From: tjroberts137@sbcglobal.net   
      
   On 9/14/22 1:19 AM, Bruchpilot Aki wrote:   
   > Into which form of energy is the energy actually converted, which   
   > photons lose through the cosmic redshift?   
      
   Nothing "happens" to the energy of the EM radiation known as the CMBR,   
   because energy is not a "thing", it is a quantity used for bookkeeping.   
      
   A) Remember that the energy of a given object depends on the coordinates   
   used to measure it.   
   B) Remember also that Noether's theorem relates conservation of energy   
   to time-translation invariance, which in GR requires that a timelike   
   Killing vector is present   
      
   If there is no such invariance (no timelike Killing vector) one should   
   not expect energy to be conserved, and in the FLRW manifolds used to   
   model cosmology there is neither such a Killing vector nor conservation   
   of energy. [#]   
      
       [#] In a "small" local 4-volume in which the metric can be   
       considered to be constant within measurement resolutions,   
       one can construct locally-inertial coordinates in which   
       the time coordinate is indistinguishable from a timelike   
       Killing vector. So in this region, using such coordinates,   
       energy is conserved to within measurement resolutions.   
       This applies to every classical experiment that was used   
       to establish conservation of energy. And it most definitely   
       does not apply to the CMBR between emission and now.   
       Similarly for conservation of momentum and angular momentum.   
       (I have omitted some details and caveats for clarity.)   
      
   IOW: To calculate the energy of the CMBR when it was emitted requires   
   one to use locally-inertial coordinates at rest relative to the emitting   
   particles, and the dot product used to calculate it involves the metric   
   at that place and time. To calculate the energy of that same radiation   
   now requires one to use locally-inertial coordinates at rest relative to   
   the earth, and the dot product involves the metric here and now. Both   
   the coordinates and metric involved are different, which explains why   
   different values are calculated.   
      
   Tom Roberts   
      
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