Pages

Tuesday, 29 May 2018

What does it mean to say that the universe is expanding?

Q: The universe is expanding, so what is it expanding into?

A: The question is wrong.

OK, that answer could use some expansion of its own.

Newtonian Cosmology

Here's one version of the story.

Careful measurements (and careful interpretation of the measurements) tell us that the stuff in the universe is getting further apart. What's more, if we look around us, then to a good degree of approximation, everything is travelling away from us, and the farther away it is, the faster it's getting away from us. In fact, the speed at which distant galaxies are receding from us is proportional to how far away they are. We know this because of red shift in the spectra of these galaxies, from which we can calculate how fast they are travelling in order for a Doppler shift to cause that much red shift.

This seems to suggest that we are (surprisingly or not will depend on your religious and philosophical framework) at the centre of the universe; it is perhaps less flattering that everything is trying to get away from us.

Let's start off by thinking about how special our position is. We think of the universe as being full of a homogeneous dust (for scale, the dust particles are clusters of galaxies). In the simplest model, we are at the origin, and at time \(t\) the the dust particle at position \(\pmb{x}\) has velocity \(\pmb{v}=H(t) \pmb{x}\). Then \(H(t)\) is the scale factor which relates distance to velocity of recession, and we call it \(H\) the Hubble parameter, to honour Hubble who made the initial observations.

But what about somebody who isn't at the origin? What about somebody living in a different dust particle, far, far away, say at position \(\pmb{x}'\)? What would they see?

Let's use \(\pmb{v}'\) to denote the velocity as seen by this observer at position \(\pmb{x}'\). We have to subtract their velocity from the velocity we see at each point to get the velocity relative to them at each point. Then we get \[ \pmb{v}' = H(t) \pmb{x} - H(t) \pmb{x}' = H(t)(\pmb{x}-\pmb{x}') \] But \(\pmb{x}-\pmb{x}'\) is just the position as seen by the observer at \(\pmb{x}'\); this other observer also sees everything in the universe recede at a rate proportional to distance, and it's even the same constant of proportionality. Rather suprprisingly, this suggests that we aren't anywhere special, and that the cosmological principle (or principle of mediocrity) that we are nowhere special is compatible with the observations.

The first thing to notice is that the universe is infinite - it's the whole of three dimensional Euclidean space - forever. And it's always full of this cosmological dust: but if we go back in time, we see that the dust particles are getting closer together, so the dust must be getting denser, and if we go forward in time, it is getting more rarefied. But there's no 'edge' to the material, no expanding of anything, even though the dust particles are getting further apart. This is one of the joys of having infinite space: it can be full of stuff, but there's nothing to stop the contents spreading out.

There's a slight problem when we try to push back to \(t=0\), when the density would have to be infinite: but I avoid that by saying that I have a model for a universe that seems compatible with Hubble's observations for all \(t \gt 0\), and I've no idea what kind of Big Bang happened at \(t=0\); that's not part of my model. There's certainly no sense in which the universe started off as a single 'infinitely dense point' and exploded outward.

So, great news! As long as the universe just happens to be full of stuff exploding outwards, we have a rather neat mathematical model compatible with the observations, and all we need is some physics compatible with this model to make it plausible.

Physics.

Aye, there's the rub.

Problem One

The first problem is a kinematic one. No matter how small \(H(t)\) is, objects sufficiently far away have to be moving faster than light. This is definitely hard to reconcile with special relativity, and special relativity is very well confirmed these days, so don't just want to abandon it and go back to Newtonian mechanics. This cosmological model really is only good for Newtonian mechanics.

We could try to fix it by slowing down the expansion sufficiently far away, so that Hubble's law is only valid over a certain scale: but that gets hard to reconcile with the cosmological principle.

Or we might try to fiddle with special relativity by arguing that it's great at small scales, but needs correction at larger ones.

Problem Two

But even if we only look for a Newtonian model, there is still a serious problem.

We are at the origin of the universe, and we see everything receding from us. But as thing get farther away, they speed up: everything is not only getting farther apart, but is acceleration. So our distant observer, living in a dust particle far, far away, is accelerating. That's an issue because it means that we are, after all, in a special place. Though all observers see a universe expanding in the same way, only we don't feel a force pushing on us to make us accelerate.

This is actually two problems.

