# The scale factor in flat FRW model

1. Nov 22, 2014

### TrickyDicky

In the FRW model with Euclidean 3-space, k=0 in the first Friedmann equation makes the density(times a constant) directly related to the Hubble parameter. The Hubble parameter is the time derivative of the scale factor divided by the scale factor itself.
My question is: does a Euclidean geometry of the space, being scale invariant, put any constraint on the nature of the scale factor function a(t)?
Wouldn't Euclidean scale invariance indicate that the a(t) function for the flat case could only be linear in t, since the scale invariance of the space(not only of the Friedmann equations like in the general case) and thus of the proper distance invariant makes the scale function of t indifferent to any power of t?

DISCLAIMER:I know a linear scale factor ("coasting universe") has been discarded observationally, and if my reasoning was sound it would seem to imply the flat FRW model has been observationally discarded, but I'm not asking about that, so please just concentrate on refuting mathematically my reasoning about distance invariance and how it determines or not a linear scale factor.

2. Nov 22, 2014

### Orodruin

Staff Emeritus
How the density evolves with $a$ depends on the type of matter. For a matter dominated universe it is diluted by $1/a^3$, while for a radiation dominated universe, it dilutes by $1/a^4$. Assuming $p = w\rho$ gives
$$a \propto t^{\frac{2}{3(1+w)}}.$$
Here, $w = 0$ corresponds to matter domination and $w = 1/3$ to radiation domination.

3. Nov 22, 2014

### TrickyDicky

Right and for w=-1/3 (cosmic strings)you get the linear scale factor but that is an ulterior cosmological consideration and I'm not asking that, I'm not entering into the equations of state or any specific model with defined values for pressure etc, I'm only trying to clarify a specific question about the FRW flat universe and its spatial geometry. Is it not true that the scale factor function relates proper distances and what happens with that function if those distances have length scale invariance, it seems logical that it has to be a linear function.
BTW, did you read the disclaimer?

4. Nov 22, 2014

### Chalnoth

Orodruin answered your question: your assumptions are off. Basically, the lack of a constant means that in the spatially-flat the scale factor will have a power-law relationship with time (given constant $w$), but the specific power varies depending upon the contents of the universe.

5. Nov 22, 2014

### Garth

The 'coasting model' may not be discarded observationally so lightly.
Introducing the Dirac-Milne universe
There are three ways of producing a coasting model:

1. the empty universe - obviously this can be observationally discarded, however it may be the asymptotic limit to a monotonically expanding universe.
2. the Dirac-Milne universe in which an equal amount of matter and 'repulsive' antimatter overall cancel out each other gravitationally.
3. A model with a Dark Energy EoS of $\rho = -\frac{1}{3}p$ produced by some sort of 'quintessence' suggested by Kolb (A coasting cosmology ). Such a model is suggested a string theory as TrickyDicky said or by Self Creation Cosmology - if I can get it to work!

Garth

6. Nov 22, 2014

### TrickyDicky

What assumption are off? I only assumed length scale invariance of Euclidean geometry.
It would be great if someone addressed my questions directly, like using the quote button.

Last edited: Nov 22, 2014
7. Nov 22, 2014

### TrickyDicky

Thanks Garth, but the disclaimer was aimed at restricting the discussion to the mathematical question, not at debating cosmological models.
For whatever reason nobody is addressing my questions directly.

8. Nov 22, 2014

### Orodruin

Staff Emeritus
Perhaps because your questions are non-questions with an answer that you seem unwilling to accept? You have seen the solutions to the Friedmann equations in a flat space-time, the scale factor is obviously not scaling linearly with t.

No.

No, and it is unclear why you would think it does. That the comoving space is Euclidean does not mean space-time is flat.

9. Nov 22, 2014

### TrickyDicky

My questions are non-questions(how does one do that anyway?) so I guess that's why you delivered non-answers?

Wait there, you are not even understanding what I'm asking. A constant Hubble parameter doesn't imply flat space-time, it is a perfectly expanding curved spacetime cosmology, it is not Minkowski spacetime, are you aware of this?

10. Nov 22, 2014

### Orodruin

Staff Emeritus
Perfectly aware, it is what I just said. Comoving space is not space-time, it is a space-like surface at a fixed cosmic time t, which seems to be what you are referring to. And why are you bringing a constant Hubble parameter into the game? A constant Hubble parameter means the Universe is dominated by a cosmological constant and is expanding exponentially - definitely not linearly.

It is quite clear from the solutions to the Friedmann equation that there are scale factors which are not linear in t that satisfy them, also with k = 0, so no, I do not understand why you are asking if something that is manifestly false is true. It is still unclear how you want to use Euclidean scale invariance to prove that a = t2/3 is not a valid solution to the Friedmann equation.

