# What’s the matter?

Posted by on Apr 10, 2013 in Blog, Science | 0 comments

In a previous post, I derived a simple form of the Friedmann equations using an argument based upon the total energy of a galaxy in a cosmological field of matter of uniform density. In that derivation, I made the working assumption that the total energy was $0$. That was a restrictive assumption that I want to lift now. Let me write that equation for the energy down again:

$E = 1/2 m V^2 - \frac{m M G}{D}$

where I am continuing with the same notation that I set up in that previous post: $m$ is the mass of some test galaxy under consideration, $M$ is the total mass within a sphere of radius $D$ from the origin of coordinates, $G$ is Newton’s gravitational constant, $V = \dot{D}$ is the velocity of the galaxy, and $E$ is the total energy. Assuming all is well with the universe and that energy is conserved, this is a constant. It doesn’t matter right now whether it’s greater than, less than, or equal to $0$; it’s just some number.

## If the energy is not 0

Some rearrangement of this last equation yields

$( \frac{ \dot{a}(t)}{a(t)})^2 - \frac{2 M G}{ a^3(t)} = \frac {\text{const}}{a^2(t)}$

in which the constant involves the total energy and the mass of the test galaxy. By substituting for the matter density $\rho$ in this equation, and then by recalling that $\rho = \nu / a^3$, we can find

$( \frac{ \dot{a}(t)}{a(t)})^2 = \frac{8 \pi G \nu}{ 3 a^3(t)} + \frac {\text{const}}{a^2(t)}$

in which I’ve just accounted for the volume of a sphere of radius $a(t)$. [You may want to go back to that previous derivation to get a feel for this equation.]

### Positive total energy

A quick glance at this equation reveals some simple facts. Assuming that the terms on the RHS are positive, then the LHS must also be positive (it is after all, the square of some real number), and this would imply a universe that would continue to expand forever, if it was expanding at any given point in time. The expansion might slow down or pick up, but it would have to stay positive.

I already considered the case in which the $\text{const}$ pertaining to the total energy was $0$. In the very early stages of the universe, assume that $a$ was small. In this case, the $1/a^3$ term will trump the $\text{const}/a^2$ term; and so this case looks the same as the one in which the total energy is $0$.

The complete solution to this differential equation is rather nasty in the extreme. Here is Mathematica’s representation of the solution:

Solution

What this is saying is that the small piece in square brackets at the end of the string, having to do with $i t$, is the argument to a certain inverse function. In that piece, $c1$ is a constant of integration. The $C$ is the $\text{constant}$ in the formulation of the differential equation. That argument should be substituted in wherever you see the $\text{\#1}$. However, the actual function is not what is shown: it is, instead, the inverse of this function. Very nasty. Far better to do a numerical solution for particular values, if you wanted to get a sense for what is going on.

However, we can make sense of the other extreme without too much effort. At the other end, if $a$ becomes very large, then the $1/a^3$ term becomes less relevant than the $1/a^2$ term. So, we could write as an approximation to this case

$\dot{a}(t) = \text{constant}$

where this constant is the square root of the previous one. This tells us that $a(t) = \text{constant} t$ and the universe expands uniformly. So, this limit is quite simple also. Apparently, the solution begins out looking like something proportional to $t^{2/3}$ and crosses over to something just proportional to $t$ somewhere along the way.

### Negative total energy

If we let the total energy be less than $0$, we get instead

$( \frac{ \dot{a}(t)}{a(t)})^2 = \frac{8 \pi G \nu}{ 3 a^3(t)} - \frac {\text{const}}{a^2(t)}$

In this case, it is possible for the sign of $\dot{a}$ to change and for the expansion to reverse itself and become a contraction. This is just like the case of a particle in a gravitational field that begins with less than the escape velocity. Just as in the previous model, the universe begins to expand proportional to $t^{2/3}$; but at some point, the velocity of the scale factor reverses itself and in the limit becomes proportional to $-t$.

This set of solutions represents what is called the “matter-dominated universe” because all that we have accounted for so far is a uniform density of matter. This matter is assumed to very slowly moving (non-relativistic) and more or less fixed to a comoving cosmological lattice. We now need to consider energy, in the form of photons.

## An energy-dominated universe

We can see that we have been playing with a form of Einstein’s Field Equations with the Friedmann equations that we’ve been pushing around here lately. If I recast the Friedmann equation into

$( \frac{ \dot{a}(t)}{a(t)})^2 + \frac {\kappa}{a^2(t)} = \frac{8 \pi G \rho}{3}$

we see this structure more plainly. The first term on the LHS will be part of a metric tensor. The second term on the LHS is a curvature. I have introduced a common convention of calling this constant $\kappa$ and using another convention that the sign of $\kappa$ is negative if the total energy is positive. More of this later. The term on the RHS is an energy-mass density, or basically the 00 term of the energy-stress tensor. So far, I’ve just accounted for, as I say, a matter density.

