From: relativity@paulba.no   
      
   Den 15.10.2025 09:44, skrev Thomas Heger:   
   > Am Montag000013, 13.10.2025 um 21:59 schrieb Paul B. Andersen:   
      
   >>   
   >> What the display on a 'clock' shows is a 'point in time'.   
   >>   
   >> I can say that the 'time' now is October 13, 2025, 14:34:32 CET   
   >> That's 'a point in time'.   
   >>   
   >> If I left my home at   t1 = October 13, 2025, 7:05:10 CET   
   >> and arrived at work at t2 = October 13, 2025, 7:59:50 CET   
   >> Then I can say that the 'time' I used to walk to work today   
   >> was 54 minutes and 40 seconds.  (Δt = t2 - t1)   
   >> That's a duration (or 'interval').   
   >>   
   >> So "time" can be both 'a point in time' and a duration.   
   >   
   > No: a point in time is also a duration, but where the reference point (=   
   > the beginning of the interval) is assumed to be known already.   
   You _know_ that we in the western world specifies "points in time"   
   with the Gregorian calendar and the time zones based on UTC.   
      
   >>   
   >> If we are only interested in durations, we may choose to   
   >> use a clock we can set to zero at any time.   
      
   We are free to choose our time scale any way we want.   
   But it must be precisely defined, it is not 'given by nature'.   
      
   Gregorian calendar and time zones based on UTC is   
   what 'everybody' uses in their everyday life.   
      
   You use your wristwatch to meet at work at the right   
   'point in time', don't you?   
      
   But for scientific work, we will probably use other definitions of   
   our time.   
      
   Se example below.   
      
   ---------------------------------   
      
   Is there any particular reason why you haven't   
   responded to the following?   
      
   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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