11. Nov 22, 2014

### Chalnoth

Yes it can. In particular, it doesn't come close to fitting the CMB evidence. With the present value of the Hubble parameter in an empty universe, the CMB has a comoving distance of about 74 billion light years. The actual comoving distance is about 46 billion light years.

Sure, the two models are somewhat close when looking purely at the expansion history for the nearby universe. But they diverge wildly as you go further away.

12. Nov 23, 2014

### TrickyDicky

So why are you mentioning Minkowski flat spacetime? It is not related to anything I posted.

My mistake, I meant to say constant expansión.
I'm not trying to prove that.

13. Nov 23, 2014

### Orodruin

Staff Emeritus
I am not, you are the only one who has mentioned Minkowski space-time. Part of your problem seemed to be an assumption of the FRW space-time being flat, which is why I mentioned that it is not.

Yes you are, it is exactly what your first post is saying.

14. Nov 23, 2014

### TrickyDicky

Ok, I think I see where your misunderstanding lies, a flat spacetime doesn't have a scale factor function with nonzero constant first derivative, so I`m not assuming any spacetime being flat, you can confirm this in any cosmology textbook. When I used the term flat before I was referring to the spatial hypersurface.

Let's take a step back so we can reach some agreement. I made an effort to put the discussion in a previous stage to assigning the EoS "w", so let's concentrate on the FRW metric with k=0 even before we use the Friedmann equations that are deduced from it(plus the EFE and a perfect fluid Stress-Energy tensor).

So we have this spatially flat FRW metric and we try to deduce something about the scale factor a(t) from the fact that D(t)=a(t)Dt0, where D(t) and D(t0) are proper distances in a Euclidean space at times t and t0, and a(t) is the scale function depending on time t. The only thing I'm trying to get right (before bringing in any EoS and applying the Friedmann equations) is wheter this equation D(t)=a(t)Dt0 for proper distances D(t) and D(t0), that being in a scale invariant space must be related by a linear scale, implies or not that the scale function relating them must be a linear function.
This is the only step that needs to be clarified before proceeding with any implications for the Friedmann equations and the EoS that may be used in such case.

15. Nov 23, 2014

### Orodruin

Staff Emeritus
Again, exactly what I said and what you were arguing against. I am just responding to what seemed to be your argument.

And the answer is that it does not, there really is not much else to it. If you disregard the Friedmann equation, you can pick any (smooth) a(t) that you would like, the relation between the proper distances is linear in a(t), not in t. It is still unclear how you would attempt to deduce a constraint on the scale factor without imposing the Friedmann equation or something similar.

16. Nov 23, 2014

### TrickyDicky

I don't understand the difference between a function of t a(t) being linear versus a linear function of t, can you elaborate?

Well, the scale factor is in the metric and the Friedmann equations are derived from the metric not the other way around.

17. Nov 23, 2014

### Orodruin

Staff Emeritus
If a(t) is not a linear function of t, then any function which is linear in a(t) is not going to be linear in t. Let us take d(a(t)). While it is true that d(a(t1)+a(t2)) = d(a(t1)) + d(a(t2)) (d is linear in a), it is not necessarily true that d(a(t1+t2)) = d(a(t1)+a(t2)) (a is not linear in t).

No, the Friedmann equation is derived from imposing the Einstein field equations onto a homogeneous isotropic universe. There is nothing about the metric itself that puts bounds on a(t). The form of the metric with the scale factor (the form of which is still to be determined) is imposed by the assumption of the Universe being homogeneous and isotropic.

18. Nov 23, 2014

### TrickyDicky

My point was that for a scale invariant geometry it can't be any a(t), it should be linear in t, since the scale invariance renders any power of t equivalent to t.

19. Nov 23, 2014

### TrickyDicky

We are saying the same thing,a homogeneous isotropic universe determines the form of the metric including having a scale factor, it is true that it doesn't determine the form of the function a(t) in general, but I think I'm giving a plausible argument for why it would determine it in the Euclidean case.

20. Nov 23, 2014

### TrickyDicky

Sure, just to be clear, I'm not saying the equation D(t)=a(t)D(t0) is linear in t in general, I know that in general that equation only says it is linear in a. Only for proper distances in a Euclidean geometry I'm guessing it would be linear in t, it is a purely geometric effect of linear spaces.
And of course I'm imposing that a(t) must not be itself a constant, as that would indeed give Minkowski flat spacetime, maybe I should have made this clear before but I thought it was implicit since we are discussing GR and therefore only spacetimes with curvature.

Last edited: Nov 23, 2014