The scale factor in flat FRW model

[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.

 

[Orodruin]

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.

 

[TrickyDicky]

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.

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?

 

[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.

 

[Garth]

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.

The 'coasting model' may not be discarded observationally so lightly.
Introducing the Dirac-Milne universe

The main feature of this cosmology is the linear evolution of the scale factor with time, which directly solves the age and horizon problems of a matter-dominated universe. We study the concordance of this model to the cosmological test of type Ia supernovae distance measurements and calculate the theoretical primordial abundances of light elements for this cosmology. We also show that the acoustic scale of the cosmic microwave background naturally emerges at the degree scale despite an open geometry.

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.
   [*]the Dirac-Milne universe in which an equal amount of matter and 'repulsive' antimatter overall cancel out each other gravitationally.
   [*]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

 

[TrickyDicky]

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.

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.

 

[TrickyDicky]

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.
   [*]the Dirac-Milne universe in which an equal amount of matter and 'repulsive' antimatter overall cancel out each other gravitationally.
   [*]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

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.

 

[Orodruin]

For whatever reason nobody is addressing my questions directly

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.

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)?

No.

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?

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

 

[TrickyDicky]

Perhaps because your questions are non-questions...

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

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

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?

 

[Orodruin]

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?

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 = t^2/3 is not a valid solution to the Friedmann equation.

 

[Chalnoth]

The 'coasting model' may not be discarded observationally so lightly.

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.

 

[TrickyDicky]

Perfectly aware

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

why are you bringing a constant Hubble parameter into the game?

My mistake, I meant to say constant expansión.

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 = t^2/3 is not a valid solution to the Friedmann equation.

I'm not trying to prove that.

 

[Orodruin]

So why are you mentioning Minkowski flat spacetime?

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.

I'm not trying to prove that.

a(t) function for the flat case could only be linear in t

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

 

[TrickyDicky]

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.

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.

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

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.

 

[Orodruin]

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.

That the comoving space is Euclidean does not mean space-time is flat.

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

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

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.

 

[TrickyDicky]

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.

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

It is still unclear how you would attempt to deduce a constraint on the scale factor without imposing the Friedmann equation or something similar.

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

 

[Orodruin]

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

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).

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

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.

 

[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.

 

[TrickyDicky]

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.

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.

 

[TrickyDicky]

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).

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.

 

[PeterDonis]

does a Euclidean geometry of the space, being scale invariant, put any constraint on the nature of the scale factor function a(t)?

Why would it? Why would scale invariance of a spacelike hypersurface of constant time have anything to do with how the scale factor changes with time? $a(t)$ could be any arbitrary function of $t$ and each hypersurface of constant time could still be scale invariant.

 

[PeterDonis]

the Dirac-Milne universe in which an equal amount of matter and 'repulsive' antimatter overall cancel out each other gravitationally

I'm not sure why the authors of that paper think that repulsive antimatter is somehow less of an "unsatisfactory situation" than dark matter and dark energy.

 

[PeterDonis]

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.

No, it doesn't. The scale invariance of Euclidean geometry of a spacelike hypersurface is spatial: it talks about variation with $x, y, z$ (or $r, \theta, \phi$ or whatever spatial coordinates you want to use). It says nothing at all about variation with $t$.

 

[TrickyDicky]

;

Why would it? Why would scale invariance of a spacelike hypersurface of constant time have anything to do with how the scale factor changes with time? $a(t)$ could be any arbitrary function of $t$ and each hypersurface of constant time could still be scale invariant.

No, it doesn't. The scale invariance of Euclidean geometry of a spacelike hypersurface is spatial: it talks about variation with $x, y, z$ (or $r, \theta, \phi$ or whatever spatial coordinates you want to use). It says nothing at all about variation with $t$.

Thanks, this is the kind of counter-argument I was looking for, and yet the scale invariance of Euclidean space is indifferent to what you make it with respect to, any re-scaling performed between Euclidean proper distances, like dilatations (expansions) makes no change.
So the math appears to be saying that the function relating them should either be constant, or linear if we want to keep the universe expanding like in this case.
Perhaps this is not your understanding of scale invariance?

