# Redshift Academy

Wolfram Alpha:

Last modified: January 26, 2018
```Expansion of the Universe
--------------------------

Mass Dominated Period
-------------------------

.   .
.         .
.     M       .
.               .
.       +x = 0  .m
.               .
.             .
.         .
.   .
<-- D -->

Consider the case of a mass, m sitting on the surface of a sphere
with mass, M, that is being ejected from M at velocity, v, we can
use the conservation of energy and Newton's law that states:

1.  A spherically symmetric body affects external objects
gravitationally as though all of its mass were concentrated
at a point at its centre.

2.  If the body is a spherically symmetric shell (i.e., a hollow
ball), no net gravitational force is exerted by the shell on
any object inside, regardless of the object's location within
the shell.

We can write:

mv2/2 - GMm/D = constant

where mv2/2 is the KE of m and GMm/D is the gravitational PE due
to M.  Note when the RHS is equal to 0 we get the familiar formula
for the escape velocity.

mv2 - 2GMm/D = constant * 2

v2 - 2GM/D = constant * 2/m = constant

Now D = ax where a is the scale factor and x is a distance.
.   .
v = D = ax

So,
.
a2x2 - 2GM/ax = constant

Rewrite in terms of mass density, ρ
.
∴ a2x2 - (2G/ax)(4/3)πρD3 = constant
.
∴ a2x2 - (2G/ax)(4/3)πρa3x3 = constant
.
∴ a2x2 - 2G(4/3)πρa2x2 = constant

To make sense the constant on the RHS must be ∝ x2. Therefore
we can divide LHS and RHS by x2
.
∴ a2 - 2G(4/3)πρa2 = constant

By convention we write:
.
∴ (a/a)2 = (8/3)πρG - k/a2

where k relates to the curvature of the universe.

This is the FRIEDMANN-ROBERTSON-WALKER equation.

ρ is a function of t (as volume expands density will decrease).
We can get around this by replacing with M/a3.  Thus,
.
(a/a)2 = (8/3)πMG/a3 - k/a2

This is the FRW EQUATION for the MASS DOMINATED PERIOD

Consider k = 0 (just at the escape velocity).
.
(a/a)2 = (8/3)πMG/a3

Assume a solution a = βtp where p is a power.
.
∴ (a/a)2 = p2/t2

∴ p2/t2 = 8πMG/3β3t3p

∴ p = 2/3 and a = βt2/3

∴ p2 = 8πMG/3β3

∴ 4/9 = 8πMG/3β3

∴ β3 = 6πMG

∴ a = (6πMG)1/3t2/3
.
∴ H = (a/a) = 2/(3t)

Therefore, during the mass dominated period, the universe grew
according to a power law of 2/3.

Radiation Dominated Period
---------------------------

This occurred before the mass dominated period.

Universe has 109 photons for every proton.  Photons are massless
but have energy by virtue of their motion.  Slow one down and
it disappears.

Ep = hν = c/λ (de Broglie)

As universe expands λ gets stretched and so the energy (and
temperature) diminish by a factor of 1/a. There will now be
an extra a in the denominator of the FRW equation.  Thus,
.
(a/a)2 = (8/3)πMG/a4

Solve this equation in the same way as before to get.

p = 1/2.  So,

a = (6πMG)1/3t1/2
.
∴ H = (a/a) = 1/(2t)

Therefore, during the radiation dominated period, the universe grew
according to a power law of 1/2.

Vacuum Dominated Period
-----------------------

First, we need to talk about pressure and energy density for radiation
dominated period.

Consider many photons sloshing backwards and forwards inside a 1D box

|---------------------|<- wall
<------ L ------>

F = dp/dt

The change in photon momentum is 2p as it bounces off the wall.
The bouncing creates a pressure on the wall.

Time between collisions = 2L/c

F = 2p/(2L/c) = pc/L = E/L = ρ in 1D.

Move to a cube with side a:

-->x
------
/     /|
-----/ |-> P (pressure)
|     |A|
|     | /
|     |/
-----
<- a ->

F = E/L Therefore P = E/LA => P = ρ  But this only one direction.
