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## WHAT IS THE RADIUS OF A STATIC EINSTEIN UNIVERSE THAT COULD SUSTAIN A 2.7 K COSMIC BACKGROUND RADIATION ACCORDING TO GENERAL RELATIVITY?

We play the following game.  The input are (a) the empirical energy density of the universe (2.72 K) of the cosmic background radiation and (b) the Friedmann equations which are Einstein’s equations for metrics in the Friedmann-Robertson-Walker metrics

$-c^2 dt^2 + R(t)^2 ds^2$

and (c) the empirically claimed radius of the observable universe.

The Friedmann equations are:

$(\frac{\dot{R}}{R})^2 = \frac{8 \pi G}{3} \rho - k \frac{c^2}{R^2}$

$\frac{\ddot{R}}{R} = -\frac{4\pi G}{3}(\rho + \frac{3p}{c^2} )$

where $\rho$ is the MASS density and not the energy density and $p$ is the pressure.  We had found the ENERGY density in the last post $\rho = 4.14 \times 10^{-14} J/m^3$.  We could not understand the possibility of a STATIC universe of radius 24 Gpc (which is the size of the OBSERVED universe.  The static universe corresponds to $\rho + 2p/c^2=0$ which determines which is not a problem.  The real problem is the first equation

$\frac{8\pi G}{3} \rho - k \frac{c^2}{R^2} =0$

So now let us set $\mu=4.14\times 10^{-14}$ as the energy density and let $\rho=\mu/c^2$ using the famous energy equivalence $\latex E=mc^2$.  Then we find, for $R=24 Gpc$,

$\frac{8\pi G}{3} \rho \frac{R^2}{c^2} = 0.0015$

This seems to produce too low of a curvature but the order of magnitude is not too great.  In fact we can find that setting $R= 623.644 Gpc$ produces

$\frac{8\pi G}{3}\rho \frac{R^2}{c^2} = 1$

This is encouraging because it tells us that if we have nothing in the universe except the mass density equivalent to the cosmic background radiation mean, approximately 2.7 K, then the Einstein/Friedmann equations for a STATIC universe would be satisfied for a radius of 623.644 Gpc.  This is a universe where the mass density is being determined ONLY by the cosmic background radiation.  If we increase $\rho$ in order to retain 1 on the right hand side we would have to decrease the radius.  This is nice because it tells us that the radius of a static universe will be at most 623.644 Gpc and is then completely consistent with the observed radius of 24 Gpc.

The conclusion would be that if are careful to interpret $\rho$ as the MASS density rather than the energy density, we have a consistent model for a STATIC Einstein model.

Of course we should have some reasonable explanation for the Hubble redshift distance relation in a static universe.  The idea that I had discovered some years ago applies here.  The Hubble relation is explained as an ARTIFACT of spherical geometry having nothing to do with any OTHER physical phenomena.  The spherical geometry enforces a deviation from the relationship between frequency and wavelength from $\lambda = c/\nu$ to (essentially) $\lambda = c/\sqrt{\nu(nu+2)}$.  More precisely, on a sphere, frequency for light waves are quantized according to the eigenvalues of the Laplacian which occur as $\sqrt{n(n+2)}/R^2$ so there is a discrepancy between the frequency determined by the observed wavelength and the frequency at the source which grows in distance traveled by the light according to $c(1/n - 1/sqrt{n(n+2)})$.  We can implement this idea easily to match the Hubble graph qualitatively.

Here’s a quick and dirty model that produces Hubble ‘velocity’ as a function of distance based on the spherical geometry.  The $H-\alpha$ wavelength is $\lambda_{H\alpha}=6.56\times 10^{-7}m$.  In an $S^3(R)$ this corresponds to the $n$-th eigenvalue of the Laplacian where $n \sim R/\lamda_{H\alpha}$.  The following function (which includes a fudge factor to which I will return later) allows us to produce Hubble ‘velocities’ as a function of the radius of the universe.

> hubble_slope_for_universe_Gpc
Gpc<-3e25
c<-3e8
Halpha<-6.56281e-7
>
[1] 26251.24