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essay: A SEMI-CLASSICAL CALCULATION RE PROTON-RADIUS

A SEMI-CLASSICAL CALCULATION RE PROTON-RADIUS

by Mark Creek-water Dorazio, Amateur-Physics-Enthusiast;  MARK.CREEKWATER@gmail.com

SUMMARY:          Following  DR. ERNEST STERNGLASS [Refs. 1 + 2],  one can visualize the PROTON as composed of four ELECTRON-POSITRON PAIRs [ep-pairs]  and  an UNPAIR’D POSITRON AT THE CENTER … using this model, and Sternglass’s “semi-classical” math approach, one can calculate a THEORETICAL numeric value which agrees close ly with the results of experiments which have been done to determine the proton’s radius;  {Most of these have given a numeric value of approx. 8.7 x 10^(-14) cm for proton-radius [e.g., Ref. 3]} …

ADDITIONALLY,  one can propose that this numeric value is NOT a measure of the entire proton’s radius, but of the radius of one of the four [4] ep-pairs in Sternglass’s proton-model …

TEXT:       Sternglass [Ref. 1] has developed a model which accounts for the origin of PROTONs + NEUTRONs in our universe, and describes their structure …

Using the “PRIMEVAL ATOM” hypothesis of GEORGES LeMAITRE [Refs. 4 + 5] he describes a scenario in which the rotating electromagnetic field of this hypothetical “primeval atom” —(WHOSE EM FIELD INITIALLY CONTAINED ALL THE MASS/ENERGY IN OUR UNIVERSE)— divided in half, and each of the pieces divided in half, and so on and so forth …

AFTER ONLY 270 GENERATIONs of such a process, there would be 2^270 tiny pieces, each with the mass of APPROX. 5 NEUTRONs … At this point, or soon after, there was a “PHASE TRANSITION” [Sternglass’s words, P.11, Ref. 1] in which many zillions of the tiny pieces re-configured, in a way which led to the production of zilions of NEUTRONs, SOME of which quickly “DECAYED” —producing PROTONs … Many zillions of protons + neutrons emerged from this process, during the last “stage” [Sternglass’s word] of the long divide-in-half scenario, with the release of very large amounts of ENERGY, in the form of hi-energy PHOTONs [“gamma rays”];  enough energy to power a “BIG BANG”, if in fact such a “big bang” ever happened …

… THUS STERNGLASS EXPLAINs both the “BIG BANG” and the formation of all the PROTONs + NEUTRONs which now exist …

STERNGLASS’s “TABLE 1”:  Using data from Sternglass’s “Table 1” [p.234, Ref. 1], one can derive a math formula (below) for the radius of each of the many differently-sized “COSMOLOGICAL SYSTEMs” which participate in the divide-in-half scenario:  {Sternglass calls these “cosmological systems”, regardless of their size [p.234, Ref. 1], and says that “For every system, the mass is proportional to the square of the radius” [p.225, Ref. 1]} …

Here is the math formula:  (Rs) = [ 2G / c^2 ] x [ ( Mu x Ms )^(1/2) ] x [ 1 / 137.036 ] … [EQN. #1],    where “Rs” is the radius of the system’s torus-[donut]-shaped, rotating, electromagnetic field, which I will call the “SYSTEM-RADIUS”;  “G” is Newton’s gravitational constant;  “c” is speed-of-light;  “Mu” is mass-of-our-universe;  “Ms” is the mass of the system;  and [1/137.036] is the so called “fine-structure constant”…

Please note that this is a modified “Schwarzschild-formula”, in which the “local gravity” is greatly increased, according to Sternglass’s theory, which he details in his BOOK [Ref.#1]

{ Sternglass says that the tiny systems near the last stage of the long divide-in-half scenario experience a “RELATIVISTIC SHRINKAGE” by a factor of approx. (137.036), which explains the presence of that number in the formula } …

Re the mass of our universe, Sternglass uses PAUL DIRAC’s so called “LARGE NUMBERs HYPOTHESIS” [p.224, Ref. 4; pp.73-76, Ref. 5] to derive an ingenious way to calculate it … More regarding this, below …

COMPTON WAVE-LENGTH AND COMPTON-RADIUS
The physicist ARTHUR COMPTON, during the first half of the 20th century, popularized the idea that there is a so called “COMPTON WAVE-LENGTH” associated with every object which a physicist might want to study;  defined as the wave-length of a PHOTON whose ENERGY CONTENT is equivalent to that of the object, and given by a simple math formula:
(WL-compton) = (h) / ( ),
where h IS Planck’s-constant, c is speed-of-light, and “M” is the object’s mass …
Likewise, the COMPTON-RADIUS is just simply the Compton wave-length divided by [2x(pi)]:                           (Rc) = ( h-BAR ) / ( M ),
where “Rc” is Compton-radius and “h-BAR” is (Planck’s constant) / (2x(pi)) …

In Sternglass’s model there is initially only ONE “cosmological system” — the “PRIMEVAL-ATOM” … Because its mass/energy content so large —(BEING THAT OF OUR ENTIRE UNIVERSE !!)— its  Rc  is ridiculously small:  because smaller photons contain more energy … As the divide-in-half scenario proceeds, the masses of the systems, and the sizes of their em-fields, DECREASE, while their Compton-radii INCREASE:  after 270 divide-in-half generations, there are many trillions of tiny systems, and the SYSTEM-radius  [Rs]  of each tiny system is almost the same as its COMPTON-radius  [Rc]  …

