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Why is the non-zero value of spatial curvature +/- 1?

  1. Nov 3, 2015 #1
    Going from the Newtonian to relativistic version of Friedmann's equation we use the substitution
    [itex]kc^{2} = -\frac{2U}{x^{2}}[/itex]

    The derivation considers the equation of motion of a particle with classic Newtonian dynamics. I can sought of see that if space is flat the radius of curvature will be infinite, so spatial curvature will vanish, but why exactly is |k| = 1 when space is not flat ?

    Also I'm guessing that k is actually a ratio of something over that things absolute value, e.g.
    [itex]k = \frac{thing}{\|thing\|}[/itex], because why else would it be +/- 1 or 0?
     
  2. jcsd
  3. Nov 3, 2015 #2

    Chalnoth

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    It's purely a convention. By this convention, when [itex]k = \pm 1[/itex], the scale factor is the radius of curvature.

    An alternative convention is to set the scale factor equal to 1 today. With this convention, the curvature parameter can take on any value, and is proportional to the inverse of the radius of curvature squared.
     
  4. Nov 4, 2015 #3
    Thanks, do you know where I can learn more about his? The introductory cosmology books just state this without justifying it and that bugs me.
     
  5. Nov 4, 2015 #4

    Chalnoth

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  6. Nov 4, 2015 #5
    So the radius of curvature is a function of the scale factor, but why? That's what I'd like to know. It isn't obvious to me. Thanks
     
  7. Nov 4, 2015 #6

    Chalnoth

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    If the universe is described by a sphere, then the radius of curvature is the radius of the sphere. If you increase the distances between every two objects by a factor of two, you increase the radius of the sphere by a factor of two, and therefore increase the radius of curvature by a factor of two.
     
  8. Nov 8, 2015 #7

    Chronos

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    That only works for a finite universe. But, then again, Einstein strongly favored that option having realized his field equations are not well behaved in the presence of infinities.
     
  9. Nov 9, 2015 #8

    Jorrie

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    Actually, it is a function of the curvature constant [itex]\Omega_k= 1-\Omega_{total}[/itex]. The scale factor is just what it says, something to scale the effect of all the Omegas to earlier times. For a small positive curvature (i.e. a small negative curvature constant, unfortunately, due to legacy) the radius of spatial curvature is [itex]R_k \approx R_{Hubble}/\sqrt{\Omega_{total} - 1}[/itex]. For spatially flat it blows up to infinity and for a spatially open universe it comes out as an imaginary number.
     
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