Re: Time dilation
From: John Kennaugh (JKNG_at_kennaugh2435hex.freeserve.co.uk)
Date: 09/29/04
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Date: Wed, 29 Sep 2004 19:58:28 +0100
Androcles writes
>
>"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
>news:TbBgzjSMXuVBFwVe@kennaugh2435hex.freeserve.co.uk...
>| 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:
>
>
>Duh...
>I suppose I'll have to go over it with you.
Having read your comments I wish you had. You have largely ignored what
I wrote.
>
>|
>| 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.
>
>Why use one second ticks?
It was your original problem
Androcles wrote in "Re: Is time a measurement?(actually in search of
edification)":
--------------------- quote ----------------------------------------
>Perhaps amusingly, we can consider what happens to a moving oscillator that
>has been correctly coupled with it's synchronised Earth counter as it leaves
>Earth, travels to (say) Jupiter and then returns.
>For simplicity, we'll make the journey occur at 0.866c in each direction,
>and choose a convenient distance of 1 light-hour.
>Since we are doing the measuring, the oscillator reaches the planet after
>1/0.866 = 1.155 hours, still sending its 1 second pulses by radio, returns
>and arrives after 2.31 hours.
>The Earth counter will have indicated 60 * 60 * 2.31 = 8314 seconds for the
>entire trip.
>Sanity check:
>If the moving oscillator counted half the number of ticks as the Earth bound
>oscillator, it ticked at a rate of 1 tick every 2 seconds. That is 1/2 Hz.
>
>Let's see what Einstein has to say about that, but we'll need to be careful,
>we expect red shift going and blue shift returning.
>
>f' = f. sqrt[(1-v/c)/(1+v/c)]
> = 1.sqrt[ (1-0.866)/ (1+0.866) = sqrt(0.134/1.866) = 0.268 Hz,
>and since this lasted for 1.155 hours, 1114 ticks.
>Returning,
>f' = 1.sqrt[ (1+0.866)/ (1-0.866) = sqrt(1.866/0.134) = 3.732 Hz,
>and since this too lasted for 1.155 hours, 1.155 *60 * 60 *3.732 = 15516
>ticks.
------------------ end quote -----------------------------------------
Both of your calculations were wrong and I corrected them then. Since
then I have done the rest of the maths which is what I presented here.
It was you who worked out what frequency a 1 second tick presents when
affected by Doppler. As I have presented your own problem back to you I
will delete your 'new problem'. The question is the same. The original
has better resolution.
[.....] unnecessary variation of the problem/red herring
>|
>| 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.
>Good. It saves us from going a full light year.
It is your problem I am solving
>| 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.
>
>Nah... I don't believe it.
Your statement simply says you don't accept relativity. Fine! Neither do
I. But you cannot disprove relativity by showing a fault in the maths
which you yourself have created by not accepting one of the predictions
of SR.
>"as has already been shown to the first order of small quantities, the same
>laws of electrodynamics and optics will be valid for all frames of reference
>for which the equations of mechanics hold good."-- Albert "Idiot" Einstein.
>Seeing something shorter than it really is would violate the same laws.
No if there was a measuring rod stationary w.r.t the moving observer the
earth bound observer would see that as half as long. i.e. the earth
bound observer will see the space ship as half the length. Both
therefore have the same laws.
e.g. Two identical space ships passing each other at 0.866c.
A sees B's space ship half as long as his and B's clock going half as
quick.
B sees A's space ship half as long as his and A's clock going half as
quick.
Laws are the same for both. Absurd, granted, but the same.
If you like you can ask 'who's clock is actually slower' and you will
get a verity of answers which have changed over the years. One of which
is 'they both are'.
If you like you can say "That's silly" and I will agree.
The one answer which simply won't do, is that they are actually going at
the same rate and it is an illusion as in "It's like looking at
something at an angle rather than straight on. It looks shorter" -
because there is no way an illusion would produce a real difference in
age.
I think the modern answer is to attack you for asking the question. The
bilge technique. "You obviously don't understand relativity. You make no
attempt to understand relativity, I have explained it to you numerous
times. If you did understand relativity then you wouldn't ask silly
questions........" - been there, done that, not convinced.
>| His trip distance is
>| therefore 1 light-hour total so he also sees his clock tick 4157 times.
>
>Nah... can't happen. The Earth is 8 light minutes from the sun, Saturn would
>be
>about 1 light hour from Earth. You are claiming the traveller can get to
>Saturn
>and back in an hour. He'd have to be travelling at 2c to do that, in round
>figures.
No If you're on earth you will see him take 2/0.866 = 2.30947 hrs to
cover the two light years distance. I have already calculated that.
For the moving observer the distance is only 1 light year round trip.
That is what the theory says. That is what relativists believe. I'm only
trying to stop you bashing your head against a brick wall.
>This is where you attempt a shortcut, and that is a fallacy. You've
>forgotten the doppler shift.
No I haven't I did the maths in full below.
[......]
>|
>| 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,
>
>Yep.
>|
>| 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.
>
>Right. Same for the flashes,
bugger the flashes.
>so the returning clock is runing fast.
No. If a train whistle is coming towards you and the frequency appears
higher you don't assume that time in the train is going faster do you?
The reason I quoted
f' = f(1-v/c)/sqr(1-vv/cc)
which simplifies to
= f. sqrt[(1-v/c)/(1+v/c)]
which is what you used is that the first equation has two terms (1-v/c)
which is what would normally be described as Doppler shift, simply a
change of frequency due to motion and the other term sqr(1-vv/cc) time
dilation. In relativity both cause a shift of frequency. The time
dilation term is the same whether the train is going towards you or away
from you because v is squared which again is less obvious from the
simplified equation. In the more general case the v in the first term is
the component of the velocity in a line through the observer and the v
in the second term is the velocity w.r.t the FoR. So if it is travelling
at right angles to you the first term = 1 and the second is still
sqr(1-vv/cc). Because the time dilation component is included in the
Doppler equation you can apply it on its own and get exactly the same
number of ticks if you count ticks outward and inbound as you do if you
simply take the total time and assume time dilation. i.e.
[(2.154735 x 0.268) + (0.1547344 x 3.731)] x 60 x 60 =
sqrt(1-0.866^2/1^2) x 60 x 60 x 2/0.866
>You have the correct outbound frequency (albeit calculated by a
>different method, and the correct inbound.
I am reproducing your own maths which I checked were correct from
previous thread. Your calculation of the frequency was correct, but the
time it is received for was in error.
>|
>| 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.
>Right. For how long does he count them?
[.......]
>|
>| 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.
>There is no time dilation. Somewhere you've got you numbers screwed
>up. I'll let you look for it,
No if it is wrong you find it. It was the problem posed in the first
place. I found your error. If you think I have made an error then you
find it.
>| 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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