AT SOME POINT, the size of the Compton-radius must equal that of the system-radius — the Rs in EQN. #1 …
One can use easy maths to calculate both the mass (Ms) and radius (Rs) of the system whose system-radius (Rs) equals its compton-radius (Rc):  by definition, one has:
Rc = ( h-BAR ) / ( c x Ms ),   [EQN. #2],
where Rc is Compton-radius and Ms is mass-of-system …
Equating this (Rc) to the (Rs) IN EQN.#1, one has:
(h-BAR) / c x Ms ) = [ 2c^2 ] x [ ( Mu Ms ]^(1/2) ] x [1/137.036];
Solving this equation for (Ms):
(Ms) = { [ (c) x (h-BAR) / 2] x [ (137.036) / (Mu)^(1/2) ] }^(2/3) … [EQN. #3];

MASS-OF-UNIVERSE ??
Sternglass’s formula for the mass of our universe, (neglecting a TYPO in the the book, and using the identity  e^2 = K x Qe x Qe), appears on p.265, Ref. 1, as:

(Mu) = [ [ K x Qe x Qe ]^2 ] / [ G^2 x (Me)^3 ],     where “Me” is the mass of an electron;  “Qe” is the electric-charge of an electron;  “K” is Coulomb’s electrostatic constant;  and “G” is Newton’s gravitational constant …

Re-arrangeing this, to make it more beautiful + “elegant”:   (Mu) = [ { ( K x Qe x Qe ) / ( G x Me x Me  ) }^2 ] x [Me] … [EQN. #4];
[NOTE: this gives a numeric value of (Mu) = approx. 1.581 x 10^58 grams] …

Using this expression for (Mu) In EQN. #3, one has:
 (Ms)  =  { [ x (h-BAR)  / 2G ] x [ (137.036) / (Me)^(1/2) ] x [ ( G x Me x Me ) / ( K x Qe x Qe ) ] }^(2/3);

NOTE:  the “^(2/3)” at the end means that one squares the entire expression and then calculates the cube-root of the result …

Using numeric values,  = 2.9979 x 10^(10) cm/sec,  h-BAR = 1.0546 x 10^(-27) gram.cm.cm/sec,  G = 6.673 x 10^(-8) cm^3 / gram.sec^2,  (Me) = 9.1094 x 10^(-28) gram,  and (K x Qe x Qe) = 2.3071×10^(-19) (gram.cm^3)/sec^2 ;
one calculates that (Ms) = 4.0543 x 10^(-25) gram …
NOTE:  this mass is somewhere between that of two [2] pi-mesons and one [1] pi-meson:  it’s the theoretical mass of the “COSMOLOGICAL SYSTEM” in Sternglass’s model whose SYSTEM-RADIUS [Rs] is equal to the radius of a photon which contains the same amount of energy;   i.e., the system’s so called “COMPTON-RADIUS” [Rc] …
One can now use EQN. #2 to calculate this radius:
(Rc) = (Rs) =  (h-BAR) / ( (c)x(Ms) )  =
[1.0546×10^(-27) gram.cm.cm/sec] / [ (2.9979×10^(10) cm/sec) x (4.0543 x 10^(-25) gram) ] = 8.677 x 10^(-14) cm …
NOTE #1:  this is very close to the MEASURED “radius of the proton”, which experiments have determined, by a variety of methods, to be somewhere in the neighborhood of between approx. 8.42 x 10^(-14) cm AND 8.97 x 10^(-14) cm [Ref. 3]
NOTE #2:  by this method, one calculates the numeric value  [8.677 x 10^(-14) cm] in a “SEMI-CLASSICAL” way, FROM STERNGLASS’s MODEL, using none of the fiendishly difficult maths for which quantum mechanics is famous …
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As for the calculated mass, (Ms) = 4.0543 x 10^(-25) gram: WHAT MIGHT THIS BE ??
Well, in Sternglass’s model, THE PROTON consists of four [4] electron-positron pairs, and an unpaired positron at the center [p.250, Ref. 1] …
Perhaps each of these 4 [four] pairs has a mass of approx. 4.0543 x 10^(-25) gram, and a radius of approx. 8.677 x 10^(-14) cm ??
CONSIDER: 4 x [4.0543 x 10^(-25) gram] = 16.217 x 10^(-25) gram = 1.6217 x 10^(-24) gram, WHICH IS ALMOST THE KNOWN MASS OF THE PROTON !!   Perhaps the positron at the center provides the remaining mass ??
PERHAPs the “PROTON-RADIUS” which experiments determine to be approx. 8.7 x 10^(-14) cm  { CODATA VALUE IS GIVEN AS APPROX. 8.768 x 10^(-14) cm [Ref. 4] }  might actually be measurements of the radius of this particular {[pun intended]} STERNGLASS-COSMOLOGICAL-SYSTEM, whose COMPTON-radius [Rc] just happens to be equal to its SYSTEM-radius [Rs] ??

PERHAPs THIS IS WHY PROTONs ARE SO STABLE ???

REFERENCEs
1) Sternglass, Ernest; 1997, Before the Big Bang, NewYork, Four Walls Eight Windows;
2) Sternglass, Ernest; 1961, “Relativistic Electron-pair Systems and the Structure of Neutral Mesons”, Phys Rev (v.123) pp. 391-398, (July 1, 1961);

3) Antognini, A., + others; 2011, “The Proton Radius Puzzle”, Journal of Physics: Conference Series, (v.312, n.3);
4) Kragh, Helge; 1990, Dirac: a Scientific Biography Cambridge, Cambridge University Press;
5) Dirac, Paul; 1978, Directions in Physics, NewYork, John Wiley & Sons;

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This entry was posted on December 27, 2014 by .
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