# Cosmology FAQ with Equations

## Eric V. Linder

This FAQ aims to explore several classic conclusions in cosmology with a bit more mathematics than the general audience level FAQ so the reader with some undergraduate training in physics can delve deeper into the specifics of the early and present universe. For an introduction to cosmology on the upper undergraduate to beginning graduate level, see my textbook First Principles of Cosmology.

### 1. The Early Universe

Here we investigate the time evolution of the matter density of the universe and show how it implies an early dense phase. We also discuss the flatness problem and how inflation can solve it.

The easiest way to explore the conditions in the early universe without calculations involving general relativity is to consider the second law of thermodynamics. For equilibrium situations this simply balances energy changes with the work done to cause those changes. If the work is expansion at some pressure $p$ then the change is

$$dE=-p\,dV,\eqno(1)$$

where $V$ is the volume and the energy is the energy density $e$ times that volume. So $dE=d(eV)=V\,de+e\,dV$. Recall that the volume goes as the cube of the expansion factor $a$ so $dV/V=3da/a$. Equation (1) then ends up looking like

$${de\over e+p}=-3{da\over a}.\eqno(2)$$

Finally we need a relation between energy density and pressure, called the equation of state. For ordinary matter the pressure is very small so we can set it to zero and solve equation (2) to find $e\sim a^{-3}$. That is, in the early universe when the size and hence $a$ was much smaller than it is today, the density must have been extremely high. As $a$ gets smaller the density gets higher until it approaches infinity as $a$ approaches zero. This singularity of infinitely small size and infinitely high density is referred to as the Big Bang. In Exploration 2 we will see that extremely high temperatures are also expected.

If we consider radiation instead of matter then the equation of state is $p=e/3$ so the solution is $e\sim a^{-4}$. The radiation energy density dies off even faster than that of the matter, and so at some time in the past must have been greater. The early universe was therefore dominated by high temperature radiation not matter, and it is its relic glow that we see today as the cosmic microwave background, discussed further in exploration 3. Note that a very peculiar type of energy called vacuum energy has negative pressure, $p=-e$, so its density remains always constant. For this reason among others it is therefore sometimes called the cosmological constant.

To investigate further the evolution of energy in the universe we use the equation of energy conservation. This says that the total kinetic energy plus potential energy is a constant,

$$(1/2)mv^2+(-GMm/R)=K\,.\eqno(3)$$

Using Hubble's Law to write the velocity as $v=HR$ and writing the mass of the universe $M$ in terms of its density and size $R$, $M=(4\pi/3)eR^3$, we can substitute into equation (2) then divide by $mR^2/2$ to get $$H^2=(8\pi G/3)e-k/R^2.\eqno(4)$$ If the first term on the right hand side is much bigger than the second then $H^2\sim e$.

Since the Hubble parameter $H$ describes the rate, or inverse timescale, for the expansion this looks somewhat familiar. In problems where one is concerned with the reverse of expansion, collapse, the time for gravitational collapse depends on density as $t\sim e^{-1/2}$, just what we have above. In fact this equation is used in movie special effects to make scenes look more realistic. In old movies, scenes such as when Godzilla rises out of the ocean immediately look very fake because our brains are keen calculators of time scales and realize that the waves are not rising and falling properly. This timescale depends on a combination of the collapse time above, and the free fall time $t\sim h^{1/2}$ where $h$ is the height of the wave. Since in the Godzilla scene the camera is filming a miniature set of waves in a bathtub, rather than huge breakers in the ocean, the height and timescale are all wrong. By adjusting the other timescale, however, changing the density of the waves by using a liquid other than water, special effects wizards can obtain the right behavior and fool us into thinking we are watching giant waves.

The second term on the right hand side of equation (4) describes how flat the universe is. Since the first and second terms depend differently on time through the size $R$, one expects the universe to become more curved as time goes on. Today, however, observations show the universe seems fairly close to flat, i.e. the first term alone is almost equal to the left hand side. This puzzle of why the second, curvature term is apparently so small today is called the flatness problem. This is one of the things the theory of inflation solves. Inflation depends on vacuum energy which you recall had constant density. During the time of inflation therefore, the expansion timescale was also constant. Since this is defined by $H=(1/R)dR/dt$ then the implied behavior for $R$ is exponential expansion, $R\sim e^{Ht}$. This huge growth in $R$ drove the second term to such extremely small values that it still hasn't recovered yet, so our universe appears almost perfectly flat.

### 2. Hubble Expansion

We relate the Hubble law of velocities to the expansion and discuss the temperature evolution of the universe and lack of a need for the concept of a center to the universe.

Any object showing a uniform expansion (or contraction) will describe the relation between velocities and distances by the Hubble law

$$\vec v=H\vec R,\eqno(1)$$

where the arrows signify vectors, i.e. directional quantities. Expansion is uniform if it only depends on time, not position, so we can simply scale distances by the expansion factor $a(t)$: distances can be related to their earlier values by $R=R_0a(t)$. Since velocity is the time derivative of distance then

$$v=dR/dt=R_0\,da/dt=[R/a(t)]\,da/dt=[1/a(t)](da/dt)R\equiv HR.\eqno(2)$$

All the expansion information, depending only on time not position, is put into the quantity $H$, and we find that the two position dependent variables, the velocity $v$ and the distance $R$ are simply proportional to each other by this factor $H$. Thus all objects seem to recede at velocities that grow linearly with the distance away from us - Hubble's Law is a natural consequence of the expansion.

