Sunday, March 18, 2018

How old is the Universe?

The Universe is about fifteen billion years old, and started off with a Big Bang, as any fule kno. Well, kind of, but that's more than a slight over-simplification.

There are two fairly obvious questions: first, how do we know? and second, what does that actually mean? Advance warning: there will be mathematics. But if you've got as far as knowing what a differential equation is you should have all you need to cope.

Modelling the Universe

The Mathematics Contribution

Unfortunately, there's no apparent way of directly observing the age of the universe. Any measurement of its age is going to be pretty indirect, and rely quite heavily having on some kind of theory or model of the universe.

First, we'll find a mathematical framework for the model, then incorporate physics into it later.

To start off with, here are some fairly plausible assumptions about the nature of the universe.

  1. The universe is homogeneous
  2. The universe is isotropic
The first assumption is that the universe looks (at any given time) much the same no matter where you are; the second is that it is much the same in every direction. We know that the universe looks (pretty much) isotropic from where we are; assuming that we are nowhere special and the universe is much the same at every point implies that it is also isotropic everywhere.

From these two, we can deduce that (in terms of some very well-chosen coordinates) the geometry of the universe is given by the metric \[ ds^2 = c^2 dt^2 -a(t)^2\left(\frac{dr^2}{1-kr^2} + r^2( d\theta^2+\sin^2(\theta)d\phi^2) \right). \] where \(k\) is either \(-1\), \(0\) or \(1\), and \(a(t)\) is some as yet undetermined function called the scale factor.

The metric tells us about the nature of the separation between two nearby points. There's a lot packed up in that, so it's worth taking a closer look.

Consider a couple of nearby points in the universe, where nearby means that the differences in their coordinates are small, and we call those differences \(dt, dr, d\theta\) and \(d\phi\) . There are a couple of interesting cases.

If the points have the same \(r, \theta, \phi\) coordinates, then \(ds^2=c^2 dt^2\), and \(dt\) is just how much time passes between the two.

On the other hand, if the points are at the same time, then \[ -ds^2 = a(t)^2\left(\frac{dr^2}{1-kr^2} + r^2( d\theta^2+\sin^2(\theta)d\phi^2) \right) \] is the square of the distance between them. We can see how we can regard this distance as the product of \[ \sqrt{\frac{dr^2}{1-kr^2} + r^2( d\theta^2+\sin^2(\theta)d\phi^2))} \] which we call the coordinate distance and \(a(t)\) which we call the scale factor.

The different values of \(k\) tell us about the spatial geometry of the universe.

  1. If \(k=-1\) the universe has negative spatial curvature. This means that if we consider a sphere of radius \(R\), the volume of the enclosed space grows faster than \(4\pi R^3/3\).
  2. If \(k=0\) we have flat space. Then the volume enclosed by a sphere of radius \(R\) is the familiar \(4 \pi R^3/3\).
  3. If \(k=1\) then the universe has positive spatial curvature. In this case the volume enclosed in a sphere of radius \(R\) is less than \(4 \pi R^3/3\).
There is nothing to choose between these possibilities from the point of view of the mathematics. Observationally though, as far as we can tell the universe is spatially flat, i.e. \(k=0\).

It's also worth making a note that although the coordinates make it look as if there's a centre of the universe at at \(r=0\), this isn't true. This \((r,\theta,\phi)\) system of coordinates are just the familiar spherical polar coordinates: we can equally well choose any point to be at \(r=0\).

So our two assumptions mean the the universe can be described in terms of just one dynamical quantity, called the scale factor. This function \(a(t)\) described the entire history of the universe. But the mathematics doesn't tell use anything about it. For that, we require some physical input.

The Physics Contribution

In order to figure out how \(a(t)\) behaves, we need to know what kind of material the universe is full of. Then the Einstein Field Equations tell us the relationship between what the universe is filled with and its geometry (as expressed by the metric), and can be used to find the metric.

So we need a good candidate for the stuff that fills the universe.

One plausible approximation is to thing of the universe as full of a fluid, whose constituent 'atoms' are clusters of galaxies. (Obviously we are working on a very large scale here for this approximation to be sensible.) It is also reasonable to take this fluid to have negligible viscosity, so we model the content of the universe as an ideal fluid.

