To Kimbyd,
Thanks again and I meant what I said. There's nothing wrong with what you've written. I'm just a stubborn student trying to understand why things have to be a certain way.
kimbyd said:
The curvature in the FLRW universe is a constant value set by the initial conditions
Why? This is an assertion not a proof.
You say something similar again later:
kimbyd said:
But General Relativity does say that a mismatch between the rate of expansion and density manifests as spatial curvature. Which can't change because it's just a constant set by the initial conditions.
Yes to the first, my understanding of what G.R. says is summarised below. The second sentence is just an assertion not a proof. The purpose of the OP is to test that and understand why.
What does determine the curvature parameter?
kimbyd said:
It's determined by the relationship between the density of the universe and how fast it's expanding.
Yes exactly. If I can get LaTeX working properly then these expressions apply.
The density parameter is given by [Equation 1]:
$$\Omega = \frac {8 \pi G} {3 H^2} \rho = \frac {\rho} {\rho_{crit}}$$
and then the curvature*, ##\kappa## is found from [Equation 2]:
$$\kappa = ( \Omega - 1 ) (H^2a^2)$$
[Reference: Page 337, Spacetime and Geometry, Sean Carroll]
Where: ##\rho## = Density ; ##\rho_{crit}## = Critical density ; H = Hubble parameter ; a = Scale factor
* This "Curvature" in [Equation 2] has been described as the "un-normalised curvature parameter" in previous posts. It is the curvature of a spatial hypersurface at fixed co-ordinate time.
The Hubble parameter and the density are involved, so your statement makes sense. In previous posts I said the curvature is determined by comparing the density to the critical density and we can use [Equation1] and [Equation 2] to describe the full set of relationships between these things that we are interested in.
We have the following relationships:
##\rho < \rho_{crit} \iff \Omega < 1 \iff \kappa <0 \iff \text {The universe is Open}##
## \rho = \rho_{crit} \iff \Omega = 1 \iff \kappa =0 \iff \text {The universe is Flat}##
## \rho > \rho_{crit} \iff \Omega > 1 \iff \kappa >0 \iff \text {The universe is Closed}##
The density, ##\rho## depends on the scale factor. You gave the example of matter density ~ a
-3 and there are other components of the cosmological fluid with different power laws relating their density to the scale factor, as usual.
In the general case, the scale factor varies with time. You can construct exceptions (like the Einstein static universe) if you want but the case with a constant scale factor is uninteresting and won't be pursued here.
In the general case, we can be confident that the total density evolves and by choosing a mixture of species in our cosmological fluid we even have considerable control over that change.
The critical density also depends on time.
$$\rho_{crit} = \frac {3 H^2} {8 \pi G}$$
and the Hubble parameter is seen to depend on time (in the general case).
So the comparison of density to critical density is dynamic. Their ratio, ##\Omega##, varies with time. This doesn't seem to be in dispute in the texts I have read.
This is a slightly more simplified version of [Equation 2]:
$$\kappa = ( \Omega - 1 ) \dot a^2$$
Looking at the above or [Equation 2] we expect ##\kappa## to evolve because just about everything on the RHS does vary with time.
That the un-normalised curvature parameter can change with time doesn't seem to be in dispute from any of the previous posts in this thread, although I must state that it is not directly mentioned in any of the Cosmology texts I've seen. However, it's not that interesting
provided the sign of the curvature doesn't change. The magnitude of the curvature always was arbitrary depending on the co-ordinates or units of length you choose to use, if you want to look at it that way.
Summary:
I believe that ##\Omega## and ##\kappa## certainly
can change with time as the universe evolves. Therefore, I am seeking the proof that ##\Omega## cannot cross that boundary at ##\Omega = 1##, or equivalently that the curvature cannot change its sign.
@PeterDonis has suggested a proof may be found in Hawking and Ellis but isn't sure where. I haven't had access to that book yet.
You went on to say:
kimbyd said:
All that said, the effect of the spatial curvature on the universe does change over time. Specifically, it scales as a-2, where a is the scale factor.
That all seems reasonable, a quantity ##\rho_c = \frac {- 3 \kappa} {8 \pi G a^2}## can be introduced and is usually described as a fictitious energy density for the curvature. Re-writing the (first) Friedmann equation with this density contribution, the effect of curvature ##\kappa## falls off as ~a
2. That's what I have understood by your statement.
That describes the effect of curvature on the expansion of the universe. I'm not too concerned about the expansion of the universe, only on what the curvature is. So your statement, while correct, doesn't seem to relate to the OP.
An old "folk belief":
There was a time when the curvature parameter was synonymous with the ultimate fate of the universe. So talking about the expansion of the universe was exactly the same as talking about the curvature. However, modern cosmology has adapted since a vacuum energy component has been more widely accepted. This point was made earlier by
@PeterDonis.
PeterDonis said:
but our best current model of our universe has a small positive cosmological constant. For that case, the ultimate fates are somewhat different, and in particular the case k > 0 (spherical) will not necessarily recollapse to a Big Crunch.
I believe it is actually far more general then just applying to the case k>0. If I had a scanner I would copy figure 8.4 from Sean Carroll. I don't have a scanner, here's a quote:
Traditional disdain for the cosmological constant has lead to a folk belief that this [..the old correspondance between curvature and ultimate expansion fate..]
is a necessary correspondance; once the possibility of vacuum energy is admitted, however, any combination of spatial geometry and eventual fate is possible.
[Page 343, Spacetime and Geometry, 3rd printing, Sean Carroll]
Of course, I understand that there is some interest in the expansion of the universe and its ultimate fate and I enjoyed reading your description of the situation. However, for the moment I'm just being a purist and wondering if the intrinsic geometry of the universe can change.
You also went on to say:
kimbyd said:
Given our inability to measure a non-zero spatial curvature, this means that the spatial curvature of our observable universe has never been a significant factor in the way the universe expands (at early times, matter and radiation densities overwhelmed the effect of curvature, and now the cosmological constant prevents it having an impact in the future either).
Yes, I agree with the essence of what is said. Just to clarify:
1. There ARE ways to measure the spatial curvature BUT it does seem to be very close to zero (flat) at the current time. An experiment that I'm interested in is a proposal to "draw" a triangle in space using lasers from a triangle of satellites.
2. I'm not too concerned about how important curvature is or was for expansion. I'm interested in the evolution of curvature just for the impact it has on geometry. For example, is a real triangle drawn in space going to have more or less than 180 degrees of internal angles and can that change with time? More-over, it doesn't even have to be our real universe but just any FRW universe.
Back to the OP:
I'm making good progress answering my own question by looking at the process of normalising the curvature parameter carefully. The continuity requirements seem to be the key. I've bored everyone else enough already and probably won't write anymore about that. I'm very grateful for all the time, attention and replies I have received on Physics Forums.
Best wishes, bye for now.