But we derived this equation from energy considerations alone, not curvature of space or space-time. In that model, the explanation involved a summation of terms involving kinetic and potential energy. The potential energy term has become a matter density on the RHS here. The total energy term is now cast in the role of the curvature and metric tensor term. The kinetic energy is now standing in for the Ricci tensor. We see a unification between the Ricci tensor for the curvature of space-time and a kinetic energy of space. Since the elements of the Ricci tensor may be either positive or negative, we begin to form the notion of a kinetic energy of space that may also either be positive or negative. Of course, this kinetic energy term originally pertained to the motion of a given “test galaxy” in the usual form of $1/2 m V^2$; but if a uniform density of such objects acquire a motion because of the time dependence of the metric of space-time, the explanation of the source of the kinetic energy must account for this effect.

However, if we recall good old $E = m c^2$, or equivalently $m = E/c^2$, then we can introduce an effective mass density that corresponds to some arbitrary energy density of photons into this Friedmann equation. To get to thinking about this, go back to our cosmological lattice:

Lattice-with-distance

The volume of a unit cube in this lattice is $a^3$. Recall that the energy of a photon is $E = h f = (h c)/ \lambda$ where $h$ is Planck’s constant, $f$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength. If we have a wave in some medium and we gradually change the dimensions of the medium, then the wavelength of the wave will change proportionately. This implies that the energy of photons is inversely proportional to the scale factor: $E \propto \lambda ^{-1}$. This effect is due to the fact that $a(t)$ is a component of the metric of space. As the space stretches, so does the wavelength of each photon. Note that this is not due to any standing-wave aspect of the photons.

### A different density for energy

Here we find that there is an essential difference between the total energy in a cube of the grid for ordinary matter and for photons. For ordinary matter, the total, $M$ within a volume, does not depend on the scale factor. For photons, the energy decreases with increasing $a(t)$. So, for matter, we had that the density, $\rho_m = \nu_m / a^3$. For photons, the energy density has one more factor of $a$: we find that the density is $\rho_e = \nu_e / a^4$. This is a consequence of a property called adiabatic invariance, which states that certain physical parameters will remain constant if some other quantity is very slowly varied. In this case, what is being varied is the metric of space-time; that is, the scale factor $a$. The parameter that is remaining constant is the number of nodes in the wave character of the photon or wave-packet of light, if you prefer. Since, by Fourier analysis, any complex waveform can be seen as a linear superposition of component waves, we can simply think about this behavior in terms of the properties of infinitely long uniform waves. As the size of the universe expands, the number of zero-crossings of these waves are constant; hence, the wavelengths increase. If the wavelengths of all of the component waves increase, then so will the size of any wave-packet.

For a radiation-dominated universe then, we can redo the Friedmann equation in the form

$( \frac{ \dot{a}(t)}{a(t)})^2 = \frac{8 \pi G \nu_e}{ 3 a^4(t)}$

in the specific case where the total energy is $0$. The solution to this differential equation, ignoring various constants yields that $a(t) \propto t^{1/2}$. So, in a radiation-dominated universe (at a total energy of $0$ anyway) the universe expands as $t^{1/2}$.

To compare then, at a net $0$ of energy, a matter-dominated universe would expand with a scaling exponent of $0.\dot{6}$ and a radiation-dominated universe would expand with a scaling exponent of $0.5$. These two relations are not wildly different, but the difference in the scaling exponents would yield significant and measurable differences over long intervals of time.

Going back to what I said earlier concerning kinetic energy of matter being affected by the time dependence of the metric of space-time, in a general relativity point of view, the same comments apply here. We begin to see this balance between the time-dependence of the Ricci tensor, which corresponds to (or yields) variations in the kinetic energy of matter or radiation, and the densities of these elements of the stress-energy tensor. As we continue to work our way through these equations, we’ll see different forms of energy in the form of pressures and tensions introduced by matter, radiation and what amounts to the cosmological constant.

One way to think about the case of light energy is to begin with the idea of a box of dimensions equal to one cube of the comoving cosmological lattice, but with reflecting walls. In such a hypothetical case, there would be a pressure on the walls of the box due to the impulse-momentum theorem as photons reflect off of the walls. For each photon striking a wall, there is a photon reflected back. In the real case of imaginary boxes, the result is the same in a homogeneous universe: for each photon leaving the box, there is another one entering. The net effect of a pressure is identical: the momentum leaving equals the momentum entering.

Let’s now roll both matter and radiation together in a Friedmann equation assuming a net $0$ of energy:

$( \frac{ \dot{a}(t)}{a(t)})^2 = \frac{8 \pi G \nu_e}{ 3 a^4(t)} + \frac{8 \pi G \nu_m}{ 3 a^3(t)}$

For small $a$ the radiation term dominates and for large $a$ the matter term dominates. We could envisage a universe in its early stages being dominated by radiation and then, as time progresses, matter becoming more significant. In this model, the early universe would expand as $t^{1/2}$ and then later, it would expand as $t^{2/3}$. This claim makes a fairly big assumption that there is no conversion between radiation and energy; or that any conversion is consistent with the claim. This claim is therefore in need of some verification; but hold that thought for now.

