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## SUCH A CHILDISH PLAY

Sometimes it does seem like my entire life is a childish play because I was simply lacking the genius that led Shakespeare’s characters to be immortalized by names of Jupiter’s moons.  On the other hand the great Titan Prometheus only had a little moon of Saturn, and the universe is very large.  I am definitely closer to salvaging Eternity from Oblivion by overthrow of Big Bang even though I did not seriously enter the arena of science in a proper way except in brief positions at Biospect/Predicant where I had reached the rank of Scientist II ingloriously ejected by the ambitions of some punk kid to return to New York as a quant who failed to produce anything better than a prediction model for commodities based on inflation which was a result of note.  It is clear to me now that EVERY true result is of a great deal of interest.  My life is a disaster in too many ways but such a childish play indeed.

## REMINISCES OF LEHMAN BROTHERS FIXED INCOME RESEARCH 1995

My first job was at Lehman Brothers Fixed Income Research in Andy Morton’s group.  I was hard-working and naive about what was required then for vertical ascent.  I came with mathematical background but untrained in science and I spent my time coding.  My perspective has matured regarding science since, so now I realize the hard work of getting nature to follow the quantitative models now attempting to get a static Einstein universe to produce the CMB anisotropy.  What is clear is that the hard part of numerical work remains as much drudgery as I had in the trading floor of Lehman, 13 hours in the neon light punctuated by dark skies.  What has not changed when making Octave models for CMB anisotropy (as opposed to the derivative pricing models) is that the annoyance of the coding problems is dwarfed by the difficulty of producing a model that can fit the observed anisotropy.  Tonight, I can’t solve the problem: find a distribution of point masses on a three-sphere such that the gravitational redshifts alone may reproduce the anisotropy.  So much for the coding problem.

The sample-frequency-concentrated-distribution looks quite good but not in a realistic FREQUENCY SCALE.

pkg load statistics

global n=10000
global q0=100
global points=zeros(n,2)
global q0
global n
global X = mvnrnd( [0,0,0,0],diag([1,1,1,1]),n);

for i=1:n,
X(i,:)=X(i,:)/norm(X(i,:),2);
endfor

# stereographic projection + normalize

function z=pwrrnd(k,n)
u=unifrnd(0,1,n)
z=(1-u)^(1/(1-k))*0.001
endfunction

global masses = lognrnd(0,4,n,1)*50;

function d=direction(x1,x2)
d=x2-x1;
d=d/norm(d,2);
endfunction

function d=distance(x1,x2)
d=acos(dot(x1,x2));
endfunction

function f=gforce( x1,x2, m1,m2)
f=m1*m2*direction(x1,x2)/distance(x1,x2)^2;
#f=0
endfunction

timestep=10

if 0,
for a=1:n,
force_on_x = 0;

x1 = X(a,:);
m1 = masses(a);
for b=1:n,
x2 = X(b,:);
d=distance(x1,x2);
m2=masses(b);
if d>0 && d<pi,
force_on_x = force_on_x + gforce(x1,x2,m1,m2);
endif
endfor
x = x+0.5*force_on_x*timestep^2/m1;
x = x/norm(x,2);
X(a,:)=x;
endfor
endif

global z = zeros(n,3);
for i=1:n,
y=X(i,:);
y1=y(1);
y2=y(2);
y3=y(3);
y4=y(4);
zp=[ y1/(1-y4), y2/(1-y4),y3/(1-y4) ];
z(i,:) = zp/norm(zp,2);

# select from log-normal
#u = lognrnd(0,1);
#if u>5,
#z(i,:) = [0,0,0];
#endif
endfor

function t=fitangle(x)
t=x;
while t>2*pi,
t=t-2*pi;
endwhile
while t<0,
t=t+2*pi;
endwhile
endfunction

function b=bin_s2point( z )
if abs(z)<0.001,
b=[0,0];
return
endif
global q0;
theta = atan( z(2)/z(1));
theta=fitangle(theta);
phi = atan( sqrt(z(1)^2+z(2)^2)/z(3));
phi = fitangle(phi);
thetan = floor(theta*q0/(2*pi));
phin = floor(phi*q0/(2*pi));
b=[thetan,phin];
endfunction
# create a spherical grid in longitudes/latitudes
global g = zeros(q0,q0);

global refpoint = [1,0,0,0]

function dox()
global n;
global g;
global X;
global masses;
for i = 1:n,
global refpoint;
global z;
global q0;
global points;
dg = zeros(q0,q0);

cds = bin_s2point(z(i,:));
cds(1)=mod(cds(1),q0);
cds(2)=mod(cds(2),q0);
disp([cds(1),cds(2)])
points(i,:)=cds;
fflush(stdout)
#sdir = sign(direction(refpoint,X(i,:));
dist = distance(refpoint,X(i,:));
#g(cds(1)+1,cds(2)+1) = g(cds(1)+1,cds(2)+1)+1;
m1=masses(i);
dg(cds(1)+1,cds(2)+1) = m1/dist;
dg = imsmooth( dg, "Gaussian", ceil(2),ceil(2));
#disp(sum(sum(dg)))
#fflush(stdout)
dg(1,1)=0;
g = g+dg;
endfor
endfunction

dox()
sum(sum(g))
#g = (g-min(min(g)));
g=g/sum(sum(g));

