Re: Time dilation
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 09/27/04
- Next message: shevek: "Re: What is luminiferous ether made of?"
- Previous message: Eugene Shubert: "Re: Derivation of Lorentz from basic assumptions"
- In reply to: John Kennaugh: "Re: Time dilation"
- Next in thread: John Kennaugh: "Re: Time dilation"
- Reply: John Kennaugh: "Re: Time dilation"
- Messages sorted by: [ date ] [ thread ]
Date: Mon, 27 Sep 2004 18:29:52 GMT
"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.
|
| 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 makes for larger numbers... I'll use
flashes of light, delivered once every every 20 minutes for a duration
of 10 minutes.
0 10
|___________|==========|___________| <30 mins
|==========|___________|==========| < 60 mins
That's one hour, or 3600 ticks, drawn on a linear time axis but
I need two lines to show it.
|
| He sees the moving clock time dilated
| t1 = t2 sqrt(1-0.866^2/1^2) = t2/2 so will receive 4157 ticks.
Maybe you think he won't see three flashes of light for the outward trip,
but we'll use relativity to see if he does. The frequency, f, is 3 per
hour.
the observed frequency, f' is
f' = 3.sqrt( [1-0.866] / 1.866) = 0.268*3 = 0.804
so you are correct, only 0.8 flashes of light have been sent when the
turnaround has been reached.
If the traveller stops, it will take some time for rest of the light flashes
to reach the Earth, and likewise for the traveller to see similar flashes
from the Earth, they have to catch up.
This time is easily calculated, t = 1/f = 1/0.804 = 1.244 hours per flash,
3 flashes takes 3.732 hours to arrive.
Well, that sounds reasonable. The start of the 4th flash has to be at least
one hour after the experiment started and it is from 0.866 light hours
away, so the soonest it can arrive is 1.866 hours. When we consider
that the clock is running at half speed, 1.866 * 2 = 3.732 hours.
Whoopee, the numbers work out. So it must be right! Wow!
|
| 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.
| 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
Sure, and he'll know to turn the light on and off every 600 ticks.
| but will see the 'measuring rod'
| length dilated such as to be 1/2 light-hour.
Nah... I don't believe it.
"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.
| 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.
This is where you attempt a shortcut, and that is a fallacy. You've
forgotten the doppler shift.
It is blue on the return trip.
f' = 3.sqrt( 1.866 / [1-0.866]) = 11.195 flashes per hour
We didn't quite get the full flash for the outward trip, that was 0.804,
so we'll add that, totalling 11.999.
Oh dear, it looks like a rounding error, I was sure it should be 12.
Anyway the trip took a distance of 2 light hours and lasted for
3.734 hours and both see a total of 3.734 * 3 = 11.2 flashes,
nearly all of which are observed on the return trip, according to Einstein's
doppler.
Why the first flash wasn't seen when it was sent from only 10
light minutes away will forever remain a mystery, but that's relativity
for you.
|
| 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.
Nope. Both agree that the number of flashes of light each sees of the
other is the same.
|
| 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, so the returning clock is runing fast.
You have the correct outbound frequency (albeit calculated by a
different method, and the correct inbound.
|
| 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?
Answer : he counts a frequency of 0.268 for 3.731 seconds, or 1 tick
and a frequency of 3.731 for 0.268 seconds =1 tick.
So does the Earth observer.
|
| 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, I prefer values less than 10, it's easier to
see.
Androcles
|
| 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
- Next message: shevek: "Re: What is luminiferous ether made of?"
- Previous message: Eugene Shubert: "Re: Derivation of Lorentz from basic assumptions"
- In reply to: John Kennaugh: "Re: Time dilation"
- Next in thread: John Kennaugh: "Re: Time dilation"
- Reply: John Kennaugh: "Re: Time dilation"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
