From: relativity@paulba.no   
      
   Den 27.11.2025 22:12, skrev Thomas 'PointedEars' Lahn:   
   > Paul B. Andersen wrote:   
   >> Den 26.11.2025 12:52, skrev J. J. Lodder:   
   >>> Summary: The GPS sat clocks are steered such that they are synchronous   
   >>> with the GPS master clocks at USNOP.   
   >>> (applying the appropriate corrections, Doppler and relativity)   
   >>   
   >> There is no correction for the Doppler shift due to the velocity   
   >> of the SVs relative to the receiver.   
   >> The gravitational Doppler shift is included in the (1-4.4647e-10)   
   >> rate correction of the SV clock.   
      
   I should have said:   
   "The gravitational blue shift is included in the (1-4.4647e-10)   
     rate correction of the SV clock."   
      
   >   
   > Gravitational redshift is NOT a Doppler shift   
      
   It's not unheard of to call it a Doppler shift, but you are right.   
   In this case it is a gravitational blue shift.   
      
   < (it arises from spacetime   
   > curvature, not relative motion¹), but my professor in General Relativity has   
   > confirmed to me that the interpretation of gravitational redshift as   
   > successive relativistic Doppler shifts within a infinite family of observers   
   > moving relative to each other due to different gravitational acceleration   
   > (as explained by Veritasium) is valid.  (Still I think that it is an   
   > unnecessarily complicated interpretation of the phenomenon.)   
   >   
   > [By contrast, *cosmological* redshift cannot be properly understood as a   
   > relativistic Doppler shift; cf. Davis & Lineweaver, "Expanding Confusion",   
   > 2003.]   
   >   
   > ___   
   > ¹ For the Schwarzschild metric in Schwarzschild coordinates, with sign   
   >    convention (−+++) and in natural units,   
   >   
   >      ds² = −dτ²   
   >          = −(1 − 2m/r) dt² + (1 − 2m/r)⁻¹ dr² + r² dΩ²   
   >          ≡ −A(r) dt² + A(r)⁻¹ dr² + r² dΩ²,   
   >   
   >        m ≡ G M/c²,   
   >      dΩ² ≡ dθ² + sin²(θ) dφ²,   
   >   
   >    for a stationary observer (dr² = dΩ² = 0) we have   
   >   
   >      dτ = dt √[A(r)].   
      
           dτ/dt = √[1 − 2m/r] ≈ (1 − m/r)   
      
           a very good approximation for all r ≥ Earth's radius   
      
   >   
   >    For two observers at the radial coordinates r₁ and r₂, r₂ > r₁,   
   >    we have (trivially)   
   >   
   >              dτ₁ = dt √[A(r₁)]   
   >              dτ₂ = dt √[A(r₂)],   
      
            dτ₂/dτ₁ ≈ (1 − m/r₂)/(1 − m/r₁) ≈ 1 + m(1/r₁ -   
   1/r₂)   
      
            another very good approximation for all r ≥ Earth's radius   
   >   
   >    therefore   
   >   
   >          dτ₂/dτ₁ ~ f₁/f₂ = c λ₂/(c λ₁) = λ₂/λ₁,   
      
               dτ₂/dτ₁ = f₁/f₂   
   >   
   >    where f₁ and λ₁ are the frequency and wavelength of the EM radiation   
   >    emitted by the observer at r₁, and f₂ and λ₂ are the frequency and   
   >    wavelength of the same radiation as received and observed by   
   >    the observer at r₂.   
   >   
   >      ==>      λ₂ ~ λ₁ dτ₂/dτ₁   
   >                  ~ λ₁ √[A(r₂)/A(r₁)]   
   >                  ~ λ₁ √[(1 − 2m/r₂)/(1 − 2m/r₁)]   
   >                  > λ₁. ∎   
   >   
   >      [r₂ > r₁ ⇔ 2m/r₂ < 2m/r₁   
   >               ⇔ 1 − 2m/r₂ > 1 − 2m/r₁   
   >               ⇔ (1 − 2m/r₂)/(1 − 2m/r₁) > 1.]   
      
       f₁ ≈ (1 + m(1/r₁ - 1/r₂))·f₂   
      
   if   r₁ = 6.378e6 m,  radius of Earth   
   and  r₂ = 26.56e6 m,  radius of GPS orbit   
      
   then f₁ = (1 + 5.2867e-10)·f₂   
      
   The observer on a non rotating spherical Earth would see   
   the frequency of the stationary emitter over his head   
   blue shifted.   
      
   --   
   Paul   
      
   https://paulba.no/   
      
   --- SoupGate-Win32 v1.05   
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