Cosmology Introduction ...or... Physics as a Creation Myth

Cosmology 1: Doesn't it make you feel humble!

Redshift: Slipher-Hubble-Humason found light from most galaxies is redshifted. The Doppler effect gives z = (λ-λ₀)/λ₀ = Δλ/λ₀ Velocity of recession: v = zc = Δλc/λ₀

Hubble found vel. of recession ∝ distance zc = Hd = v H ~ 65kms-1/Mpc 1 Mpc (megaparsec) = 3x1022 m

Note although all galaxies are receding from us, does not imply we are at the centre: in the currant cake model all currants see all the others as recediing

Big Bang (once over lightly)

RULE 1 in Physics 100: Never mix your units!)

H = 65x10³ = 1.8x10-18 (m s-1)/m 3.1022 We can invert this to give H-1 = 5.4x1017 s = 1.7x1010 yr.

What does this time represent? Must be age of universe: if expansion does not change i.e. 17x109 yr ago, all the galaxies were in the same place. Universe had a beginning, implied by the big bang. Can run Hubble expansion back: we would like to use this to predict what will happen in the end

A 2-D analog is the surface of a balloon: Note the following:

• It has no centre in 2-D space.
• Deflating it reduces it to zero size: i.e. at the moment of the big bang, not only matter was created, but also space and time
• The galaxies are not receding from us: space is expanding
• We require a curved 2-D surface embedded in a 3-D volume.
• We can define coords on the balloon that will expand along with it: this is a "co-moving" system
• This is a positively curved universe: we can also construct negatively curved ones (harder to visualise)

If we measure from now (t = t₀) then R = 0 when t = t₀ - 1/H₀ independent of R₀.

Gravitational attraction would have slowed expansion since the early universe. So Hubble's constant is important: we had better be sure of what it is!

(Incidentally, it isn't a constant...when the universe was smaller, R was less; if v was constant H must have been bigger. Better "the Hubble parameter"

What's going to happen in the end?

As a model, consider this as an escape velocity problem. How hard do we need to throw a galaxy on the "outside" so that it escapes? Note: our calculation had better not depend on r! \color{red}{ \frac{1}{2}mv^2 - \frac{{GMm}}{r} = 0} but \color{red}{v = Hr} and the total mass of the universe inside \color{red}{M = \frac{{4\pi }}{3}\rho r^3 }

• \color{red}{H^2 r^2 = 2G\frac{{4\pi }}{3}\rho r^2 } (we got lucky: the r cancels out!).
• We can turn this round and write it as an equation for the density ρ
• \color{red}{ \rho _0 = \frac{{3H^2 }}{{8\pi G}}}
• Hence the "critical density"

• ρ₀ ~ 6 x 10^-27 kg m^-3 ~ 3.6 Hydrogen Atoms m^-3 (Number is flaky: we'll use 4).
• Instead of the actual density, we'll use
  \color{red}{ \Omega = \frac{\rho }{{\rho _0 }}}
  because some errors cancel out.

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  So if Ω > 1 Universe come to nasty end in ~ 50 x 109 yr. Ω = 1 Universe expansion slows down asymptotically : "critical universe" Ω < 1₀ Universe expands forever More important:we live forever if Ω ≤ 1, (well maybe).

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  • Note that this implies that the rate of expansion must change Gravity will slow down expansion in the early stages, so Hubble's constant isn't a constant... when the universe was smaller, v was larger so H must have been bigger. Better "the Hubble parameter"

  • From this (incredibly naive) model we can deduce:
  • Density
  • Expansion time
  • Temperature of the universe
  • Corrections if the density is not critical
  • But we now need to do the calculation more honestly: i.e. using General Relativity