From: relativity@paulba.no   
      
   Den 13.10.2025 08:31, skrev Thomas Heger:   
   > Am Samstag000011, 11.10.2025 um 21:43 schrieb Paul B. Andersen:   
   >> Den 11.10.2025 09:18, skrev Thomas Heger:   
   >>   
   >>> Einstein used (secretly) an absolute time and an absolute   
   >>> space in SRT.   
   >>   
   >> Consider the following scenario:   
   >> Given an Euclidean space where the axes are called x, y and z.   
   >> A is a point at x = 0, y = 0 and z = 0.   
   >> B is a point at x = L, y = 0 and z = 0.   
   >> Clock Ca is placed at A and another, equal clock Cb is at B.   
   >> The two clocks simultaneously show the same; they are synchronous.   
   >> An object is moving at constant speed from A to B.   
   >> When the object is at A clock Ca shows t = ta.   
   >> When the object is at B clock Cb shows t = tb.   
   >>   
   >> The time the object used to travel from A to B is Δt = tb - ta.   
   >> The object was moving at the speed v = L/Δt = L/(tb - ta).   
   >>   
   >> Note that the above is equally true according Newtonian Mechanics(NM)   
   >> and according to The Special Theory of Relativity (SR).   
      
   One would expect that everybody would understand the above,   
   but voalà:   
      
   >   
   > The clock at point 'A' should show 'A-time', which is the local time at   
   > point A.   
   >   
   > Same with point B and 'B-time'.   
   >   
   > But those time measures are local to A and B (and do not necessary run   
   > into the same direction at all possible points).   
      
   Which planet do you live on? :-D   
      
   On Earth, say in your living room, we have two stationary   
   clocks a distance L from each other.   
      
   These two clocks are equal, which means that   
   they run at the same rate.   
      
   These two clocks simultaneously show the same,   
   they are synchronous.   
      
   Do you seriously claim that this is impossible because   
   "those time measures are local to A and B (and do not necessary   
     run into the same direction at all possible points).   
      
      
   >   
   > That remote location uses the same time, which runs into the same   
   > direction and makes clocks tick at the same rate, that is an assumption.   
   >   
   > But this assumption also means, that all clocks in all the universe   
   > would run into the same direction and make clocks tick at the same rate.   
   >   
   > THAT is actually 'Newton's absolute time', which is 'external' (kind of   
   > 'God's clock').   
   >   
   > But if time isn't external, then time had to be restricted to the   
   > location in question.   
   >   
   > This would allow the observer at point 'A' (for instance) to declare   
   > himself to be at rest and everything else as moving.   
   >   
   > But that observer would need to consider, that all other observers could   
   > do the same, but with other time-measures.   
   Why are you stating all this irrelevant nonsense?   
      
   Not even you can be so ignorant and stupid that you don't   
   understand the following scenario:   
      
   Remember, this scenario happens in the real world.   
   Read it again and allow your self think!   
      
   Consider the following scenario:   
   Given an Euclidean space where the axes are called x, y and z.   
   A is a point at x = 0, y = 0 and z = 0.   
   B is a point at x = L, y = 0 and z = 0.   
   Clock Ca is placed at A and another, equal clock Cb is at B.   
   The two clocks simultaneously show the same; they are synchronous.   
   An object is moving at constant speed from A to B.   
   When the object is at A clock Ca shows t = ta.   
   When the object is at B clock Cb shows t = tb.   
      
   At t = ta, object O at A   
      
      Ca = ta                     Cb = ta   
      A                           B   
     -|---------------------------|---------> x   
      0                           L   
      O   
      
      
   At t = tb, object O at B   
      
      Ca = tb                     Cb = tb   
      A                           B   
     -|---------------------------|---------> x   
      0                           L   
                                  O   
   In case your editor clutters up the figure:   
   https://paulba.no/temp/Fig1.pdf   
      
      
   The time the object used to travel from A to B is Δt = tb - ta.   
   The object was moving at the speed v = L/Δt = L/(tb - ta).   
      
   Note that the above is equally true according Newtonian Mechanics   
   and according to The Special Theory of Relativity (SRT).   
      
   So is there no difference between NM and SR?   
      
   Of course it is.   
   In NM the Δt is 'absolute' in the sense that it is independent   
   of frames of reference.   
      
   So what about SR?   
      
   The metric below defines SR. So what can be deduced from this   
   metric is what SR predicts. This is indisputable.   
      
      (c⋅dτ)² = (c⋅dt)² − dx² − dy² − dz²                    (1)   
   or:   
      dτ² = (1 − (1/c²)⋅[(dx/dt)² + (dy/dt)² + (dz/dt)²])dt² (2)   
   or:   
      dτ = √(1 − v²/c²)dt                                    (3)   
      where v² = (dx/dt)² + (dy/dt)² + (dz/dt)²   
      
   (1), (2) and (3)  are the same metric.   
      
   Let's calculate the proper time of a clock that is moving from   
   the event    E1: t =  0, x = 0, y = 0, z = 0   
   to the Event E2: t = Δt, x = 0, y = 0, z = 0   
      
   According to (3): v = 0   
     τ = ∫(from t=0 to t=Δt)dt = Δt   
      
   But what would it be in the frame of reference K' (t',x',z') ?   
      
   Note that the metric (1) is true for any frame of reference   
   So:   
     (c⋅dτ)² = (c⋅dt')² −dx'² −dy'² −dz'²   
   and:   
     (c⋅dτ)² = (c⋅dt)² −dx² −dy² −dz² = (c⋅dt')² −dx'²   
   −dy'² −dz'²   
   note:   
     this does _not_ mean that t = t', x = x', y = y' and z = z'   
      
   Let the origin of K' move along the positive x-axis of K   
   with the speed v.   
   The events E1 and E2 will then have the coordinates in K':   
   E1: t' =   0, x' = 0,    y' = 0, z' = 0   
   E2: t' = Δt', x' = v⋅t', y' = 0, z' = 0   
      
   Note that v = dx'/dt' and v² = (dx'/dt')²   
   So according to (3):   
   dτ = √(1 − v²/c²)dt'   
   τ = ∫(from t'=0 to t'= Δt') √(1 − v²/c²)dt' = √(1 − v²/c²)Δt'   
      
   The proper time between two events is invariant   
   so: τ = Δt = √(1 − v²/c²)Δt'   
      
   The temporal interval between E1 and E2 is Δt in K   
   The temporal interval between E1 and E2 is Δt/√(1 − v²/c²) in K'   
   The temporal intervals between the same two events are   
   different in K and K'.   
      
   Temporal intervals between events are not absolute,   
   they are frame dependent.   
   (Note that Δt and Δt' are _not_ proper times.)   
      
   --   
   Paul   
      
   https://paulba.no/   
      
   --- SoupGate-Win32 v1.05   
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