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CHAPTER 4: A SEMI-CLASSICAL CALCULATION REGARDING PROTON-RADIUS

CHAPTER 4:  A SEMI-CLASSICAL CALCULATION REGARDING PROTON-RADIUS

by  Mark Creek-water Dorazio, ApE (amateur physics enthusiast);

email:  MARK.CREEKWATER@gmail.com

“It seems obvious to me … that having two incompatible theories of nature is intellectually intolerable”  —–LEONARD SUSSKIND [book: The Blackhole War (2008)]

Following Dr. ERNEST STERNGLASS [Ref.#1],  one can visualize the PROTON as composed of four ELECTRON-POSITRON PAIRs [ep-pairs]  and  an UNPAIRED POSITRON-AT-THE-CENTER … Using this model, and Sternglass’s “semi-classical” math approach, one can calculate a theoretical numeric value which agrees closely with the results of experiments which have been done to determine the proton’s radius … [approx. 8.7 x 10^(-14) cm] …

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

Sternglass 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. #18 + #19] he describes a scenario in which the electromagnetic field of this hypothetical “primeval atom” —(WHOSE EM-FIELD INITIALLY CONTAINED ALL THE MASS/ENERGY IN OUR UNIVERSE !!)— divides in half, and each of the pieces divides in half, and so on and so on …

After only 270 generations of such a process, there would be 2^270 tiny pieces, each with the mass of approximately 5ive [5] neutrons …  Note:  2^(270) is a very very large number:  more than a trillion trillion trillion trillion trillion trillion !!

Shifting to the present tense:  at the end of the divide-in-half scenario (which I call “the  count-down to the Big Bang”), there is a “PHASE TRANSITION” [Sternglass’s words, P.11, Ref.#1] in which many trillions of the tiny pieces of the primeval atom re-configure, in a way which leads to the production of many trillions of neutrons, most of which quickly “decay” —producing protons … Many trillions of protons + neutrons emerge 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” …

Thus Sternglass explains both the “Big Bang” and the formation of all the protons + neutrons which now exist …

NOTE:  regarding the above mentioned “phase transition”:  Sternglass mentions this phrase ({“phase transition”}) several times in his book [Ref.#1]:  he observes that when water freezes and forms ice, this, too, is regarded as a “phase transition”:  because the molecules of water rearrange [i.e., “re-configure”] themselves in a way which enables them to lose energy without getting any colder:  because water and ice co-exist at exactly the same temperature:  32 degrees Fahrenheit, equivalent to zero degrees Centigrade …
{[ Personally, I remember in my high school physics class doing a simple physics experiment, which thousands, perhaps millions, of high school students have also done:  in this experiment, one chills some water, while using a thermometer to measure its temperature, until it freezes, and then continues to chill it:  one observes that the temperature decreases to 32 F, then stops decreasing while water and ice are both present, then continues to decrease after all the water has changed to ice … The point of this simple experiment is to show that, while the water + ice are co-existing together, in contact with each other, their ENERGY-CONTENT continues to decrease, while their TEMPERATURE stays the same:  this is to show that the stuff is losing energy, tho it’s not getting any colder … The energy which it loses is binding energy:  when the molecules of water re-configure and form the bonds which enable them to change from a LIQUID to a SOLID, they “automatically” give up this binding energy ]} …
 
Likewise, the phase transition which happened at the start of the so called “Big Bang”, as the little rascals formed the bonds which enabled them to “create” neutrons + protons, released a very very large amount of “binding energy” —– enough to power a “Big Bang” … 
 
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” [“cosmo.systs”] which participate in the divide-in-half scenario:  { Sternglass calls these objects “cosmological systems”, regardless of their size:  he says that “For every system, the mass is proportional to the square of the radius” [p.225, Ref.#1] } …

Here is the math-formula, derived from Sternglass’s “Table 1”:  

(Rs) = [ 2G / c^2 ] x [ ( Mu Ms )^(1/2) ] x [1/137.036]     [EQN.#1],

where Rs is the radius of the system’s torus-[donut]-shaped 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-call’d “fine-structure constant”…

