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Newsgroups: fr.sci.physique
Date: Sun, 11 Feb 2024 01:29:39 -0800 (PST)
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Subject: =?UTF-8?Q?Re=3A_Le_probl=C3=A8me_de_l=27anisochronie_relativiste=2E?=
From: Richard Verret <rverret97@gmail.com>
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Le 10/02/2024 =C3=A0 23:24, Richard Verret a =C3=A9crit :
> Le 10/02/2024 =C3=A0 20:55, Richard Hachel  a =C3=A9crit :
>> Le 10/02/2024 =C3=A0 16:31, Richard Verret a =C3=A9crit=C2=A0:
>>> un temps Tr pour parcourir la distance D=E2=80=99, tel que D=E2=80=99 =
=3D Vr Tr.=20
>>  Non, non.
>>  Un temps Tr tel que D=3DTr.Vr=20
>>  Restons simple..
> Oui, tu as raison, c=E2=80=99est plus rationnel ainsi d =3D v.t et d=E2=
=80=99 =3D v=E2=80=99.t=E2=80=99.
Dans le cas du paradoxe de Langevin, la distance d parcourue en m.r.u. est =
=C3=A9gale, pour le jumeau terrestre =C3=A0 d =3D v.t, et =C3=A0 d =3D v=E2=
=80=99.t=E2=80=99 pour le jumeau voyageur, comme v=E2=80=99 =3D v, il vient=
 t=E2=80=99=3Dt. La distance d1 parcourue lors d=E2=80=99une acc=C3=A9l=C3=
=A9ration ou d=E2=80=99une d=C3=A9c=C3=A9l=C3=A9ration est =C3=A9gale =C3=
=A0 d1 =3D a t1^2/2 pour le jumeau immobile et =C3=A0 d1=3Da=E2=80=99t=E2=
=80=991^2/2 pour le voyageur. Comme a=E2=80=99=3Da, t=E2=80=991 =3D t1. La =
dur=C3=A9e du voyage pour le jumeau immobile t+4t1 est la m=C3=AAme que pou=
r le voyageur t=E2=80=99+4t=E2=80=991 si l=E2=80=99on consid=C3=A8re que le=
s distances d=E2=80=99acc=C3=A9l=C3=A9ration et de d=C3=A9c=C3=A9l=C3=A9rat=
ion sont toutes identiques. Les jumeaux ont donc le m=C3=AAme =C3=A2ge quan=
d ils se retrouvent.=20
Et voili!