**CHAPTER 4: A SEMI-CLASSICAL CALCULATION RE PROTON-RADIUS**

by Mark Creek-water Dorazio, amateur physics enthusiast

mark.creekwater@gmail.com

**“It seems obvious to me … that having two incompatible theories of nature is intellectually intolerable” Leonard Susskind [Ref.#35]**

Following Dr. Ernest Sternglass **[Ref.#1],** one can visualize the proton as being composed of four electron-positron pairs [ep-pairs] with an unpaired positron at the center. Using this model, and Sternglass’s “semi-classical” math approach, one can calculate a numeric value which agrees closely with the results of experiments which have been done to determine the proton’s radius. Most of these have found a value of approximately 8.7*10^(-14) cm for the proton’s radius **[e.g., Ref.#20]**. Plus, one can propose that this is **not **the radius of the entire proton, but the radius of one of the four ep-pairs which compose the proton in Sternglass’s model.

Sternglass has developed a model which accounts for the origin of protons and neutrons in our universe, and describes their structure. Starting with the “primeval atom” hypothesis of Georges Lemaitre **[Refs.#18, #19, #47] **he describes a scenario in which the rotating electromagnetic field of this hypothetical “primeval atom” —(which initially contained all the mass and 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 this divide-in-half process, there would be 2^270 tiny pieces, each with the mass of approx. 5 neutrons.

{ Note: 2^270 is a very large number: more than a trillion trillion trillion trillion trillion !! }

At this point, or soon thereafter, there was a **“phase transition” [**Sternglass’s words, p.11, Ref.#1] in which many zillions of the tiny pieces of the primeval atom re-configured, in a way which led to the production (one wants to say “creation”) of many zillions of protons and neutrons, 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 high-energy gamma rays, enough to power a “big bang.”

Sternglass compares this phase transition to the phase transition which happens when water freezes, a process which releases energy (binding energy), the same amount needed to melt the resulting ice. Likewise the phase transition which happened at the start of the Big Bang released a very very large amount of energy, as zillions of sub-atomic-sized objects re-configured, forming neutrons, most of which quickly “decayed” — producing protons.

Thus the model explains both the Big Bang and the formation of all the protons and 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” [cosmo.systs] which participate in the divide-in-half scenario. Sternglass calls these objects “cosmological systems” regardless of their size, saying that **“For every system, the mass is proportional to the square of the radius” [p.225, Ref.#1].**

Here is the math-formula: ** Rs =**** [ ****2*G**** /**** c^2 ****]*[ ( Mu*Ms )^(1/2) ]*[ 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; [1/137.036] is the so-called “fine-structure constant”; and “^(1/2)” means that one calculates the square root of [*Mu*Ms*]. This is a modified “Schwarzschild formula” — in which the “local gravity” associated with the system is larger than Newton’s gravity, according to Sternglass’s theory {More details in Chapter 10}. 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. * *

{Note: after I “discovered” Simhony’s model **[Ref.#2] **I realized that the reason **why** the little rascals shrink is because they **“want” **to go inside an epola-cell. This coincides with Sternglass’s idea that the cosmo.syst at the place in Table 1 which he calls “stage 27” is very special. With a mass of approx. 8.33*10^(-24) gram, (almost exactly that of five protons), it’s mid-way between the mass of a **J/psi meson** and that of an** Upsilon-meson, **discovered during 1974 and 1977 respectively, as Sternglass details in Ref.#1. Please google these mesons if you want or need to}

Table 1 gives the radius of the cosmo.syst at “stage 27” as approx. 5.4*10^(-11) cm: if one shrinks this by a factor of 137.036 [the inverse of the fine-structure constant] then the radius becomes approx. 4*10^(-13) cm, just right to fit inside a single epola-cell.

Inspired by Paul Dirac’s so-called “large numbers hypothesis” **[p.224, Ref.#18, pp.73-76, Ref.#19], **Sternglass derived an ingenious way to calculate the mass of our universe, theoretically. 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 / c*M, (Eqn.2), **

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 2*(pi):

**Rc = ****h-bar / c*M (Eqn.2a), **

where “*Rc*” is Compton radius and “h-bar” is Planck’s constant divided by 2*(pi).

Note that, in general, objects with greater energy-content have smaller Compton radii, because it’s not a measure of the object’s actual size, but a measure of the size of a photon with an equal energy-content. In this essay we look at an object whose actual size is equal to that of its Compton radius.