There's the actual physical issue of just what is supposed to be doing the pushing. We need a pretty weird model of gravity to be compatible with gravity being attractive (so as to hold solar systems and galaxies together) at one scale, but repulsive (so as to make stuff accelerate apart at the bigger scale).

And there's the philosophical issue of what makes us so special: in this model we are, once again, in the actual centre of the universe.

So, can we find another way of modelling the universe which retains the good property of satisfying Hubble's law, but doesn't have either of the problems of this repulsive force which only becomes significant at large distances, or puts us in a privileged position?

Relativistic cosmology

Well of course we can. As usual, that's the sort of question that only gets asked if there is a good answer available.

But it involves a radically new way of understanding how the universe fits together.

Newtonian space-time

In the Newtonian picture of space-time, with the origin at the centre of expansion, we have spatial position, represented by a position vector \(\pmb{x}\) and time, represented by \(t\). If we have two events \(A\) and \(B\), which happen respective at position \(\pmb{x}_A\) and time \(t_A\), and at position \(\pmb{x}_B\) and time \(t_B\) then the distance between them is the usual Euclidean distance \(\|\pmb{x}_A -\pmb{x}_B\|\), and the time between them is \(|t_A-t_B|\).
This is all very persuasive, and seems to match well to experience.

But it isn't quite right, in ways which don't show up until things are moving fairly fast, at which point we have to worry about special relativity.

Fortunately, there's another way to think about distance and time, which is both compatible with special relativity (in the sense that it is very well approximated by special relativity over small regions) and with the observations of red shift (though for a somewhat subtler reason than you might expect).

Space-time and the metric

In this picture, we don't have separate notions of spatial and temporal distance, but a unified notion. Just as before, every point has a time coordinate \(t\) and space coordinates \(\pmb{x}\). But now we write \[ ds^2 = c^2 dt^2 - a(t)^2 dx^2 \] which is a way of packaging up a combination of difference between the (time and space) coordinates of two nearby events: \(dx^2\) is a shorthand for the sum of the square of the spatial coordinate differences.

So what does this new thing tell us?

If two events happen in the same place, then the time between them is \(ds/c\). If they happen at the same time, then the physical distance between them is \(a(t)\) times the coordinate distance, which is the square root of the sum of the squares of the coordinate differences (i.e. the usual Pythagoras' theorem thing). This really is the physical distance, as in it counts how many meter sticks (or yard sticks if you're old) you'd have to use to join them up.

It's worth noting that over a small enough region of space-time, this is indistinguishable from special relativity: so we really are doing what was hinted at above, by saying that special relativity (great though it is over small regions) might need to be adjusted to deal with large regions.

\(a(t)\) is called the scale factor, and it relates the coordinate distance between two simultaneous events to the spatial distance. It's a function of time, so if we have two objects, both of whose positions are fixed (i.e. their spatial coordinates are unchanging), the distance between them is nevertheless changing as time passes. This isn't because they're moving apart (or together, for that matter), but because the notion of distance is now something that depends on time.

Let's take a look at this: suppose we have two objects, at fixed spatial locations, so the coordinate distance between them, \(d_C\) is unchanging. The physical distance between then is \(d_P = a(t) d_C\). Now we can think about how fast \(d_P\) is changing. We have \[ \dot{d}_P = \dot{a}(t) \times d_C = \frac{\dot{a}(t)}{a(t)} \times a(t) d_C = \frac{\dot{a}(t)}{a(t)} \times d_P. \] So the rate of change of the distance between these fixed objects is proportional to the distance between them, and the Hubble parameter is \(\dot{a}/a\).
When \(a(t)\) is increasing, we loosely (and maybe misleadingly) call this expansion of space, or even worse, expansion of the universe.

OK, we have Hubble's law again, in the sense that the dust particles are all moving apart with a rate proportional to separation. And again, the universe is always full of these dust particles. But now the particles are stationary (meaning that their spatial coordinates don't change) and geometry is evolving so that the distance between them is changing.

But now we get more.

We also find that particles at fixed coordinates are not accelerating, so no force is required to get this effect of everybody seeing all the material in the universe obeying Hubble's law.

Red shift and recession

But there's something fishy about all this.

The original observation that gives rise to Hubble's law is obtained from measurements of red shift, interpreted as relative velocity. What does all that mean if everything is 'staying still' while 'geometry evolves'?

At this point we have to take into account that when we look at things a long way away, we are seeing them as they were a long time ago. In fact, we're seeing them when the scale factor was a bit different.

