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Twin clocksis it acceleration? 
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#37
Feb2613, 10:46 AM

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The explanation must be in terms of some asymmetry between the twins. If there is no asymmetry then their ages must be identical. The reason students are confused by the twin scenario is that they have just been introduced to this idea of symmetry under boosts so they believe that the twins are symmetrical. Saying that "it's just geometry" is not a good answer, after all, you can have symmetrical geometric figures. Even geometrically, you must identify the asymmetry. In the standard twin paradox, the geometrical asymmetry is the bend in the worldline, i.e. the acceleration. So that is an accurate and not misleading answer. In other scenarios there are other asymmetries, the topology or the curvature or whatever. 


#38
Feb2613, 10:57 AM

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#39
Feb2613, 12:22 PM

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Drawing curves, comparing angles, sliding figures, measuring lengths with a string, etc. [In SR, measuring time with a clock is equivalent to measuring length with a string in plane geometry]. do not involve coordinates. You can then draw on the plane many different cartesian coordinates, or polar coordinates, then do calculations in terms of those. In SR you have a particular type of flat geometry. You can draw on it without coordinates. It has, objectively, straight lines. If you draw curvilinear coordinates on it, your curves are still curves  you haven't made them straight in a geometric sense. As for frames, I think a key thing that is hard to convince people is that there is actually no such thing as a global noninertial frame in SR (just as there is no such thing as a global frame at all in GR  inertial or otherwise). There are many possible accelerated coordinates (equivalent to curvilineary coordinates on a plane), each of which will represent the intrinsic flat geometry superficially differently. 


#40
Feb2613, 12:39 PM

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http://physicsforums.com/showthread....670653&page=10 post #135 http://physicsforums.com/showthread.php?t=644948&page=6 posts #92 and #93 http://physicsforums.com/showthread.php?t=671398&page=3 posts #35 and #36 In each of these three cases I provided an inertial spacetime diagram for the stayathome twin's inertial rest frame and a noninertial spacetime diagram for the traveling twin's noninertial rest frame and yet none of the OP's for which I provided these diagrams responded with any statement to the effect that they understood the diagrams. I just don't know if they are effective. 


#41
Feb2613, 12:40 PM

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#42
Feb2613, 12:54 PM

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"When you draw a space time diagram or integrate the path element, you are assuming a frame" for which I agree computing an integral requires coordinates, but drawing a spacetime diagram does not. I was not making any statement about pedagogy. On that score, I find different arguments work for different people. 


#43
Feb2613, 02:10 PM

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#44
Feb2613, 02:20 PM

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#45
Feb2613, 02:54 PM

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#46
Feb2613, 03:11 PM

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One interesting pedagogical trick: Draw up a spacetime diagram without the axes, then draw up a bunch of transparencies, each with their own set of t and x axes  and then drop the various transparencies onto the diagram to show how coordinates work. This has the nice property of making it clear that the proper intervals belong to the coordinatefree picture, and suggests why "frames" are called that. (There still is one downside  some people fall into the trap of believing that one particular frame, namely the one in which the x and t axes are at a 90degree angle, is somehow special and distinguished). 


#47
Feb2613, 03:17 PM

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I don't understand the phase where they have constant separation. In reality the blue signals send during that phase should arrive blueshifted, but there is no movement of the source in the diagram. If the signals are simply a picture seen through telescope, the angular size of blue twin, as seen by black twin would be increasing after turnaround. But in the diagram the viewing distance is constant for a few years. I think it is very difficult to make a smooth and correct diagram for the sharp turnaround version. The simpler case might be the one with constant proper acceleration, then you have just one coordinates chart (Rindler) for the rest frame of the accelerating twin, not two that you have to merge somehow. See Fig 7: http://cds.cern.ch/record/497203/files/0104077.pdf 


#48
Feb2613, 03:43 PM

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#49
Feb2613, 04:05 PM

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#50
Feb2613, 04:18 PM

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#51
Feb2613, 05:31 PM

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http://physicsforums.com/showpost.ph...80&postcount=5 From the wikipedia article, the metric for Rindler coordinates is: ds^2 =  g^2 x^2 dt^2 + dx^2 ... I don't know of a good online reference for FermiNormal coordinates of a uniformly accelerated observer in SR. However, section 6.6 of MTW derives this (without calling it that). This section is all flat spacetime. The metric becomes: ds^2 = (1+ g x)^2 dt^2 + dx^2 ... Note how this form approaches Minkowski metric for x=0. For Rindler coordinates x=0 represents the horizon. Also note that if you just transform Rindler to have its x=1 line become new x=0, you still don't get Fermi Normal; you get: ds^2 = (g+gx)^2 dt^2 + dx^2 ... What you need is x' = x  1/g to get from Rindler to FermiNormal. [Edit: On further thought, it appears that FermiNormal coordinates as above, for a uniformly accelerating observer would still have the same simultaneity surfaces as Rindler and Radar  just labeled differently. Thus, it seems, you need change in acceleration  e.g. the classic twin with slightly rounded turnaround  to expose the difference between Radar simultaneity and Born Rigid simultaneity (which is what FermiNormal uses).] 


#52
Feb2613, 06:53 PM

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#53
Feb2613, 07:20 PM

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#54
Feb2613, 07:33 PM

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