THE AGE AND SIZE OF A STEADY STATE UNIVERSE
By Clarence L. Dulaney and Patrick L. Dulaney
e-mail cldtx1@sbcglobal.net dulaneyp@comcast.com
Abstract: The current model of the Universe espoused by the majority of the physics establishment is the “big bang”. This theory is based primarily on the redshift of stellar spectra, and on the 2.7K background radiation.
It is shown in this paper that the redshift is
not due to a “Doppler Shift” but due to interaction of the light rays from the stars
with the particles of the aether
The 2.7K temperature of the background is also due to this
interaction. If there were not large
numbers of these particles (neutrinos) present, there would be no “temperature
of space”
The Universe is at least 1010 years
old. It has neither expanded nor contracted
during this period, so it may be called “steady state”. It probably has a volume of at least 1062 m3. It may be larger, but the redshift
would be so large that light could not reach the earth.
BIG BANG THEORY
In 1922, Friedman [1] showed that Einstein’s General Theory of Relativity could lead to an expanding (or contracting) universe.
In 1927, Lemaitre [2] proposed, based on the redshift of light from the stars, and on Friedman’s expansion idea, that the Universe began as a “primordial atom”. This entity then expanded to eventually become the Universe. This idea eventually became the “Big Bang” Theory.
The theory was reinforced in 1927 by Hubble and Humason [3] who showed that the amount of redshift was proportional to the distance of the star or nebula from the solar system. This effect was taken (by others) as as Doppler effect and an indication that the stars move faster the further they are away from earth.
(Actually, Hubble was more inclined to believe a “tired light” concept than the Doppler Effect [4].)
At any rate, the current Physics Establishment view is that the redshift is due to the Doppler Effect, and that the Universe is expanding. This, along with the 2.7K background radiation is taken as conclusive evidence that the Universe began with a “Big Bang”.
(Note that it has been assumed, without any physical proof, that light from a moving object follows the same type of law as does sound, that is that the frequency of light from a source that is moving away is decreased.)
HUBBLE’S LAW
Hubble’s Law states that V = H0D. V is the speed of the emitter in km/sec, D is the distance from earth in megaparsecs and H0 is Hubble’s constant, probably 50 km/sec/megaparsec. (A parsec is 3.26 light years, so a megaparsec is 3.086 x 1019 km. Thus, one way of expressing H0 is as 1.6202 x 10-18 sec-1. The age of the Universe is taken as the reciprocal of Hubble’s constant, or 1.95 x 1010 year. (Note that there are 3.15 X 107 seconds in a year.)
In addition, Dl/l = H0D. Dl is the redshift for a given l. For example, if D is 1 megaparsec and l is 4500 Å, Dl = 0.75 x 10-9 cm.
“TIRED LIGHT”
There are a number of “tired light” theories that are usually linked with steady-state, or at least non-expanding Universe theories. Hubble was in favor of “tired light” as was Millikan [5] who agreed in a letter to Hubble.
Others include,but are not limited to, Kieren and Sharp [6], Marik, et al [7], Reber [8], Marmet and Reber [9], and Assis [10].
Interaction of the light with “something” in outer space was proposed. This something ranged from the aether to electrons to H atoms to space dust to reaction with other photons, to something in the “virtual vacuum” of space. The most recent speculations are based on the deflection of light by electrons or hydrogen atoms by means of Compton Effect Scattering. This is countered by the argument that the light ray would be deviated from its path by the scattering, and thus should lead to blurring of the source. No blurring occurs. (A possible explanation of no blurring based on a “random walk” analysis is given by Reber [8].)
In 1923,
Marmet
[12], noting that there could be as many. as 0.01 atom
of H per cm3
of space proposed
Suppose that light is scattered by electrons, the effect per collision would be, (assuming light of 4500 Å ) 1.09 x 10-18cm(1-cosq). If q = 1°, Dl = 1.66 x 10-14cm. For a 1 megaparsec light path, this would mean 7.5 x 10-9/ 1.66 x 10-18 = 4.52 x 109collisions for each light ray, each with a deflection of 1°. It is difficult to imagine that such would not lead to blurring of the light.
Even with smaller deflecting angle, there would be many more collisions and opportunities to blur the light. For example, with q = 0.1°, Dl would be 1.66 x 10-20cm per collision, or 4.5 x 1011collisions in a megaparsec.
If the scatterer is an H atom, Dl = 5.95 x 10-18 (1-cosq) cm per collision. In any case, there would be more collisions than for electron scattering.
It thus appears that for the Compton Scattering to work, the scatterer would have to weigh less than an electron. There are at least two such particles, Van Flandern’s C-Graviton[13], and the neutrino aether[14].
Van Flandern only says that the C-Graviton is very small and travels at many times the speed of light. Speed is so high that the mean free path between collisions is at least a thousand parsecs. Suppose the C-Graviton has a mass of 1/100 of an electron. On the same basis as above, there would be 4.5 x 107 collisions per megaparsec, still an intolerable value.
