The scale factor in flat FRW model

[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.

 

[Orodruin]

Then I do not understand your point. There is nothing in GR prohibiting such global properties.

 

[TrickyDicky]

Then I do not understand your point. There is nothing in GR prohibiting such global properties.

Sorta, and yet:
http://en.wikipedia.org/wiki/General_covariance
"In theoretical physics, general covariance (also known as diffeomorphism covariance or general invariance) is the invariance of the form of physical laws under arbitrary differentiable coordinate transformations. The essential idea is that coordinates do not exist a priori in nature, but are only artifices used in describing nature, and hence should play no role in the formulation of fundamental physical laws."

I guess you'll argue the global properties we are talking about are not "physical laws" but either "initial conditions" like homogeneity& isotropy, or local curvature, and there is a point in that because a metric is not a global physical law in GR, it is a "local law" so to speak. Still that local prescription has important physical consequences that seem to go counter the idea of coordinates playing no role "in the formulation of fundamental physical laws".

 

[Orodruin]

I guess you'll argue the global properties we are talking about are not "physical laws" but either "initial conditions" like homogeneity& isotropy, or local curvature, and there is a point in that because a metric is not a global physical law in GR, it is a "local law" so to speak. Still that local prescription has important physical consequences that seem to go counter the idea of coordinates playing no role "in the formulation of fundamental physical laws".

Local curvature is local, so this will be a local (and coordinate independent). Otherwise, something like that. Homogeneity and isotropy are global conditions that we are putting on the FRW universe. What local prescription are you suggesting has important physical quantities?

 

[TrickyDicky]

Local curvature is local, so this will be a local (and coordinate independent). Otherwise, something like that. Homogeneity and isotropy are global conditions that we are putting on the FRW universe. What local prescription are you suggesting has important physical quantities?

Curvature of the 4-manifold. Cosmology is a bit peculiar in this respect from other applications of GR, since one tries to infer from local curvature consequences for the whole universe and its evolution and origin, and despite the formal background independence of the theory.