**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 R^{4} to A4, connected to a boost B**v**
of velocity **v** in the following manner

Let B

**v**be a boost (see A1-2) and let A**0**be an Aristotle frame.Let Z = (T,

**R**) e A4 be an event.Let Z0 = B

**v**(Z) = (T0,**R0**) be its image by the boost B**v**.Let

__z__= (t,x,y,z) be the coordinates of Z in frame A**0**, e.g. Z = A**0**(__z__)Let

__z0__= (t0,x0,y0,z0) be that of Z0 =A**0**(__z0__) in frame A**0**so that

Z = (T0(t),**R0**(x, y, z)) and Z0 = (T0(t0),**R0**(x0, y0, z0)) = B(Z)

The space-time coordinates __z__ of Z in A**0** are
called space-time coordinates of Z0 in the inertial frame A**v** moving at velocity **v**
in Aristotle space-time[1].

The application b**v** : IR^{4}
®
IR^{4}

__z__ = (t, x, y,z)
®
__z0__ = (t0, x0, y0, z0)

is the so-called expression of boost B**v** in Aristotle
frame A**0**.

Application A**v** :
IR^{4}
®
A4

__z__ = (t, x, y,
z) ®
B**v**(Z) = Z0 = (T0(t0),**R0**(x0,y0,z0))

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

A**v**(t,**r**) = B**v**(T0(t),**R0**(**r**))
e.g. A**v**(t,**r**)= B**v** o A**0**(t,**r**)

Of course, if F belongs to the restricted Aristotle group
SE(1)xSE(3), then FoA**v** is an other inertial frame moving at velocity **v**. So,
a moving inertial frame A**v** is a coordinate system which localizes event Z0 thanks
to coordinates __z__ = (t,x,y,z) where (t,x,y,z) = A**v**^{-1}(Z0) = A**0**^{-1}(B**v**^{-1}(Z0)),
e.g. before boost B**v** has been applied to event Z = A**0** (t,x,y,z)

A**v** is said to have same space-time origin than A**0**
if

A**v**(0,**0**) = A**0**(0,**0**), e.g. B**v**(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 B**v**, the spatial origin E^{1}x{**R0**(**0**)}
of A**0** moves at constant velocity **v** along a straight line,

e.g. $ **K** e E^{3},
such that " t e IR, $ ! t0 e IR such that

A**v**(t,**0**) = B**v** (T0(t),**R0**(**0**)) = (T0(t0), **v** t0 + **K**)
where **v** t0 + **K**

is the spatial location of the spatial origin of A**v** 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. dT^{2} = dt^{2} and dL^{2}
= dx^{2} + dy^{2} + dz^{2} (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 A**v** (steadily moving at velocity **v**)
to a motionless frame A**0** is the same than switching from a motionless frame A**0**
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 B**v** which satisfy the additional mathematical
properties

- shear boosts B
**v**are linear (e.g. space-time origin (0,**0**) of A4 is unchanged under the action of shear boosts). Hence, no space-time translation is embedded in a shear boost. - " Z
**v**® B**v**(Z) is continuous and B**0**= identity of A4 so that no spatial rotation is embedded in a shear boost. - They don't reverse the time arrow (no embedded time reversal)
- They don't reverse the parity (no embedded space reversal)

Consequently, the combinations FoB**v** of shear boosts
with a transformation F belonging to Aristotle group are shear boosts if and only if F = i_{A4}.

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

Indeed, A**0**^{-1}(0,**0**)= A**v**^{-1}(0,**0**)
and A**v** is steadily moving at velocity **v** with regard to A**0** along **x0**
direction

- condition
**5/**reads B**v**^{-1}= B-**v**

**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 B**v** 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 B**v** and
its image by F are considered as pairs (Z, **v**) and (F(Z), F(**v**)) where Z is an
event such that B**v**(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(B**v**) in the case when F is an Aristotle group action. The
assumed covariance of the effects of a boost B**v** with regard to Aristotle group
reads then
B**v** = F^{-1} . F(B**v**) . F

**Translation TDZ of
a boost Bv**

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

Hence, thanks to the linearity of shear boosts, when a shear
boost B**v** is involved (e.g. when Zc = (0,**0**) ) we have

[TDZ(B**v**)](Z) = DZ + B**v**(Z-DZ)
e.g. [TDZ(B**v**)](Z) = [TDZ.B**v**.TD
Z^{-1}](Z) so that B**v** = TDZ^{-1}.TDZ(B**v**).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(B**v**) of a shear boost B**v** is defined as F(B**v**)=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 p**x** around **x** axis on the matrix of the
linear transformation b**v**.

As R p**x** (**v**)
= **v** we have b**v** = R-p**x**
b**v** R p**x**

where R p**x**
matrix =

R p**x** right
action reverses the sign of y and z columns of b**v** and R-p**x** left action reverses the sign of y and z lines
of b**v**. Hence above equality proves that y and z off-diagonal terms of b**v**,
but the yz terms, vanish.

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

b**v** = R-p**x**/2
b**v** R p**x**/2
where R p**x**/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 P**y** = P R p**y**
the spatial symmetry with regard to the **xz** plane.

As P**y**(**v**)
= **v** and P**y**^{ -1} = P**y** we have b**v** = P**y** b**v** P**y**

where P**y**
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 :

B

**v**^{-1}= B-**v**, e.g. the observer moving at velocity**v**believes he is at rest and that's the "railway station" which is moving at velocity -**v**[3].The maximum Lorentz invariant wave propagation speed is covariant with regard to Aristotle group action, hence this maximum value is isotropic and has same norm c in A

**0**as in A**v**.

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 b**v**^{-1} = b-**v**
where b-**v** = P**x** . b**v** . P**x**,
hence

and e^{-1}= e e.g. e^{2}
= 1, so that

aa' - bb' = 1, a = a' moreover e = 1 (indeed, b**v** is assumed to tend
towards the identity of IR^{4} when **v** tends towards **0**).

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

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 a^{2} - b'^{2}
= 1

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

a = (1-v^{2}/c^{2})^{-1/2}.

**Finally we get the classical Lorentz transformations**

ct0 = (ct + v x/c)/(1-v^{2}/c^{2})^{1/2}

x0 = (vt + x)/(1-v^{2}/c^{2})^{1/2}

y0 = y and z0 = z