Posted by on Apr 5, 2013 in Blog, Science | 0 comments In the past couple of posts, I applied the shell theorem to the case of the Earth as a giant bead with a hole drilled through the middle of it and a large assemblage of diffuse matter (such as hydrogen gas or dust). In these two cases, the location of a center of mass was more or less apparent. In the following derivation, there is going to be a pretty obvious question about a real center of mass for the system. I propose that you hold your nose about how this smells until I finish. It will turn out that the actual center of mass will not be particularly relevant in the end.

## A Galactic Chunk

So, where are we starting and going? Let our test mass or chunk of matter be a galaxy. We go back to our cosmological lattice and imagine that our galaxy under study is at some location on the lattice with respect to its center of coordinates. In the general vicinity of the center of coordinates and our test galaxy is some uniform distribution of matter.

By extending this lattice as necessary, we can establish that the distance between the origin and our test galaxy is just $D = a(t) \sqrt{x^2 + y^2 + z^2}$, where $a(t)$ is the scale factor and the $x, y, z$ coordinates are the comoving coordinates of our space. If I just define $R = \sqrt{x^2 + y^2 + z^2}$, then I can set $D = a(t) R$, and $R$ is a constant. Then the velocity of this test galaxy is $V = \dot{a}(t) R$ and the acceleration is $A = \ddot{a}(t) R$. [Note that I’m using capitals for $D, V,$ and $A$ to avoid confusion with the scale factor $a$.]

Our picture so far is identical to the one in my last post, except that I have changed the scale significantly. The test chunk of matter is now a galaxy and the distances are on the order of tens or hundreds of megaparsecs.

## Force on the test galaxy

So far, we’ve just made some simple definitions of terms. Now, let’s assume that we can apply the shell theorem with a center of mass at the center of coordinates (hold your nose!). We assume that the total mass within a sphere of radius $D$ centered at the origin of coordinates is $M$. If we define the mass of the test galaxy as $m$, then the force on it will be $F = -\frac{G m M}{D^2}$

where the minus sign implies that the force is attractive. By dividing out the mass of the galaxy, $m$, we can get an expression for the acceleration, $A$, and setting this equal to $\ddot{a}(t) R$, and substituting for $D = a(t) R$, we find $\ddot{a}(t) R = -\frac{G M}{a^2(t) R^2}$

Then, dividing both sides by $D = a(t) R$, we get $\frac{\ddot{a}(t)}{a(t)} = -\frac{G M}{a^3(t) R^3}$

But the volume of the sphere centered at the origin of coordinates and with radius $D$ out to our test galaxy is just $V = 4/3 \pi D^3$, and the density of matter in the sphere is $\rho = M/V$. So we can rewrite this last equation as $\frac{\ddot{a}(t)}{a(t)} = -\frac{4 \pi G \rho}{3}$

## No center, everywhere center

Here is an interesting result. Assuming that we have a homogeneous universe with a uniform density at some grand scale, the actual scale $R$ becomes irrelevant; it has been removed from the equation. We could have picked any point for an origin of coordinates and any test galaxy at any distance from the origin, and we’d have come to the same result.

What does it mean that there would be an attractive force towards any arbitrary location? We could have picked the origin of coordinates to be at any arbitrary location in the previous derivation. Any location could be the center of mass. The result is a differential equation for the scale factor, $a(t)$, which tells us that the acceleration of $a(t)$ can only be $0$ if the universe has no matter in it. Since the scale factor is component of a metric of space, this equation tells us that this metric cannot be static if space contains matter.

## A preferred scale

There are other issues are pertinent to our working scale. As we saw in previous posts, for scales of the order of planets, solar systems and galaxies, where the distribution of matter is not uniform and where we can pick out the center of mass of a system otherwise isolated by distance from more remote objects, the equations of motion yield harmonics, Kepler orbits, Lissajous orbits, and other such classical motions. As we begin to work up to scales of the order of many galaxies, we encounter something completely different, which has been called variously a clustering heirarchy or a fractal [See for example, The Fractal Galaxy Distribution, by P.J.E. Peebles, in Fractals in Physics, ed. J. Feder & A. Aharony, North-Holland, (1990), p.273-278, aka Physica D 38.] At these intermediate scales, the distribution of matter appears to be a fractal whose correlation function scales with a power of -1.77±0.04 [See H. Totsuji, & T. Kihara, Publ. Astron. Soc. Japan 21 (1969) p.221, and P.J.E. Peebles, Astrophys. J 189 (1974) L51.] This scaling appears to hold up to about twice a clustering length of roughly $r_0 = 0.002 H_d$, where $H_d$ is the Hubble distance, estimated at about 14 billion light-years. So, the cut-off for fractal scaling is around 56 million light-years. For scales larger than this, the distribution of matter becomes homogeneous; and we can begin to think about a uniform density of matter.

