Newtonian Universe

The equations which govern the evolution of the Universe can be derived from a study of the Newtonian dynamics of a large gas cloud, that is, we can get the equations which describe how the scale factor of the Universe, R(t), evolves as the Universe ages from Newtonian physics. This leads to essentially the same results one would derive using Einstein's much more complex General Theory of Relativity (GR). And, with a little creativity on our part, we describe what the shape of the mathematical universe has to do with the physical Universe -- we relate the curvature of the Universe (i.e., whether it has positive, flat, or negative curvature) to a more physical quantity. Let us proceed

Force Law

Imagine that we have a large (but not infinite) Universe with a uniform and isotropic distribution of material (does this mean that there is a center?). Now, hollow out a sphere in this Universe.

We can show for uniform and isotropic distributions of matter that it is only the mass contained interior to the radius, r, that contributes to the gravitational force on a mass located at r. Everything outside of the sphere of size r does not contribute to the force on r! The forces all cancel out due to symmetry.

If the sphere has uniform density, then the amount of mass in each segment of the sphere is ~ density x pi x D2 x A x T and ~density x pi x d2 x A x T, where A is angle subtended by the chunk of the sphere at r, D and d are the distances from r to the sphere, and T is the thickness of the shell.
Since the force due to gravity is -GMm/distance**2, the two forces are ~ G x density x pi x A x m and ~ G x density x pi x A x m -- the distances cancel out and the two forces are equal in magnitude but opposite in direction -- they cancel each other (as advertised)!

There is no gravitational force inside of the hollowed-out sphere due to the exterior mass. Therefore, there is no force inside of the sphere, the universe is flat inside of the hollowed-out sphere and one can use Newtonian ideas on how physics works to describe how things move in this region.

Now, add a little bit of mass to the inside of the sphere. Add enough mass to cause a force but not enough to cause a significant deviation from a flat space-time.

So, for a galaxy of mass m and speed v at at distance r from the center of the sphere, Newton's law of motion, F = ma -- the force on the object is equal to the mass of the object, m, times the change in its velocity, dv, over a time interval, dt,
F = ma ===>m dv/dt = - GMm/r2 ===>m dv/dt = - 4 pi G density m r / 3
===> dv/dt = - 4 pi G density r / 3

Since we assumed that the universe is homogeneous and isotropic, we must have v = H(now) x distance for all observers in their own little spheres which means that r = R(t) x [some constant r = r(0)]
===> v = r(0) dR(t)/dt and density = [some constant density = rho(0)] / R(t)3

Newton's Law ===> R(t)**2 d[dR(t)/dt]/dt = - 4 pi G rho(0) / 3
Note that there is always a force unless the density is 0 ===> we cannot have a static universe which contains matter. The universe must be doing something.

Energy Conservation

It can be shown using Newton's law of motion that

-k/2 = 1/2 [dR(t)/dt]2 - 4 pi / 3 G rho(t) R(t)2

this expression is quite familiar. Recall that the energy of a particle in a gravitational field is E = K.E. + P.E. = 1/2 mv2 -GMm/r ===> E/m = 1/2 v2 - G [4 pi/3 density r**2]
Hmmm, this is quite suggestive. The constant k in the above theory is simply a measure of the total energy (kinetic plus potential) of the mass m. A GR calculation leads to the same relationship given above, however, the constant -k/2 defines the curvature of the universe (and one can add a term characterized by the cosmological constant). This suggests that the curvature of the Universe is similar to the energy of our Newtonian universe.

An object is bound to a star (or planet) if Energy < 0 ===> our Universe is bound (closed, positive curvature) if the constant k > 0, i.e., positive curvature. The Universe is free (open, negative curvature) if the constant k is <0, and the Universe can just escape to infinity (is flat) if k = 0.

The curvature of the Universe is telling us whether the Universe has positive, 0, or negative energy.

Fate of the Universe

This notion that an unbounded Universe has postiive energy, leads us to a natural test to determine the ultimate fate of the Universe. Recall:

- k = [dR(t)/dt]2 - 8 pi / 3 G rho(t) R(t)2

v = H(now) distance (Hubble's Law) ===> R(0) dR(t)/dt = H(now) R(0) R(t) ===> dR(t)/dt = H(now) R(t)

Combining the above relations leads to

-k = [H(now) R(t)]2 - 8 pi / 3 G rho(t) R(t)2

An open Universe has positive energy so the critical universe has k = 0 and the critical Universe has ===> rho(t) = H(now)2 / [8 pi G / 3]

Note that the LHS of the above equation is the current density in the Universe, so that if
rho(now) < H(now)2 / [8 pi G / 3]
then the Universe has exceeded its escape speed and will expand forever. This is a nice test because all we need to do is measure the current density and expansion rate of the Universe. Much of our current observational efforts are directed toward this seemingly simple task. I will much more to say about this exercise later.