
#1
Mar2112, 08:04 PM

P: 61

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
Mar2112, 08:26 PM

Emeritus
Sci Advisor
PF Gold
P: 5,198

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. 



#3
Mar2212, 07:15 AM

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P: 1,551





#4
Mar2212, 07:18 AM

Sci Advisor
P: 4,721

The meaning of the curvature termWith 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]. 



#5
Mar2212, 09:55 AM

Emeritus
Sci Advisor
PF Gold
P: 5,198

Now the book I have then takes a couple of other extra steps. First, you can apparently replace your comoving radial distance coordinate "r" with comoving angular diameter distance r_{1} 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 r_{2}^{2} = κr_{1}^{2}. 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" R_{1} 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. 



#6
Mar2212, 01:14 PM

Sci Advisor
P: 4,721





#7
Mar2812, 05:52 PM

P: 61

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 . 


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