XPost: rec.arts.tv, alt.fan.rush-limbaugh   
   From: relativity@paulba.no   
      
   Den 29.10.2025 05:09, skrev The Starmaker:   
   >   
   > Michelson’s analysis of errors in his 1881/87   
   > experiments of the two swimmers contest;   
      
   You obviously meant Reinhard A. Schumacher's analysis of errors in   
   Michelson's 1887 MMX experiment.   
      
   Michelson never made an experiment with two swimmers.   
      
   >   
   > https://iopscience.iop.org/article/10.1088/1742-6596/2197/1/012012/pdf   
      
   The idea of using two swimmers to analyse the MMX is quite stupid.   
      
   Schumacher claims that he with this analogy can prove that   
   if light were a wave in the ether as assumed by Michelson,   
   then a Michelson interferometer would not be able to measure   
   the speed of the ether relative to the interferometer.   
      
   This is a ridiculous claim, and I will not bother to explain why.   
      
   However, Schumacher is correct when he claims that there is   
   an error in Michelson's derivations, so let's go through them.   
      
   (I am sure that this error is well known.)   
      
   https://paulba.no/paper/Michelson_1887.pdf   
      
   In the following is "c" the speed of light relative to the ether,   
   and "v" is the speed of the ether relative to the interferometer.   
   "D" is length of the interferometer arms.   
      
   Analysis of the interferometer arm which is parallel to   
   the velocity of the ether:   
   =======================================================   
      
   In the rest frame of the interferometer, the speed of the light   
   going with the speed of the ether is (c+v) while the speed   
   of the light going in the opposite direction is (c-v).   
      
   So the time for the light to go forth and back the arm is:   
     t₁ = D/(c+v) + D/(c+v) = 2D⋅c/(c²−v²)   
   The distance the light have travelled through the ether is:   
     D₁ =  t₁⋅c = 2D/(1-v²/c²) ≈ 2D⋅(1+v²/c²)    (v/c << 1)   
      
   This is in  accordance with Michelson's calculation.   
      
   Analysis of the interferometer arm which is transverse to   
   the velocity of the ether:   
   =======================================================   
      
   The correct calculation:   
   -----------------------   
            .   
           / \   
          /θ|θ\   
         /  |  \   
      ct/   |   \ct   
       /    |D   \   
      /     |     \   
     /______| _____\   
        vt     vt   
      
     (vt)² + D² = (ct)² => t = D/√(c²−v²)   
      
   So the time for the light to go forth and back the arm is:   
      t₂ = 2D/√(c²−v²)   
   The distance the light has travelled through the ether is:   
     D₂ = t₂⋅c = 2D/√(1−v²/c²) ≈ 2D/(1-v²/2c²) ≈    
   D⋅(1+v²/2c²)   
      
   Note: sin θ = v/c   
      
   Michelson's calculation:   
   ------------------------   
            .   
           / \   
          /θ|θ\   
         /  |  \   
      D'/   |   \D'   
       /    |D   \   
      /     |     \   
     /______| _____\   
        vt     vt   
      
   Michelson (wrongly) consider it as evident that   
   the angle θ is the classical aberration.   
   (Aberration according to the Galilean transform.)   
   That is, tan θ = v/c   
      
   So vt/D = v/c  => vt = D⋅v/c   
      
   D'² = D² + (vt)² = D² + D²⋅v²/c² = D²(1+v²/c²)   
   D' = D⋅√(1+v²/c²)   
      
   The distance the light has travelled through the ether is:   
     D₂ = 2D' =  2D⋅√(1+v²/c²) ≈ 2D⋅(1+v²/2c²)   
      
   Note that this approximation of the wrong D₂ is equal   
   to the approximation of the correct D₂.   
      
   So Michelson's calculation that (D₁-D₂) ≈ D⋅v²/c²   
   remains correct.   
      
   Michelson's error has no consequence for the validity of the MMX.   
      
   --   
   Paul   
      
   https://paulba.no/   
      
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