NOTE1:  ( Mu x Ms )^(1/2)  means  “the square-root of ( Mu x Ms )” …

NOTE2:  this is a modified “Schwarzschild equation” (google it if you need to), in which the “local gravity” inside a system is much stronger than “Newton’s gravity” … More re “local gravity” in CHAPTER 10 …

NOTE3:  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:  inspired by PAUL DIRAC’s so called “large numbers hypothesis” [p.224, Ref.#18; pp.73-76, Ref.#19] Sternglass uses a similar hypothesis to derive an ingenious way to calculate, theoretically, the mass of our universe, and details how he did this in his book [Ref.#1] … More regarding this, below …

COMPTON WAVELENGTH 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 WAVELENGTH” associated with every object which a physicist might want to study;  defined as the wavelength 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 wavelength divided by (2x(pi)):
(Rc) = ( h-BAR ) / ( x 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 is so large —(BEING THAT OF OUR ENTIRE UNIVERSE !!)— its  Rc  is ridiculously small:  because more energetic photons are smaller … As the divide-in-half scenario proceeds, the masses of the systems, and the sizes of their EM-fields, [Rs], DECREASE, while their Compton radii INCREASE: after 270 divide-in-half generations, there are many zillions of tiny systems, and the SYSTEM radius  [Rs]  of each tiny system is almost the same length as its COMPTON radius  [Rc]  …

At some point in the divide-in-half scenario, the size of the Compton radius must equal that of the system radius — the Rs in equation #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]:  This involves solving two math-equations simultaneously:

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];  as above, Mu” is mass of our universe …

MASS OF OUR UNIVERSE ??

Sternglass’s formula for mass of universe appears on p.265 {careful:  there’s a TYPO (typographical error) in the book} in [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 …

Rearranging this, to make it look more 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, which is approx. 100x greater than the mass of our universe which one usually sees in books + papers + essays:  this is “consistent with the evidence that only about one percent of the mass of the universe is in visible form” [p.210, Ref.1]} …

Using this expression for Mu in EQN.#3, one has:
 (Ms)  =  { [ c 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 whole thing, then calculates the cube-root of that …

Using numeric values  c = 2.9979 x 10^(10) cm/sec,  h-BAR = 1.0546 x 10^(-27) gram.(cm/sec).cm,  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 x 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 pi-mesons and one 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., equal to its so called “Compton radius” [Rc] …

One can now use EQN.#2 to calculate this radius:
(Rc) = (Rs) =  (h-BAR) / ( (c)x(Ms) )  =
[1.0546 x 10^(-27) gram.(cm/sec).cm] / [ (2.9979 x 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 …
NOTE #2:  the numeric value  (8.677 x 10^(-14) cm)  was calculated theoretically, in a “semi-classical” way, from Sternglass’s model, using none of the fiendishly difficult maths for which quantum mechanics is famous …

======================================

Regarding 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] ep-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.#20] }  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 ???

$$$$$$$$$$$$$$ << END OF CHAPTER 4 >> $$$$$$$$$$$$$$

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6 comments on “CHAPTER 4: A SEMI-CLASSICAL CALCULATION REGARDING PROTON-RADIUS

  1. Pingback: REGARD-ING DR.STERNGLASS’s PROTON-MODEL | markcreekwater

  2. Pingback: REFERENCEs + APPENDIXs | markcreekwater

  3. Pingback: BOOK-TITLE: ?? WHAT ARE “QUARKs” ?? | markcreekwater

  4. Pingback: INTRODUCTIONs | markcreekwater

  5. Pingback: BOOK-TITLE: HOW PROTONs WORK: ESSAYS RE THE WORK OF DR. ERNEST STERNGLASS + DR. MENAHEM SIMHONY | markcreekwater

  6. Pingback: essay: PRIMER-FIELDs; + some STERNGLASS-GOLD | markcreekwater

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