From: hees@fias.uni-frankfurt.de   
      
   On 30/05/2020 16:17, Nicolaas Vroom wrote:   
   > On Thursday, 28 May 2020 20:14:03 UTC+2, Tom Roberts  wrote:   
   >> On 5/25/20 8:38 AM, Nicolaas Vroom wrote:   
   >>   
   >>> What is the reason that you cannot use one reference frame for the   
   >>> whole experiment from start to finish i.e. when both clocks again   
   >>> can be compared at point O?   
   >> There is no such reason, and my discussion explicitly showed using the   
   >> inertial frame of A for the entire analysis.   
   > In fact, there are two issues:   
   > 1) Is it wise to select only one reference frame?   
   > 2) Which is the best reference frame to select?   
   > Apparently you agree with issue 1.   
   > I would prefer point O, as indicated in the text.   
   Ad 1) It is wise to formulate a problem in a manifestly covariant   
   way and then use the most convenient reference frame to do the   
   calculations.   
      
   The obviously Lorentz-invariant quantity to be calculated here is   
   the proper time tau of the moving twin as a function of the   
   proper time of the twin staying in his inertial frame at rest, which   
   is the coordinate time, t, of this inertial frame.   
      
   Ad 2) In this case of course the inertial frame (the restframe of   
   the twin staying at home all the time).   
      
   The unique Lorentz-invariant solution is   
   \tau = \int d t \sqrt{1-\beta^2(t)},   
   where \beta=|\vec{v}|/c, where \vec{v} is the three-velocity of the   
   moving twin as measured in the inertial frame of the resting twin.   
      
   That completely solves the task and shows that \tau   
   >>>> T = \integral sqrt(1-v^2/c^2) dt   
   >>> My understanding is that t represents the time (clock count) of a   
   >>> clock at rest and T the (time) clock count of a moving clock.   
   >> Not quite. t is the time coordinate of the inertial frame,   
   > This raises a new question: What is the difference my 't represents the   
   > time of a clock in a reference frame' versus 'the time coordinate of   
   > the inertial frame'   
   t is the proper time of the resting twin, which is of course identical   
   with the inertial   
   restframe's coordinate time.   
   >   
   >> So it is clear that the time coordinate   
   >> of the inertial frame is the same as the elapsed proper time of a clock   
   >> at rest in it (reset the clock to zero at t=0). They are conceptual   
   >> different but numerically equal.   
   > My philosophy is to use as many concepts as possible   
   My philosophy is to use as simple as possible concepts, and in   
   relativity the most simple   
   concept is to describe a physical situation in terms of   
   Lorentz-invariant quantities. Here   
   we compare two Lorentz invariant proper times: the proper time of the   
   twin at rest and the   
   proper time of the traveling twin the round-trip of the traveling twin   
   takes.   
      
   This resolves all paradoxes with different frames of reference, because   
   the results do not depend   
   on any frame of reference, and it's allowed to use any frame of   
   reference to evaluate the quantities.   
   For that I choose the most convenient reference frame.   
   >   
   >> 	A clock can only indicate its proper time, and only   
   >> 	along its worldline; the time coordinate of an inertial   
   >> 	frame can indicate the frame's coordinate time anywhere   
   >> 	the frame is valid. Of course, such coordinates are   
   >> 	models, and to implement them requires physical clocks   
   >> 	located where events of interest occur; all such clocks   
   >> 	must, of course, be synchronized to the coordinate time.   
   > That means IMO that all the clocks should be synchronized with   
   > one standard clock. IMO at the origin O   
   Of course the clocks have to be synchronized at one instant, i.e., in   
   this case on the   
   beginning of the trip, where both twins are at the location 0. That's   
   another important   
   concept: use as long as you can simple local events to describe a   
   situation in relativity.   
   >   
   >>> The question is how is v (the speed of the clock) measured?   
   >> In the usual way: v=dx/dt, where x(t) is the position of the object at   
   >> time t, and t is the time coordinate of the inertial frame.   
   > In practice, this means that t is the time of the nearest physical clock.   
   There is not one "nearest physical clock". As this example shows, what a   
   clock once   
   synchronized with another clock, when both clocks have been at the same   
   place, shows after   
   some time when you compare these clocks (again at a common place!)   
   depends on   
   the motion of these clocks.   
      
   In SR it's pretty easy to synchronize a set of clocks in one inertial   
   frame, because you can   
   use light signals and the univerality of the speed of light. That's how   
   Einstein's clock synchronization   
   works for an inertial frame.   
      
   In accerated reference frames or even relativity, clock synchronization   
   and the definition of "distance"   
   becomes a complicated issue, but that's not related to the   
   special-relativistic twin paradox at all.   
      
      
   --   
   Hendrik van Hees   
   Goethe University (Institute for Theoretical Physics)   
   D-60438 Frankfurt am Main   
   http://fias.uni-frankfurt.de/~hees/   
      
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