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I'm interested to find out if we can say anything useful about spacetime at the singularity in the FLRW metric that occurs at $t = 0$.

If I understand correctly, the FLRW spacetime is a combination of the manifold $\mathbb R^{3,1}$ and the FLRW metric $g$. The metric $g$ has a singularity at $t = 0$ because at that point the proper distance between every pair of spacetime points is zero. Presumably though, however the metric behaves, the manifold remains $\mathbb R^{3,1}$ so we still have a collection (a set?) of points in the manifold. It's just that we can no longer calculate distance between them. Is this a fair comment, or am I being overly optimistic in thinking we can say anything about the spacetime?

I vaguely recall reading that the singularity is considered to be not part of the manifold, so the points with $t = 0$ simply don't exist, though I think this was said about the singularity in the Schwarzschild metric and whether it applies to all singular metrics I don't know.

To try to focus my rather vague question a bit, I'm thinking about a comment made to my question Did the Big Bang happen at a point?. The comment was roughly: yes the Big Bang did happen at a point because at $t = 0$ all points were just one point. If my musings above are correct the comment is untrue because even when the metric is singular we still have the manifold $\mathbb R^{3,1}$ with an infinite number of distinct points. If my memory is correct that the points with $t = 0$ are not part of the manifold then we cannot say the Big Bang happened at a point, or the opposite, because we cannot say anything about the Big Bang at all.

From the purely mathematical perspective I'd be interested to know what if anything can be said about the spacetime at $t = 0$ even if it has no physical relevance.

John Rennie
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    Also related https://physics.stackexchange.com/q/148838/ –  Dec 07 '14 at 16:58
  • to quote a comment I made yeterday: a GR singularity is not necessarily topological: possibly, it's 'just' a metric degeneracy; @CristiStoica probably has something to say about that...; the gist: the FLRW singularity is 'quasi-regular', the densitized stress-energy-momentum tensor remains smooth and the Weyl curvature hypothesis holds – Christoph Dec 07 '14 at 17:06
  • I don't think that's the manifold. It should be $\mathbb R^3\times \mathbb R_{>0}$. – MBN Dec 07 '14 at 22:20
  • @MBN: entirely possible, my grasp of the maths involved is very shaky. – John Rennie Dec 08 '14 at 06:13
  • Yes, actually I think there is a bit of confusion: the metric is a structure you put on the manifold. The manifold is $\mathbb{R}^4$ topologically, and nothing changes that. When you consider a singular metric, you may exclude some point and restrict the maifold to a submanifold like $\mathbb{R}^4/{0}$. – Oscar Mar 16 '15 at 09:32
  • @Oscar: that suggests my final comment is true i.e. that the Big Bang is not part of the manifold and therefore the question "did the Big Bang happen at a point" is meaningless. – John Rennie Mar 16 '15 at 09:40
  • The underlying manifold where the FLRW metric is put on is not necessarily $\mathbb R^3\times \mathbb R_{>0}$. It depends on the three dimensional topology. The manifold is $\mathbb{R}_{>0}\times\cal{S}$ in general where $S$ is a three dimensional manifold. http://en.wikipedia.org/wiki/Shape_of_the_universe – yess Mar 18 '15 at 17:45
  • This paper might help, @JohnRennie –  Mar 22 '15 at 16:38

6 Answers6

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The nature of singularities in GR is a delicate issue. A good review of the difficulties presented to define a singularity are in Geroch's paper What is a singularity in GR?

The problem of attaching a boundary in general to a spacetime is that there is not natural way to do it. for example, in the FRW metric the manifold at $t=0$ can be described by two different coordinate systems as: $$\{t,r\cos\theta,r\sin\theta \cos\phi,r\sin\theta \sin\phi\}$$ or $$\{t,a(t)r\cos\theta,a(t)r\sin\theta \cos\phi,a(t)r\sin\theta \sin\phi\}$$ In the first case we have a three dimensional surface, in the latter a point.

