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FRW - why is k constant?

  1. Mar 1, 2009 #1
    Hi, just a quick question.....in the Friedman equation, k takes a value of either 1,0,-1.

    But why is it a constant?

    Thanks you for your time!
     
  2. jcsd
  3. Mar 2, 2009 #2

    Chalnoth

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    Well, there are a few ways to think about it. I'll do it by writing down the FRW metric.

    First, consider the assumptions: we're assuming that there are observers for which the universe is homogeneous and isotropic. First, let's write down a completely general metric:

    [tex]ds^2 = dt^2 + f(x,y,z,t)(dx^2 + dy^2 + dz^2)[/tex]

    This is completely general because if there were a function in front of the time term, I could just perform a coordinate transformation to get rid of it. Now, let's start making use of those assumptions. The first assumption we can make use of is isotropy. This means that this function, whatever it is, cannot be a function of direction, so I can rewrite the metric as:

    [tex]ds^2 = dt^2 + f(r, t)(dx^2 + dy^2 + dz^2)[/tex]

    ...where r is shorthand for [tex]\sqrt{x^2 + y^2 + z^2}[/tex], and not an actual coordinate here.

    What remains, then, is to see what sorts of forms are valid for f(r,t) by enforcing the second condition, that of homogeneity. There are a couple of ways of doing this. First, we could make sure that if I perform a change of coordinates by setting [tex]\vec{r}' = \vec{r} + \delta\vec{r}[/tex], and leaving time unchanged, then the metric I should get must be identical. Or, alternatively, I could compute the spatial terms of the curvature tensor and ensure that they are independent of the spatial coordinates. Either will do.

    For simplicity's sake, the first thing that we do here is re-write the dependence on time in terms of the expansion factor a(t). Again, this is completely general: all I'm doing is specifying that there exists an invertible function a(t) for some range in t which we can use to determine the coordinate t within that range. Then the function can be cast as f(a, r). It turns out that one choice that works here that is completely general within our assumptions is the following:

    [tex]f(a, r) = \frac{a(t)}{1 - kr^2}[/tex]

    So, there you have it. Note that I could get a different form if I'd made a(t) a different function of time, but it turns out that since the function is separable in a and r, this makes for the simplest way to write it.

    (disclaimer: I haven't actually done this derivation in a while, so I might be misremembering the specifics. But I'm pretty sure this is accurate)
     
  4. Mar 2, 2009 #3

    Jorrie

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    Chalnoth has offered you a sophisticated view; here is my very down-to-Earth view.

    If I look at this form of the first Friedmann equation,

    [tex] H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3} \rho - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3} [/tex]

    it tells me that [itex]k[/itex] is the present curvature parameter of the universe, since at present [itex]a=1[/itex]; hence [itex]k[/itex] is constant for a given solution, i.e. closed, open or flat.
     
    Last edited: Mar 3, 2009
  5. Mar 3, 2009 #4

    Chronos

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    More importantly, the value of 'k' is irrelvant unless you reject the entire equation! I dont think you can do that without assuming the burden of some seriously 'out there' theories.
     
  6. Mar 3, 2009 #5

    Jorrie

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    Your statements do not seem to make sense to me. :confused: To which post/statement are you replying?
     
  7. Mar 3, 2009 #6
    It can be any value of a real number.But there's a degeneracy between k and redefinition of spatial coordinates.So we set k to be constant for simplicity.
     
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