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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.

The cosmic tug of war

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.

Critical density of the universe

As we discussed, the critical density is that combination of matter and energy that brings the universe coasting to a stop at time infinity. Einstein’s equations lead to the following expression for the critical density ( ρ crit ):

ρ crit = 3 H 2 8 π G

where H is the Hubble constant and G is the universal constant of gravity (6.67 × 10 –11 Nm 2 /kg 2 ).

Solution

Let’s substitute our values and see what we get. Take an H = 22 km/s per million light-years. We need to convert both km and light-years into meters for consistency. A million light-years = 10 6 × 9.5 × 10 15 m = 9.5 × 10 21 m. And 22 km/s = 2.2 × 10 4 m/s. That makes H = 2.3 ×10 –18 /s and H 2 = 5.36 × 10 –36 /s 2 . So,

ρ crit = 3 × 5.36 × 10 –36 8 × 3.14 × 6.67 × 10 –11 = 9.6 × 10 –27 kg/m 3

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 ).

Check your learning

  1. A single grain of dust has a mass of about 1.1 × 10 –13 kg. If the average mass-energy density of space is equal to the critical density on average, how much space would be required to produce a total mass-energy equal to a dust grain?
  2. If the Hubble constant were twice what it actually is, how much would the critical density be?

Answer:

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

V = E grain ρ crit = 1.1 × 10 –13 kg 9.6 × 10 –26 kg/m 3 = 1.15 × 10 12 m 3 = ( 10,500 m ) 3 ( 10.5 km ) 3

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 ρ crit = 4 × 9.6 × 10 –27 kg/m 3 = 3.8 × 10 –26 kg/m 3 .

Practice Key Terms 3

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Source:  OpenStax, Astronomy. OpenStax CNX. Apr 12, 2017 Download for free at http://cnx.org/content/col11992/1.13
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