (aⁱ_+- + aⁱ_-+)… Par un abus de langage sans aucune conséquence, la reparamétrisation admissible de xⁱ(t⁺,t^-) en xⁱ(s/c) nous fait passer de Fⁱ[x(t⁺,t^-),t⁺,t^-] à Fⁱ[x(s/c),s/c]. Les équations (28) coïncident donc avec les équations de mouvement habituelles m₄Aⁱ(s/c) = Fⁱ[x(s/c),s/c] à l’intérieur du cône de lumière.

La différence, ensuite, est flagrante : si la 4-accélération ne pose pas de problème de signe, en revanche, la 4-vitesse Vⁱ en a un sérieux, qui… restreint la dynamique à 0 =< v < c.

Aucune restriction de ce genre ne se retrouve plus dans vⁱ₊ ou vⁱ_-. Le mouvement peut donc être prolongé à l’extérieur du cône.

Où se trouve l’apparent paradoxe ?

Si l’on décrit le mouvement en termes de courbes xⁱ(s/c) = [ct,x(t)] dans l’espace-temps, on aboutit inévitablement à une frontière physique : les trajectoires de points matériels dans l’espace-temps ne peuvent sortir du cône de lumière, sous peine de sortir du domaine physique. C’est ce que l’on constate, parce qu’on raisonne en termes de trajectoires. Ce n’est donc pas faux.

Si l’on décrit le mouvement en termes de surfaces xⁱ(t⁺,t^-), toujours dans l’espace-temps, on ne trouve plus de frontière physique infranchissable, on trouve un lieu singulier, le cône de lumière, et encore, uniquement sur la composante vⁱ_- de la vitesse surfacique. A l’intérieur du cône, les deux mouvements coïncident, moyennant un changement de paramétrisation, puisque les équations de mouvement sont les mêmes. Par contre, (28) mais surtout vⁱ₊ et vⁱ_- se laissent prolonger sans aucune difficulté, ni technique, ni de principe, au-delà de c.

Sauf erreur de ma part :

 

CONTRAIREMENT A L’APPROCHE PAR TRAJECTOIRES, L’APPROCHE PAR SURFACES EST CAUSALE PARTOUT : ON NE TROUVE PLUS AUCUNE VIOLATION DE LA CAUSALITE NULLE PART. CELA SIGNIFIE QUE TOUT EFFET RENVOIE A UNE CAUSE, MAIS QUE CELLE-CI NE PRECEDE L’EFFET QU’A L’INTERIEUR DU CÔNE DE LUMIERE. A L’EXTERIEUR DU CÔNE, L’EFFET PRECEDE SA CAUSE PARCE QUE L’INFORMATION ARRIVE APRES LE PROCESSUS.

 

En effet, la cause de mon accélération est la résultante des forces extérieures appliquées : cette résultante est bien présente partout, sauf peut-être en des points singuliers isolés, pôles éventuels du champ de forces. L'effet est l'accélération. Si ma vitesse de déplacement devient > c, mon effet précède bien ma cause.

On interprète ça assez logiquement comme une violation de la causalité. C’est correct dans la représentation en trajectoires, ça ne l’est plus dans la représentation en surfaces. Le signal, qui véhicule l’information, se déplace toujours à c. Le processus tachyonique est plus rapide, l’information qu’il contient « reste donc en arrière ». C’est la même chose qui se produit, dans un contexte complètement différent, avec le régime supersonique. Ça ne devrait donc pas surprendre outre mesure. Dérouter, peut-être. Surprendre, non.

Ne pas perdre de vue qu’on dispose désormais de deux paramètres temporels. Ainsi, le fait de dire « à l’extérieur du cône – zone d’éloignement absolu – un événement peut se produire en même temps à deux endroits différents » (ubiquité) est une interprétation par courbes (un seul paramètre temporel). En surfaces, on peut évoluer suivant l’axe t⁺, l’axe t^- ou le plan (t⁺,t^-) : il ne devrait plus y avoir de situations « exotiques ».