The critical density is the dividing line between two types of Big Bang cosmological models; the low-density, negative curvature, hyperbolic geometry, and the high-density, positive curvature, closed spherical geometry. A universe at exactly the critical density has a flat, zero curvature, euclidean geometry. Both the hyperbolic and euclidean models are infinite, only the spherical geometry with a density greater than the critical density has a closed, finite geometry, and is destined to recollapse.
Because the critical density corresponds to a flat, euclidean geometry for space-time, it is possible to derive it from simple Newtonian physics. The escape velocity from a spherical volume with a mass density of 'rho' is found below:
Kinetic energy = potential energy
1 2 G m M
___ m V = -------
2 R
<\pre>
where m = the mass of your test particle, and we see that this cancels on
both sides of the equation. Now, M, is just the total mass of the material
within the spherical volume of radius R, and since density = M/volume
where volume = 4/3 pi R^3 for a sphere, we can obtain:
3
1 2 G Rho x (4/3) pi R
___ V = --------------------
2 R
or with a little algebra:
2
2 8 pi G Rho R
V = ---------------
3
<\pre>
Now, from Hubble's Law, the recession velocity of a galaxy is just
V = H R so that we get:
2 8 pi G Rho
H = -----------
3
or with a little algebra:
2
3 H
Rho = ----------
8 pi G
<\pre>
And if we use G = 6.67 x 10^-8, pi = 3.14159 and H = 75
kilometers/sec/megaparsec which in cgs units = 2.5 x 10^-18 sec^-1, you get
a critical density of Rho = 1.1 x 10^-29 grams/cc. For H = 50 you get Rho =
5.0 x 10^-30 grams/cc, and for H = 100, you get Rho = 2.0 x 10^-29 grams/cc.
Currently, as of 2000, the best data seems to suggest H = 75 kilometers/sec/mpc, so this
means that astronomers have to be able to find a critical density of
Rho = 5.0 x 10^-30 grams/cc which defines an 'Omega' of 1.000. So far,
stars and luminous matter only account for about 10 - 15 percent of this at
best, and the remaining is called 'dark matter' and it seems to account for about 0.30. There also seems to be a cosmological constant term that contributes Omega = 0.70 or there abouts.
Copyright 1997 Dr. Sten Odenwald Return toAsk the Astronomer.