It might be tempting to define a singularity following other physical theories as the points where the metric tensor is undefined or below $C^{2}$. However, this is troublesome because in the gravitational case the field defines also the spacetime background. This represents a problem because the size, location and shape of singularities can't be straightforward characterize by any physical measurement.

The theorems of Hawking and Penrose, commonly used to show that singularities in GR are generic under certain circumstances have the conclusion that spacetime must be geodesically incomplete (Some light-paths or particle-paths cannot be extended beyond a certain proper-time or affine-parameter).

As mentioned above the peculiar characteristic of GR of identifying the field and the background makes the task of assigning a location, shape or size to the singularities very delicate. If one thinks in a singularity of the gravitational potential in classical terms the statement that the field diverges at a certain location is unambiguous. As an example, take the gravitational potential of a spherical mass $$V(t,r,\theta,\phi)=\frac{GM}{r}$$ with a singularity at the point $r=0$ for any time $t$ in $\mathbb{R}$. The location of the singularity is well defined because the coordinates have an intrinsic character which is independent of $V$ and are defined with respect the static spacetime background.

However, this prescription doesn't work in GR. Consider the spacetime with metric $$ds^{2}=-\frac{1}{t^{2}}dt^{2}+dx^{2}+dy^{2}+dz^{2}.$$ defined on $\{(t,x,y,z)\in \mathbb{R}\backslash \{0\}\times \mathbb{R}^{3}\}$. If we say that there is a singularity at the point $t=0$ we might be speaking to soon for two reasons. The first is that $t=0$ is not covered by our coordinate chart. It makes no sense to talk about $t=0$ as a point in our manifold using these coordinates. The second thing is that the lack of an intrinsic meaning of the coordinates in GR must be taken seriously. By making the coordinate transformation $\tau=\log(t)$ we obtain the metric $$ds^{2}=d\tau^{2}+dx^{2}+dy^{2}+dz^{2},$$ on $\mathbb{R}^{4}$ and remain isometric to the previous spacetime defined in $\{(t,x,y,z)\in \mathbb{R}\backslash \{0\}\times \mathbb{R}^{3}\}$. What we have done is find an extension of the metric to $\mathbb{R}^{4}$. The singularity was just a coordinate singularity, similar to the event horizon singularity in Schwarzschild coordinates. The extended spacetime is of course Minkowski spacetime which is non-singular.

Another approach is to define a singularity in terms of invariant quantities such as scalar polynomials of the curvature. This are scalars formed by the Riemann tensor. If this quantities diverge it matches our physical idea that and object approaching regions of higher and higher values must suffer stronger and stronger deformations. Also, in many relevant cosmological models like FRW and Black Holes metrics one can show that this indeed happen. But as mentioned the domain of the gravitational field defines the location of events so a point where the curvature blow up might not be even in the domain. Therefore, we must formalise the following: statement "The scalar diverges as we approach a point that has been cut out of the manifold.". If we were in a Riemann manifold then the metric define a distance function $$d(x,y):(x,y)\in\cal{M}\times\cal{M}\rightarrow \inf\left\{\int\rVert\dot{\gamma}\rVert \right\}\in\mathbb{R}$$ where the infimum is taken over all piecewise $C^{1}$ curves $\gamma$ from $x$ to $y$. Moreover, the distance function allows us to define a topology. A basis of that topology is given by the set $\{B(x,r):y\in{\cal{M}}| d(x,y)\le r \forall x\in \cal{M}\}$. The topology naturally induce a notion of convergence. We say the sequence $\{x_{n}\}$ converges to $y$ if for $\epsilon> 0$ there is an $N\in \mathbb{N}$ such that for any $n\ge N$ $d(x_{n},y)\le \epsilon$. A sequence that satisfies this conditions is called a Cauchy sequence. If every Cauchy sequence converges we say that $\cal{M}$ is metrically complete Notice that now we can describe points that are not in the manifold as a point of convergence of a sequence of points that are. Then the formal statement can be stated as: "The sequence $\{R(x_{n})\}$ diverges as the sequence $\{x_{n}\}$ converges to $y$" where $R(x_{n})$ is some scalar evaluated at $x_{n}$ in $\cal{M}$ and $y$ is some point not necessarily in $\cal{M}$. In the Riemannian case if every Cauchy sequence converges in $\cal{M}$ then every geodesic can be extend indefinitely. That means we can take as the domain of every geodesic to be $\mathbb{R}$. In this case we say that $\cal{M}$ is geodesically complete. In fact also the converse is true, that is if $\cal{M}$ is geodesically complete then $\cal{M}$ is metrically complete.
So far, all the discussion has been for Riemann metrics, but as soon as we move to Lorentzian metrics the previous discussion can't be used as stated. The reason is that Lorentzian metrics doesn't define a distance function. They do not satisfy the triangle inequality. So we only have left the notion of geodesic completeness.

