🚀 go-pugleaf

> 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