Re: Time dilation
From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 09/29/04
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Date: Wed, 29 Sep 2004 22:06:58 GMT
"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
news:iFUMkmRUXwWBFwqL@kennaugh2435hex.freeserve.co.uk...
| 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.
No I didn't. I simplified the problem.
|
| >
| >|
| >| 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.
Yes, you are correct. They ARE wrong.
As I've since pointed out, the 0.268 Hz and the 3.732 Hz are observed
for 3.732/(3.732 + 0.268) * 2.31 hours and
0.268/(3.732 + 0.268) * 2.31 hours
In other words, time must be allowed for the ticks to reach the observer,
the show isn't over when the traveller reaches turn around.
| 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.
You've forgotten something. 4157*0.866 = 3600 seconds = 1 hour,
so for a journey time of 1.155 hours 4157 ticks will be sent. However,
most are still strung out in space on their way to the observer,
and the last will not arrive until ??? (by either clock).
So where is the time dilation?
|
| >|
| >| 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
'Fraid not. Let's go on.
|
| >| 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.
It's easy to jump the gun without looking at the reason, and to snip my
simplified version without looking at it, isn't 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.
|
| 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.
Nope. The distance contraction applies to the entire frame. Just put a long
pitot tube on the nose of the vehicle to see why. Make it long enough to
reach Saturn. We would then see the tube shrink away from Saturn as the ship
accelerated, instead of going past it. That violates the laws of optics.
| e.g. Two identical space ships passing each other at 0.866c.
[.....] unnecessary variation of the problem/red herring
|
| >| 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.
Correct, but you are also claiming he can make the trip in half that, ship
time.
That also means it includes the entire frame.
|
| For the moving observer the distance is only 1 light year round trip.
Exactly. The entire universe shrinks. And since light enters the ship from
the side has the same c as the light entering from the front...
But anyway, according to relativity's Lorentz equations, the distance is
one light year round trip, and according to relativity's doppler equations,
it isn't.
|
| 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.
Yeah, we know... 4157 missing ticks.
| >
| >|
| >| 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.
NO!
| >|
| >| 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.
That's the outgoing frequency, not the tick count.
|
| >|
| >| 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.
Right. And it took how long to sent them?
| >|
| >| 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.
bugger your calculations.
|
| >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?
I wouldn't, of course, and only a relativist would assume it was going
slower.
| 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 reason I used f. sqrt[(1-v/c)/(1+v/c)]
is because that is what Einstein gives in section 7 of "Electrodynamics",
and because YOU haven't bothered to read that paper you are unaware of my
analysis
given in the post "The Seven Deadly Sins of Relativity"
| 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.
Crap.
Einstein's entire basis for that ridiculous conjecture is his statement:
"IF we place x' = x-vt ...", my capitialization.
We then have
xi = (x')/sqrt(1-v^2/c^2)
but IF we place x' = x+vt (in other words change direction)
we then have length expansion, and the faster you go the further away
the destination.
Learn to read!
| 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.
Bugger the previous thread. You are wrong in thinking there is time dilation
and length contraction in both directions.
| >|
| >| 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?
| [.......]
Likewise.
What is strange here is that we have two sane engineers arguing over
something neither of us accept, yet one of them isn't interested in WHY it
is wrong. Maybe there is only one sane engineer after all.
Androcles
- Next message: TMG: "Re: God=G_uv proves 40k B.C. Creation"
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- In reply to: John Kennaugh: "Re: Time dilation"
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