Derivation of Lorentz transformations in Aristotle space-time A4 (6/07/04)

(Appendix A1 of Proposition d'un test de transmission instantanée d'information par effet EPR )

A1-1 Mathematical definition of inertial frames

Boosts are diffeomorphisms from A4 to A4 that will actually happen to be affine transforms owing to the physical requirement that free particles propagate at constant speed along straight lines. The mathematical properties that boosts are required to satisfy will be stated precisely in paragraph A1-2. Now, a so called inertial frame is a diffeomorphim from R4 to A4, connected to a boost Bv of velocity v in the following manner

The space-time coordinates z of Z in A0 are called space-time coordinates of Z0 in the inertial frame Av moving at velocity v in Aristotle space-time[1].

The application bv :    IR4                         ®                   IR4

                                   z = (t, x, y,z)          ®      z0 = (t0, x0, y0, z0)

is the so-called expression of boost Bv in Aristotle frame A0.

Application Av :         IR4                         ®                  A4

                                  z = (t, x, y, z)          ®    Bv(Z) = Z0 = (T0(t0),R0(x0,y0,z0))

is a so-called inertial frame moving at velocity v.

Av(t,r) = Bv(T0(t),R0(r))     e.g.     Av(t,r)= Bv o A0(t,r)

Of course, if F belongs to the restricted Aristotle group SE(1)xSE(3), then FoAv is an other inertial frame moving at velocity v. So, a moving inertial frame Av is a coordinate system which localizes event Z0 thanks to coordinates z = (t,x,y,z) where (t,x,y,z) = Av-1(Z0) = A0-1(Bv-1(Z0)), e.g. before boost Bv has been applied to event Z = A0 (t,x,y,z)

Av is said to have same space-time origin than A0 if

Av(0,0) = A0(0,0), e.g.  Bv(T0(0),R0(0)) = (T0(0),R0(0))

A1-2 Mathematical properties expected from boosts

The expected physical properties of boosts read mathematically as follow

1/ When boosted by Bv, the spatial origin E1x{R0(0)} of A0 moves at constant velocity v along a straight line,
e.g. $ K e E3, such that " t e IR, $ ! t0 e IR such that
Av(t,0) = Bv (T0(t),R0(0)) = (T0(t0), v t0 + K)     where v t0 + K
is the spatial location of the spatial origin of Av at time T0(t0)

2/ Boosts are affine transformations. This is a necessary and sufficient condition for a diffeomorphism of affine space-time to transform straight lines into straight lines [2].

3/ Boosts give rise to a comotive spatial metric and a comotive temporal metric satisfying covariance requirement of proper lengths and durations, e.g.  dT2 = dt2    and   dL2 = dx2 + dy2 + dz2 (conservation of the proper length of steadily moving bodies and the proper duration of cyclic phenomena).

4/ The physical effects of a boost are covariant with regard to the complete Aristotle group actions (they don't depend on motionless observer space-time location and spatial orientation)

5/ Steadily moving observers using only phenomena satisfying all the relativist symmetries aren't up to detect their absolute velocity v. Hence, switching from an inertial frame Av (steadily moving at velocity v) to a motionless frame A0 is the same than switching from a motionless frame A0 to an inertial frame A-v moving at velocity -v.

Now, we will state the mathematical expression of shear boosts, that's to say boosts Bv which satisfy the additional mathematical properties

Consequently, the combinations FoBv of shear boosts with a transformation F belonging to Aristotle group are shear boosts if and only if F = iA4.

We will achieve the calculations of expression bv of Bv in a frame A0 which space-time origin is located at Z0=(0,0) (e.g. such that A0(0,0)=(0,0)) and axis x0 (defined by (0,x0)= A0(0,1,0,0)) has same direction than velocity v. Then, in A0, condition 1/ reads " t e IR, $ ! t0 e IR such that bv (t,0) = (t0,vt0,0,0)

Indeed, A0-1(0,0)= Av-1(0,0) and Av is steadily moving at velocity v with regard to A0 along x0 direction

 

A1-3 Significance of the covariance of physical effects of a boost on a body with regard to the Aristotle group actions in Aristotle space-time

The effects of a boost Bv on a physical system S are supposed to be covariant with regard to actions F of the complete Aristotle group

In the above illustration, the observed boost Bv and its image by F are considered as pairs (Z, v) and (F(Z), F(v)) where Z is an event such that Bv(Z) = Z (the so called space-time origin of the boost). When translated, rotated and possibly submitted to P or T symmetry with whatever he is surrounded by, the motionless Aristotle observer of the effects of a boost isn't aware of the Aristotle group action he undergoes together with the boosted system he is observing. Now, let us define F(Bv) in the case when F is an Aristotle group action. The assumed covariance of the effects of a boost Bv with regard to Aristotle group reads then                               Bv = F-1 . F(Bv) . F

