On Saturday, September 5, 2020 at 3:36:26 AM UTC-5, PengKuan Em wrote:   
   > a)	Material clock   
   >   
   > What is time? This question is tricky because in relativity time-rate   
   > changes when frame of reference changes. Time-rate changing is puzzling   
   > because it is in conflict with our intuition that time is the flow of   
   > ticks of clocks of which the mechanical structure does not change. Then   
   > how to clearly explain the contradiction between the constant flow of   
   > ticks delivered by clocks and the relativistic time dilation?   
      
   We can fix that. Make it both. A coordinate time (t) and historical   
   time (s), which we'll identify as proper time. Throw it in as another   
   coordinate too. The Minkowski line element for proper time   
      
   ds^2 = dt^2 - (1/c)^2 (dx^2 + dy^2 + dz^2)   
      
   then becomes   
      
   ds^2 - dt^2 + (1/c)^2 (dx^2 + dy^2 + dz^2) = 0.   
      
   Now ... let's regularize it by transforming the coordinate to the   
   *difference* of proper time and coordinate time, defining   
      
   u := c^2 (s - t).   
      
   Then, the line element can be rewritten as the quadratic invariant:   
      
   dx^2 + dy^2 + dz^2 + 2 dt du + (1/c)^2 du^2 = 0   
      
   and, in addition, we have a linear invariant   
      
   ds := dt + (1/c)^2 du.   
      
   which may be interpreted as a "soldiering form" which ties the   
   historical "flowing" time onto the space-time geometry.   
      
   In the resulting geometry, it's *all* *four* *dimensions* which   
   flow in time, not just space! It's both "block time" and "flowing   
   time" at the same time. The entire block, itself, is flowing in   
   time!   
      
   To find what these things mean, consider first that these both have   
   non-relativistic limits:   
      
   dx^2 + dy^2 + dz^2 + 2 dt du = 0,   
   ds = dt.   
      
   In non-relativistic theory, by this account, it is justified to   
   treat coordinate time as the "flowing" time. On account of this,   
   the two may be safely confused in non-relativistic theory.   
      
   But by stepping forward into relativity and then moving back into   
   non-relativistic form, an extra item has cropped up that wasn't   
   there before and now, suddenly, you also actually have a space-time   
   metric for non-relativistic theory ... and one that morphs continuously   
   into the metric for relativity by adjusting a parameter alpha from   
   0 to (1/c)^2, with the generic case being   
      
   dx^2 + dy^2 + dz^2 + 2 dt du + alpha du^2 and ds = dt + alpha du.   
      
   The quadratic invariant identifies a metric that *always* has a 4+1   
   signature, for *all* values of alpha.   
      
   Second, consider what the most general linear transforms are that   
   leave these invariants intact; denoting infinitesimal transforms   
   by D(...):   
      
   dx D(dx) + dy D(dy) + dz D(dz) + (dt + (1/c)^2 du) D(du) + du D(dt) = 0,   
   0 = D(ds) = D(dt) + (1/c)^2 D(du).   
      
   Substituting the second equation D(dt) = -(1/c)^2 D(du) into the   
   first yields:   
      
   dx D(dx) + dy D(dy) + dz D(dz) = -dt D(du).   
      
   The general solution is:   
   D(dx, dy, dz) = omega x (dx, dy, dz) - upsilon dt   
   D(du) = upsilon . (dx, dy, dz)   
   D(dt) = -alpha upsilon . (dx, dy, dz)   
   where alpha is the above-mentioned parameter and the two new vectors are:   
      
   omega = infinitesimal rotations,   
   upsilon = infinitesimal boosts,   
   () . () denotes 3-vector dot-product,   
   () x () denotes 3-vector cross-product.   
      
   This is the 5-dimensional representation of the Lorentz group (when   
   alpha = (1/c)^2) and of the Galilei group (when alpha = 0) ... which   
   *not* the Galilei group, but the Bargmann group!   
      
   All cases are restrictions of the symmetry group SO(4,1), for the   
   4+1 metric, that correspond to the little group for the linear   
   invariant ds.   
      
