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PC1672 Advanced dynamics


3.13 Cosmology

Matter in the universe as we see it today has a low density and so the gravitational fields acting over large distances are weak. Although a full treatment of the ``geography'' of the universe requires general relativity, we can understand many of its important features using simple Newtonian gravity.

We assume that the universe is uniform on large distance scales, with a density $\rho(t)$. In addition we assume that this matter is now in the form of ``dust'' (consisting of particles as large as stars or even galaxies!) which means that it does not exert any pressure. In the absence of pressure, the only force acting on matter is gravity.

A typical galaxy at some distance $r$ from us feels a gravitational attraction towards us exerted by all the matter inside a sphere of radius $r$ centred on us:

\begin{displaymath}{\bf g}=-{GM(r,t)\over r^2}\end{displaymath}

where

\begin{displaymath}M(r,t)={4\pi\over 3}\rho(t) r^3\end{displaymath}

If we imagine this sphere to expand or contact as the galaxies on its surface move with respect to us, then no mass escapes from it. The mass inside the sphere is then constant in time. $M(r(t),t)=M$ The density of the universe must vary as $r(t)^{-3}$ where $r(t)$ is the distance to our ``test'' galaxy.

The total energy of our test galaxy is a constant of its motion. In an isotropic universe the velocity of the galaxy must point radially with respect to us. The energy (per unit mass) of the galaxy is

\begin{displaymath}\eta={1\over 2}\dot r^2-{GM\over r}\end{displaymath}

Solving for the speed gives

\begin{displaymath}\dot r=\pm\sqrt{{2GM\over r}+2\eta}\end{displaymath}

which just like the case of an orbit with zero angular momentum. This shows that we cannot have $\dot r=0$ for all galaxies at all times; the universe must be expanding (or contracting). From observations of the redshifts of galaxies, we know that it is expanding and so we want the positive root.

An important special case is that of a universe which is just unbound, $\eta=0$. In general relativity this corresponds to a universe where space is flat. In this case the speed of our test galaxy is

\begin{displaymath}\dot r=+\sqrt{{8\pi\over 3}G\rho(t)}\,r\end{displaymath}

in terms of the density. This shows that the speeds satisfy Hubble's law, velocity is proportional to distance from us, with a Hubble ``constant'' of

\begin{displaymath}H(t)=\sqrt{{8\pi\over 3}G\rho(t)}\end{displaymath}

This can also be written in terms of the constant mass $M$ as

\begin{displaymath}\dot r=\sqrt{2GM\over r}\end{displaymath}

which can be integrated to give

\begin{displaymath}r=\left({9GM\over 2}\right)^{1\over 3}t^{2\over 3}\end{displaymath}

where we have taken $r(0)=0$. In other words, the big bang occurs at time $t=0$. This result also implies that the Hubble ``constant'' varies in time as

\begin{displaymath}H={\dot r\over r}={2\over 3}t^{-1}\end{displaymath}

Despite its name, the Hubble ``constant'' decreases in time as gravity slows the expansion of the universe. Our best estimates of its current value are in the region of $0.6-0.7\times 10^{-10}\ \hbox{\rm yr}^{-1}$. (See the March 1998 Scientific American article by Wendy Freedman and NASA's Hubble web site.) Since the expansion of the universe slows down, the universe cannot be older than $1/H$. Estimates of the age of the universe lie in the range $12-15\times 10^9$ years, which is (just) consistent with these values for $H$.

For $\eta\neq 0$ things are a bit more complicated but the basic features are analogous to those of orbits in a gravitational field. If $\eta<0$ the universe is bound; any galaxy reaches a maximum distance from us and the universe eventually collapses in a ``big crunch''. If $\eta>0$ then the speed at which a galaxy is moving away from us tends to a constant at large times.

Up to a couple of years ago the standard picture of the universe was that the density of matter was equal to the critical value that makes $\eta$ exactly zero. This would also give a flat space (in which a straight line is the shortest distance between two points). The idea that space is flat is supported by theorists' prejudices and, more importantly, by observations of fluctuations in the cosmic microwave background by COBE and other experiments, some of which involve Jodrell Bank. (See also the web sites for MAP and Planck Surveyor.) Very recently the BOOMERANG and MAXIMA experiments have produced beautiful data on fluctuations in the microwave background on angular scales of the order of a degree. These provide strong evidence that space really is flat.

However there are two problems with this picture.

The solution is to reintroduce the cosmological constant. This was first proposed by Einstein as way to cancel the gravitational forces in the universe which, as we have seen, require an expanding universe. He then rejected the idea as his ``biggest blunder'' when Hubble showed that the universe is indeed expanding.

The cosmological constant can be though of as a mass density of the vacuum (``empty'' space)

\begin{displaymath}\rho_0={\Lambda c^2\over 8\pi G}\end{displaymath}

where the constant $\Lambda$ can be adjusted so that this provides the missing 60% of the density needed to flatten the universe. The idea of a mass (or energy) density in the vacuum is not so strange in the context of quantum field theory where the vacuum can be thought of as filled with ``virtual'' particles which arise through a combination of the uncertainty principle and Einstein's $E=mc^2$.

If a mass density of empty space isn't weird enough, consider what happens as space expands. This density is a constant and so as the volume of some region of space expands so does the total mass contained in it. From special relativity we know that this extra mass is equivalent to extra energy. If we are to hold onto the idea of conservation of energy, this energy cannot just be created. It come from work done on the region of space as it expands. The vacuum must therefore exert a negative pressure

\begin{displaymath}p_0=-\rho_0c^2\end{displaymath}

Although we really need general relativity to calculate the energy of our test galaxy in the presence of the combined forces of gravity and vacuum pressure, the final result is deceptively simple:

\begin{displaymath}\eta={1\over 2}\dot r^2-{GM\over r}-{4\pi\over 3}G\rho_0r\end{displaymath}

The final term looks like the additional potential energy generated by a constant mass density $\rho_0$, until we notice that it has the wrong sign! The negative pressure wins out over the extra gravitational attraction of the vacuum and the cosmological constant gives rise to a repulsive force driving the galaxy away from us.

The net effect of the cosmological constant is to slow the deceleration of the universe. At the present day its effect is almost exactly cancelled by the gravitational attraction between the matter in the universe. However, as the universe continues to expand, matter will get less and less dense and the repulsion will dominate, leading to an ever accelerating expansion. (For more on the cosmological constant, see: Ned Wright's cosmology tutorial, Sean Carroll's encyclopedia article, Eli Michael's web document and John Norbury's course on general relativity.)

Finally, a cosmological constant $\Lambda>0$ may provide the extra mass density needed to make space flat as well as the repulsive pressure to overcome gravity, but it also brings its own problems. Why is it there? Why is it so very small (up to 120 orders of magnitude smaller than estimates of its natural size in field theories)?

Textbook references


Home: PC 1672 home page | Up: 3 Gravity | Weekly plan | Help: Guide to using this document |
Next: 4 Rigid-body motion | Previous: 3.12 Tides |

Mike Birse
17th May 2000