 

[TrickyDicky]

Peter, I think what you are referring to above is translational invariance but that is a property of the three spatial geometries possible in the FRW metric, it is not something that only Euclidean geometry has, however the scale invariance I'm talking about is privative of Euclidean geometry.

 

[Orodruin]

I still do not see how you are going to use the scale invariance to impose that a(t) must be a linear function of t. It is possible to just put an arbitrary a(t) that results in a smooth manifold and you will have a 4-dimensional pseudo-riemannian manifold which is distinctly different from that where a(t) is linear in t (unless you choose a(t) = kt). Rescaling to make a(t) linear in t for the spatial part is going to introduce cross terms between dt and the spatial differentials into the metric and so is distinctly different from if you had just put a(t) linear in t from the beginning.

 

[TrickyDicky]

I still do not see how you are going to use the scale invariance to impose that a(t) must be a linear function of t. It is possible to just put an arbitrary a(t) that results in a smooth manifold and you will have a 4-dimensional pseudo-riemannian manifold which is distinctly different from that where a(t) is linear in t (unless you choose a(t) = kt). Rescaling to make a(t) linear in t for the spatial part is going to introduce cross terms between dt and the spatial differentials into the metric and so is distinctly different from if you had just put a(t) linear in t from the beginning.

There are no drdt cross terms in FRW metrics, being hypersurface orthogonal.

 

[Orodruin]

There are no drdt cross terms in FRW metrics, being hypersurface orthogonal.

Again, exactly my point. If you do a scale transformation from an arbitrary a(t) to a linear one of the spatial part, you will introduce cross terms and so no longer be dealing with a FRW metric on the standard form. It is therefore distinctly different from a FRW universe where you have the scale factor proportional to t.

 

[TrickyDicky]

Again, exactly my point. If you do a scale transformation from an arbitrary a(t) to a linear one of the spatial part, you will introduce cross terms and so no longer be dealing with a FRW metric on the standard form. It is therefore distinctly different from a FRW universe where you have the scale factor proportional to t.

But an arbitrary a(t) doesn't give you just a smooth pseudoriemannian manifold, it is a FRW metric already and besides my point is valid only when the metric of the 3-space is Euclidean, not in a general FRW metric.

 

[Orodruin]

But an arbitrary a(t) doesn't give you just a smooth pseudoriemannian manifold, it is a FRW metric already and besides my point is valid only when the metric of the 3-space is Euclidean, not in a general FRW metric.

You are choosing to ignore the point entirely. The point is that the manifolds with a(t) = t and a(t) = sqrt(t) are fundamentally different even if the 3-space is Euclidean and not related by a simple scale transformation of the 3-space.

You still have not provided your argumentation for how you explicitly would perform the rescaling you are talking about. Since you are obviously not getting your message through with words, how about just showing us what you mean in terms of equations? Otherwise this is just an exercise in semantics. I therefore give you the following task:

Consider the flat (i.e., 3-space flat) FRW universes with the metrics $ds^2 = dt^2 - t^2 \, d\vec x^2$ and $ds^2 = dt^2 - t\, d\vec x^2$, respectively. Show us the transformation you want to make in order to make the second universe equivalent to the first.

 

[TrickyDicky]

You are choosing to ignore the point entirely. The point is that the manifolds with a(t) = t and a(t) = sqrt(t) are fundamentally different even if the 3-space is Euclidean and not related by a simple scale transformation of the 3-space.

You still have not provided your argumentation for how you explicitly would perform the rescaling you are talking about. Since you are obviously not getting your message through with words, how about just showing us what you mean in terms of equations? Otherwise this is just an exercise in semantics. I therefore give you the following task:

Consider the flat (i.e., 3-space flat) FRW universes with the metrics $ds^2 = dt^2 - t^2 \, d\vec x^2$ and $ds^2 = dt^2 - t\, d\vec x^2$, respectively. Show us the transformation you want to make in order to make the second universe equivalent to the first.

Since I was trying to avoid a constant function I was thinking more in terms of a function like $a(t)=ct^n$ with $c$ a factor that would be rescaled so that $t^n=t$ and one scale function is a scale transformation of the other.