The three kinds of vectors available in any Lorentzian metric define three nonequivalent notions of geodesic completeness depending on the character of the tangent vector of the curve: spacelike completeness, null completeness and timelike completeness. Unfortunately, they are not equivalent it is possible to construct spacetimes with the following characteristics:

  • timelike complete, spacelike and null incomplete
  • spacelike complete, timelike and null incomplete
  • spacelike complete, timelike and null incomplete
  • null complete, timelike and spacelike incomplete
  • timelike and null complete, spacelike incomplete
  • spacelike and null complete, timelike incomplete
  • timelike and spacelike complete, null incomplete

Moreover, in the Riemannian case if $\cal{M}$ is geodesically complete it implies that every curve is complete, that means every curve can be arbitrarily extended . Again, in the Lorentzian case that is not the case, Geroch construct an example of a geodesically null, timelike and spacelike complete spacetime with a inextendible timelike curve of finite length. A free falling particle following this trajectory will accelerate but in a finite amount of time its spacetime location would stop being represented as a point in the manifold.

Schmidt provided an elegant way to generalise the idea of affine length to all curves, geodesic and no geodesics. Moreover, the construction in case of incomplete curves allows to attach a topological boundary $\partial\cal{M}$ called the b-boundary to the spacetime $\cal{M}$.

The procedure consist in building a Riemannian metric in the frame bundle $\cal{LM}$. We will use the solder form $\theta$ and the connection form $\omega$ associated to the Levi-Civita connection $\nabla$ on $\cal{M}$. Explicitly,

\begin{equation} G_{ab}(X_{a},Y_{a})= \theta(X_{a}) \cdot \theta(Y_{a})+\omega(X_{a})\bullet \omega(Y_{a}) \end{equation}

where $X_{a},Y_{a}\in T_{p}\cal{LM}$ and $\cdot,\bullet$ are the standard inner product in $\mathbb{R}^{n}$ and $\mathfrak{g}\cong\mathbb{R}^{n^{2}}$.

Let $\gamma$ be a $C^{1}$ curve through $p$ in $\cal{M}$ and a basis $\{E_{a}\}$. Now choose a point $P$ in $\cal{LM}$ such that $P$ satisfies $\pi(P)=p$ and the basis of $T_{p}$ is given by $\{E_{a}\}$. Using the covariant derivative induced by the metric we can parallel propagate $\{E_{a}\}$ in the direction of $\dot{\gamma}$. This procedure defines a curve $\Upsilon $ in $\cal{LM}$. This curve is called the lift of the curve $\gamma$. The length of $\Upsilon$ with respect to Schmidt metric, $$l=\int_{\tau}\|\dot{\Upsilon} \|_{G} dt$$ is is called a generalised affine parameter. If $\gamma$ is a geodesic $l$ is an affine parameter. If every curve in a spacetime $\cal{M}$ that has finite affine generalised length has endpoints we call the spacetime b-complete. If it is not $b$- complete we call the spacetime b-incomplete. A classification of singularities in terms of the b-boundary (See Chapter 8, The large Scale Structure of spacetime) was done by Ellis and Schmidt here.