Given this, the state of the fluid is given by two quantities, the pressure \(p\) and density \(\rho\). Because of the homogeneity and isotropy assumptions, these only depend on \(t\). With quite a lot of calculation, one eventually finds a pair of coupled ordinary differential equations for \(a\), \(\rho\) and \(p\): \[ \begin{split} \left( \frac{\dot{a}}{a} \right)^2 &= \frac{8\pi G}{3}\rho +\frac{\Lambda c^2}{3} - \frac{kc^2}{a^2}\\ \left(\frac{\ddot{a}}{a}\right) &= -\frac{4 \pi G}{3}\left( \rho + \frac{3p}{c^2} \right) + \frac{\Lambda c^2}{3} \end{split} \] In these equations, \(G\) is the gravitational constant, and \(\Lambda\) is the cosmological constant, which appears naturally in the Einstein Field Equations. For a long time it was believed to be \(0\), but recent measurements suggest is is a very small positive quantity.

OK, so this gives us two equations: but there are three quantities to be dealt with, \(a\), \(p\) and \(\rho\). We're missing some information.

To fill the gap, we need to know a bit more than just that the universe is full of an ideal fluid. We need to know what kind of ideal fluid it is. And one way of saying that is by giving a relationship between \(p\) and \(\rho\), which relationship we call the equation of state. Equipped with this we have, at least in principle, all the information required to specify the history of the universe in terms of \(a\), \(p\) and \(\rho\) as functions of \(t\).

A Brief Digression on the Hubble Parameter

In passing, it's worth noting that that quantity \(\dot{a}/a\) which appears in the first equation above is our old friend the Hubble parameter, usually denoted \(H(t)\). (I don't like calling it the Hubble constant, since it's a function of time.) At any time \(t\), the Hubble parameter \(H(t)\) tells us us the relationship between the distance between two objects at fixed spatial coordinates (i.e. fixed \((r,\theta,\phi)\), and the rate at which that distance is changing.

You might briefly wonder how things at fixed position can be getting further apart: but remember, the distance between things is the product of the scale factor \(a(t)\) and the coordinate distance, so even though the coordinate distance is fixed, this physical distance between them changes with \(a(t)\).

The Age of the Universe

Now that we have a model of the universe, we can at least try to work out how old the universe is according to that model. And then we can try to deduce from that something about the actual universe.

There's a useful bit of notation worth introducing here. I decide to call the current time \(t_0\), and then just put a subscript \(0\) on any quantity to indicate that it is the value at \(t_0\): so \(a_o = a(t_0)\), \(H_0 = H(t_0)\) etc.

We should note that that the units of \(H\) are those of inverse time, so on dimensional grounds one might hope that the age of the universe is given by (some multiple of) \(1/H_0\). (Thanks to Leo Stein, @duetosymmetry, for pointing this out.) We'll see how this hope plays out for one simple case just below.

A Simple Dust Filled Universe

As an illustration of how we can extract an age of the universe from this mathematical model, we'll consider the simplest (non-trivial!) model.

First we take \(k=0\). There is no observational evidence that this is wrong, and it makes the equations simpler.

Next, we take \(\Lambda =0\). This isn't quite right, but we're going to be pushing the evolution backwards, which will mean that \(\rho\) will be increasing, so the contribution from \(\Lambda\) will become less important rather than more important as we push further. With luck this won't be too much of a problem.

Finally, we take \(p=0\), i.e. we assume that the universe is full of a fluid that does not resist compression. Since there is a lot of space between galaxies, and they aren't moving very fast, this seems plausible too. In this context such a fluid is called dust.

Now, with all these simplifications, our differential equations reduce to \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8\pi G}{3}\rho, \qquad \left(\frac{\ddot{a}}{a}\right) = -\frac{4 \pi G}{3} \rho \] so we immediately get \[ \left( \frac{\dot{a}}{a} \right)^2 = -2 \left(\frac{\ddot{a}}{a}\right) \] or \[ \dot{a}^2 = -2a \ddot{a}. \]

Faced with a differential equation like this - nonlinear, and not one I've seen a special method to solve - I resort to guessing. It looks as if taking \(a(t)\) to be some power of \(t\) might work, since the powers on each side of that equation would match, so I guess that \(a(t)\) is proportional to \(t^\alpha\), and try to find a value of \(\alpha\) that works.

