Experiment to test mutual time dilation
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From: kenseto on 10 Aug 2010 09:28 On Aug 9, 9:02 pm, "Inertial" <relativ...(a)rest.com> wrote: > "Sam Wormley" wrote in message > > news:ftSdncUnv6Dr-v3RnZ2dnUVZ_qOdnZ2d(a)mchsi.com... > > > > Note that to measure the clock rate of some other clock, you need to take > (at least) TWO readings of that other clock (at different times) and look at > the difference, and compare to the difference in readings on own correct > clock. > > If the other clock is moving, then you'll need TWO synchronised correct > clocks (A1, A2 say) at rest in your frame at different location, and each > can take a reading from the moving clock (B) as it passes > > Let set up such an experiment (as a gedanken) as above > > When B passes A1, A1 records its own time, and the time on B, and when B > passes A2, A2 records its own time, and the time on B > > Then you can compare the difference between the time on B that A1 recorded, > and the time on B the A2 recorded, and compare that to the difference in > times on A1 and A2 themselves. > > SR says that if you do that, you will find the difference in time on B is > *less* than the difference in time between A1 and A2. That is that clock B > is ticking *SLOWER* than clocks in A Right....SR predicts that the passage of an interval of clock time on the B clock corresponds to the passage of a lesser interval of clock time on the A clock....that means that the B clock is running slower than the A clock. Delta(T_B)=gamma[Delta(T_A)] The problem is: Sr predicts the same for an observer on B. > > Now lets have a second gedanken. > > In this case we have a single clock A at rest in your frame, and two > co-moving moving clocks (B1, B2 say) that are synchronised in their own > frame and are moving past A (with the same speed as B in the previous > experiment) Why is B1 and B2 moving past A...why not A moving past B1 and B2 ??? > > AS in the previous experiment, clock A takes a set of two readings, one of > its own time, and the other of the time shown on the moving clock. > > So we get two sets of readings from A .. one set when B1 passes and another > set when B2 passes. So you are assuming that the B clocks are doing the moving while A remains stationary....right? > > If you compare them, you'll find, in this case, a *longer* elapsed time > between B1 and B2 than is shown on A (the opposite results to the previous > experiment) This is a bogus assumption based on the faulty SR math that all clocks moving wrt the observer (A or B) are running slow. > > How can this be? > > The result of the second experiment implies that either clocks B1 and B2 > were in sync in your frame but ticking *FASTER* -- OR -- that the clocks are > *NOT* in sync in your frame. This is wrong....B is the observer and B1 and B2 are in sych in the B frame. Therefore B must predict that A is running fast as follows: Delta(T_A)=Delta(T_B)/gamma Ken Seto > > As the previous experiment shows that a moving clock, B, is measured as > ticking slower .. that must mean that the second interpretation (that moving > clocks are NOT in sync in your frame) is the correct one. Otherwise you > would have a contradiction between the experimental results.
From: Michael Moroney on 10 Aug 2010 11:34 "Paul B. Andersen" <something(a)somewhere.no> writes: >On 10.08.2010 09:21, Koobee Wublee wrote: >> The little professor at Trondheim is utterly asinine in the past few >> days. He is totally ignoring logic and embracing his mystic and very >> stupid view points as always.<shrug> >Giving up, Koobee? :-) >Your surrender graciously accepted. Aww geez, it's over? I just popped more popcorn.
From: Michael Moroney on 10 Aug 2010 12:09 kenseto <kenseto(a)erinet.com> writes: >On Aug 9, 9:02 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> Let set up such an experiment (as a gedanken) as above >> >> When B passes A1, A1 records its own time, and the time on B, and when B >> passes A2, A2 records its own time, and the time on B >> >> Then you can compare the difference between the time on B that A1 recorded, >> and the time on B the A2 recorded, and compare that to the difference in >> times on A1 and A2 themselves. >> >> SR says that if you do that, you will find the difference in time on B is >> *less* than the difference in time between A1 and A2. That is that clock B >> is ticking *SLOWER* than clocks in A >> Now lets have a second gedanken. >> >> In this case we have a single clock A at rest in your frame, and two >> co-moving moving clocks (B1, B2 say) that are synchronised in their own >> frame and are moving past A (with the same speed as B in the previous >> experiment) >Why is B1 and B2 moving past A...why not A moving past B1 and B2 ??? Geez, you've been kooking in here for 15 years now but still don't understand the basic fundamentals of relativity? All inertial frames in SR have the same math, so of course an observer in A's frame can do measurements with B1 and B2 moving past A by assuming A is stationary. >> AS in the previous experiment, clock A takes a set of two readings, one of >> its own time, and the other of the time shown on the moving clock. >> >> So we get two sets of readings from A .. one set when B1 passes and another >> set when B2 passes. >So you are assuming that the B clocks are doing the moving while A >remains stationary....right? All motion is relative, so of course A will see the B clocks as moving. (and yes, an observer in B will see the A clocks as moving, but that was already discussed as the first gedanken, so it's not relevant here) You got to keep the A and B frames separate. This is where you always get mixed up. >> If you compare them, you'll find, in this case, a *longer* elapsed time >> between B1 and B2 than is shown on A (the opposite results to the previous >> experiment) >This is a bogus assumption based on the faulty SR math that all clocks >moving wrt the observer (A or B) are running slow. No, just do the logic that was just shown to you! >> How can this be? >> >> The result of the second experiment implies that either clocks B1 and B2 >> were in sync in your frame but ticking *FASTER* -- OR -- that the clocks are >> *NOT* in sync in your frame. >This is wrong....B is the observer and B1 and B2 are in sych in the B >frame. But not the A frame. And that is the point.