To get the correct answer for 3D we need to divide by 3, Thus,

P = ρR/3

Now consider the above box again.  Let's move the face A of the
cube by an amount dx.  The work done is PAdx or PdV.

The decrease in the internal energy of the box is -PdV. Now E
inside box is:

E = ρV Thus,

dE = ρdV + Vdρ = -PdV

Vdρ = -(P + ρ)dV

Now introduce the parameter, w, as P = ρw

Vdρ = -(w + 1)ρdV

dρ = -(w + 1)ρdV/V

dρ/ρ = -(w + 1)dV/V

lnρ = -(w + 1)lnV   after integration

= ln{1/Vw + 1} + c'   c' is the constant of integration

ρ = c'/Vw + 1

= c'/a3(w + 1) since V = a3

Rewrite the FRW equation (a/a)2 + k/a2 = (8/3)πρG, as follows:
.
(a/a)2 + k/a2 = (8/3)πGc'/a3(1 + w)

We can now use w to describe any combination of periods.

w = 1/3 => (8/3)πGc'/a4 => (8/3)πGρR   ... RADIATION DOMINATED

w = 0 => (8/3)πGc'/a3 => (8/3)πGρM    ... MASS DOMINATED

w = -1 => c' => (8/3)πGρo   ... VACUUM DOMINATED

The quantity 8πGρo is the COSMOLOGICAL CONSTANT, Λ

Summary
-------

(a).   ρm = mc2/V

m is constant as V increases.  Therefore, ρ decreases.

(b).   ργ = hf/V = hc/λV

λ increases as V increases.  Therefore, ρ rapidly
decreases.

(c).   ρV = constant.

Today, ρV is approximately 70% of the total energy density.
ρm is approximately 30% (including Dark matter) and ρR is
negligible.   If we consider just the vacuum energy we get:

.
(a/a) = √(8/3)πGρo

∴ a = c'exp{(√(8πGρo/3)t}

Period  | Density  |       Expansion     | Pressure |    w
----------+----------+---------------------+----------+---------
Radiation | ∝ 1/a4   | ∝ t1/2               | P = ρR/3 | w = 1/3 |
Mass      | ∝ 1/a3   | ∝ t2/3               | P = 0    | w = 0   |
Vacuum    | constant | ∝ exp{(√(8πGρV/3)t} | P = -ρV  | w = -1  |

The Universe grew fairly rapidly during the radiation period,
slowed down during the mass period and grows exponentially
during the vacuum period.  This is illustrated below.

Image courtesy of Wikipedia.

de SITTER SPACE
---------------

A universe where the only source of energy comes from the vacuum
(Λ ≠ = 0, ρM = ρR = 0) is referred to as de SITTER SPACE.  Thus,
.
(a/a)2 = H2 = Λ/3 - k/a2

For the k = 0 case we get:
.
(a/a) = H = √8πGρ0

With the solution:

∴ a = Aexp(√(8πGρ0)t)

Alternatively,

a = Aexp(√H t)

Thus, in the vacuum dominated phase the universe is growing
exponentially.  Recall that:

v = HD = D√8πGρ0

∴ DH = c/√8πGρ0

Since Λ is a constant,  DH is fixed.  This distance corresponds to
the distance at which we can't see beyond because the universe is
expanding at the speed of light.  DH is referred to as the COSMIC
HORIZON.

The implication of this is that as the universe continues to expand,
the objects that we see in the night sky today will eventually slip
through the cosmic horizon and become invisible.  Another way of
saying this is that the light emitted at the horizon is effectively
redshifted to infinite wavelength.  Observers in the distant future
will look up at the heavens and see a different universe than we see
today.  Ultimately, the night sky will become completely black!

The radius of the cosmic horizon is approximately 46.6 billion light
years.  The radius of the surface of last scattering is about 45.7
billion light years.  Eventually the SOLS will pass through the cosmic
horizon and the CMB will disappear.

Note:  The age of the universe is estimated to be 13.8 billion years.
However, the radius of the observable universe is not this number.
In the real universe, spacetime is expanding and the distances
obtained by multiplying the speed of light multiplied by this interval
has no direct physical significance.

It is worth noting that while the universe as a whole is expanding,
it can be contracting locally due to gravity.  A good example is a
black hole. ```