 

[Orodruin]

So your c will have to depend on t to be rescaling in the correct manner. Since you do not want to provide the actual transformation, let me try.

Assume you have $ds^2 = dt^2 - a(t)^2 d\vec x^2$ with $a(t) = t^2$. There is a transformation that will transform the spatial part to $(t)^2 d\vec X^2$, namely:
$$
\vec X = \frac{\vec x}{t}.
$$
Is this the transformation you want to do?

Or perhaps you are more into rescaling time and want to do $T = t^2$ to obtain the spatial part $(T)^2 d\vec x^2$?

 

[TrickyDicky]

So your c will have to depend on t to be rescaling in the correct manner. Since you do not want to provide the actual transformation, let me try.

Assume you have $ds^2 = dt^2 - a(t)^2 d\vec x^2$ with $a(t) = t^2$. There is a transformation that will transform the spatial part to $(t)^2 d\vec X^2$, namely:
$$
\vec X = \frac{\vec x}{t}.
$$
Is this the transformation you want to do?

Or perhaps you are more into rescaling time and want to do $T = t^2$ to obtain the spatial part $(T)^2 d\vec x^2$?

I was thinking along the lines of the second one, but rescaling the constant that multiplies $t$, i.e in the example doing $\frac{c}{t}t^2=ct$, (otherwise it could be thought we are taking t as a constant and that leads to flat spacetime) in the end it is the scale invariance of the $\vec x$, spatial part that allows this rescaling, this kind of scale transformation of a(t) wouldn't be possible for k=+/-1.

 

[Orodruin]

I was thinking along the lines of the second one, but rescaling the constant that multiplies $t$, i.e in the example doing $\frac{c}{t}t^2=ct$, (otherwise it could be thought we are taking t as a constant and that leads to flat spacetime) in the end it is the scale invariance of the $\vec x$, spatial part that allows this rescaling, this kind of scale transformation of a(t) wouldn't be possible for k=+/-1.

So, the second one is not preserving the FRW structure of the metric as $dt = \frac{1}{2\sqrt T} dT$, leading to the line element $ds^2 = \frac{1}{4T}dT^2 - T^2 d\vec x^2$. It therefore represents a metric which is physically different from the FRW metric with $a(t) = t$ as the coordinate $T$ is no longer the proper time for a stationary observer.

You simply cannot define a scaling which is time independent that gives you a linearly growing scale factor with a FRW metric unless you already have a linearly growing scale factor.

 

[TrickyDicky]

So, the second one is not preserving the FRW structure of the metric as $dt = \frac{1}{2\sqrt T} dT$, leading to the line element $ds^2 = \frac{1}{4T}dT^2 - T^2 d\vec x^2$. It therefore represents a metric which is physically different from the FRW metric with $a(t) = t$ as the coordinate $T$ is no longer the proper time for a stationary observer.

You simply cannot define a scaling which is time independent that gives you a linearly growing scale factor with a FRW metric unless you already have a linearly growing scale factor.

So I'm not rescaling t but c; and is your point that your first transformation is a valid one? I would say it is equivalent to my rescaling.

 

[Orodruin]

Yes, my first transformation is valid, although there is a typo, it should be $\vec X = t \vec x$ in order to get the scale factor $t^2$ in front of $d\vec X^2$ in the metric, otherwise it would be scaling in the wrong direction.

This results in the following:
$$
d\vec x^2 = d(\vec X/t) = (d\vec X)/t - (\vec X/t^2) dt.
$$
The line element is therefore given by
$$
ds^2 = dt^2 - t^4 [(d\vec X)/t - (\vec X/t^2) dt]^2 = \left(1 - \vec X^2\right) dt^2 + 2t (\vec X \cdot d\vec X)\, dt - t^2 d\vec X^2,
$$
which has the spatial part scaling linearly with $t$ but definitely is not a metric on the FRW form due to the cross terms and the altered time-time component and therefore also represents a universe distinct from that of a linearly expanding one as I stated earlier:

Rescaling to make a(t) linear in t for the spatial part is going to introduce cross terms between dt and the spatial differentials into the metric and so is distinctly different from if you had just put a(t) linear in t from the beginning.

