<< Chapter < Page | Chapter >> Page > |
In an empty universe (the dashed line [link] and the coasting universe in [link] ), neither gravity nor dark energy is important enough to affect the expansion rate, which is therefore constant throughout all time.
In a universe with dark energy, the rate of the expansion will increase with time, and the expansion will continue at an ever-faster rate. Curve 4 in [link] , which represents this universe, has a complicated shape. In the beginning, when the matter is all very close together, the rate of expansion is most influenced by gravity. Dark energy appears to act only over large scales and thus becomes more important as the universe grows larger and the matter begins to thin out. In this model, at first the universe slows down, but as space stretches, the acceleration plays a greater role and the expansion speeds up.
We might summarize our discussion so far by saying that a “tug of war” is going on in the universe between the forces that push everything apart and the gravitational attraction of matter, which pulls everything together. If we can determine who will win this tug of war, we will learn the ultimate fate of the universe.
The first thing we need to know is the density of the universe. Is it greater than, less than, or equal to the critical density? The critical density today depends on the value of the expansion rate today, H _{0} . If the Hubble constant is around 20 kilometers/second per million light-years, the critical density is about 10 ^{–26} kg/m ^{3} . Let’s see how this value compares with the actual density of the universe.
where H is the Hubble constant and G is the universal constant of gravity (6.67 × 10 ^{–11} Nm ^{2} /kg ^{2} ).
which we can round off to the 10 ^{–26} kg/m ^{3} . (To make the units work out, you have to know that N , the unit of force, is the same as kg × m/s ^{2} .)
Now we can compare densities we measure in the universe to this critical value. Note that density is mass per unit volume, but energy has an equivalent mass of m = E / c ^{2} (from Einstein’s equation E = mc ^{2} ).
a. In this case, the average mass-energy in a volume V of space is E = ρ _{crit} V . Thus, for space with critical density, we require that
Thus, the sides of a cube of space with mass-energy density averaging that of the critical density would need to be slightly greater than 10 km to contain the total energy equal to a single grain of dust!
b. Since the critical density goes as the square of the Hubble constant, by doubling the Hubble parameter, the critical density would increase by a factor a four. So if the Hubble constant was 44 km/s per million light-years instead of 22 km/s per million light-years, the critical density would be ${\rho}_{\text{crit}}=4\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}9.6\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{\u201327}}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}=3.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{\u201326}}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}.$
Notification Switch
Would you like to follow the 'Astronomy' conversation and receive update notifications?