## What is the matter?

The items that are comprised by the matter term include all the heavier particles that are essentially at rest with respect to our cosmological lattice. These include the ordinary matter particles with which we are all familiar: anything made out of quarks (like protons & neutrons) and at least some of the leptons (the electron and positron). However, there is another component to this, at least in theory; this component is so-called dark matter.

The evidence for dark matter is varied. One of the most fundamental aspects of this evidence comes from the rotational speeds of galaxies. From Kepler’s third law, we can write for the orbital angular frequency (in radians per second)

$\omega^2 = \frac{G M}{R^3}$

where $R$ is the distance of the orbiting object to the large mass $M$ at the focus of the ellipse. Notice that the angular frequency $\omega$ is proportional to the square root of the large mass, $M$. For a galaxy, this implies that stars will orbit the galactic center in a way that depends on the total mass. By adding up the mass of all of the stars and dust, it is possible to get an estimate of what the mass $M$ should be. Of course, by the shell theorem, the relevant mass is just that within a sphere of radius $R$. For surveys of hundreds of galaxies, it turns out that the orbital frequencies are higher by almost an order of magnitude that what would be calculated by just the visible matter. Other evidence for dark matter comes from gravitational lensing effects that exceed what would be expected on the basis of visible matter alone.

Assuming that we are not wrong in our assumptions about Newtonian mechanics and the general theory of relativity, this evidence suggests some other form of matter that is not accounted for in the standard model of particle physics, that has mass, and that does not interact with photons. A working assumption is some subatomic particle that has, as yet, not been discovered; probably because it does not interact with quarks or leptons; that is, with ordinary observable matter.

There is some more direct evidence as recently as this month (April 2013) from the Alpha Magnetic Spectrometer in terms of a high count of positrons (anti-electrons) that could be explained by the annihilation of dark matter particles in space.

Another set of experiments designed to search for dark matter involve the placement of detectors deep in underground mines around the world. This placement is intended to reduce the detection of cosmic rays, high energy photons from outer space. Assuming that dark matter interacts only weakly with ordinary matter, it is to be expected that it would easily penetrate into these deep mines much more easily than even the highest energy photons. By accumulating a sufficient number of detection events, it is hoped that a unique signature for dark matter particles would become evident.

A variety of candidates have been proposed for dark matter, one being the lightest of the so-called neutralinos, a particle predicted by supersymmetry. There are other candidates broadly known as WIMPs (weakly interacting massive particles). I’ll come back to these ideas in later posts.

At present, dark matter is estimated to be around 85% of the matter in the universe, with only 15% being the ordinary matter that we are familiar with.

## Faster than light?

It is worth touching on the fact that the Hubble law, $V = H D$, tells us that the more distant an object is from us, the greater its velocity relative to us. This velocity can exceed the speed of light, and this could be taken as a violation of special relativity, which states that nothing can move faster than light. It also seems to raise some vexing questions about the proper form for an expression of kinetic energy; that is, $1/2 m V^2$ or $m (\gamma - 1) c^2$.

First note that those distant galaxies that are receding with a velocity greater than $c$ are no longer visible to us: they have passed a kind of horizon not unlike a Rindler horizon. So these distant objects can not send us any energy or information. On their side of the horizon, everything seems perfectly normal locally, just as it does to us on our side of the horizon. We have no sense of hyper-speed motion, because locally we are not traveling at speeds exceeding $c$.

So, recession at great distances at relative velocities that exceed $c$ is perfectly feasible, with the addition of a horizon through which information cannot flow. Of course, that horizon, from our point of view, is very far away, about 14 billion light years. Someone who was on a planet 7 billion light years away from us would presumably see their horizon at a different place than we do. They would still have a horizon about 14 billion light years away from them. This suggests that the universe could be much larger than we can see from our vantage point. More of this later, too.

## What’s wrong with this picture?

The picture that I have drawn so far is, in many ways, false. At present, we do not believe that the universe is either expanding at some rate like $t^{0.5}$ or $t^{0.667}$ or collapsing back after a previous phase of expansion. Rather, we find that the universe is expanding at an accelerating rate. Something has been missed in the equations I’ve presented so far, since nothing in them would provide for an accelerating expansion.

If the picture I’ve painted so far were accurate, then the velocity of recession of distant objects would gradually reduce, and perhaps even reverse itself. In this case, objects that are presently beyond the horizon would become visible again as their velocity slowed to less than $c$. However, our current understanding suggests that this is not that case and that galaxies that are not yet beyond the horizon will gather speed and disappear from our point of view. And this will continue into the foreseeable future until perhaps only our galaxy or local cluster remains visible.

However, Houston, we have a problem. We still have nothing in our equations that comes close to what is actually observed. On to this problem in my next post.