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I Why does an Empty Universe have to obey Negative Curvature?

  1. Jan 13, 2019 #1
    Its stated that empty universe should have a hyperbolic geometry (Milne Universe) but I dont understand how its possible.
    $$H^2=\frac {8\pi G\epsilon} {3c^2}-\frac {\kappa c^2} {R^2a^2(t)}$$


    For an empty universe when we set ##\epsilon=0## we get
    $$H^2=\frac {-\kappa c^2} {a^2(t)}$$

    $$\ddot{a}(t)=-\kappa c^2$$



    However, $$\frac {\ddot{a}(t)} {a(t)}=-\frac {4\pi G} {3c^2}(\epsilon+3P)$$

    and for the acceleration equation, we get ##\frac {\ddot{a}(t)} {a(t)}=0##,

    for ##\epsilon=0## non-trivial solution happens only ##\ddot{a}(t)=0## for ##\kappa=0##

    So whats the problem here ?
     
    Last edited: Jan 13, 2019
  2. jcsd
  3. Jan 13, 2019 #2

    Jorrie

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    Without going into the technical stuff, as I understand it, a Milne universe is spatially flat, but has negative spacetime curvature, because the (empty) space is expanding at a constant rate (##\ddot a = 0)##.
     
  4. Jan 13, 2019 #3
    Hmm, so thats a different metric then the FLRW since FLRW rerpesents only spatial metrics. But why do we need a negative spacetime curvature ? Why it cant be just flat ?

    So ##\kappa=0## is true but space-time in negatively curved ?
     
  5. Jan 13, 2019 #4

    George Jones

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    Actually, it is the other way 'round: the Milne universe has zero spacetime curvature, and negative spatial curvature.


    How did you get the second equation? ##H^2 = \left( \dot a /a \right)^2##, which gives
    $$\left(\frac{\dot a}{a} \right)^2 = \frac{-\kappa c^2} {a^2}.$$
    Consequently, ##\kappa## must be negative.
     
  6. Jan 13, 2019 #5

    Jorrie

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    Nope, FLRW is a spacetime metric, because H has time in it: ##H = \frac{\dot {a}}{a}##.

    You must distinguish between curved space and curved spacetime. Minkowski spacetime is flat, because it does not expand: ##\kappa=0## refers to zero spatial curvature, but you must also have ##\dot {a} = 0## to get flat spacetime.
     
  7. Jan 13, 2019 #6
    oh yes theres also ##-c^2dt^2##
    I see now. I dont know why I said its not a space-time metric..
    oh I see cause theres square. I didnt notice that. So for a non trivial solution ##\kappa=-1## but it can be also ##\kappa=0## right ?
     
  8. Jan 13, 2019 #7
    If we say ##\kappa=0## then ##a(t)=C##

    and for
    ##\kappa=-1##

    ##\dot{a}(t)=c/R## or ##a(t)=tc/Rt_0## ?

    Whats the unit of ##a(t)## its unitless right ? From ##s_p=a(t)r##, but in ##V=HD##, ##H##has a unit of 1/s so, if ##a(t)## is unitless its derivative "gains" unit ??

    Are above equations true ?

    I also noticed that I missed ##R^2## term in the Friedmann Equation
     
    Last edited: Jan 13, 2019
  9. Jan 13, 2019 #8

    Jorrie

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    Oops yea, I goofed the words. Thanks for correction.
     
  10. Jan 13, 2019 #9
    I edited my post..It seems I made a mistake
     
  11. Jan 13, 2019 #10

    Orodruin

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    The Milne universe is Minkowski space, just with different coordinates. It is similar to using Rindler coordinates, but in the future light-cone of the Mikwoski space origin rather than the spatially separated region. You certainly do not need ##\dot a = 0## to get flat spacetime, as the Milne coordinates clearly show. It is similar to how Euclidean space is still flat regardless of whether you use spherical coordinates or not.
     
  12. Jan 13, 2019 #11

    kimbyd

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    Nit: negative spatial curvature, but that's an artifact of the coordinate system used. The total space-time curvature is identically zero.

    Basically, there's positive curvature from the expansion that is matched by the negative spatial curvature.

    And as Orodruin mentioned, it's the same exact space-time as the non-expanding Minkowski space-time. It's just a different coordinate system that makes it look curved.
     
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