There simply is no transformation that will make your FRW universe with an arbitrary scale factor equivalent to one where the scale factor is linear with the cosmological time t.

 

[TrickyDicky]

You seem to be right about that, I'll think about it some more but the fact that I have not been able to find any reference that backs it up and I would think that such a mathematical transformation would be known is rather telling.

 

[TrickyDicky]

Since there is no transformation that allows a linear a(t) without changing the form of the metric, but you seemed to implicitly endorse the interpretation I made about Euclidean scale invariance, how would one refute mathematically the implications about a constant scale factor?

 

[Orodruin]

The coordinate surfaces with constant t are Euclidean and scale invariant. But it is only a subspace of the manifold you are considering and you are making statements about the properties of the full 4-dimensional FRW universe when you are discussing a(t). Those properties cannot be inferred from the properties of a coordinate surface.

 

[TrickyDicky]

The coordinate surfaces with constant t are Euclidean and scale invariant. But it is only a subspace of the manifold you are considering and you are making statements about the properties of the full 4-dimensional FRW universe when you are discussing a(t). Those properties cannot be inferred from the properties of a coordinate surface.

This argumentation was my first guess, but then it is evident that properties from the constant t hypersurfaces like k or p are used in the Friedmann equations to infer properties of the 4-manifold.Not to

 

[Orodruin]

You are now going into properties implied by the Einstein field equations. These do not a priori have to hold for an arbitrary manifold with a metric of the particular form $ds^2 = dt^2 - a(t)^2 d\vec x^2$ as you were insisting before. It is just when we impose the Einstein field equations and an equation of state we obtain constraints. The point here was that a(t) does not have to be (and indeed are not) linear in t just because the coordinate surfaces are flat.

 

[TrickyDicky]

You are now going into properties implied by the Einstein field equations. These do not a priori have to hold for an arbitrary manifold with a metric of the particular form $ds^2 = dt^2 - a(t)^2 d\vec x^2$ as you were insisting before. It is just when we impose the Einstein field equations and an equation of state we obtain constraints. The point here was that a(t) does not have to be (and indeed are not) linear in t just because the coordinate surfaces are flat.

Yes, I agree. What is really more surprising and I actually find harder to get is that we can obtain in part these constraints on the scale factor and the FRW cosmology after imposing the equations based ultimately on the geometry of coordinate hypersurfaces, a particular slicing of the manifold; after all one of the basis of GR is the notion that coordinates have no physical significance, so any other slicing should be equivalent physically, even if it were less practical computationally. But it is obvious that the coordinate surfaces are identified with our physical 3-space in the theory, so that other slicings of the 4-manifold would be physically incompatible with what we observe.

 

[PeterDonis]

the scale invariance of Euclidean space is indifferent to what you make it with respect to

Sure, but FRW spacetime is not the same as Euclidean space. Each spacelike hypersurface of constant time (in the flat FRW model) is a Euclidean space, but the spacetime as a whole is not. So any argument about Euclidean space can only apply to a particular spacelike hypersurface.

 

[TrickyDicky]

Sure, but FRW spacetime is not the same as Euclidean space. Each spacelike hypersurface of constant time (in the flat FRW model) is a Euclidean space, but the spacetime as a whole is not. So any argument about Euclidean space can only apply to a particular spacelike hypersurface.

This was already cleared up. Now I was reminding the issue that the scale factor a(t), from which many physical consequences are derived, is coordinate-dependent and in GR(and in physics in general) it is postulated that physical properties are not coordinate-dependent. I think it is important to be aware of this for whatever is worth.

 

[Orodruin]

The coordinate independence is mainly a local issue and that will still be fine. In general, the evolution of the Universe will of course depend on what you put inside it. The presence of a fluid will introduce a special frame (also in special relativity), i.e., the rest frame of the fluid. This does not break the principle of relativity in any way.

 

[TrickyDicky]

The coordinate independence is mainly a local issue and that will still be fine.

Locally inertial frames are preferred per the equiavalence principle. Coordinate independence is clearly a global issue in GR by diffeomorphism invariance.

In general, the evolution of the Universe will of course depend on what you put inside it. The presence of a fluid will introduce a special frame (also in special relativity), i.e., the rest frame of the fluid.