Obviously, the same sort of thing can be said for the neutrino, which has a mass of 1/3 that of an electron [14]. Based on this, and the objections of Neves and Assis [15], it is concluded “tired light” cannot be explained by Compton Effect Scattering.
SPIN-FLIP OF NEUTRINO AETHER
One possibility that would simply slow the light down rather than scatter it could consist of a spin-flip of the neutrino of the aether. Neutrinos have 2 different spins, either +½ or -½. Suppose a light ray strikes a neutrino with a +½ spin, stays with it for an infinitesimal length of time (possibly 10-10 sec.) and causes an inversion of the spin to -½ in this period (or from -½ to +½). How much energy is required? The spin energy is roughly ½ h/2p, so that converting from +½ to -½ requires 2h/4p or 4.217 x 10-27 ergsec. If the flip occurs over 10-10 sec, then 4.217 x 10-17 erg changes hands.
The total redshift is 0.75 x 10-8cm, so in a megaparsec, the total energy transferred would be hc/l , or 2.65 x 10-8 ergs. Thus, 6.28 x 108 would occur for each parsec., or 5.77 x 1010km/spin-flip.
The primary advantage of a “spin-flip tired-light” mechanism is that no deflection of the light ray would occur, thus no blurring.
There is a question of how many neutrinos are present in the universe. Marmet [12] asserted that the density of outer space is such that 0.01 H atoms could be present per cm3. Since a neutrino weighs approximately 1/5500 part of an H atom, there could be 55 neutrinos per cm3 of space.
VOLUME OF THE UNIVERSE
The age of the Universe was estimated as 2 x 1010 year. This would indicate that the furthest star should be 2 x 1010 light years away, or 1.892 x 1028 cm. Assuming this is the radius of a spherical Universe, the volume would be 2.837 x 1088 cm3. There would be a total of 1.56 x 1090 neutrinos.
THE 2.7K BACKGROUND RADIATION.
COSMIC RAY THEORY OF REGENER
E. Regener in 1933 [16], proposed that cosmic rays could account for the background radiation. We believe this to be incorrect, because the cosmic rays supply their energy not to “outer space”, but to the earth, and other massive bodies they contact. Cosmic rays are zero charged ions [14] that do not react with the uncharged aether, and only supply their energy to the atmosphere or to the earth, which they penetrate to several thousand feet in some cases.
Assume the energy of the spin-flip is given to the neutrino. This is 4.2 x 10-17 ergs to be spread over the surface area of the neutrino (the radius is 0.25 x 10-8 cm, so the surface area is 7.854 x 10-17cm2 ). There would be 5.37 x 10-1ergs/cm2., which would be R in the Stefan-Boltzmann equation. R = sT4. So, T4 = (5.37 x 10-1 ergs/cm2)/(5.669 x 10-5 ergs/cm2K4sec). T = 9.91K. The question is, what would the energy have to be to make T= 2,7K? The value would have to be 2.367 x 10-19 ergsec.. For this to be so, the time of contact of the light ray and the neutrino would have to be 1.78 x 10-8 sec. rather than the 10-10 sec previously estimated. Thus the resultant energy of the spin-flipped neutrino is 2.36 X 10-19 erg, and the “black body” temperature is 2.7K.
TEMPERATURE OF “OUTER SPACE”
Something should be said about the “Temperature of Space”. Most physicists consider outer space to be virtually a vacuum, with no more than the mass equivalent of 104 H atoms/m3.
What does temperature mean in such a system? Generally, the temperature of a gas is proportional to the mean speed of the molecules. If there is a (very) large number of molecules, the square of the mean velocity (ideal gas) is a statistical function of the absolute temperature [16].
According to P. Marmet [12], there may be the mass equivalent of about 104 H atoms/m3 of space. From kinetic theory [17], it is known that Z11 = (2½p/2) (d2ca(n*)2). Here Z11 is the number of collisions/m3/sec, d is the particle diameter or about 3 x 10-10meter, ca is the average velocity of the particle and n* = 104 particles per cubic meter. Ca = (8kT/pm)½. For a hydrogen atom m = 1.67 x 10-27kg, and T=2.7K. Thus ca = 2.367 x 102 m/sec. For a neutrino with m = 3 x 10-31 kg, ca = 1.77 x 104 m/sec. Calculation for H gives Z11 = 4.73 x 10-9 collisions per ses/m3. This extremely small number indicates it would be impossible to measure a “temperature” in such a dilute system.
To have a large number of collisions would require a large number of particles such as is postulated for the neutrino aether. The neutrino has a mass about 1/5500 that of a H atom. Calculation of Z11 for 5500 x 104 neutrinos/m3 gives 7.41 collisions/sec, a much more reasonable number as far as “temperature” is concerned..
SUMMARY
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