Plugging this into the equation gives \(\alpha^2=-2\alpha(\alpha-1)\), or \(\alpha(3\alpha-2)=0\). The \(\alpha=0\) solution is not interesting: it would mean that \(a(t)\) was constant, so the \(\alpha = 2/3\) solution is the relevant one.

This gives us \[ a(t) = a_0\left( \frac{t}{t_0} \right)^{\frac{2}{3}} \]

Now we're getting somewhere. We can see that this is fine for any value of \(t>0\), but when \(t=0\) it all goes to pieces. \(a(0)=0\), and if you look at the first equation you'll see that the expression for \(\rho\) involves a division by \(0\), which is never a good sign.

But let's gloss over that for the moment. What I'd like to know is the value of \(t_0\), because that's how long the universe (according to this model) has been around.

We can do something a little bit cunning here, and compute \(H(t)=\dot{a}/a\). This gives us \[ H(t) = \frac{2}{3t} \] or, more to the point, \[ t_0 = \frac{2}{3H_0} \] and \(H_0\) is something that we can try to find from observation. Admittedly, the measurements are quite tricky, but at least it's some observational data that tells use the age of the universe.

Well, no.

That doesn't tell us the age of the universe. What it tells us is that if we push the model back in time, the very longest we can push it is this \(t_0\). It doesn't tell us that the universe sprang into existence that long ago and then evolved according to these equations subsequently.

In fact, we can even say that this value can't be quite right.

The Age of the Universe?

If we push this back then eventually we are in a regime where the density is very large. At this point, it is almost certainly a bad approximation to assume that \(p=0\), and we need to use a more accurate model that takes better care of how pressure and density are related in some kind of hot plasma.

Well, of course, we have reasonable models for this. We plug in the appropriate equation of state, and work out how old the universe is given that for a sufficiently early 'now' that equation of state is relevant.

But then the problem recurses. For sufficiently early times, the conditions become so extreme that we really don't know how to model the matter.

And of course, in all the above I was working with various simplifications. We need to work with rather more complicated evolution equations to do something more realistic.

And when we do all that, we find that the 'age of the universe' is a little under fourteen billion years.

But this age is not really the age of the universe. It is how long it has been since the start of the time when we have some physical understanding of how the stuff filling the universe behaves.

Before that? Since we don't know how the stuff behaves, and it is very likely that classical physics will stop being the relevant model in any case, it's hard to push earlier.

There are lots of speculative theories, some more fun and some more plausible than others. But we certainly can't deduce from what we know that the universe itself has only existed for this fourteen billion years, only that our ability to describe it fails when we try to get further back than that.

The Age of the Universe As We Know It

This is the bottom line. When we talk about the age of the universe in cosmology, we aren't really talking about the age of the universe as such. We're talking about the length of time for which we have some understanding of what is going on in the universe, i.e the time since which content of the universe was something we have reasonable models for. The Big Bang is an idealization of the time at which the universe became accessible to our theories and models.

Before that, it may have spent an undetermined, or even infinite, length of time full of some kind of very high energy quantum field which underwent a phase transition to the hot dense plasma that looks to us like a Big Bang beginning to the universe. Or maybe something even weirder, such as a state in which the notion of time itself doesn't make sense, so that the 'age of the universe' is the length of time for which time has been a useful notion. There are various notions out there of what might have preceded the Big Bang, but by the very nature of things, they tend to be relatively unconstrained by observations.

Addendum

Nalini Joshi (@monsoon0) has pointed out that \[ \dot{a}^2 = -2a \ddot{a} \] can be solved without resorting to guessing the right answer, by using separation of variables (twice). Dividing both sides by \(a \dot{a}\) gives \[ \frac{\dot{a}}{a} = -2 \frac{\ddot{a}}{\dot{a}} \] which can be integrated immediately to give \[ \ln(a) = -2 \ln(\dot{a}) + C_0 \] so that \(a = C_1\dot{a}^{-2}\).

This rearranges to \[ \dot{a}^2 = \frac{C_1}{a} \] and so \[ \dot{a}=\frac{C_2}{\sqrt{a}} \] which can then be separated to give \[ \sqrt{a}da = C_2 dt \] Integrating this (and choosing the constant of integration so that \(a(0)=0\)) yields the solution given above.

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