 

Translation TDZ of a boost Bv

Let Zc be the fix event of Bv, e.g. Bv(Zc) = Zc. By definition, the translation TDZ(Bv) of a boost Bv will be the composition of Bv with a translation (following Bv action) which will translate the origin event Zc of Bv up to event Zc+DZ, e.g.  [TDZ(Bv)](DZ+Zc) = DZ+Zc , so that [TDZ(Bv)](Z) = DZ+Zc + Bv(Z)-Bv(DZ+Zc)

Hence, thanks to the linearity of shear boosts, when a shear boost Bv is involved (e.g. when Zc = (0,0) ) we have
[TDZ(Bv)](Z) = DZ + Bv(Z-DZ)   e.g.  [TDZ(Bv)](Z) = [TDZ.Bv.TD Z-1](Z)  so that Bv = TDZ-1.TDZ(Bv).TDZ

This proves the covariance of boosts effects with regard to space-time translations TDZ, e.g. motionless Aristotle observers observe the same boosts effects whatever the moment and the location where they are looking a boosted system. This covariance, with regard to space-time translations, stems from the linearity of boosts (e.g. boosted free particles are free particles).

 

Spatial rotation and P symmetry of a shear boost Bv

Now, if the considered action F of Aristotle group E(1)xE(3) is a rotation or a P symmetry (so that it doesn't change the shear boost origin Zc = (0,0)), the image F(Bv) of a shear boost Bv is defined as F(Bv)=BF(v)

 

A1-4 Covariance of boosts effects with regard to spatial rotations and P symmetry proves that off-diagonal terms of bv vanish (but the t x ones)

A1-4-1 Let us consider the action of a 180° rotation R px around x axis on the matrix of the linear transformation bv.

As R px (v) = v we have     bv = R-px  bv  R px

where R px matrix =

R px right action reverses the sign of y and z columns of bv and R-px left action reverses the sign of y and z lines of bv. Hence above equality proves that y and z off-diagonal terms of bv, but the yz terms, vanish.

 

A1-4-2 Now, let us consider the action of a 90° rotation R px/2 around x axis

bv = R-px/2  bv  R px/2            where R px/2 matrix =

We get the equality of the y and z diagonal terms and yz term = -zy term

 

A1-4-3 Now, to prove that the yz term vanish, we consider Py = P R py the spatial symmetry with regard to the xz plane.

As Py(v) = v and Py -1 = Py we have bv = Py bv Py

where Py matrix = . Consequently, y off-diagonal terms vanish.

Thanks to the assumed covariance of boost effects with regard to rotations and P symmetry, we have proven that yz and zy terms vanish, e.g. the physical effects of a boost don't cause any rotation around the direction of boost's velocity v.
Hence, $ a, a', b', b'' and e e IR such that

      t0    =  a t + a' x
      x0   =  b' t + b'' x
      y0   =  e y    z0 = e z

 

A1-5 Boost-symmetry

To end the settlement of Lorentz transforms, we write :

For convenience we will introduce this maximum speed c in the A1-4 equations, so that $ a, a', b, b' and e e IR such that

        ct0 = a (ct) + b x
        x0  = b' (ct) + a' x
        y0  = e y    z0 = e z

Now bv-1 = b-v            where b-v = Px . bv . Px, hence

and     e-1= e     e.g.    e2 = 1, so that

aa' - bb' = 1, a = a'   moreover e = 1 (indeed, bv is assumed to tend towards the identity of IR4 when v tends towards 0).

As the origin of Av located at x = y = z = 0 moves at velocity v = v x0 in A0,

x0 = vt0 when x = y = z = 0, so that x0 = b'(ct) = vt0       and ct0 = a (ct),

hence   a (v t0)  = a b'(ct) = b'(ct0)    so that b' = av/c

 

Now, let us express the covariance of the relative speed c of light

x0 = c t0 ̃ x = c t  so that b' (ct) + a' x  = a (ct) + b x   ̃  b' + a' = a + b

Now, as a = a' we get b = b' hence aa' - bb' = 1 provides a2 - b'2 = 1

e.g. a2 - (av/c)2 = 1 , now as a>0 (because we have excluded time reversal)

a = (1-v2/c2)-1/2.

 

Finally we get the classical Lorentz transformations

     ct0  = (ct + v x/c)/(1-v2/c2)1/2

     x0   = (vt + x)/(1-v2/c2)1/2

     y0  = y     and      z0 = z