With one clock stationary at the common origin for 10 time units and the other rocket clock(adsbygoogle = window.adsbygoogle || []).push({});

traveling out and back at v=.6c. When both clocks are compared at the origin the elapsed

time on the traveling clock will show less time by a factor of (.8) from the Lorentz Transformation.

The spacetime interval for the traveling clock will be S=8, sqrt( 10^2-6^2)

Compared to the stationary clocks spacetime interval S=10.

Note, it is understood if the rocket clocks coordinates were transformed to the inertial frame

by: t=gamma(1.25) (t' + (v)(x')/cc and x=gamma(1.25)(x'+(v)(t') then S=8

=sqrt(17^2-15^2).

So the question: Is the 8 units of elapsed time in the rocket clock because the physical mechanism was moving at a slower rate (time between ticks) in the rockets frame.

Or in the case of the spacetime interval, was the mechanism in the rockets clock ticking at the same rate (in rocket frame) as the rate of ticking in stationary frame, but the rocket clock just traveled a lesser distance in spacetime, S=8

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# Less elapsed time: Lorentz Transformation or Spacetime Interval ?

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