In the case of the FRW the b-boundary $\partial\cal{M}$ was computed in this paper The result is that the boundary is a point. However, the resulting topology in $\partial\cal{M}\cup\cal{M}$ is non-Hausdorff. This means the singularity is in some sense arbitrary close to any event in spacetime. This was regarded as unphysical and attempts to improve the b-boundary construction were made without any attempt having a particular acceptance. Also the high dimensionality of the bundles involved make the b-boundary a difficult working tool.

Another types of boundaries can be attached. For example:

  • conformal boundaries used in Penrose diagrams and in the AdS/Cft correspondance. In this case the conformal boundary as seen here at $t=0$ is a three dimensional manifold.

  • Causal boundaries. This constructions depends only on the causal structure, so it doesn't distinguish between boundary points at a finite distance or at infinity. (See chapter 6, The large scale structure of spacetime)

  • Abstract boundary.

I am unaware if in the two last cases explicit calculations have been done to the case of the FRW metric.

yess
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    still, t=0 isn't part of the manifold. The transformation R{0} x R^3 -> R^{4}; (t,x,y,z) -> (log(t),x,y,z) is no diffeomorphism and therefore it's not suprising that the singularity disappears. Transformations that are no diffeomorphisms change the manifold. – image357 Mar 17 '15 at 17:17
  • There is no singularity to start with. The difficulty is to know what is the maximal domain of the solution. This also make contact with the PDE point of view of GR and the smoothness of the defined solution. – yess Mar 17 '15 at 17:30
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    If the results turns out non-Hausdorff, it is clear that the attempt (to sensibly attach a boundary) must have failed because you no longer have a manifold, right? – Danu Mar 21 '15 at 13:00
  • That is correct in general $\cal{M}\cup\cal{\partial M}$ is not going to be a manifold. But,as I mentioned there were some attempts to fix the situation, nevertheless no attempt fix the situation entirely. – yess Mar 22 '15 at 15:10
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Just for clarification: the manifolds used in general relativity are locally (in the sense of diffeomorphisms) $\mathbb R^{3+1}$. They are not $ \mathbb R^{3+1}$ in general, which is Minkowski space with zero curvature.

In this sense, the points at $t=0$ do not belong to the manifold, as there is no neighborhood which is diffeomorphic to $\mathbb R^{3+1}$ as @Ali Moh already stated. This means that with the FLRW metric alone, one cannot make predictions for the "big bang" (although one can make predictions around $t=0+\epsilon$ for every $\epsilon > 0$).

image357
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I'm assuming you are not talking about the modern view of the history of the universe where the big bang (reheating) was preceded by inflation, and we don't know what happened before inflation or whether there is even a beginning of time.

So if you want to just look at the FRW metric then I think you're comment that the singularity does not belong to the manifold is correct because the limit $t \rightarrow 0$ is not well defined. For example looking at the volume $$\lim_{t\rightarrow 0^+} \int d^4 \sqrt{-g} = \infty$$ whereas $$\int d^4 \sqrt{-g}\,\,\Big|_{t=0} = 0$$

So you cannot smoothly add the singularity point to the manifold.

Ali Moh
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  • where the big bang (reheating) was preceded by inflation Huh? The big bang is the singularity. Inflation happened after the big bang. –  May 07 '19 at 19:20
  • @BenCrowell to be fair, the term Big Bang is sometimes used to mean the reheating after inflation ended. This isn't the way I understand the term as I, like you, would take it to mean the singularity but nevertheless I do see it used in this other sense. – John Rennie May 08 '19 at 04:28
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Just as an example, you can think of the degenerated metric $ds^2=dx^2$ on $\mathbb{R}^2$. Certainly the underlying manifold is $\mathbb{R}^2$ instead of $\mathbb{R}$, but from a purely Riemannian geometrically point of view, you will see no difference if you treat it as $\mathbb{R}$. Mathematically, this is because the metric can be pushed forward along the projective map $\pi:\mathbb{R}^2\to \mathbb{R} \,\,(x,y)\mapsto(x,0)$ (Recall that in general metric can only be pulled back), and we say that the plane shrinks to a line.

