From: relativity@paulba.no   
      
   Den 29.11.2025 15:15, skrev Thomas 'PointedEars' Lahn:   
   > 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.   
   >   
      
   https://paulba.no/pdf/Clock_rate.pdf   
      
   >> [...]   
   >>>     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-)   
   >   
   >   
   > \\//,   
      
      
   --   
   Paul   
      
   https://paulba.no/   
      
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