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Newtonian limit

  1. Jan 30, 2006 #1

    hellfire

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    According to Sean Carroll’s Lecture Notes on General Relativity, equation (4.20) and text below, the Newtonian limit is obtained when [itex]g_{00} = -(1+2\phi)[/itex]. However, in Schutz’s “A first course in General Relativity” it is additionally required that [itex]g_{11} = g_{22} = g_{33} = (1+2\phi)[/itex]. I do not understand the need for this. Are both metrics a correct newtonian limit? Why does Schutz put this additional requirement?
     
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  3. Jan 30, 2006 #2

    Stingray

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    To be complete. But the spatial components of the metric aren't needed to recover the Newtonian equations of motion. Look at the geodesic equation for objects moving at slow speeds.
     
  4. Jan 30, 2006 #3

    hellfire

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    Thanks Stingray. So both are correct Newtonian limits. What do you mean with "to be complete"?
     
  5. Jan 30, 2006 #4

    Stingray

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    I mean that the "Newtonian metric" satisfies both of those conditions. They are not separate things. It's just that the spatial components happen to drop out of the equations of motion as long as they're close to [tex]\delta_{ij}[/tex].
     
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