# The universe appears to have a lower bound in the time dimension, why not an upper bound?

The Big Bang looks like a lower bound to the "size" of the universe in the time dimension. Could it also have an upper bound, some furthest point in time from the Big Bang?

• A few words of caution: I think the words 'upper and lower bound' are not very appropriate here; furthermore, it is not clear what was happening right at the moment of the big bang - whether time was even a concept that made sense then, or whether time ever really 'started' are not obviously well-defined and/or answerable questions.
– Danu
Aug 12 '14 at 15:43
• ...that being said, the answer to your question is: No, as far as we know there is no 'latest possible time' or anything like that.
– Danu
Aug 12 '14 at 15:44
• this is also coordinate dependent. In conformal time, for inflationary physics and early universe cosmology, the big bang happened at $t=-\infty$, which removes the lower bound
– Jim
Aug 12 '14 at 15:45
• ...and of course the same coordinate-dependence makes the question itself ill-defined...
– Danu
Aug 12 '14 at 15:50
• Because there are some systems where there is a last possible moment of time, such as those used to make penrose diagrams
– Jim
Aug 12 '14 at 15:51

It's certainly possible, though on current evidence it looks unlikely.

The past bound isn't really a bound in the usual sense of the word, but instead it's a singularity. If we solve Einstein's equations for the universe with a few apparently plausible assumptions we find that the universe is described by a scale factor, normally written as $a(t)$, and as this notation suggests the scale factor is a function of time, $t$. If we take any two points in the universe currently separated by some distance $d_0$ then the distance between those points varies with time as:

$$d = a(t)d_0 \tag{1}$$

As the universe ages $a(t)$ gets bigger and $d$ increases, and this is why the universe is expanding. If we wind time backwards towards the Big Bang then $a(t)$ decreases and the universe contracts.

The problem is that as $t \rightarrow 0$ then $a(t) \rightarrow 0$, and therefore from equation (1) we find $d \rightarrow 0$. This means at time zero the spacing between every point in the (possibly infinite) universe was zero. As a side effect the density of the universe goes to $\infty$. This point is the Big Bang.

The Big Bang is singular because at that point we cannot use Einstein's equations to tells us what happened before it, so the singularity places a bound on our ability to calculate the behaviour of the universe. In principle time could extend backwards before the Big Bang to negative values, but we cannot calculate anything about the behaviour of the universe at those negative times.

As an aside, few physicists believe there really was a singularity at the Big Bang. Most of us believe that some form of quantum gravity becomes important at very high densities and this will prevent the density becoming infinite. For example Loop Quantum Cosmology predicts there was a Big Bounce. This is all wildly speculative, but if something like this did happen it means there was no past singularity and time extends smoothly backwards to $-\infty$.

The point of all the above waffling was that the past boundary (if it exists) is due to a singularity, and likewise if there is a future boundary it too must be due to a singularity. In the early days of general relativity it was widely believed that the universe was closed and would recollapse in a Big Crunch. This would be a future singularity and would represent a future boundary of the sort you describe.

However it looks as if the universe is flat and won't recollapse so there is no Big Crunch to put an end to our timekeeping. About the only even remotely possible future singularity would be if dark energy has a particularly pathological equation of state, in which case there could be a Big Rip. This is a singular point and would create a future boundary. However you should appreciate that while the Big Rip is a fun idea there is absolutely no evidence to suggest it's likely to happen.

So the answer to your question is that no, there is (almost certainly) no future boundary to time.

• As I stated in the comments above, this is entirely coordinate dependent. In conformal time, the scale factor is often written as $a(\tau)=e^{H\tau}$ and so you see that the big bang ($a=0$) happens at $\tau=-\infty$. Thus removing the lower bound. There are also coordinate systems like those used to make penrose diagrams that turn infinity into a finite number. In these systems, time has both a minimum and maximum value. But the basic point is that the correct answer to this question heavily depends on how one chooses to measure.
– Jim
Aug 12 '14 at 17:11
• @Jim: the point is that there was (probably) a past singularity, and there probably isn't a future singularity. This seems to me the answer to the question the OP meant to ask. Aug 12 '14 at 17:18
• It is physically not clear, that the past is actually a singularity. It's only a singularity in general relativity, which is an essentially classical theory. You have the same problem in Newtonian mechanics with point masses, and much more so, in classical electromagnetism with charges. These theories assume that these objects are singularities, and they produce a number of artifacts as a consequence. In reality, of course, point masses don't exist, at all, and charge carriers are subject to the laws of quantum mechanics, which completely change their behavior as we approach small scales. Aug 12 '14 at 19:06
• @Jim, CuriousOne: You can avoid the definitional problems with coordinate time and singularities by asking whether there are past-directed or future-directed geodesics of infinite (proper) length, or some variant of that criterion. Aug 12 '14 at 19:32
• @benrg: Sorry for the misunderstanding. My real question is, at what point does our extrapolation of time as the order parameter lose physical significance (i.e. where does GR break down) and what does (if anything!) replace it? It seems intuitive that the disappearance of spacetime should dissolve all order (since, from our perspective, it looks like a high temperature case). That, however, is not obvious, since there is e.g. plenty of structure in high temperature plasmas, it's just different from the low temperature gas. There may therefor very well be an order parameter "before" time. Aug 12 '14 at 19:54