🚀 go-pugleaf

"Wolfgang G. Gasser" wrote:

> > Wolfgang G. Gasser wrote:
>
> >> Two distant observers with each a (transversal) LIGHT CLOCK orbit
> >> at a constant distance around their common center (assumed at rest).
> >> Special relativity postulates that the light beams in both light
> >> clocks propagate at c relative to both (almost inertially moving)
> >> observers. This requires that for each observer the clock rate of
> >> the opposite clock CAN be lower than the clock rate of his own clock.
> >> Because of symmetry and unchanging distance, clock rate CANNOT differ.
>
> Tom Roberts in news:dSymh.10793$yC5.9487@newssvr27.news.prodigy.net :
>
> > For approximately inertial motion, the basic equations of SR are only
> > APPROXIMATELY valid. You ignored the approximation you, yourself, made.
>
> Let us assume a relative speed of v = 0.99995 c (with gamma = 100)
> of each observer with respect to the other. Deviation from inertial
> motion can be made as small as we like by choosing a large enough
> radius of the orbit in relation to the height of the light clock.
> Applying the simple and transparent LIGHT CLOCK reasoning, we get:
>
> With respect to each observer the light beam in the opposite clock
> can only propagate at c, if the rate of this clock is gamma = 100
> times lower than the observer clock rate*. On the other hand, if
> the clock rates of both light clocks are equal, then with respect
> to each observer the velocity of the beam in the opposite clock is
> composed of the two orthogonal components c and v, resulting in a
> speed of light of c_v = sqrt(v^2 + c^2) = 1.41c.
>
> So we have a real contradiction and not something caused by an
> approximate calculation.
>
> > At a deeper level, however, such orbiting observers cannot be treated
> > as inertial observers, because a circle is different topologically from
> > a line.
>
> Also this attempt to remove the contradiction leads to an untenable
> consequence: Let us assume that at a given moment the observers start
> moving inertially instead of further orbiting. When still orbiting,
> the time of each observer runs identically with respect to the other.
> Just after the start of inertial movement however, the time of each
> observer runs one hundred times slower with respect to the other!
>
> Cheers, Wolfgang
>
> * This gamma time dilation leads to a gamma reduction in the speed
>   of the opposite clock beam from c to c/gamma. This opposite clock
>   is moving at v with respect to the observer, and this movement
>   is orthogonal to the light beam direction. Therefore the velocity
>   of the beam with respect to the opposite observer results in
>   the hypotenuse of the right-angled triangle with the legs v and
>   c/gamma_v, and we get: c_v = sqrt(v^2 + (c/gamma_v)^2) = c

   Si tu postes en fran�ais (ou si quelqu'un veut bien traduire),
   je te dirais o� se trouve l'erreur de raisonnement
   (j'ai d�j� mis le doigt sur des erreurs d'Hawkins et de Penrose,
   c'est pas pour craquer devant un petit bol de Sangria)    :)))

  R.H.