# value at a spherical harmonic

function v=sphCoeff( l,m, g)
v = 0;
q0=20;
for j=1:q0,
for k=1:q0,
w1 = cos(2*pi*j/q0);
w2 = exp(1i*2*pi*k/q0);
fnval = gsl_sf_legendre_sphPlm( l, m, w1 )*w2;
if ~isnan(fnval*g(j,k)),
v = v + fnval*g(j,k);
endif
endfor
endfor
disp([l,m,v])
fflush(stdout)
endfunction

function v=sumSphCoeff( l, g)
v=0;
for m=0:1
v = v+sphCoeff(l,m,g);
endfor
v=abs(v)^2
endfunction

valsByL = zeros(1,1000);

for k=1:1000,
valsByL(1,k)=sumSphCoeff(200*k,g);
endfor

plot( 1:1000, valsByL);

## SIMPLEST EXPLANATION OF UNIFORMITY OF CMB

The simplest explanation of the uniformity of the cosmic background radiation is thermal equilibrium in a STATIC universe, in fact the Einstein static universe model is infinitely more plausable than inflation theories where things happened during the beginning of the universe (whatever that means).  The Standard Model of Cosmology is completely incredible in the sense that it makes very little sense.  I bet that a relatively simple model where the anisotropy of the CMB is explained by things like gravitational redshift due to mass clusters could fit the anisotropies with less work than what went into fitting inflation models.  Unlike the Standard Model of Particle Theory the cosmological models are completely not based on experiments.

## RADIUS OF EINSTEIN-STATIC UNIVERSE FOR REALISTIC VALUES OF MATTER DENSITY

In my last blog, I considered radius of static Einstein universes with only CMB radiation of 2.7K.  We are interested in radius of static Einstein universes with realistic values for matter density.  Published estimates of matter density (for references see this) are $\rho=10^{-30} g/cm^3$ (Tipler 1987), $\rho=4-18\times 10^{-30} g/cm^3$ (Guth 1987), $\rho=5\times 10^{-30} g/cm^3$.  We need to convert to kg/m^3 so the order becomes $10^{-30} g/cm^3 = 10^{-27} kg/m^3$ and solve the equation

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

• $\rho=10^{-27} kg/m^3$ corresponds to a radius of 13.38 Gpc
• $\rho=5\times 10^{-27} kg/m^3$ corresponds to radius 6 Gpc
• $\rho=18\times 10^{-27} kg/m^3$ corresponds to radius 3.154 Gpc

Here we have to worry about the negative pressure issue, i.e. the second Friedmann equation for static universe is satisfied.

$\frac{\ddot{R}}{R} = -4\pi G(\rho+3p/c^2) =0$

Unlike radiation for which $\rho_{rad}+3p/c^2=0$ we don’t automatically get the cancellation for pure matter where $p=0$.  So here we need a cosmological constant to produce the negative pressure.  Here we can invoke quantum field theory for a static Einstein spacetime to obtain a cosmological constant.  The Friedmann equations with a cosmological constant are

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

$0 = -\frac{4\pi G}{3}(\rho + 3p) + \Lambda/3$

In the case of interest to us, where $\rho \sim 10^{-27}$ the pressure due to radiation will be much smaller, say due to CMB 2.7K $p \sim 10^{-31}$ (which is obtained by the Stefan-Boltzmann law and then dividing by $c^2$).  We fix $\rho=5 \times 10^{-27}$ and $p = 10^{-31}$ and solve for $R$ in the first equation for the sake of clarity

$kc^2/R^2 = \frac{4\pi G}{3}( 3\rho+3p)$

In this case we get the following radii for static universes:

• $\rho= 1\times 10^{-27}$ corresponds to 10.92 Gpc
• $\rho= 5\times 10^{-27}$ corresponds to 4.88 Gpc
• $\rho= 18\times 10^{-27}$ corresponds to 2.57 Gpc

This is a purely CLASSICAL picture that does not include quantum field theory effects.  When pressure $p$ increases the radii will increase as well.  For static Einstein universes the vacuum energy for example for massless neutrinos are reasonable so these could explain slightly larger radii.

## 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
2*(xMpc*Gpc)*1e-3/(sqrt(n^2+2)+n)
}

Although chi-by-eye is a bad habit, the issue here is to just ensure that we are not doing something completely absurd by checking that we get some reasonable numbers for ‘Hubble velocity’ by spherical geometry alone.  So we can check that for a 30 Gpc universe at 400Mpc and a fudge factor of 1e-4 we obtain something between 25000 and 30000 as in the empirical graph.

>
> hubble_slope_for_universe_Gpc(30,400,1e-4)*c
[1] 26251.24

Of course this is not very tight.  What this shows is that we can expect to constrain the radius of the universe between 24 Gpc and 624 Gpc which arise from considerations of empirical energy density of 2.7 K CMB temperature and still be able to explain the empirical redshifts without any expansion.  Stability of the Einstein model dominated by radiation is known (with a speed of sound lower bound).  Therefore we can have viable static models that does not violate Einstein’s equations and explains redshift, and can remove dark energy and dark matter.  There are enormous benefits to such models since the cosmological constant problem is resolved in the model.  Also Euclidean quantum field theory have no problems by work of Jaffee-Ritter.  There is no singularity finite time in the past which I believe is the truth obviously.  The problem thus is to fit the CMB anisotropy in a static Einstein universe.  Recall that Einstein’s first reaction to Lemaitre was that the physics was atrocious and he was right about that.  Quantization phenomena are probably GLOBAL geometric phenomena rather than phenomena that are purely microscopic.  This is something that makes a great deal more sense in a static universe.

## 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.