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## CMB AND STATIC RADIUS: NOT TRIVIAL FIT AND AN ODD NUMEROLOGICAL OBSERVATION

In FRW models $ds^2 = -dt^2 + R(t)^2 dz^2$ where the $dz^2$ is the line-element of a unit $S^3$ the Einstein equation reduce to Friedmann equations

$(R')^2 - \frac{8\pi G}{3} \mu R^2 = -k$

$\frac{d}{dt}(\mu R^3) = - p dR^3/dt$

$\frac{2 R''}{R} = - (\frac{R'}{R})^2 - k R^{-2} - 8\pi G\rho$

where $\rho$ is the energy density and $p$ is the pressure.  I wanted to see if there was a way that we could get something simple and stupid: can we take the 2.7 K cosmic background radiation temperature and find the radius of the universe?  The answer seems to be no.  First, look at the paper of Barrow et. al. 0310233 for an analysis of this model (where they show that there is not recollapse for these closed models etc.)

Now let’s compute the energy density of 2.7K.  Stefan-Boltzmann law $\mu = a T^4$ with $a=7.5657\times 10^{-16} Jm^{-3}K^{-4}$ gives us $\mu = 4.14 \times 10^{-14} J/m^3$.  Set $R' = R'' = 0$.  Take the radius of the universe to be 24 Gpc which in units with speed of light $c=1$ $R = 2.4 \times 10^{18}$.  Then

$\frac{8\pi G}{3} \mu R^2 = 4\times 10^{14} \not = 1$

So for the observed temperature we don’t get the observed radius for the static model.  Purely numerologically (which is not meaningful) if we take $\mu^2$ instead of $\mu$ we do get something more reasonable order-of-magnitude although this is physically not meaningful:

$\frac{8\pi G}{3} \mu^2 R^2 = 16.56$

Anyway, assuming that the universe is static with the 2.7 K does not immediately give us the correct radius of order 24 Gpc using only Einstein equations.  I am convinced that there is no expansion at all and that the blackbody radiation is a steady state phenomenon.