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The meaning of the curvature term

by Loro
Tags: curvature, meaning, term
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Loro
#1
Mar21-12, 08:04 PM
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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?
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cepheid
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Mar21-12, 08:26 PM
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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
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Mar22-12, 07:15 AM
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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
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Mar22-12, 07:18 AM
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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
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Mar22-12, 09:55 AM
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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
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Mar22-12, 01:14 PM
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Quote Quote by cepheid View Post
I don't think I said anything wrong,
No, I was just trying to clarify.
Loro
#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 .


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