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A question about the equivalence principle. 
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#73
Jun212, 07:33 AM

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Another approach is to speed up the rear clock (as mentioned in an earlier post) so that it appears to be running at the same rate as the front clock. Now we have an unequivocal method of defining simultaneous in the accelerating rocket reference frame and can synchronise clocks using the usual Einstein clock synchronisation convention. Now when we send signals from the front and back using the on board rocket reference clocks, the signals arrive simultaneously at the back and front respectively according to the rocket clocks and the elapsed time between sending and receiving is equal according to both the front and back accelerating rocket observers. At no point here have we had to refer to the MCIRF's but if we compare results, simultaneous emission and simultaneous reception of signals as measured by the accelerating rocket observers, agrees with simultaneous as measured in the MCIRF's. Now the rocket observers know they had to speed up the rear clock to make it run at the same rate as the front clock so they know the real proper time rate of the rear clock must be slower than the proper time rate of the front clock. If they consider themselves to be stationary in their own reference frame they would ascribe this differential clock rate to a gravitational field and this coincides with the proper acceleration they can feel and and measure. The observers in a given inertial reference frame outside the rocket would ascribe the differential clock rates observed by the rocket observers to classical Doppler shift. 


#74
Jun212, 10:59 AM

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√(1 (g_{rear} δt)^{2}/c^{2}))/√(1 (g_{front} δt)^{2}/c^{2})) but the actual ratio of rates is g_{front} /g_{rear} those aren't close at all. The first goes to 1 as δt → 0, but the second doesn't. 


#75
Jun212, 12:09 PM

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Some web pages that seem to me to be relevant are Greg Egan's page on the Rindler horizon: http://gregegan.customer.netspace.ne...erHorizon.html And the Usenet Physics FAQ pages on SR and acceleration (which links to the page on the relativistic rocket equation, another good resource), and on the clock hypothesis: http://math.ucr.edu/home/baez/physic...eleration.html http://math.ucr.edu/home/baez/physic.../SR/clock.html 


#76
Jun212, 12:31 PM

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#77
Jun212, 12:48 PM

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The formula 1 + gL/c describes something a bit different than what I would call the "ratio of instantaneous clock rates", but I'd rather not get into that until I'm sure I correctly understand the scenario you're talking about. 


#78
Jun212, 01:46 PM

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e_{1} = the rear clock shows time 0. Let F = the instantaneous rest frame of the rear clock at this event. e_{2} = some event at the front clock that is simultaneous with e_{1} in frame F . Let T_{2} = the time showing on the front clock at event e_{2}. e_{3} = the rear clock shows time δT Let F' be the instantaneous rest frame of the rear clock at this event. e_{4} = some event at the front clock that is simultaneous with e_{3} in frame F'. Let T_{4} = the time showing on the front clock at event e_{4}. So from the point of view of an observer in the rear of the rocket, the rear clock advanced by δT, and the front clock advanced by: T_{4}  T_{2} Then the ratio of clock rates, as measured by the rear clock is: R_{front}/R_{rear} = (T_{4}  T_{2})/δT As δT → 0, this ratio does not go to 1, but goes to 1 + gL/c^{2}, where L is the length of the rocket, and g is the acceleration of the rear. 


#79
Jun212, 01:57 PM

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#80
Jun212, 02:09 PM

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#81
Jun212, 02:28 PM

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In fact it's even weirder than that. Consider the entire worldlines of the two clocks; I'll write their equations in the "launch frame" as follows: Rear clock: x^2  t^2 = R^2 Front clock: x^2  t^2 = (R + L)^2 Now draw any straight line through the origin, in the "launch frame", t = vx, where v >= 0. This is either a horizontal line (for v = 0) or a line sloping up and to the right at less than 45 degrees (for v > 0). Call the event where the line intersects the rear clock's worldline Rv, and the event where the line intersects the front clock's worldline Fv. Then all of the following are true: (1) For any v, the line t = vx is a "line of simultaneity" in the instantaneous rest frame of the rear clock at Rv *and* of the front clock at Fv. Therefore, Rv and Fv are simultaneous as seen by both the front and the rear clocks. (2) For any v, relative to the launch frame, the rear clock at Rv and the front clock at Fv are both moving at velocity v. Therefore, the front and rear clocks both see each other as at mutual rest at these events, and they both have the same "time dilation" factor at these events (because of the clock postulate). (3) For any v, the proper time experienced by the front clock between F0 and Fv is greater, by the ratio (R+L)/R, than the proper time experienced by the rear clock between R0 and Rv. So we have two clocks, which remain a constant distance apart at mutual rest, and have the same "time dilation factor" at any pair of corresponding events, and yet experience different proper times between corresponding events. I realize all this is true; I just point it out to explain why one has to be careful using the term "clock rate" without more explanation. Many people will think "time dilation factor", as in 1/sqrt(1  v^2), when they see "clock rate" (after all, that's what I initially thought when I saw it), but that's not what you mean by the term. In particular, I think this may be one source of confusion in some of Austin0's posts. 


