Experiment to test mutual time dilation
in [Relativity]
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From: Paul B. Andersen on 9 Aug 2010 14:37 On 09.08.2010 18:52, Koobee Wublee wrote: > As usual, you have challenged me to find your error, and I have done > so.<shrug> See the link below. > > http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8deab2d?hl=en > > Unable to face your own short comings, you are once again becoming > asinine with silly and childish denials. You need to calm down and > stop babbling and repeating the same nonsense before I can continue to > guide you out of mysticism.<shrug> > > Do you think you can behave like a responsible adult for a change? Repeating a posting where you failed to point out an error won't do. Here is the derivation again: A wave propagating in the positive x direction in the unprimed frame can be written: E cos(phi(t,x)) where phi(t,x) = wt - (w/c)x Let the primed frame be moving at v along the positive x-axis. The same wave transformed to the primed frame can be written: E' cos(phi'(t',x')) where phi'(t',x') = w't' - (w'/c)x' Applying the LT transform: t = g(t' + vx'/c2) x = g(x' + vt') g = 1/sqrt(1-v2/c2) yields: wt - (w/c)x = wg((t' + vx'/c2) - (w/c)g(x' + vt') = wg(1-v/c)t' - (w/c)g(1-v/c)x' = w't' - (w'/c)x' Thus: w' = wg(1-v/c) = w sqrt((1-v/c)/(1+v/c)) The wave is red shifted in the primed frame. Can you point out an error, Koobee? -- Paul http://home.c2i.net/pb_andersen/
From: Androcles on 9 Aug 2010 18:40 "Paul B. Andersen" <someone(a)somewhere.no> wrote in message news:4C604AFE.1020507(a)somewhere.no... | Applying the LT transform: | t = g(t' + vx'/c2) | x = g(x' + vt') | g = 1/sqrt(1-v2/c2) Bwahahahahahahahaha! Classic schoolboy error! Standard algebra: a = (b-c)*d a/d = b-c b = a/d+c Norwegian algebra: a = (b-c)*d a +c = b * d b = (a+c) * d
From: Inertial on 9 Aug 2010 21:02 "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 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) 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. 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) 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. 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: BURT on 9 Aug 2010 23:17 On Aug 9, 6:02 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 > > 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) > > 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. > > 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) > > 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. > > 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. Theory can be tested by thought experiment. If a near light speed train passes the station and observes the station's clock as slow how does the train age less? Mitch Raemsch
From: Koobee Wublee on 10 Aug 2010 03:21
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> |