The meaning of the curvature term

  1. I just wanted to make sure whether I've understood something correctly

    In the FRW equation:

    [itex](\frac{ \dot a}{a})^2 = \frac{8 \pi G}{3} \rho - \frac{k}{a^2}[/itex]

    ...there is this curvature term. I'm confused about the meaning of this k. Sometimes they say it can ONLY be -1 , 0 or +1. Sometimes they say it's smaller, bigger or equal zero. So can it or can it not be fractional? If it can - what does it mean?

    My understanding so far is, that this whole term is the Gaussian curvature:

    [itex] \pm \frac{1}{a^2} [/itex]

    Where a is the radius of curvature - and it changes with time as the universe expands;

    And so k is there just to provide an appropriate sign for the three cases: flat, spherical or hyperbolic geometry.

    Am I right, or can it be fractional?
  2. jcsd
  3. cepheid

    cepheid 5,189
    Staff Emeritus
    Science Advisor
    Gold Member

    a is not the radius of curvature, it is the scale factor.

    There are two different versions of k that can appear in the RW metric. One has dimension and can be either >0, <0, or =0. This k is equal to 1/R^2 where R is the radius of curvature.

    The other version of k is dimensionless. It has been normalized somehow (can't remember exact details). Therefore it is either 1, -1, or 0.

    I think that, with the Friedmann equation in the form that you gave, the k has to be the dimensional one. Therefore it is the spatial curvature

    One book I have uses kappa for the dimensional one and k for the dimensionless one.
    Last edited: Mar 21, 2012
  4. bapowell

    bapowell 1,982
    Science Advisor

    In a closed universe, the scale factor *is* the radius of curvature.
    I think you are making this more confusing than needed. The term [itex]k/a^2[/itex] gives the spatial curvature -- it is the Gaussian curvature of spatial slices of constant time. In the equation that Loro has written, [itex]k[/itex] is clearly a constant. It is equal to 1, 0, or -1 depending on the geometry.
  5. Chalnoth

    Chalnoth 5,672
    Science Advisor

    There are a couple of ways of doing it. One is to simply redefine [itex]a[/itex] so that the entire term, [itex]k/a^2[/itex] takes on the correct value.

    With this definition, the first Friedmann equation as written by Loro remains accurate. But by convention we usually take [itex]a=1[/itex] at the present time, and we don't have the freedom to pick the overall scaling of [itex]a[/itex] if we make that choice.

    Another way of doing it is to add a separate "radius of curvature" term, which requires replacing [itex]k[/itex] with, for example, [itex]kR^2[/itex].
  6. cepheid

    cepheid 5,189
    Staff Emeritus
    Science Advisor
    Gold Member

    I don't think I said anything wrong, you are just talking about the details of the normalization that I couldn't remember. The way I learned it was that[tex]\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho - \frac{1}{a^2\mathcal{R}^2}[/tex]where [itex] \mathcal{R} [/itex] is the radius of curvature at the present day. You can also write this term as [itex]\kappa/a^2[/itex] where [itex] \kappa = 1/\mathcal{R}^2[/itex] and κ is either > 0, or < 0, or = 0. This κ is what I think of as the "spatial curvature." This [itex]\mathcal{R}[/itex] is the thing that appears in the RW metric, i.e. [tex] ds^2 = dt^2 - a^2(t)[dr^2 +\mathcal{R}^2 \sin^2(r/\mathcal{R})(d\theta^2 + \sin^2\theta d\phi^2)][/tex]

    Now the book I have then takes a couple of other extra steps. First, you can apparently replace your co-moving radial distance coordinate "r" with co-moving angular diameter distance r1 instead, where [itex] r_1 = \mathcal{R}\sin(r/\mathcal{R}) [/itex]. With this substitution, the metric apparently becomes[tex] ds^2 = dt^2 - a^2(t)\left[\frac{dr_1^2}{1 - \kappa r_1^2} +r_1^2(d\theta^2 + \sin^2\theta d\phi^2)\right][/tex]The final substitution that the book mentions is that you rescale your radial distance coordinate so that r22 = κr12. Then the metric becomes [tex] ds^2 = dt^2 - R_1^2(t)\left[\frac{dr_2^2}{1 - k r_2^2} +r_2^2(d\theta^2 + \sin^2\theta d\phi^2)\right][/tex] where k = +1, 0, or -1 for universes with spherical, flat, and hyperbolic geometries respectively. The book points out that under this rescaling, [itex]R_1(t) = \mathcal{R}a(t) [/itex] so that at the present day, the value of your "scale factor" R1 is [itex]\mathcal{R}[/itex] rather than unity. So I can understand what you mean by the scale factor representing the curvature after this normalization has been done. The stuff I outlined above was the basis for what I said in my first post.
  7. Chalnoth

    Chalnoth 5,672
    Science Advisor

    No, I was just trying to clarify.
  8. Thank you all,

    The explaination of Cepheid clarifies that a lot. When I look at my notes now, that's actually exactly what my lecturer did, but then I have in my notes that [itex]\frac{1}{R^2}[/itex] is either ±1 or 0, which is obviously wrong...

    So in one form of the metric (one using the same units for all coordinates) there's [itex]\kappa = \frac{1}{R^2}[/itex] , and when we for some reason rescale our radial coordinate we get the other k = ±1 or 0 .
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?