  • a degenerate metric makes no sense since the imposed topology is not hausdorff. in your example, take a number x, then every point (x,y) couldn't be distinguished from a point (x,y') with y!=y'. one can try make that manifold hausdorff again by identification (x,y)~(x',y') if x=x' but then you would end up with R and not R^2 – image357 Mar 22 '15 at 00:25
  • @MarcelKöpke that's what I have said. – Alice Akitsuki Mar 22 '15 at 05:53
  • I didn't like notion of a non-hausdorff manifold, since manifolds are per definition hausdorff spaces – image357 Mar 22 '15 at 11:41
  • @but when dealing with Ricci flow or the Einstein eq. people always say "shrink" to something, that's what I wanna explain – Alice Akitsuki Mar 23 '15 at 06:22
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    @MarcelKöpke you can think of the example of Minkowski manifold, where the topology doesn't match with the metric. – Alice Akitsuki Mar 23 '15 at 13:22
  • @MarcelKöpke generally we don't expect the underlying topology be induced by the not necessarily positive definite metric. – Alice Akitsuki Mar 23 '15 at 13:24
  • I'm not sure but for such cases one usually embedds the tangent space at a point into the topological space induced by the metric which has eigenvalues $|e_i|$ if the tangent space metric has eigenvalues $e_i$. For local minkowski-spaces: $\mathbb R^{3+1} \rightarrow \mathbb R^4; x \rightarrow x$. One pulls back that topology, such that a set A is open by definition if and only if it's diffeomorphic image at all points of that set is open in that embedding. This will not work with a degenerate metric tensor since the metric of embedding is sill degenerate leaving the hausdorff problem. – image357 Mar 23 '15 at 18:01
  • I think the above definition of open doesn't work. take: open if and only if for all points of A there exists a open neighborhood of the image point, that when pulled back is completely in A. – image357 Mar 23 '15 at 18:10
  • @MarcelKöpke I still think you shouldn't expect the topology of the manifold be relevant with the metric. though in Riemann case there exists always a metric compatible with the topology, in Lorentz case it may not. And there even exists topological manifold that doesn't admit any smooth structure, in that case we still have open sets. – Alice Akitsuki Mar 24 '15 at 10:39
  • " in Lorentz case it may not" I mean there maybe no Lorentzian metric on some manifold. – Alice Akitsuki Mar 24 '15 at 10:41
  • But such cases are not about generel relativity which is essentially differential geometry with a local minkowski-metric. This only works with smooth manifolds that are locally minkowsky-space beforehand. sure you can look at some strange non-smooth stuff but this is not the scope of general relativty. – image357 Mar 24 '15 at 11:38
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Wikipedia has a webpage on Big Bang Nucleosynthesis, covering from ~1 second to perhaps as long as 20 minutes. In Mangiarotti and Martens's paper "A review of electron–nucleus bremsstrahlung cross sections between 1 and 10 MeV" they discuss screening, the effect of confinement on bremsstrahlung and the influence of multiple scattering.

[I'm currently reading "Spin densities in 4f and 3d magnetic systems by Ian Maskery" to determine if I can improve this answer.]

"I'm interested to find out if we can say anything useful about spacetime at the singularity in the FLRW metric that occurs at t=0.".

Wikipedia's webpage titled "Scale Factor" says: "The relative expansion of the universe is parametrized by a dimensionless scale factor "a". Also known as the cosmic scale factor or sometimes the Robertson-Walker scale factor, this is a key parameter of the Friedmann equations.

In the early stages of the big bang, most of the energy was in the form of radiation, and that radiation was the dominant influence on the expansion of the universe. Later, with cooling from the expansion the roles of mass and radiation changed and the universe entered a mass-dominated era. Recently results suggest that we have already entered an era dominated by dark energy, but examination of the roles of mass and radiation are most important for understanding the early universe.

