Geodesic Equation: Lagrange Approximation Solution for Schwarzschild Metric

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AleksanderPhy
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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 (which could be used to describe example sun) which 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.
 
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AleksanderPhy said:
that means that ∂gμν/∂xλ=0.

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