Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If the universe is spatially infinite (and assuming, if it makes a difference, that we don't have eternal inflation), what actually happened 13.7 billion years ago? Was the energy density infinite (or "very large") at each point in R3? Or did R3 itself collapse to some other structure?

share|improve this question
    
Related: physics.stackexchange.com/q/1915/2451 and links therein. –  Qmechanic Dec 5 at 18:54

3 Answers 3

up vote 4 down vote accepted

The laws of physics - field theory and/or general relativity - are local (and in string theory, they're approximately local). So it doesn't make much difference whether a slice of the Universe is compact or noncompact. Near the Big Bang, the proper distances were shrunk almost to zero. But in cosmology, it still makes sense to use the coordinates in which the geometry of the slice is $$ds^2 = a(t)^2 (dx^2+dy^2+dz^2)$$ where $a(t)$ is an overall rescaling factor that goes to zero, $a(t)\to 0$, for $t\to 0$. The parameter $t$ may also be reparameterized so that the $t=|r|$ lines are null which is useful to clarify the causal structure of the spacetime.

At any rate, near $t=0$, physics works "locally" in the coordinates $(x,y,z)$ above, regardless of the fact that $a(t)$ goes to zero. Objects moving along lines with variable $t$ but fixed $(x,y,z)$ correspond to static objects. You shouldn't imagine that objects are getting further in the $(x,y,z)$ coordinates near $t=0$ just in order to keep a proper distance fixed. No real objects are trying to keep their proper distance fixed as the Universe expands; the expansion of the proper distances is very real and you shouldn't try to deny it.

At any nonzero $t$, the $R^3$ stays the very same $R^3$ (and $S^3$ or $H^3$ would stay the same manifolds as well), and the laws of physics can't be blindly extrapolated to the strict $t=0$ point, anyway. So it makes no physical sense to ask what was "before" the Universe was an $R^3$. There was just a singularity at $t=0$; at every moment that physics can discuss, the geometry was $R^3$.

Eternal inflation has some new things to say - which would be relevant if eternal inflation were right - about these matters but you explicitly asked this topics to be avoided.

share|improve this answer

Your question is very good. I find the idea of eternal inflation to be problematic. The vacuum which induces this frantic inflation is a scalar field which gets stretched beyond the horizon length. In this inflationary situation the vacuum energy density is about 14 orders of magnitude lower than the Planck energy density. This means any region has a Hubble volume at subatomic scales and the inflaton field is stretched across regions where it is not causally complete. It gets frozen, in a sense. It also becomes attenuated by its own inflationary action. So the eternal inflation concept has always struck me as questionable.

The potential for the inflationary potential early in the universe is a de Sitter form. The FLRW equations are $$ \Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\Lambda}{3}~-~\frac{k}{a^2}, $$ where we assume $k~=~0$ for the generally flat space we appear to observe. The early inflationary universe was driven by a scalar field which generated this vacuum energy where $V(\phi)~=~-a\times\phi$, $a$ a constant. This set the early cosmological constant for the de Sitter expansion with a vacuum energy about 13 orders of magnitude smaller than the Planck energy. The universe had more vacuum energy density than quark-gluon field density in a hadron.

The Lagrangian for a scalar field is $L~=~(1/2)\partial^a\phi\partial_a\phi~–~V(\phi)$ and in QFT we work with the Lagrangian density ${\cal L}~=~L/vol$ so the action $S~=~\int d^3xdt{\cal L}(\phi, \partial\phi)$. We run this into the Euler-Lagrange equation $\partial_a(\partial{\cal L}/\partial(\partial_a\phi))~-~\partial{\cal L}/\partial\phi~=~0$, and keep in mind $vol~\sim~x^3$. This gives a dynamical equation $$ \partial^2\phi ~-~ (3/vol^{4/3})\partial_a\phi~–~ \frac{\partial V(\phi)}{\partial\phi}~=~ 0. $$ If we assume the inflaton field is more or less constant on the space for a given time on the Hubble frame this DE may be simplified to $$ {\ddot\phi}~–~(3/vol^{4/3}){\dot\phi}~–~\frac{\partial V(\phi)}{\partial\phi}~=~0 $$ That middle term is interesting for it is a sort of friction. It indicates the inflaton field, the thing which drives the inflationary expansion, is running down or becoming diffused in the space. The potential function here is complicated and not entirely known, but it is approximately constant, or some small decrease with the value of $\phi$. What then happens, which is not entirely understood, is that the field experiences a phase transition, the potential becomes $V(\phi)~\sim~\phi^2$ with a minimum about 110 orders of magnitude smaller than it was in the unbroken phase.