Using the dimensionless scale factor to characterize the expansion of the universe, the effective energy densities of radiation and mass scale differently. This leads to a radiation-dominated era in the very early universe but a transition to a matter-dominated era at a later time and, since about 5 billion years ago, a subsequent dark energy-dominated era.".


See the Wikipedia webpage section "Idealized Hubble's Law" which says: "The mathematical derivation of an idealized Hubble's Law for a uniformly expanding universe is a fairly elementary theorem of geometry in 3-dimensional Cartesian/Newtonian coordinate space, which, considered as a metric space, is entirely homogeneous and isotropic (properties do not vary with location or direction). Simply stated the theorem is this:

  • Any two points which are moving away from the origin, each along straight lines and with speed proportional to distance from the origin, will be moving away from each other with a speed proportional to their distance apart.

Diagram of Friedmann equations

The age and ultimate fate of the universe can be determined by measuring the Hubble constant today and extrapolating with the observed value of the deceleration parameter, uniquely characterized by values of density parameters (ΩM for matter and ΩΛ for dark energy). A "closed universe" with ΩM > 1 and ΩΛ = 0 comes to an end in a Big Crunch and is considerably younger than its Hubble age. An "open universe" with ΩM ≤ 1 and ΩΛ = 0 expands forever and has an age that is closer to its Hubble age. For the accelerating universe with nonzero ΩΛ that we inhabit, the age of the universe is coincidentally very close to the Hubble age.

That diagram uses the following exact solutions to the Friedmann equations:

$${\displaystyle {\begin{cases}a(t)=H_{0}t&\Omega _{M}=\Omega _{\Lambda }=0\\{\begin{cases}a(q)={\tfrac {\Omega _{M}}{2(1-\Omega _{M})}}(\cosh q-1)\\t(q)={\tfrac {\Omega _{M}}{2H_{0}(1-\Omega _{M})^{3/2}}}(\sinh q-q)\end{cases}}&0<\Omega _{M}<1,\ \Omega _{\Lambda }=0\\a(t)=\left({\tfrac {3}{2}}H_{0}t\right)^{2/3}&\Omega _{M}=1,\ \Omega _{\Lambda }=0\\{\begin{cases}a(q)={\tfrac {\Omega _{M}}{2(\Omega _{M}-1)}}(1-\cos q)\\t(q)={\tfrac {\Omega _{M}}{2H_{0}(\Omega _{M}-1)^{3/2}}}(q-\sin q)\end{cases}}&\Omega _{M}>1,\ \Omega _{\Lambda }=0\\a(t)=\left({\tfrac {\Omega _{M}}{\Omega _{\Lambda }}}\sinh ^{2}\left({\tfrac {3}{2}}{\sqrt {\Omega _{\Lambda }}}H_{0}t\right)\right)^{1/3}&0<\Omega _{M}<1,\ \Omega _{\Lambda }=1-\Omega _{M}\end{cases}}}$$

"The comment was roughly: yes the Big Bang did happen at a point because at t=0 all points were just one point. If my musings above are correct the comment is untrue because even when the metric is singular we still have the manifold R3,1 with an infinite number of distinct points. If my memory is correct that the points with t=0 are not part of the manifold then we cannot say the Big Bang happened at a point, or the opposite, because we cannot say anything about the Big Bang at all.

From the purely mathematical perspective I'd be interested to know what if anything can be said about the spacetime at t=0 even if it has no physical relevance.".

A pointlike particle present at the center of a black hole has zero mass and size so theoretically an infinite number of zero sized particles can be compressed into a zero sized space; from a practical standpoint fewer than an infinite number of particles are available.

As much mass as is present in the Milky Way galaxy, converted to photons or other zero sized particle, could theoretically be compressed to a point of zero size. While not a whole lot can be said, and far less proven, about what happened simultaneously with the moment of the Big Bang I feel safe saying that even less could be said about 'the other side of the Big Bang'.