   Now, to determine what this means, consider how the mass (m),   
   momentum (p) and kinetic energy (H) transform in non-relativistic   
   theory:   
      
   D(m) = 0, D(p) = omega x p - upsilon m, D(H) = -upsilon.p.   
      
   Under the classical version of the correspondence rule   
   p-hat = -i h-bar del, H-hat = i h-bar @/@t   
   (@ denotes the curly-d partial derivative symbol)   
   one has the correspondence   
   p <-> -del, H <-> @/@t   
   and this leads to the consideration of the corresponding one-form:   
   p.dr - H dt.   
      
   What is the transform of this?   
   D(p.dr - H dt)   
      = (omega x p - upsilon m).dr + p.(omega x dr - upsilon dt)   
      - (-upsilon.p) dt - H (0)   
      = -m (upsilon.dr)   
      = -m D(du)   
      = -D(m du),   
   This puts the spot-light on the one-form   
   p.dr - H dt + m du   
   showing that it is actually an invariant.   
      
   If we adopt these same quantities for Relativity and *continue* to   
   assume that this is the case for the relativistic form, by turning   
   on the parameter alpha = 0 to alpha = (1/c)^2, then the assumption   
   that this be invariant leads to:   
      
   0 = D(p.dr - H dt + m du)   
     = Dp.dr + p.(omega x dr - upsilon dt)   
     - DH dt - H (-alpha upsilon.dr)   
      + Dm du + m (upsilon.dr)   
     = (Dp - omega x p + upsilon (m + alpha H)).dr   
     - (DH + upsilon.p) dt + (Dm) du   
   then we obtain the following transforms   
      Dp = omega x p - upsilon M   
      DH = -upsilon.p   
      Dm = 0   
   which singles out the "moving" mass M = m + alpha H as the mass   
   that goes with the momentum in the formula (momentum = mass times   
   velocity). Its transform, derived from those above, is   
      DM = -alpha upsilon.p   
   and, as a consequence, we find the following as the two invariants   
   under these transforms:   
      
   Linear invariant: mu := M - alpha H = m,   
   Quadratic invariant: lambda := p^2 + 2MH - alpha H^2 = 0.   
      
   The coordinate u is conjugate to m and in a quantized theory, m   
   would be represented as i h-bar @/@u. The coordinates (s,u), when   
   used in place of (t,u) produce the one-form   
      
   m du - H dt = m du - H (ds - alpha du) = (m + alpha H) du - H ds = M du - H ds.   
      
   So the corresponding operator forms would be, respectively,   
   M <-> i h-bar (@/@u)_s, H <-> -i h-bar (@/@s)_u.   
      
   The energy term H - which is the relativistic form of the *kinetic*   
   energy (not the total energy) is conjugate to the proper time s,   
   provided that it and u be taken together as the coordinates.   
      
   That generalizes the non-relativistic prescription of taking H to   
   be the generator of "flowing time".   
      
   This clean, continuous deformation from non-relativistic to   
   relativistic form is obscured because in Relativity, one normally   
   takes the *total* energy E, instead of the kinetic energy H as the   
   relevant component of momentum.   
      
   Here, that arises from the fact that the 4D sub-representation (M,p)   
   of the 5D representation (M,p,H) closes under the transforms, so   
   (M,p) forms a Minkowski 4-vector. M is, of course, converted to E   
   by way of the equation   
      
   E = M c^2.   
      
   So the corresponding transforms would read:   
      
   D(p) = -(1/c)^2 upsilon E, D(E) = -upsilon.p   
      
   and we may find the rest-mass as the square of the "mass shell"   
   invariant, which can be constructed from the quadratic invariant   
   lambda and linear invariant mu as:   
      
   mu^2 - alpha lambda   
      = (M - alpha H)^2 - alpha (p^2 - 2MH + alpha H)   
      = M^2 - alpha p^2   
      = (E/c^2)^2 - (p/c)^2 = m^2.   
      
   Strictly speaking, this construction goes BEYOND relativity, since   
   it has 5 components. The difference can be brought out clearly by   
   considering what the rest-frame form of the 5-vector (H,p,M) is in   
      
   [continued in next message]   
      
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