Re: Special Relativity is Dead! (second proof)

jan.verheul@xxxxxxxxxx wrote:
Brian Kennelly schreef:

The acceleration does not lead to burst of clock pulses. It
leads to change in the Doppler factor. The rate at which the
pulses are received increases smoothly during the acceleration,
and then remains steady when the acceleration ends.

That means that there will be a conflict between the clocks upon
arrival home of the traveling clock. The traveling clock is ahead of
the stationary clock (based on received timing pulses) and the
stationary clock is ahead of the traveling clock (based on application
of S.R., which would at least be valid for the stationary clock,
according to proponents of S.R.).

Assume that the Doppler factors on each leg of the trip are K and K', with K>1 and K'<1.

On the outbound journey (time T), the travelling clock will have received T/K pulses. On the return journey, it will receive T/K' pulses. So, while the travelling clock measures 2T, the stationary clock has pulsed T(1/K'+1/K) times.

Now, if the velocities have equal magnitude, then K'=1/K, and we find that, while the travelling clock measured 2T, the stationary clock ticked 2T(K^2+1)/2K. Now, (K^2+1)/2K >1 unless K=1, so the stationary clock ticked more times that the travelling clock, and there was no 'burst' of signals.

I ignored the pulse received during the acceleration, so we need to look at them a little closer. Assume that Y pulses are received during the turnaround. Then we need to add Y to the stationary clock time, and Y' to the travelling clock time. Now, the Doppler factor changes smoothly from K to 1/K, in a way that is independent of the distance, so Y' will be independent of the distance. Its value depends on the details of the acceleration, but in any case SR predicts that Y'<Y. We find the the travelling clock measures 2T+Y'<2T(K+1/K)+Y. We still do not need to postulate any burst of signals.

(We can turn it around as well, and measure pulses sent by the travelling clock and received by the stationary clock. Again we find that, while the travelling clock pulses 2T times, the stationary clock receives 2T(K+1/K) (>2T) pulses. It is entirely symmetric.)

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