In Sternglass’s model there is initially only **one **cosmo.syst, the primeval atom. Because its mass/energy content is 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, {which I call “the count-down to the Big Bang”}, the masses of the systems, and the sizes of their electromagnetic fields, decreases, while their Compton radii increase: after 270 divide-in-half generations, there are many trillions of tiny systems, each with a **Compton-radius [ Rc]** almost equal to its

At some point during this “count-down” 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*Ms (Eqn.2b), **

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*Ms**

Solving for *Ms *leads to:

* Ms = *{ [ c*(h-bar) / 2**G *]*[137.036 / *Mu^*(1/2) ] }^(2/3) (Eqn.3);

**MASS OF OUR UNIVERSE ??**

Inspired by Paul Dirac’s “large numbers hypothesis,” Sternglass derived an elegant way to calculate the mass of our universe. Using the identity *e***^2 = K*Qe*Qe, **and neglecting a typo in the book, the formula for this appears on p.265, Ref.#1, as:

**Mu = **** [**** K*Qe*Qe**** ]^2**** /**** [ (G^2)*****Me^3**** ****] (Eqn.4), **

where “*K*” is Coulomb’s famous electrostatic constant; “*Qe*” is the electric charge of an electron; “*G*” is Newton’s famous gravitational constant; and “*Me*” is the gravitational mass of an electron.

Re-arranging this, to make it more beautiful and “elegant”:

**Mu = ****{ [ K*Qe*Qe / G*Me*Me ]^2 }*Me (Eqn.4a);**

{Note: this expression gives a numeric value of approx. 1.581*10^58 grams, which is approx. 100x greater than the mass of our universe which one usually sees in books and essays which address this subject. As Sternglass says: this is **“consistent with the evidence that only about one percent of the mass of the universe is in a visible form” [p.210, Ref.#1]**}

Inserting this expression for *Mu *into Eqn.3, one obtains:

**Ms = ****{ [ c*(h-bar) / 2*G ]*[ 137.036 / Me^(1/2) ]*[ G*Me*Me / K*Qe*Qe ] }^(2/3);**

Note: the “^(1/2)” at the end means that one squares the entire expression and then calculates the cube root of the result.

Using numeric values: *c = *2.9979*10^10 cm/sec, h-bar = 1.0546*10^(-27) gram.cm.cm/sec, *G = *6.674*10^(-8) cm.cm.cm/gram.sec.sec, *Me =* 9.1094*10^(-28) gram, and *K*Qe*Qe *= 2.3071*10^(-19) gram.cm.cm.cm/sec.sec; one calculates that **Ms = ****4.0546*10^(-25) gram.**

Note: this mass is between that of two pi-mesons and that 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., the system’s so-called Compton-radius [*Rc*].

One can now use Eqn.2b to calculate this radius: **Rc = Rs = ****(h-bar) / c*Ms = **

**[1.0546*10^(-27) gram.cm.cm/sec] / [(2.9979*10^10 cm/sec)*(4.0546*10^(-25) gram) = **

**8.676*10^(-14) cm.**

Note #1: This is very close to the **measured **“radius of the proton” — which many teams of experimental physicists have determined, by a variety of methods, to be somewhere in the neighborhood of between approx. 8.42 and 8.97 x 10^(-14) cm **[Ref.#20].**

Note #2: By this method one calculates the numeric value [8.676*10^(-14) cm] in a “semi-classical” way, from Sternglass’s model, with no reference to any of the fiendishly difficult maths for which quantum field theory is famous.

**### ### ### ### ###**

As for the calculated mass, *Ms = *4.0546*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. 4.0542*10^(-25) gram, and a radius of approx. 8.677*10^(-14) cm ??

Consider: 4 x 4.0546*10^(-25) gram = 16.218*10^(-25) gram = 1.6218*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*10^(-14) cm —{CODATA value is given as approx. 8.768*10^(-14) cm **[Ref.#20]**}— might be actually 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 ??

**CONCLUSION**

Using only Sternglass’s model, and some of his numbers, one can solve two easy algebraic equations [Eqns. #1 and #2b, above] simultaneously, to calculate theoretical values for the mass and radius of one of the four electron-positron pairs which constitute most of the proton’s mass in Sternglass’s proton model. The calculated numeric value for radius given by this method is very near to that of the “proton radius” —– as measured by many teams of experimental physicists during the past 50 years. One suspects the possibility that the experimenters might have been measuring the radius of one of the four ep-pairs assumed by Sternglass to be inside protons, instead of that of the entire proton.

**Note that this numeric value [8.676*10^(-14) cm] is calculated with no reference to any of the fiendishly difficult mathematics for which quantum mechanics is famous.**

Plus, the calculated mass obtained by this method is slightly less than one fourth [1/4] of the known mass of the proton, which supports Sternglass’s idea that there are four electron-positron pairs in a proton (or neutron), and that these four ep-pairs carry most of the mass of a proton or a neutron.

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

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