From: PointedEars@web.de   
      
   Paul B. Andersen wrote:   
   > Den 27.11.2025 22:12, skrev Thomas 'PointedEars' Lahn:   
   >> Paul B. Andersen wrote:>>> The gravitational [blueshift] is included in the   
   (1-4.4647e-10)   
   >>> rate correction of the [GPS] SV clock.   
   >>   
   >> [...]>> ___   
   >> 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   
      
   Actually, expanding dτ/dt into a Maclaurin series gives   
      
       √(1 − 2m/r)   
     = (1 − 2m/r)^{1/2}   
     = (1 − w)^{1/2}   
     = (1 − w)^{1/2}|_{w=0} w^0 − 1/2 (1 − w)^{−1/2}|_{w=0} w^1 + O(w^2)   
     = 1 − 1/2 w + O(w^2)   
     = 1 − m/r + O[(2m/r)^2],   
     ≈ 1 − m/r   
      
   so this is a good approximation for all r ≫ 2m = 2G M/c^2 = R_S; but   
      
     R_{S,Terra} ≈ 2G M_Terra/c^2 ≈ 8.85 mm ≪ 6371 km = R_Terra,   
      
   with M_Terra ≈ 5.97 × 10²⁴ kg.  So it is a very good approximation "for   
   all   
   r ≥ [Terra's] radius" *in the case of Terra as the spacetime   
   curvature-generating object*, indeed.   
      
   > [...]   
   >>    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.   
      
   Indeed.  Well done 8-)   
      
      
   \\//,   
   --   
   PointedEars   
      
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