So nucleation bubbles, such as the observable universe with a small cosmological constant, form so long as the inflaton field is large. However, as it rolls downwards and becomes attenuated bubble nucleation may decrease. It is like soda water which loses its fizziness. So the whole $R^3$ may by now be “flat,” and our observable universe is due to one bubble out of a vast number produced spread across this $R^3$.

So what started this whole shebang? It may be due to the $Dp$-brane collisions, or due to the tunneling of some vacuum region near a black hole singularity in another spacetime, or … . There are a number of possible ways this may have been generated. So we might imagine some vacuum “blob” near a black hole singularity which tunnels out of that spacetime and generates a nascent deSitter spacetime. For the Reissnor-Nordstrom metric with $g_{tt}~=~1~-~r_0/r~-~\Lambda r^2/3$ then near the signularity the Wheeler-deWitt (WDW) euqation is approximately $$ \Big[\frac{\partial^2}{\partial a^2}~-~\frac{9\pi^2}{4G^2}\Big(a^2~-~\frac{\lambda}{3}a^4\Big)\Big]\Psi[a]~=~0, $$ where there is this pottential $U(a)~=~Ca^2~-~Da^4$ the quantum field tunnels through. This is then a barrier for the production of a cosmology. If $k~=~1$ for this blob then this is a tiny $R^3$ region which tunnels into a new spacetime. So somehow if the observable universe is flat with $k~=~0$ there is a topology change.

The WDW equation defines Hartle-Hawking states. A recent paper by Ashoke Sen http://arxiv.org/abs/1101.4254 illustrates how the $AdS_2$ boundary is a $CFT_1$ which corresponds to Hartle-Hawking states. This means there may be some “braney” correspondence between the above model with the WDW and $Dp$-brane physics. So the blob which tunnels from the neighborhood of a black hole singularity might amount to the quantum “deposition” of IIB strings onto a brane. This then kicks off the inflationary spacetime, or think of dropping the Mentos into the coke bottle, which sets off lots of bubble nucleation.

This is admittedly rather speculative. I am suggesting this as a possible way of thinking about this, and way which might get around this little issue you raise with the apparent problem of there being an infinite amount of energy on an $R^3$.

share|improve this answer
    
Another tour de force, @Lawrence +1. Your answers are getting better with a more optimal mix of math and prose. Just to mention for clarity, the Wheeler-deWitt equation is, roughly speaking, the quantization of the first FLRW equation (mentioned above), where $a$ is the configuration variable and $p_a \propto \dot a$ the momentum variable are replaced by the operators $\hat a$ and $\partial/\partial_a$. –  user346 Feb 10 '11 at 16:21
    
I suppose somebody down voted it then. Yes this is the WDW for a mini-superspace of the FLRW. What I wrote above is something I have been kicking around the last few weeks. –  Lawrence B. Crowell Feb 10 '11 at 17:32
1  
yes, it seems you have a hater. Can't be helped I suppose. Just keeping giving good answers and maybe the hate will melt away. –  user346 Feb 10 '11 at 18:36
    
Thanks for this "origins" post. My particular interest is what happens to the R^3 topology as t -> 0, and what we can say about the limit point t=0 in post-GR theories. –  Nigel Seel Feb 11 '11 at 9:57
    
The R^3 at t = 0 is critical. If it starts out as S^3 and is topologically changed to R^3 then there may be topological charges or quantum numbers involved with the structure of matter on the classical R^3. These numbers are due to the Kahler form $4\pi k~=~/int F\wedge F$, which might be related to the change in the Betti number. It is interesting to ponder. –  Lawrence B. Crowell Feb 11 '11 at 19:38

I will treat this question as primarily about General Relativity, and its Cosmological metric solution as discussed by Lubos. So I would just like to add a few points to that answer.