So, earlier I pointed out that there is a preferred frame of reference, one that is stationary with respect to the cosmic microwave background radiation. Now, we find a preferred scale, one that is large enough that the distribution of matter is homogeneous.

## Friedmann equations

If we consider this uniform distribution of matter with respect to our comoving cosmological lattice, we can think of a density $\nu$ such that $M = \nu R^3$ where $R$ is the length of a side of a cube of the comoving lattice. This gives us that $\rho = \nu / a^3$. We can then substitute for $\rho$ in our last equation to get $\frac{\ddot{a}(t)}{a(t)} = -\frac{4 \pi G \nu}{3 a^3(t)}$

This is one form of the Friedmann equations, about which we are going to have much to add in this and subsequent posts. While Friedmann derived this form from the trace of the Einstein field equations, there has been nothing fancier than good old Newtonian mechanics in this derivation so far.

Assuming that $a(t)$ must be positive, this last equation suggests that the acceleration of the scale factor should be negative; but whether the scale factor is actually increasing or decreasing depends upon initial conditions that we have not worked out yet. That is, we have not considered $\dot{a}(t)$ as yet. So, the scale factor could be increasing but at a decreasing rate, just as an object thrown from the Earth might leave the planet if it is started with something greater than the escape velocity.

## A derivation from energetics

Nonetheless, this equation is a model of a decelerating universe; but in point of fact, our universe appears to be expanding in an accelerating manner. So, something is missing. Let’s go back and start over using energy conservation for our test galaxy: $E = \frac{1}{2} m V^2 - \frac{m M G}{D}$

The total energy can be less than, equal to, or greater than $0$ depending upon whether or not the kinetic energy is greater than the gravitational potential energy. The escape velocity $V_e$ is the critical point when the total energy is $0$, and $V_e = \sqrt{2 M G / D}$. By analogy, the entire universe is similarly situated: if the total energy is $0$, it expands at a decreasing rate. If the total energy is greater than 0, it keeps on going. If the total energy is less than $0$, it eventually contracts.

Substituting for $D$ and $V$ in terms of the scale factor and $R$, we get $E = \frac{1}{2} m \dot{a}^2(t) R^2 - \frac{m M G}{a(t) R}$

Assume for now that the total energy is $0$, then after some rearranging, $\frac{\dot{a}^2(t)}{a^2(t)} = \frac{2 M G}{a^3(t) R^3}$

We’ll come back to the cases where the total energy is not $0$ later. Recognizing that the volume of our sphere is $4/3 \pi a^3 R^3$, we can rewrite this as $\frac{\dot{a}^2(t)}{a^2(t)} = \frac{8 \pi \rho G}{3}$

Recall that $H = \dot{a} / a$ is the Hubble non-constant. This is another form of the Friedmann equation, derived from energetics. Substituting for $\rho$ in terms of $\nu$, $\frac{\dot{a}^2(t)}{a^2(t)} = \frac{8 \pi \nu G}{3 a^3(t)}$

### The solution

There are several ways to solve this differential equation, and without belaboring the point, the result is $a(t) = \left(\frac{3}{2}\sqrt{\frac{8 \pi \nu G}{3}} t+1\right)^{2/3}$

assuming that $a(0) = 1$ for $t=0$ being the present time. In other words, the scale factor increases as the $2/3$rd power of the time. Note that the actual value of $\nu$ depends upon our choice of scale, so the constant on the LHS is somewhat fungible.

Such a universe with a scale factor of this form would look as follows, where I’ve just set the constant multiplying $t^{2/3}$ to be $1$:

[Click on the image for a larger version.] With the constants set up so that $a = 0$ at $t=-1$, the units of time are the age of the universe, about 14 billion years. So, in this version of the universe, it simply keeps on expanding although at a constantly reducing rate.

In coming posts, I’ll start to include various forms of energy as well as matter into the picture. Each component will add its own unique element to this dynamic picture of cosmological space and time. 