The meaning of the curvature term


by Loro
Tags: curvature, meaning, term
Loro
Loro is offline
#1
Mar21-12, 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?
Phys.Org News Partner Space news on Phys.org
Unique pair of supermassive black holes in an ordinary galaxy discovered
How far are the planets from the Sun?
Previewing the bizzare April 29th solar eclipse
cepheid
cepheid is offline
#2
Mar21-12, 08:26 PM
Emeritus
Sci Advisor
PF Gold
cepheid's Avatar
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.
bapowell
bapowell is offline
#3
Mar22-12, 07:15 AM
Sci Advisor
P: 1,567
Quote Quote by cepheid View Post
a is not the radius of curvature, it is the scale factor.
In a closed universe, the scale factor *is* the radius of curvature.
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
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.

Chalnoth
Chalnoth is offline
#4
Mar22-12, 07:18 AM
Sci Advisor
P: 4,721

The meaning of the curvature term


Quote Quote by cepheid View Post
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.
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].
cepheid
cepheid is offline
#5
Mar22-12, 09:55 AM
Emeritus
Sci Advisor
PF Gold
cepheid's Avatar
P: 5,198
Quote Quote by Chalnoth View Post
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].
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.
Chalnoth
Chalnoth is offline
#6
Mar22-12, 01:14 PM
Sci Advisor
P: 4,721
Quote Quote by cepheid View Post
I don't think I said anything wrong,
No, I was just trying to clarify.
Loro
Loro is offline
#7
Mar28-12, 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 .


Register to reply

Related Discussions
What is the physical meaning of curvature? Calculus 3
What is the meaning of the term ''refractive index of a medium''? Introductory Physics Homework 1
Meaning of complex term General Math 2
Gaussian Curvature, Normal Curvature, and the Shape Operator Calculus & Beyond Homework 0
Help wanted for quantizing a pure Pontryagin term (Theta term) General Physics 1