Big Bang Modelling

From the (incredibly naive) model we can deduce:

• Density
• Expansion time
• Temperature of the universe
• Corrections if the density is not critical
• Modifications if we have more than just matter in the universe

Connection between time, distance and red-shift:

At time t₀, a galaxy is at a distance d₀ from us, and is travelling at v We see the photon at time t1

In time t between γ emission & absorption, the distance has increased (universe has expanded) by zct. so γ has to travel further:
travel time

 
t = d1/c

so

d1 = d₀ + zct = d₀ + zd1 so that   d1 = d₀/(1-z) (non-relativistic)

i.e. if z = .25, the universe was 75% of current size when γ was emitted

(Note we have made a subtle change in the description: it is not the other galaxy which is moving away, it is the space in between which is stretching!)

We can also relate red-shift and age: At t = 0, universe had no size so

 d₀ = vt₀ = zct₀ hence  t₀/t1 = d₀/d1  = (1-z)

so that

t₀ = t1(1-z)

so again if z = .25, the universe was only 75% of its current age

(Correctly d₁ = d₀(1+z) and t₀ = t₁/(1+z), because of relativistic effects, but we also need to allow for slowing down of the expansion)

To make this look a little fancier, we will introduce the "Cosmic Scale Factor", R(t). All distances will scale by the same amount, so we can define:

r(t) = σR(t) σ is a dimensionless number, which does not vary with time. If you like, σ=r(t=0). In this case, r₀ = d, r = 2d so R= ..?

This means that instead of talking about the distance between all the galaxies varying as a function of time, we can extract all of the time variation and put it into one parameter.

Note some relations:
$$ \color{red}{ v\left( t \right) = \frac{{\partial r\left( t \right)}}{{\partial t}} = H\left( t \right)r\left( t \right) = \sigma \frac{{\partial R\left( t \right)}}{{\partial t}} = \sigma H\left( t \right)R\left( t \right)} $$ (since σ doesn't change with time) so $$ \color{red}{ \frac{{\partial R\left( t \right)}}{{\partial t}} = H\left( t \right)R\left( t \right)} $$

(Note since σ is a dimensionless initial value, can use units system in which σ = 1, and drop it now.)

Density

How does the density change in this model? ρ(t) = ρ(t₀) (R(t₀)/ R(t))³ = ρ₀ R₀³/R³ Note: we are talking about matter density here

Expansion time:

Since$$ \color{red}{ \frac{1}{2}mv^2 - \frac{{GMm}}{r} = E} $$ which can be rewritten as $$ \color{red}{ \left( {\frac{{\partial R\left( t \right)}}{{\partial t}}} \right)^2 = \alpha R^2 - k,\alpha = \frac{{8\pi G\rho _0 }}{{3R_0 ^{} }},k = \frac{{2E}}{m}} $$ This can easily be solved if k = 0: R(t) = Ct2/3 (C is an uninteresting constant)

Hubbles "constant"

In this model

∂R(t)/∂t = ²/3 C t-1/3

and hence $$ \color{red}{ H\left( t \right) = \frac{1}{{R\left( t \right)}}\frac{{\partial R\left( t \right)}}{{\partial t}} = \frac{2}{{3t}}} $$

If we take t as the time now since the Big Bang

t =  ²/3 H-1 = 12x109 yrs

History of CMBR:

The universe was originally opaque (i.e. mean free path of γ's very small) and hence CMBR was in thermal equilibrium with matter. Then the universe "condensed" (or froze) out.

Photon freezeout:

Thermal equilibrium between matter and radiation implies they are at the same mean energy, with a Black Body distribution of γ's. Expansion ⇒ Cooling.

Need the R dependence of energy density, ρ Non-Relativistic Matter (i.e. Baryons): ρ ≈ 1/R³(t) Relativistic particles (i.e. γ's, ν's) get red-shifted as well, so ρ ≈ 1/R⁴(t) We live in matter dominated universe ρmatter >> ρrad, but it was not always thus

This means that we need to be able to solve for a more general universe, where we don't have just cold matter.