#82
Jun212, 04:37 PM

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Well, what you have to take into account is that simultaneity is relative. Let e_{1} be the traveling twin celebrated his 20th birthday. Let e_{2} be the event at which the distant twin celebrated his 20th birthday. Let F be the "launch" frame of the traveling twin, and let F' be the "traveling" frame. Events e_{1} and e_{2} are simultaneous in frame F, but not in frame F'; in frame F', e_{2} took place long, long before e_{1}. So in jumping from frame F to frame F', the traveling twin changed his notion of what time "now" is on the distant planet, and so he changed his notion of how old the distant twin is. So for an accelerated observer, the time on a distant clock changes both because that clock advances, and also because the observer's notion of what is "now" for the distant clock changes. 


#83
Jun212, 05:39 PM

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#84
Jun312, 02:19 AM

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[tex]\frac{\Delta\tau_2}{\Delta\tau_1} = (1+g_1L/c^2) = \frac{R_2}{R_1} = \frac{g_1}{g_2}[/tex] To the accelerated observers on board the rocket, the clocks remain at constant distance from each other, so from their point of view, none of the observed red shift is due to classical Doppler shift caused by relative velocities. 


#85
Jun312, 10:56 PM

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so any calculation based on the MCIRF couldn't reveal relative velocity between front and back. When you say here the observer in the LF do you mean the initial launch frame or the current mCIRF ,,,,earlier you were refering to MCIrfs as LF In any case Doppler shift is as you say ,apparent dilation, so not really relevant the question was why take an approach which had these problems and the inherent ambiguity of the result due to these problems.?? It is still an accelerating system. Even disregarding the acceleration/velocity differential, a synchronization which works for one velocity cannot work for other different velocities. Yeah?? How could it? regarding measurement of relative clock rates at the front and back from the launch frame. LF accelerating system AF Simultaneity is not an issue. measurement of clock rates of course requires some interval of time a single event doesn't work. so in LF at (bx_{o},t_{0}) (fx_{0} ,t_{0}) we get observations bT'_{0} and fT'_{0} of AF .. at some later point at (bx_{1}, t_{1}), (fx_{1},t_{1}) we get observations bT'_{1} and fT'_{1} of AF (bx) t_{1}  (bx)t_{0} = (bx)dt (fx)t_{1} (fx)t_{0} = (fx)dt bT_{1}bT_{0} =dbT' fT_{1}  fT_{0}= dfT then (fx)dt/[itex]\gamma[/itex] =dfT' and (bx)dt/[itex]\gamma[/itex]=dbT' or not Would you agree that in this circumstance the simultaneity of the LF clocks at either measurement is not important because there is no direct comparison between the observations between them? The comparison is between an observation of the back clock with a later observation of the back clock.Etc. Of course it would be necessary to calculate the proper times for these events using the Rindler coordinates to make these comparisons. Also the measurement points could be widely separated say 0.7c and 0.8c More and more i suspect that the velocity dilation would be insignificant and might possibly agree with the Rindler predictions. I.e. would not increase with greater velocities. The acceleration magnitudes you used were totally unrealistic. The back of the rocket quickly passing the front and leaving it in the dust ;) so I am still unsure. thanks 


#86
Jun412, 03:11 AM

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P.S. I think Peter has a pretty good handle on it all in post #81. 


#87
Jun412, 07:09 AM

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#88
Jun412, 10:49 AM

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P.S. Some reservations about the none length contracting case have occurred to me. I will try to analyse that in detail later. 


#89
Jun412, 02:23 PM

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#90
Jun412, 03:06 PM

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The prediction is that if we allow length contraction (Born rigid acceleration) then the moving clock will be ahead of the clock that was always in the rear by a factor of 1+gL/c^{2}. So it's not simply some kind of illusion. 


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