Lecture Notes for Astronomy 321, W 2004 1 Stellar Energy ...

subject to a spherically symmetric gravitational field, for which E = 0 represents
the dividing line between m being gravitationally bound or unbound.
So we have:
• K > 0 ⇒ E < 0 ⇒ “bound” (“closed”), R returns to 0.
• K < 0 ⇒ E > 0 ⇒ “unbound” (“open”), R goes to infinity.
• K = 0 (“flat”) ⇒ E = 0 ⇒ R → ∞ and Ṙ → 0 as t → ∞.
These solutions for R(t) as a function of t are summarized in Figure 12.
The last solution, for a flat universe, corresponds in our analog to the solution
for the classical escape velocity, where the velocity of the test mass goes to
zero at infinity. We note that K = 0 is quite strongly favored by current
observations. We also note that in this case, the condition E = 0 means that
two large numbers K and U, the total kinetic and potential energies of the
universe, have to cancel to large precision. We will discuss this seemingly
unlikely situation when we discuss “inflation.”
It is interesting to evaluate the time evolution of the three cases above.
For K = 0, it is easy to show that the Friedmann equation with ρ =
M/(4πR 3 /3) becomes R 1/2 dR = √ 2GM dt, which gives
R ∝ t 2/3
and the age of the universe as 2/(3H o ). Furthermore, the critical density
which enables this E = 0 solution (see homework) can straightforwardly be
shown to be
ρ c = 3H2 o
8πG . (28)
For K > 0, one can show using a similar analysis that a “big crunch”
occurs at a time 2πGM/c 3 ∼ 10 Gyr after the big bang.
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