Doing it properly

This is all a bit of a cheat, since we have assumed non-relativistic Newtonian mechanics: we should use General Relativity. However, we can do almost everything in terms of our (naive) model
To start with, we want the Friedmann equation $$ \color{red}{ \frac{1}{2}mv^2 - \frac{{GMm}}{r} = E \to \left( {\frac{{\dot R}}{R}} \right)^2 = \frac{8}{3}\pi G\rho - k\frac{{R_0^2 }}{{R^2 }}} $$

We will choose our scale so that k is an integer: the "curvature" of universe

k = -1 means E > 0, open, negative curvature
k = 0 means E = 0, critical, zero curvature
k = 1 means E < 0, closed, positive curvature

Also need the connection between $\rho$ and R: for matter and radiation we have argued $$ \color{red}{ \rho _M \sim \frac{1}{{R^3 }}:\rho _R \sim \frac{1}{{R^4 }}} $$

In general, if we have an expanding gas, change in energy = W.D. $$ \color{red}{ d\left( {\rho V} \right) = - PdV \to \frac{d}{{dt}}\left( {\rho \frac{4}{3}\pi R^3 } \right) = - P\frac{d}{{dt}}\left( {\frac{4}{3}\pi R^3 } \right)} $$

For cold matter, P = 0, so we get back the old result. For radiation:can repeat old kinetic theory of gases derivation giving P = 1/3 ρ_R. In general we'll write P = wρ. Then $$ \color{red}{ \frac{{d\left( {\rho R^3 } \right)}}{{dt}} = - w\rho \frac{{d\left( {R^3 } \right)}}{{dt}}} $$

Then $$ \color{red}{ \frac{{d\left( {\rho R^{3\left( {1 + w} \right)} } \right)}}{{dt}} = 0} $$ or ρR^3(1+w) is a constant (prove it: hint first show $$ \color{red}{ \frac{{d\left( {R^{3\left( {1 + w} \right)} } \right)}}{{dt}} = \left( {1 + w} \right)R^{3w} \frac{{d\left( {R^3 } \right)}}{{dt}}} $$

Can now solve easily for flat universe $$ \color{red}{ \left( {\frac{{\dot R}}{R}} \right)^2 = \frac{8}{3}\pi G\rho _0 \left( {\frac{R}{{R_0 }}} \right)^n ,n = 3\left( {1 + w} \right)} $$

where n = 3 (matter) or 4 (radiation) or anything else

So we want the relation between t and T:

Expansion time:

α2 = 8/3πGρ

The Solution is............

Hubble's "constant" is $$ \color{red}{ H = \frac{v}{d} = \frac{{\dot R}}{R} = \frac{2}{{nt}}} $$

Note: For small t, v(t) ≈ 1/R(t) >> c => the horizon problem. No immediate problem with relativity, (since no info is transferred but ...

Parts of the universe have never been in contact with each other. How do they know to be at the same temperature?

For k = 0$$ \color{red}{ \rho = \rho _c = \frac{{3H^2 }}{{8\pi G}}} $$

as we expect. If we have only radiation $$ \color{red}{ \rho = \frac{{4\sigma T^4 }}{{c^3 }}} $$

which gives an exact expression for the temp: $$ \color{red}{ T^2 = \frac{1}{t}\sqrt {\frac{{3c^3 }}{{64\pi G\sigma }}} ,t = \frac{\xi }{{T^2 }}} $$

As the peak in the BB curve falls below hydrogen binding energy, ¹H forms, at ~3700 K, ie about ¹/₂eV. When did this happen?

(Note, this is less than the 13.6 eV that you would expect: need Saha equation)

τ ≈ ζ/T² ≈ 500,000 yrs, or 1+z ≈  T/T₀ ≈ 3700/2.736 ≈ 1360

What happens if the universe is not flat? we have to solve $$ \color{red}{ \int {\frac{{dR}}{{\sqrt {\frac{\alpha }{{R^{n/2 - 1} }} - k} }}} = t} $$

(see Berry for details: tedious but straightforward! Note thsi give us f(R) = t, not R = g(t) which we would like

As the universe expands, the density of γ's decreases. ρ(t) = ρ₀ R₀³/R(t)³ and also the temperature T falls as each γ gets redshifted

Hence the energy density E(t) ∝ R(t)⁴

More exactly: from the Black Body equation:

Energy density:

u = T⁴ × 8/15π5k⁴/(h³c³) = aT⁴ = 4σT⁴/c

Mean γ-energy E ≈ 2 x 10^-4 T eV actually = hc/λ = 6.58x10^-16 x 3x10⁸ /λ
Peak of BB curve is at λ = 2.898x10^-3/T
Now, the peak is at 1.05 mm: what is the temperature, N, E, ρ?

