Forums before death by AOL, social media and spammers... "We can't have nice things"
|    sci.physics.relativity    |    The theory of relativity    |    225,861 messages    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
|    Message 223,989 of 225,861    |
|    Paul B. Andersen to All    |
|    Re: The problem of simultaneity    |
|    19 Oct 25 21:02:20    |
      From: relativity@paulba.no              Den 18.10.2025 15:37, skrev Richard Hachel:       >       > To fully understand the theory of relativity, we must begin with the       > Poincaré-Lorentz transformations and visualize as precisely as possible       > what they imply in our minds.       > To research these transformations, there are two simple methods: the one       > used by Henri Poincaré and Richard Hachel (known as the French method)       > and the one used by Albert Einstein and Hermann Minkowski (known as the       > Mittleuropa method).       > The first involves observing a train car moving from left to right; the       > second involves imagining oneself in the train car and observing the       > landscape moving from right to left.              The Lorentz transform is a coordinate transformation       from one frame of reference to another.       So we must obviously start with defining the frames       of references, and how they move relative to each other.       Let the frames be called K(t,x,y,z) and K'(t',x',y',z').              If we let the axes be parallel (x||x',y||y',z||z'),       let the origins be aligned at t = t'= 0, and let       the motion be along the x axes, we will have two variants:              In the following is γ = 1/√(1−v²/c²)              #1: The origin of K' is moving along the positive x-axis        of K with the speed v.        t' = γ(t - (v/c²)⋅x)        x' = γ(x - v⋅t)        y' = y        z' = z              The inverse transform:        t = γ(t' + (v/c²)⋅x')        x = γ(x' + v⋅t')        y = y'        z = z'              #2: The origin of K is moving along the positive x'-axis        of K' with the speed v.        t' = γ(t + (v/c²)⋅x)        x' = γ(x + v⋅t)        y' = y        z' = z              The inverse transform:        t = γ(t' - (v/c²)⋅x')        x = γ(x' - v⋅t')        y = y'        z = z'              It is nonsense to call #1 and #2 "different methods".       The difference is a trivial change of direction of the motion.              When you call them "different methods" you haven't understood       that the difference is only a change in direction of the motion.                     > The transformations are then said to be positive or negative. In       > positive form (Poincaré, Hachel), we have:       > x'=(x+v.To)/sqrt(1-Vo²/c²)       > y'=y       > z'=z'       > t'=(t+x.Vo/c²)/sqrt(1-Vo²/c²)              The above is a coordinate transformation from       the frame (t,x,y,z) to (t',x',y',z')              > In negative form, we have:       > x'=(x-v.To)/sqrt(1-Vo²/c²)       > y'=y       > z'=z'       > t'=(t-x.Vo/c²)/sqrt(1-Vo²/c²)              The above is a _different_ coordinate transformation       from the frame (t,x,y,z) to (t',x',y',z')              > It's just the same thing, said differently.              No. It's simply wrong!       You can't have two different coordinate transformations       between the same two frames of reference.              Your failure to define the frames of reference properly       shows that you don't know what you are doing.              --       Paul              https://paulba.no/              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
(c) 1994, bbs@darkrealms.ca