Re: Time dilation

From: John Kennaugh (JKNG_at_kennaugh2435hex.freeserve.co.uk)
Date: 09/26/04 

Date: Sun, 26 Sep 2004 16:52:44 +0100

Androcles

I followed up a posting of yours to see if there is indeed a discrepency
to be found. I'm afraid there isn't:

Consider what happens to a moving clock that has been synchronised with
an Earth clock as it leaves Earth, travels to a point a fixed distance,
1 light year (as measured by the earth bound observer) from earth and
then returns. For simplicity, we'll make the journey occur at 0.866c in
each direction. The moving clock transmits by radio its clocks ticks
which are received by the earth and counted. Note that a clock tick not
only represents time, can be represented as a frequency it is also an
event. With telemetry each tick could be uniquely identified. An event
either takes place or it doesn't. No ticks can be lost or gained
although we may be permitted to lose the odd one in the maths if we do
not calculate to sufficient accuracy.

The Earth observer measures the time for the 2 way trip as
2/0.866 = 2.30947 hrs and this works out at 8314 ticks at 1 tick/sec for
his clock.

He sees the moving clock time dilated
t1 = t2 sqrt(1-0.866^2/1^2) = t2/2 so will receive 4157 ticks.

The 'measuring rod' between earth and 'the fixed point' is stationary in
earths FoR and is 1 light-hour long. From the point of view of the
observer travelling with the moving clock, he will see his clock as
accurately ticking 1 tick per second but will see the 'measuring rod'
length dilated such as to be 1/2 light-hour. His trip distance is
therefore 1 light-hour total so he also sees his clock tick 4157 times.

Both observers therefore agree that the moving observer's clock has only
ticked 4157 times compared to the earth bound observer's 8314 because
the moving clock is going slower, the moving observer because the
distance he travels is shorter.

What would the earth observer actually receive on his radio when
listening for the ticks? They will be subject to Doppler shift. One can
apply the Doppler equation and get the same answer.

Outward journey
f' = f(1-v/c)/sqr(1-vv/cc)
    = f. sqrt[(1-v/c)/(1+v/c)]
    = 1.sqrt[ (1-0.866)/ (1+0.866) = sqrt(0.134/1.866) = 0.268 Hz,

Earth will start receiving this frequency at t=0 it will stop when the
tick which left the clock at the turnaround point reaches earth. The
last tick (at this frequency) was transmitted at t = 1/0.866 =
1.154735hrs it will take an hour to reach earth ( 1 light hour distance)
so 0.268Hz will be received for t = 2.154735 hrs.

Number of Ticks = 2.154735 x 60 x 60 x 0.268 = 2079 ticks

Returning,
f' = 1.sqrt[ (1+0.866)/ (1-0.866) = sqrt(1.866/0.134) = 3.731 Hz,

This frequency will be received for the remainder of the trip starting
at t = 2.154735 and ending at 2.30946hrs a total of 0.154725hrs.
The number of ticks = 0.1547344 x 60 x 60 x 3.731 = 2078 ticks.

The total number for both go and return is the same and the grand total
= 4157 the same as previous figure (give or take a tick). Doppler shift
will effect the interval between ticks but cannot effect the total
number.

What if the Moving observer can receive, by radio, ticks from the earth
bound clock. On the outward journey he will receive ticks at 0.268Hz and
on the return at 3.731Hz.

Outward trip he starts receiving ticks at 0.269Hz at t = 0. At the turn
around point he will have been receiving them for 0.5/0.866 = 0.5774 hrs
so will have received 0.5774 x 60 x 60 x 0.269 = 559 ticks.

On the return 0.5774 x 60 x 60 x 3.731 = 7755
a total of (559 + 7755) = 8314 which is the same as the earth bound
observer counted.

Summery
Earth bound observer sees his clock tick 8314 times and the moving
observers clock tick 2079 times at 0.268Hz on the outward journey and
2079 times at 3.731Hz for the return - a total of 4158 - half that of
his own. He puts the difference down to time dilation.

The moving observer sees his clock tick 2079 on the outward journey
which is for him half a light hour distance, and another 2079 on the
return. The same as the earth clock received from him. On the outward
journey he receives 559 ticks at 0.269Hz and on the return 7755 at
3.731Hz from the earth bound clock. Total received from earth = 8314
which agrees with that of the earth bound observer. 

-- 
John Kennaugh
to email convert the number from hex to decimal

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