When did the universe stop being radiation dominated?
We said: ρ_c ≈ 3 H/m³. We see less: ρ_M ∼≈ .01 H/m³. !!!

Better: energy in protons = (number/unit volume)×mc² Say

ΩM∼ .01 ⇒ EM ≈ 0.01×10MeV/m³ ∝ 1/R³

Energy in CMBR photons

Erad = 4σT⁴/c ≈ 0.4 MeV/m³ ∝ 1/R⁴

So these were equal when:

R(t)/R(t₀) = Ematter/Erad ≈ 1250 = (1+z)

ie, at about the same time that the universe became transparent. (Note: we began with a very large uncertainty, setting Ω ≈ 0.01, based on .03 proton per cubic metre: depending on the nature of the DM we can change this by a factor of 10)

This "light" is now at 2.736°K, and almost uniform in every direction. It was emitted just 100000 years after the Big Bang and has been travelling round the universe ever since. At time t₀, the universe is full of γ's at a temperature T₀.

Inflationary Universes

One more special case: how about if vacuum itself has an energy ρ_v=Λ/3 ≠0? Λ is the cosmological constant "Einstein's greatest blunder"! Introduced as a fudge factor in 1919 to stop the universe from collapsing

so (setting ρ_m= ρ_R=0 for simplicity) we can use $$ \color{red}{ \left( {\frac{{\dot R}}{R}} \right)^2 = \frac{\Lambda }{3}} $$

which is our previous results with w = -1: this gives$$ \color{red}{ \left( {\frac{{\dot R}}{R}} \right)^2 = \frac{\Lambda }{3},R(t) = R_0 e^{\sqrt {\frac{\Lambda }{3}} t} } $$

i.e. the universe expands exponentially. Why would we want this?

Why the Hell?

6) Fortunately there is a special place for us

And God said " Let us make man in our own image, after Our likeness, and let them have dominion....over all the earth

Genesis I.26

Why is Ω = 1 so important? Combining

• If k = 0, Ω = 1 for all time
• f k > 0, then Ω decreases away from 1.
• If k < 0, then Ω increases away from 1.

Why is Ω = 1 so important?

Since we now measure Ω ~ 0.1, this means that at the time of the BB it must have been ~ 1 - 10-60 i.e. Ω = 1 is an unstable critical point The flatness problem is worse than you would think: If the universe started out at 10-44 s with (say) Ω = 3, it would last ≈ 10-35 s (!!!!)

The inflationary universe

Need to separate the 3 contributions to Ω: ΩM=ρM/ρ ΩR=ρR/ρ ΩΛ=ρΛ/ρ If the universe expands, the value of ρΛis static, but the other two fall. If we start off with any values of ΩM and ΩR but ΩV=1, the other two will disappear during inflation

consider a very small volume of the (original) universe. e.g. at t = 10-34 s, the size of the horizon is d = ct = 10-23 cm (≈ 10-8 size of a proton) The inflationary phase leads to an increase in the size of universe by 1080 (effect is very like continuous creation, in that the energy density remains the same).	(diagram not exactly to scale!)
An inflationary phase "smooths out the wrinkles". The temperature in this volume was already in equilibrium. Now, after inflation, it is much larger and smoother than before.

At 10^-35 s, the universe could not have deviated very much from Ω = 1.

After inflation, the density hasn't changed (remember steady state model?) but the universe is 10⁴³ times larger. Need to turn off inflatio at thsi time: introduce some new, very heavy particle which will decay, both stopping inflation and creating matter and radiation at the critical density.

This means that it can never deviate much from Ω = 1.

Before finishing this off, we really need to the calculation more honestly: i.e. using General Relativity