Re: Time dilation

From: Androcles (androc1es_at_nospamblueyonder.co.uk)
Date: 10/01/04 

Date: Fri, 01 Oct 2004 21:21:28 GMT

"John Kennaugh" <JKNG@kennaugh2435hex.freeserve.co.uk> wrote in message
news:6UQagbCDwVXBFwQU@kennaugh2435hex.freeserve.co.uk...
| Androcles writes
| >
| >"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.
| >|
| >| >|
| >| >| 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?
|
| There are two methods of calculating how many ticks the earth observer
| will get from the moving observer.

Crap. Counting is the only method, no matter how long it takes.
Androcles.

|
| Method 1 Total journey time method
| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
| By the time the traveller gets back to earth all ticks he has sent will
| have been received so your statement about being strung out in space is
| wrong.
| Time dilation equ t1 = t2 sqrt(1-v^2/c^2) .
| Note that v is squared so it doesn't matter which direction he travels.
| Total elapse time according to earth observer 2/0.866 = 2.30947 hrs
| Speed squared = (0.866c)^2
| t1 = t2 sqrt(1-0.866^2/1^2) = t2/2
| Time goes at half the earth clock rate for 2.3097 hrs. All ticks having
| reached earth by the time the traveller has. The earth clock ticks 8314
| times, the travellers half that 4157.
|
| Method 2 Doppler method
| ~~~~~~~~~~~~~~~~~~~~~~~
| Here the trip, still from the point of view of the observer on earth is
| viewed not as a whole but as individual ticks calculated using the
| Doppler equation. The number at the 'outward journey rate' (0.268Hz)
| plus the number at the 'return journey rate' (3.731Hz).
| [(2.154735 x 0.268) + (0.1547344 x 3.731)] x 60 x 60 = 4157
|
| >
| >|
| >| >|
| >| >| 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?
|
| Why should I look at another version of the same problem when the
| problem I am using is your own? There is only one problem which set of
| numbers you use doesn't matter.
|
| >
| >| >"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.
|
| That is far more complicated. To start with if you move a long rod at
| one end you elastically compress it and that compression travels along
| the rod at the speed of sound (in rod). If you try to move one end at
| faster than the speed of sound in the rod the rod will deform and you
| will have a permanently shorter rod. If you could instantly accelerate
| to 0.866c then the rod would stay stationary in the earths FoR and you
| would drive along it crumpling it as you went...... shall we stick to
| the original problem.
|
| >| >| 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.
|
| don't understand your comment.
|
| >
| >
| >|
| >| For the moving observer the distance is only 1 light year round trip.
| >
| >Exactly. The entire universe shrinks.
|
| No only that part of the universe which is travelling relative to him at
| 0.866c. Granted that is most of it but that is because you have chosen
| an extreme difference in speed.
|
| >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,
|
| Yes 1 for the moving observer. 2 for the earth observer
|
| > and according to relativity's doppler equations,
| >it isn't.
|
| Doppler equations give you frequency not distance.
|
| >
| >
| >|
| >| 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.
|
| No the moving observer says his clock ticked 4157 times during his 1
| light year trip and he received 8314 from the earth. The earth
| observers says his clock ticked 8314 times and he received 4157 ticks
| from the moving observer's slow clock during its observed 2 light year
| round trip. None missing. all accounted for.
|
| >
| >| >
| >| >|
| >| >| 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!
|
| Yes absolutely
|
| >
| >
| >| >|
| >| >| 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.
|
| Agreed
|
| >
| >|
| >| >|
| >| >| 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?
|
| According to the Earth observer the moving observer was sending them for
| 1.54735 Hrs 1 hours worth were in transit when he turned around because
| he was 1 light year away. He was not sending them at 0.268Hz that is the
| frequency they are received at. He was sending them at one every 2
| seconds (0.5Hz) according to the earth observer but he was further away
| each time. 0.268 x 2.154735 = 0.5 x 1.154735
|
| >
| >
| >
| >| >|
| >| >| 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
|
| On the contrary you are the one who failed to study that paper. If you
| go to section 7 to the equation above the one you quoted it says:
|
| ------------------------------ quote -----------------------------
| f' = f (1- Cos(Fi)v/c)/Sqr(1-v^2/c^2) ----- [1]
|
| "This is Doppler's principle for any velocities whatever. When Fi = 0
| the equation assumes the perspicuous form"
|
| f' = f sqrt[(1-v/c)/(1+v/c)] ----------[2]
| --------------------- end quote ----------------------------------
|
| Which is the equation you used. Einstein hasn't shown how he got from
| [1] to [2]
|
| If you do put Fi = 0 into [1] i.e. if the source is travelling in a
| straight line going through you Cos(fi) = 1 and you get:
|
| f' = f (1- v/c)/Sqr(1-v^2/c^2) ----- [1a]
|
| Which is the form I used because f' = f(1 - v/c) is the same Doppler
| shift as in acoustics and is direction dependent and the other term is a
| correction for time dilation which isn't because it has v^2 in it. His
| working would then have continued
|
| f' = f sqr[ (1-v/c)^2 /((1-v/c)(1+v/c))] --- [1b]
| f' = f sqr[ (1-v/c)/(1+v/c)] --------------- [2]
|
| Which is the form you used.
|
| > you are unaware of my
| >analysis
| >given in the post "The Seven Deadly Sins of Relativity"
| >
|
| I am aware of it. I see no reason to study it. I don't accept relativity
| for a host of reasons I don't need any more. If what you say is right
| fine. If not there are plenty who will tell you without me.
|
|
| >| 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!
|
| Unless I have missed something you have made a very basic mathematical
| error
|
| (-v) x (-v) = (+v) x (+v) = v^2
|
| You can of course argue that a square root has two valid roots and he
| has selected one of them but that would give you a shorter negative
| length not a longer length.
|
| >| 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.
|
| I don't understand your statement. I am right in what I state relativity
| says. Relativity says that speed is relative so everything must be
| symmetrical.
|
| 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.
|
|
| >| >| 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.
|
| It does not help your credibility for me to allow you to keep arguing
| when I know you are on a hiding to nothing.
|
| --
| John Kennaugh
| to email convert the number from hex to decimal

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