That said, I would speculate that 'how things worked/still operate? there' likely bears some similarities to how things work here. Perhaps a large low density black hole captured many small and high density black holes forming a object without size (mass), yet still able to exceed critical mass (or critical energy, if you prefer), that simply ripped a hole in where it was spitting it's energy into where we are.


These are the illustrations that NASA has prepared on The Beginning:

How far can we see, and why?

Looking backwards from today to The Beginning

CMB Timeline

[Note: Images obtained from NASA's WMAP Website.]

"... what if anything can be said about the spacetime at t=0 even if it has no physical relevance.".

It was a point from which everything around us was derived.

Rob
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If the FLRW metric $dt^2- A(t) (dx^2+ dy^2 + dz^2)$ is asymptotically radiation dominant as $t → 0$ (i.e. that the curvature scalar $R ~ 0$ for $t ~ 0$), then $A(t) ~ kt$ for $t ~ 0$, where $k$ is a constant.

Such a metric has the following properties:

  1. its geometry is not Riemannian or Riemann-Cartan; which limits the applicability or relevance of many folklore results (e.g. singularity theorems)
  2. it is geodesically complete; but tangent vectors on the $t = 0$ surface do not uniquely determine a geodesic
  3. the metric is signature changing across the $t = 0$ boundary, becoming locally 4+0 Euclidean for $t < 0$
  4. the initial hypersurface is a null surface (i.e. light speed is infinite at $t = 0$)
  5. the $t = 0$ hypersurface is, thus, an instantaneous slice of Newton-Cartan geometry; the transition across the t = 0 boundary is a physical realization of both the Galilean limit (at $t = 0$) and Euclideanization for ($t < 0$); light speed goes to infinity as $t → 0$ and distances, as measured in light-speed units, therefore go to 0, though the geometry itself does not contract to a point (actually: none of the FRW geometries contract to a point, the common visualization of a universe contracting to a point is little more than a myth that's passed along in popularizations and even in professional circles as folklore)
  6. the cosmology is therefore inflationary (inflation follows as a junction condition for signature-changing cosmologies with a null initial hypersurface)
  7. the past-directed null geodesics reflect off of $t = 0$ as parabolic curves and reverse direction, becoming future-directed
  8. each point is therefore causally connected to remote areas of the universe
  9. the co-moving timelike geodesics go straight through $t = 0$ to the other side $t < 0$
  10. all other timelike geodesics reflect off of $t = 0$ as catenary curves
  11. it is acausal: your future light cone and future worldline are contained inside the past light cone
  12. the spacelike geodesics for $t > 0$ are sinusoidal, and bounded away from $t = 0$.
Kyle Kanos
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  • Curved-spacetime QFT on this background may be possible ... at least for t > 0; but subject to constraints since (a) past geodesics reverse and become future-directed at t = 0, (b) the effective Galilean limit at t = 0 and Euclideanization for t < 0 requires that the underlying field theory be formulated in a unifying framework that contains both relativistic and non-relativistic forms as special cases (as well as the Euclideanized version of field theory). – Rock Brentwood May 07 '19 at 19:10
  • its geometry is not Riemannian or Riemann-Cartan Not sure what you mean by this. The geometry is of course semi-Riemannian, as with all the spacetimes we study in GR. which limits the applicability or relevance of many folklore results (e.g. singularity theorems) "Folklore results" would normally be used to indicate something that isn't true but that is widely believed to be true. The Penrose and Hawking singularity theorems are not folklore results. They are rigorous mathematical theorems. The assumption of an energy condition is a problem for cosmological spacetimes, but this [...] –  May 07 '19 at 19:25
  • [...] is not really an issue for the cosmological parameters we observe (i.e., there is no "bounce"), as shown by Borde, Guth, and Vilenkin. –  May 07 '19 at 19:26
  • it is geodesically complete No, the FLRW spacetimes that we normally study are geodesically incomplete. the metric is signature changing across the t=0 boundary, becoming locally 4+0 Euclidean for t<0 No, the geodesic incompleteness means that you can't carry out an analytic extension through t=0. –  May 07 '19 at 19:27