It needs to be understood that the Big Bang was the creation of Space-Time in General Relativity: thus in particular Space was created at the Big Bang. The use of R$^3$ in the question I take to be a synonym for "Space". Looking at the wording in the question:

"the universe is spatially infinite" - this might be interpreted as a FLRW solution (a Cosmological solution like Lubos's) which has negative or zero curvature. The other FLRW case is positive curvature which would not be considered "spatially infinite". EDIT I shall assume that the question refers to the zero curvature case, in which case the FLRW metric simplifies to

$ds^2 = dt^2 - a(t)^2(dx^2+dy^2+dz^2)$

Like any GR solution there is a question as to what this means, especially Cosmologically. There is clearly a Euclidean R^3 metric here and it is associated with constant times in this metric, giving a family of R^3 spaces. In this sense there is an R^3 model in GR. However from the perspective of an individual observer O it is worth noting that O's particle horizon is finite ie the Observer doesnt "see" R^3 at a given time t, just elements of their past light cone.

It is of interest to note that a transformation of the t coordinates renders this equation as:

$ds^2 = a(T)^2(dT^2 - dx^2+ dy^2+ dz^2)$

This is conformally equivalent to Minkowski space - as a model for the Universe!

The question is about t=T=0 however. In Penrose Conformal Space-Time diagrams and elsewhere the traditional notion that the Singularity is a "point" is replaced by it being a "3-surface" (drawn as a wavy line in these diagrams). From this perspective we could view the T=0 singularity as a 3-surface too ie perhaps R^3. So the notion that the Big Bang is a zero dimensional point could be misleading here.

In either case it becomes a question as to whether there was a pre-Big Bang state - perhaps one which generated the 3-surface.

Penrose has in fact pursued this idea in the Conformal Cyclic Cosmology model. Although I shall not give the full details, in this approach a mathematical technique called "Conformal rescaling" is applied at or near the Big Bang. What this does is expand (or "inflate") the metric from a point to an entire 3-surface (not necessarily R$^3$ ).

It should be noted that "Conformal rescaling" maintains the light cone and causality structure of a metric but expands (or contracts) infinite spaces into finite regions (as in the Conformal diagrams alluded to earlier). Thus in a sense it is more "inflationary" than Inflation.

This 3-surface is taken to be the end surface after a previous aeon. So the Big Bang is not a point, nor is it the beginning of the Universe, just the beginning of an "aeon". Penrose reckons that, more or less, he can deduce this from GR without much new Quantum Theory. This theory claims to be able to explain any large scale structures in CMB that even "Inflation" might not be able to explain.

The CCC idea is tentative in several respects and does not as yet include Quantum effects which it will surely need to become a full theory of pre-Big Bang.

share|improve this answer
    
In your third para you say that a non-positive curvature corresponds to an "open ball" topology for the universe (finite but unbounded). I find that surprising. Are you saying there are no universe models which are "actually infinite" for t > 0, i.e. have the underlying set-structure of R^3 for t>0?? –  Nigel Seel Feb 11 '11 at 10:18
    
The problem is with the word "is" in the quote - it refers to a "now" which doesnt exist in GR. There is the classic R^3 proper t=const Spacelike slice through each point however in the FLRW model - but its physical significance is complicated in that Space and Time cannot really be separated in GR. This subtlety may not always be significant: but for this question about t=0 it probably is. I think I need to EDIT and expand that para and maybe relate it to the CCC section. Might take some time... –  Roy Simpson Feb 11 '11 at 16:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.