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I Derivative of metric

  1. Aug 25, 2016 #1
    Hello so if we have geodesic equation lagrange
    approximation solution:
    d/ds(mgμνdxν/ds)=m∂gμν∂xλdxμ/ds dxν/ds. So if we have schwarzschild metric (wich could be used to describe example sun) wich is:ds2=(1-rs/r)dt2-(1-rs/r)-1dr2-r2[/SUP]-sin22. But that means that ∂gμν/∂xλ=0. So that means that first equation will equal to zero so that means that sun has no gravity effect to test particle. But according to my knowledge sun does pull things towards itself.
     
  2. jcsd
  3. Aug 25, 2016 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    No, it doesn't. The metric coefficients are functions of ##r##, which is one of the ##x^\lambda##.
     
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