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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.

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.

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.

hubble-line

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
function(gpc_radius,xMpc,fudge_factor){
Gpc<-3e25
c<-3e8
Halpha<-6.56281e-7
n<-gpc_radius*Gpc*fudge_factor/(Halpha)
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.

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.

 

Black holes are a theoretical concept invented by Oppenheimer-Snyder who showed that pure Einstein equations lead to a collapse of a star to a pure singularity (Oppenheimer_1939).  In 2014 Laura Mersini-Houghton shows that with additional considerations a black hole cannot form because the collapsed star must have a finite radius (1408.1449).  Her papers are published in Physics Letters B so this is considered seriously.  I think this makes sense.  The idea of black holes is based only on a very simple model whose conclusions were just extrapolated wildly.  Just like the Big Bang singularity these probably are figments of our imagination.  Singularities in spacetime have never been observed.  It’s fairly crazy to extrapolate about such things.  Of course experimenters have ‘found’ black holes.  Experimenters make their living finding things that theorists tell them to find.

This is a very simple simulation that is not meant to reproduce an exact fit to CMB anisotropy but to get some feel for whether CMB anisotropy occurs only with gravitational redshift as the only source of anisotropy which it does not try to do exactly.  A static Einstein universe is an excellent model now that we know that the original reasons for abandoning it were not quite right (Eddington’s instability result).  The six parameter Lambda-CDM model fits the anisotropy well which is the model to overthrow.  So is it possible to do this?  I believe that a static Einstein universe is an infinitely superior model (as it gets rid of the cosmological constant problem since quantum field theory on static Einstein universe is actually close to the measured value of around 10^{-52} cm and the quantum gravity problem is not as difficult here since for example the Wheeler-deWitt equation can be solved explicitly).  So here is a little simulation that shows that there is some interesting anisotropy in the simplest type of model for gravitational redshift effect.  Distribute a uniform set of points over a round three-sphere and assign them masses by an exponential distribution.  Choose a point on the three-sphere as a reference pont and draw a two-sphere around it and histogram the geodesics meeting the 2-sphere.  Assume the background radiation has 2.7 K uniform background so that the anisotropy is due only to the gravitational redshift of photons that scatter on the fixed objects (scattered uniformly in the three-sphere).  Smoothen the three-sphere histogram and extract the spherical harmonics coefficients for them and plot the power spectrum.  We ask:  do we see an anisotropy that gives any sense of the anisotropy.  The answer to this question is yes (in a relatively quick implementation).  This model can be refined to fit the actual CMB anisotropy I am confident.  If I don’t succeed, someone else should do this because the Big Bang cosmology is insanely ambitious in claiming anything about the ‘origins of the universe’.  A static Einstein universe would be infinitely better science with what we actually can know about the universe without any ability to do any experiments.  Of course I am not doing any experiments either but I am doing real work:  find a tight fit to the CMB anisotropy in a SIMPLE model.  This cannot be that hard.

The code is in octave:  The first example of anisotropy for the model is this little graph:

artificial-simple-anisotropy-static-universe-gravitational-redshiftonly

pkg load statistics
pkg load gsl
pkg load image

n=10000
q0=200
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

masses = exprnd(1,n)

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

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

function f=gforce( x1,x2, m1,m2)
f=m1*m2*direction(x1,x2)/distance(x1,x2)^2
#f=0
timestep=0.1

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 and 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)
endfor

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);
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 )
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*100/(2*pi));
phin = floor(phi*100/(2*pi));
b=[thetan,phin];
endfunction
# create a spherical grid in longitudes/latitudes
g = zeros(q0,q0);

refpoint = [1,0,0,0]
for i = 1:n,
cds = bin_s2point(z(i,:));
cds(1)=mod(cds(1),100);
cds(2)=mod(cds(2),100);
sdir = sign(direction(refpoint,X(i,:));
dist = distance(refpoint,X(i,:));
dg = zeros(q0,q0);
dg(cds(1)+1,cds(2)+1) = 1;
dg = imsmooth( dg, “Gaussian”, ceil(m1),ceil(m1));
g = g+dg;
endfor

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

h=imsmooth(g,”Gaussian”,10,10);

g=h;
# value at a spherical harmonic

function v=sphCoeff( l,m, g)
v = 0;
q0=100;
for j=1:q0,
for k=1:q0,
w1 = cos(2*pi*j/100);
w2 = exp(1i*2*pi*k/100);
fnval = gsl_sf_legendre_sphPlm( l, m, w1 )*w2;
v = v + fnval*g(j,k);
endfor
endfor
endfunction

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

valsByL = zeros(1,100);

for k=1:100,
valsByL(1,k)=sumSphCoeff(20*k,g);
endfor

plot( 1:100, valsByL)

 


 

I took Differential Geometry at Princeton in my sophomore year in 1992 and have become incredibly rusty so it is time to do some computations in General Relativity to get back in shape.  An elementary problem: find the light-cone geodesics for the metric

g = dt^2 - e^{-2Ht} dx^2

which satisfy g_{ab} \dot{x}^a \dot{x}^b=0.  The Christoffel symbols are computed from

\Gamma_{abc} = \frac{1}{2}( g_{ca,b} + g_{cb,a} - g_{ab,c})

There are too many of these to compute exhaustively so we note first that

g_{ab,c} = 0 unless a=b=2, c=1

Therefore the only nonzero \Gamma_{abc} are those with two 2 and one 1 in the indices.  For example,

\Gamma_{212} = \frac{1}{2} ( g_{21,2} + g_{22,1} - g_{12,2}) = \frac{1}{2} g_{22,1} = -H e^{-2Ht}

and

$\Gamma_{122}=\Gamma_{212}=\Gamma_{221} = -H e^{-2Ht}$

Then we have to raise the first index which has the solution

\Gamma^1_{ab} = g^{11} \Gamma_{1ab} + g^{21} \Gamma_{2ab} = \Gamma_{1ab} = -He^{-2Ht}

\Gamma^2_{ab} = g^{12}\Gamma_{2ab} + g^{22}\Gamma_{2ab} = e^{+2Ht} \Gamma_{2ab} = - e^{+2Ht} H e^{-2Ht} = -H

Now the geodesic equation is

\frac{ d^2{x}^c}{ds^2} + \Gamma^c_{ab} \frac{dx^a}{ds} \frac{dx^b}{ds} = 0

For coordinates t,x we have

\ddot{t} + (-He^{-2Ht}\dot{x}^2 = 0

\ddot{x} - 2H\dot{t}\dot{x} = 0

I don’t know how to solve this system of equations in general.  But the light cone condition g_{ab}\dot{x}^a\dot{x}^b =0 in this case is

\dot{t}^2 = e^{2Ht} \dot{x}^2

which we plug into the first equation \ddot{t} - H\dot{t}^2=0.  This is \dot{z} = Hz^2 solved by integrating \dot{z}{z^2} = H, so z=-\frac{1}{Hs} and t = -\frac{1}{H} \log(s).

The second equation now simplifies when we plug in \dot{t} = -\frac{1}{Hs}.

\ddot{x} +2/s \dot{x} = 0

Again \dot{z} = -(2/s)z and \log(z) = -\log(s^2) +C and x = \int^{s} \exp(-r^2) dr.