Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The meaning of the curvature term

  1. Mar 21, 2012 #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. Mar 21, 2012 #2


    User Avatar
    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. Mar 22, 2012 #3


    User Avatar
    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. Mar 22, 2012 #4


    User Avatar
    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. Mar 22, 2012 #5


    User Avatar
    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. Mar 22, 2012 #6


    User Avatar
    Science Advisor

    No, I was just trying to clarify.
  8. Mar 28, 2012 #7
    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 thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: The meaning of the curvature term
  1. Positive curvature (Replies: 2)

  2. Cosmological curvature (Replies: 27)