Now let's ask if there is anything special about the fact that everything seems to be receding directly from us; does this mean we are at a special position? Think of a photograph of a billiards table, undergoing an enlargement into a poster. The distance between any two balls will grow over time and every ball will get further from every other, but no ball is said to be at the center of the enlargement. Similarly Hubble's Law still holds no matter from where you look at it. Consider two galaxies, 1 and 2, each obeying Hubble's Law with respect to you. The difference in their velocities we will call

$$\vec v_{12}\equiv \vec v_2-\vec v_1=H\vec R_2-H\vec R_1=H(\vec R_2-\vec R_1)\equiv H\vec R_{12}\,.\eqno(3)$$

But this is exactly the velocity that galaxy 2 appears to have if you stand on galaxy 1, and it also obeys Hubble's Law. No matter what point you choose in the picture every other point will seem to recede according to Hubble's Law; there truly is no center to the expansion - it is uniform.

What effects does the expansion have on the matter and energy in the universe? In the first exploration we looked at the effects on the density. Now consider the temperature or energy of the radiation, such as the cosmic microwave background. When simple gases expand they cool. A demonstration of this is easy to come by. Put your hand a few inches in front of your mouth. If you blow on it with your mouth wide, your breath feels hot, but if you purse your lips it feels cold. This is from the air expanding as it leaves the narrow opening between your lips - the expansion causing cooling. In a related way the expansion of the universe cools down the radiation background. So while at present it is only about three degrees Kelvin, in the past it was much higher, so observations today of an expanding universe imply that in the early universe it was very hot.

We can figure out how hot by noting generally the number of thermal background photons making up the radiation glow is unchanging. The number is given by the number per volume, or number density, times the volume. Because of expansion the volume increases as the cube of the expansion factor, i.e. if distance increases as $a$ then volume increases as $a^3$. Thus to keep the same number of photons the number density must decrease as $a^{-3}$. However, the number density of thermal photons always goes as temperature cubed according to statistical mechanics (i.e. the energy density obeys the $T^4$ Stefan-Boltzmann Law), so as the universe expands the relation between temperature and expansion must be $T\sim 1/a$. When the universe was smaller the temperature must have been higher in exactly the same proportion. For example to get temperatures as hot as three billion degrees, the universe must have been a billion times smaller than now, an important point for Exploration 4 where we examine nuclear fusion in the early universe.

### 3. Cosmic Microwave Background

A simple version of the historic 1948 prediction of the microwave background temperature is presented as well as the derivation of the first asymmetry in its pattern - the dipole anisotropy.

### 4. Primordial Nucleosynthesis

Recreating the 1948 argument that elements could have been synthesized in the hot early universe we illustrate some important concepts in nuclear astrophysics and get a rough estimate of the predicted helium abundance.

A very simple resolution of the dark night sky question is presented in terms of basic geometric arguments, allowing prediction of what sort of universes would instead show a bright night sky.

First we need to consider what we actually observe when we talk about the darkness of the night sky. This is called the surface brightness $I$ and measures how much light energy arrives at our eyes or telescopes each second from a patch of sky covering a given angular area (solid angle). We can relate this to the number of photons (light particles) $N$ by the following:

$$I=N\cdot E/(A\cdot t)\,,\eqno(1)$$

just putting into symbols what we had in words above. This is helpful because the number of photons is a conserved quantity; it doesn't change as the light travels. The energy $E$ is always proportional to the frequency $f$ (Planck's law) while the time interval $t$ varies inversely since frequency is defined as an inverse period. The change of angular area with motion is called stellar aberration and goes as the inverse square of the frequency change. Thus, all told, $I\sim f^4$. In most situations we are used to, the frequency does not in fact change so the brightness $I$ is also conserved. However, in relativistic situations, in particular cosmology, the frequency does change, measured by the redshift $z$ as $f_o=f_e/(1+z)$ where $o$ denotes the observer and $e$ the emitter. Thus we obtain

$$I_o=I_e(1+z)^{-4}\,.\eqno(2)$$

This says that in order to have a dark night sky either $I_e$ is low, i.e. there are no bright objects in that direction, or $z$ is large, the expansion dims away the brightness. To investigate the chances of a bright object, say a star, in a certain direction we look at the optical depth, or collision probability. Think of hitting a line drive on a baseball diamond where the players are motionless. The chances of someone catching the ball depend on the density of players (how many relative to the size of the field), their reach (how close the ball must come), and the distance the ball travels (the further the more players have a chance at it).

The analogous situation for our light ray to hit a star and hence for us to see the sky bright in that direction gives a probability

$$P=n\cdot A\cdot (ct)\,.\eqno(3)$$

If stars were spread out evenly there would be about one per 10 billion cubic light years of space ($n=10^{-10}ly^{-3}$). The size of a typical star is roughly two light seconds across so the area it blocks is one hundred trillionth of a square light year ($A=10^{-14}ly^2$). So if we wanted the probability to be one half ($50\%$) that a star was in the way the light would have to travel for a trillion trillion years ($t=10^{24}y$)! Since our universe is only about 10 billion years old (and so all the stars are younger) clearly it is wildly unlikely that looking in a random direction you will see a star - so the night sky is dark.

Note that we never had to use the expansion redshift at all. When considering not discrete objects but a diffuse background though like the cosmic background radiation, the photons come from the time of last scattering at $z=1000$ and the $(1+z)^{-4}$ dimming is critical. If one did live in a vastly ancient universe, say the eternal steady state universe without a cosmic background radiation, then light could travel for long enough that every line of sight would end on a star. In this sort of universe though the redshift factor becomes exponentially small (as discussed in Exploration 1 under the name vacuum energy) and so again the night sky stays dark.

(c) Eric V. Linder 1993