Amazingly, there's a geometric effect which exactly (well, almost exactly) fits the required bill.

It turns out that if a light signal is emitted at time \(t_e\) and received at time \(t_r\), then the signal is red-shifted; the ratio of the wavelength of the emitted signal to that of the received one is the ratio of the scale factor at these two times. This is called cosmological red shift and is an entirely geometric effect.

How does this fit in with Hubble's observation, though?

If you consider what an observer sees when looking out at a distance object, the cosmological redshift matches (to a very high degree of approximation) the redshift that you would get if there were no cosmological redshift, and the redshift was due to a Doppler effect from a velocity that's just the same as the rate of change of physical distance.

This is even more amazing than the fact that Hubble's law pops out.

Actually, this isn't exactly true. It is, though, true to a first degree of approximation, and it's very hard to measure distances and redshifts accurately enough to see any deviation. More importantly, it doesn't affect the result that the dependence of rate of recession (really redshift) on distance is the same no matter where in the universe you live.

So how does the scale factor evolve?

Up at the top, I was unhappy with the Newtonian approach because it had some mysterious field of force pushing stuff apart. I seem to have just replaced that with a mysterious scale factor that does the same job.

Not so. In the same way as Newtonian gravity tells us how massive bodies pull on each other, Einsteinian gravity has an equation of motion that tell us how \(a(t)\) behaves.

In Einsteinian gravity, there is the metric (and we've restricted attention to a particular class of metric already) and there is a field equation (the Einstein Field Equation, or EFE) which relates how the metric varies in space and time to the matter filling the universe. In order to work out how \(a(t)\) evolves, we need to have a model for the material that fills the universe.

This model comes in two parts.

First, there's the general model. The simplest thing that has a chance of working is a perfect fluid: so we're considering galactic clusters as parcels of a compressible fluid which has negligible viscosity.

Then, there's the specific model. A fluid has a density and a pressure, and a functional relation between the two is called an equation of state. There are standard equations of state for describing a universe full of easily compressible dust, or full of electromagnetic radiation.

Once you've chosen an equation of state, you can work out what the EFE tells us: and this is a differential equation for \(a(t)\), which can then be solved. One part of constructing a good cosmological model is deciding just what equation of state to use for your model.
Let's not go there.

The point is that \(a(t)\) isn't arbitrary; it comes out of the physical model.

In fact, all this theory was worked out by Friedmann decades before Hubble's observational work.

It's rare in science for somebody to manage to predict something before it actually happens, so stop for a moment to be impressed.

What have we got for our effort?

By investing some effort in having a model of space-time in which the geometry itself is dynamic, we get a universe in which Hubble's law is satisfied, but most of the ingredients are now interpreted in a new way. In particular, the recession of distant galaxies is no longer due to them moving, but now due to the fact that the definition of distance itself depends on time in such a way that the distance between stationary objects can be changing in time.

Have I missed anything out?

Oh boy, have I ever.

I've missed out

  • all the controversy in the interpretation of distance/velocity measurements.
  • just about all the calculations that justify the claims I make. Nothing new there.
  • what the Einstein Field Equations say. That's a big topic just in its own right, and it's a bit technical.
  • what the matter model of a perfect fluid actually looks like.
  • any discussion of the curvature of space. (I've only considered the simplest case.)
  • and much, much more.
Fortunately, there's lots of excellent material out there. If all I've done is whet your appetite for more, I'm happy.

3 comments:

  1. Very clear thanks. If i think of equations of state ("constitutive") equations, like Hooke's say. By equating Hooke to Newton 2 I eliminate the need for the Force concept, relating acceleration to relative position. Similarly are we eliminating the need to unpick the metric (potential) when we plug in the equation of state for e-m tensor of rhs of Einstein's equations? Also - did you mean compressive perfect fluid- or incompressible?

    ReplyDelete
  2. I think you could see it that way: you're just stating a relationship between the dynamics of the geometry and the mass-energy distribution, without saying anything about some kind of force being exerted by the matter.

    It's a compressible fluid - in relativity, there's no such thing as an incompressible fluid (since that would enable you to send a signal instantaneously).

    ReplyDelete
  3. No proper paragraphs (there must be space between them, not just new lines).

    Text is not justified (text-align: justify).

    I refuse to read this.

    // b.

    ReplyDelete