Now you are introducing a distinction between the manifold as continent and its contents that is philosophical.

This does not break the principle of relativity in any way.

In fact it does and that is generally acknowledged by the mainstream, but it is considered an example of spontaneous symmetry breaking:"General relativity has a Lorentz symmetry, but in FRW cosmological models, the mean 4-velocity field defined by averaging over the velocities of the galaxies (the galaxies act like gas particles at cosmological scales) acts as an order parameter breaking this symmetry." http://en.wikipedia.org/wiki/Spontaneous_symmetry_breaking

 

[Orodruin]

Now you are introducing a distinction between the manifold as continent and its contents that is philosophical.

Not really, all I said was that how the Universe behaves depends on its contents. It does so through the Einstein field equations. If you put a fluid which behaves as radiation in the Universe, it is going to behave differently than if you put one in that behaves as matter. This of course affects the properties of the manifold itself.

You can still use any coordinates locally and get away with it.

In fact it does and that is generally acknowledged by the mainstream, but it is considered an example of spontaneous symmetry breaking:"General relativity has a Lorentz symmetry, but in FRW cosmological models, the mean 4-velocity field defined by averaging over the velocities of the galaxies (the galaxies act like gas particles at cosmological scales) acts as an order parameter breaking this symmetry."

This is exactly what I said:

The presence of a fluid will introduce a special frame (also in special relativity), i.e., the rest frame of the fluid.

It does not mean that the principle of relativity is violated. You can still find a locally inertial frame. A test particle that goes through this space-time without interacting with the background is going to feel no difference. The differences appear when you start being able to discern the global properties of the manifold or start interacting with the background (which does introduce a preferred frame).

 

[TrickyDicky]

Not really, all I said was that how the Universe behaves depends on its contents. It does so through the Einstein field equations. If you put a fluid which behaves as radiation in the Universe, it is going to behave differently than if you put one in that behaves as matter. This of course affects the properties of the manifold itself.

You can still use any coordinates locally and get away with it.
This is exactly what I said:

It does not mean that the principle of relativity is violated. You can still find a locally inertial frame. A test particle that goes through this space-time without interacting with the background is going to feel no difference. The differences appear when you start being able to discern the global properties of the manifold or start interacting with the background (which does introduce a preferred frame).

You mean any coordinates locally as long as you favor inertial frames right?
I agree local Lorentz invariance is not violated, but you are missing that the point I'm making is global, GR being strict doesn't allow introducing a preferred frame with physical properties, it is related to principles such as background independence and diffeomorphism invariance-general covariance.
You might want to argue those principles don't hold in GR but last time I looked they still appear in textbooks.

 

[Orodruin]

I maintain that from a pure metric viewpoint, there still does not exist a preferred local frame. There is nothing saying that you have to prefer locally inertial frames either. You can write down the equations in any coordinates, or do it coordinate independent if you prefer that. If you want to refer to a particular statement in a textbook, then please provide the reference as well as the statement.

Even if you do not have any connection between the metric and the matter content, there can be special directions in a manifold, e.g., implied by the Ricci tensor. There is nothing wrong with such manifolds. An ellipsoid does have special directions and is still perfectly fine within differential geometry.

 

[TrickyDicky]

I maintain that from a pure metric viewpoint, there still does not exist a preferred local frame. There is nothing saying that you have to prefer locally inertial frames either. You can write down the equations in any coordinates, or do it coordinate independent if you prefer that. If you want to refer to a particular statement in a textbook, then please provide the reference as well as the statement.

Not a local frame, we are talking about global properties. As you showed a simple coordinate transformation that should leave the manifold and its properties invariant, changes the form of the FRW metric and also its physical properties like density or pressure.

Even if you do not have any connection between the metric and the matter content, there can be special directions in a manifold, e.g., implied by the Ricci tensor. There is nothing wrong with such manifolds. An ellipsoid does have special directions and is still perfectly fine within differential geometry.

Of course, isotropy is a quite stringent condition for a manifold. And anisotropy of an ellipsoid doesn't introduce a coordinate dependence, only coordinates that are simpler.