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Date: Fri, 9 Jun 2023 03:37:00 -0700 (PDT)
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Subject: =?UTF-8?Q?Re=3A_De_la_relativit=C3=A9_des_distances?=
From: Richard Verret <rverret97@gmail.com>
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Le vendredi 9 juin 2023 =C3=A0 11:57:27 UTC+2, Richard Hachel a =C3=A9crit=
=C2=A0:
> Si tu pars sur le principe qu'il n'y a pas d'int=C3=A9r=C3=AAt =C3=A0 fai=
re des=20
> transformations entre syst=C3=A8mes de coordonn=C3=A9es en relativit=C3=
=A9 restreinte,=20
> je vois pas ce qu'il te reste =C3=A0 traiter.
Salut Richard ! Quand on prend des =C3=A9v=C3=A9nements simultan=C3=A9s dan=
s un r=C3=A9f=C3=A9rentiel,  on obtient la transformation d=E2=80=99une fig=
ure d=E2=80=99un r=C3=A9f=C3=A9rentiel =C3=A0 un autre, ce qui est le propr=
e d=E2=80=99une transformation ponctuelle. La transformation de Lorentz du =
r=C3=A9f=C3=A9rentiel R =C3=A0 R=E2=80=99 donne pour un segment L=E2=80=99 =
=3D k L; k =C3=A9tant l=E2=80=99inverse du coefficient de Lorentz. Lors du =
passage de R=E2=80=99 a R, on obtient L =3D k L=E2=80=99;=20
d=E2=80=99o=C3=B9 L =3D k^2 L. Cherchez l=E2=80=99erreur!=20
Ce que j=E2=80=99ai donc trait=C3=A9 c=E2=80=99est une transformation T qui=
 donne T o T^-1 (L) =3D L, c=E2=80=99est =C3=A0 dire telle que L =3D k^2 L,=
 soit k=3D 1. Et cette transformation n=E2=80=99est pas celle de Lorentz.