From: Koobee Wublee on 10 Aug 2010 13:09 For the record, Koobee Wublee was explaining how the Lorentz transform actually predicts both blue and red shift at the same time. See the post below. http://groups.google.com/group/sci.physics.relativity/msg/c4d248370bde9eee?hl=en Suddenly, professor Andersen, out of the blue, changed the discussion with this post without addressing Koobee Wublees previous post. http://groups.google.com/group/sci.physics.relativity/msg/077983367e3fe36a?hl=en Koobee Wublee, an ever so humble, good Samaritan, answered professor Andersens questions fair and square. The good hearted Koobee Wublee saw the same mistakes and garbage that professor Andersen had defecated all over these newsgroups in the past few years and decided not to press any further charges. See the post below. http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8deab2d?hl=en Seeing himself checkmated once again, professor Andersen decided to behave childish and defiant with the following useless post. http://groups.google.com/group/sci.physics.relativity/msg/29d192787e693974?hl=en The good hearted and ever so humble Koobee Wublee tried once again to reach out to save professor Andersen from drowning in his cesspool of fermented diarrhea of Einstein the nitwit, the plagiarist, and the liar with the following post. In doing so, Koobee Wublee even showed professor Andersen what a mathematical model of Doppler effect would look like and guided with hints on how to arrive with that. http://groups.google.com/group/sci.physics.relativity/msg/22d990f50ede9754?hl=en The asinine professor Andersen then refused to discuss any further by repeating the same bullshit from the very beginning. Professor Andersen is indeed a very small man. <shrug>
From: spudnik on 10 Aug 2010 14:47
to me, it is just a matter of acceleration, and I really don't see what the problem is, there never having been any "twin paradox," that was not immediately dyspozed-of. of course, when ever there is acceleration, one goes beyond galilean relativity-a-cruisin'. thus: of course, no-one knows whether Universe is finite (but we might have a handle on Olber's paradox .-) now, if the angular momentae of atoms are limited to the velocity (not speed) of lightwaves, then it isn't hard to see that there'd be effects of "going" at fractions of that velocity. thus: so, presumably, iodine-131 is a daughter of cesium-137; eh? anyway, the whole thing was a wash, as adequately shown by the UNSCEAR 2000 report, or in this historical treatment: http://www.21stcenturysciencetech.com/Articles_2009/Summer-2009/Fear_radiation.pdf thus: what's the difference between spacetime & a 3d movie? I don't believe that Einstein ever thought that there was any paradox of the twins, other than, "wow, Bro; you've really aged!" thus quoth: Max Born first heard of the theory through attending the lectures of Minkowski, in which 'we studied papers by Hertz, FitzGerald, Larmor, Lorentz, Poincare and others, but also got an inkling of Minkowski's own ideas'. Later, 'I went in 1907 to Cambridge', where he heard nothing of Einstein, and afterwards (how long afterwards he does not say) he returned to Breslau, 'and there at last I heard the name of Einstein and read his papers ... Although I was quite familiar with the relativistic idea and the Lorentz transformations, Einstein's reasoning was a revelation to me.'7 http://www.21stcenturysciencetech.com/Articles_2010/Gifts_de_Broglie.pdf thus: what is your problem with relativity?... I could see from the outset of Dingle's book, that he is in a quandary over "which clock is slower," and that is a very simple thing, the same as the strawman paradox of the twins. thus: it's simpler to say that, it is rational, iff the decimal part (after a finite number of places) repeats, *including* a tail of zeroes or nines. thus: you only have your toe in it. the surface of the sphere is pi*d*d, and it is four times the surface of the great circle -- a thing that Bucky apparently didn't know, oddly enough. it is pretty laughable, that you'd think that you are dysproving F"L"T, because it is clear from the available stuff that it was the key to his method (along with the fact that he basically created numbertheorie, dood .-) --les ducs d'Enron! http://tarpley.net --Light, A History! http://wlym.com |