<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">JAMP</journal-id><journal-title-group><journal-title>Journal of Applied Mathematics and Physics</journal-title></journal-title-group><issn pub-type="epub">2327-4352</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/jamp.2022.109187</article-id><article-id pub-id-type="publisher-id">JAMP-120214</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  Rotating Lepton Model of Pions and Kaons: Mechanics at fm Distances
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Constantinos</surname><given-names>G. Vayenas</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Dionysios</surname><given-names>Tsousis</given-names></name><xref ref-type="aff" rid="aff2"><sup>2</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Dimitrios</surname><given-names>Grigoriou</given-names></name><xref ref-type="aff" rid="aff3"><sup>3</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Konstantinos</surname><given-names>Parisis</given-names></name><xref ref-type="aff" rid="aff4"><sup>4</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Elias</surname><given-names>C. Aifantis</given-names></name><xref ref-type="aff" rid="aff5"><sup>5</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib></contrib-group><aff id="aff5"><addr-line>Michigan Technological Institute, Houghton, USA</addr-line></aff><aff id="aff2"><addr-line>Stanford University, Stanford, USA</addr-line></aff><aff id="aff1"><addr-line>Academy of Athens, Athens, Greece</addr-line></aff><aff id="aff4"><addr-line>Aristotle University of Thessaloniki, Thessaloniki, Greece</addr-line></aff><aff id="aff3"><addr-line>University of Patras, Patras, Greece</addr-line></aff><pub-date pub-type="epub"><day>02</day><month>09</month><year>2022</year></pub-date><volume>10</volume><issue>09</issue><fpage>2805</fpage><lpage>2819</lpage><history><date date-type="received"><day>2,</day>	<month>July</month>	<year>2022</year></date><date date-type="rev-recd"><day>27,</day>	<month>September</month>	<year>2022</year>	</date><date date-type="accepted"><day>30,</day>	<month>September</month>	<year>2022</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  The present article is a continuation of a recently published paper 
  [1] in which we have modeled the composition and structure of neutrons and other hadrons using the Rotating Lepton Model (RLM) which is a Bohr type model employing the relativistic gravitational attraction between three ultrafast rotating neutrinos as the centripetal force. The RLM accounts for special relativity and also for the De Broglie equation of quantum mechanics. In this way this force was shown to reach the value of the Strong Force while the values of the masses of the rotating relativistic neutrinos reach those of quarks. Masses computed for twelve hadrons and bosons are in very close (~2%) agreement with the experimental values. Here we use the same RLM approach to describe the composition and structure and to compute the masses of Pions and Kaons which are important zero spin mesons. Contrary to hadrons and bosons which have been found via the RLM to comprise the heaviest neutrino eigenmass 
  <em>m</em>
  <sub>3</sub>, in the case of mesons the intermediate neutrino mass eigenstate 
  <em>m</em>
  <sub>2</sub> is found to play the dominant role. This can explain why the lowest masses of mesons are generally smaller than those of hadrons and bosons. Thus in the case of Pions it is found that they comprise three rotating 
  <em>m</em>
  <sub>2</sub> mass eigenstate neutrinos and the computed mass of 136.6 MeV/c
  <sup>2</sup> is in good agreement with the experimental value of 134.977 MeV/c
  <sup>2</sup>. The Kaon structure is found to consist of six 
  <em>m</em>
  <sub>2</sub> mass eigenstate neutrinos arranged in two parallel pion-type rotating triads. The computed Kaon mass differs less that 2% from the experimental K
  <sup>&#177;</sup> and K
  &amp;#176; values of 493.677 MeV/c
  <sup>2</sup> and 497.648 MeV/c
  <sup>2</sup> respectively. This, in conjunction with the experimentally observed decay products of the Kaons, provides strong support for the proposed K structure.
 
</p></abstract><kwd-group><kwd>Pions and Kaons-Structure and Masses</kwd><kwd> Gravitational Bohr-de Broglie-Newton-Einstein Type Models</kwd><kwd> Rotating Lepton Model (RLM)</kwd><kwd> Hadronization</kwd><kwd> Neutrino Masses</kwd><kwd> Special Relativity</kwd><kwd> Gravitational Force</kwd><kwd> Quantum Mechanics</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>The Standard Model (SM) of particle physics has long provided a basis for understanding the fundamental structure of all observable matter in our Universe. Among the indivisable particles it describes are quarks and leptons, which include electrons, positrons and neutrinos. The SM also describes four forces: gravity which plays a very limited role, electromagnetism which regulates interactions between charged particles via photons, the Strong Force which acts between quarks via gluons and the Weak Force which involves exchanges of W and Z boson and plays an important role in radioactive decay. So far the SM has provided an excellent basis for researchers to explain their experimental results [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>]. In recent years, however, there is increasing evidence that the SM in its current form is not complete. In fact, the SM does not seem to be compatible with special or general relativity and, in addition, it assumes that neutrinos are massless, despite of subsequently established experimental evidence that neutrinos possess mass and play a paramount dominant role in our Universe [<xref ref-type="bibr" rid="scirp.120214-ref4">4</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref5">5</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref6">6</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref7">7</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref8">8</xref>]. This lack of completeness has led to developing the rotating lepton model (RLM) which is a Bohr-type rotating lepton model [<xref ref-type="bibr" rid="scirp.120214-ref9">9</xref>], combining gravity, special [<xref ref-type="bibr" rid="scirp.120214-ref10">10</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref11">11</xref>] or general relativity [<xref ref-type="bibr" rid="scirp.120214-ref12">12</xref>] and quantum mechanics [<xref ref-type="bibr" rid="scirp.120214-ref13">13</xref>] in a simple manner [<xref ref-type="bibr" rid="scirp.120214-ref1">1</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref14">14</xref>] - [<xref ref-type="bibr" rid="scirp.120214-ref20">20</xref>].</p><p>Following the general observation [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>] that all composite particles eventually decay to only up to five lepton types (electrons e<sup>−</sup>, positrons e<sup>+</sup> and the three neutrinos) the RLM considers these particles as the only truly undivisable, thus fundamental, elementary particles which can synthesize all composite particles [<xref ref-type="bibr" rid="scirp.120214-ref9">9</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref20">20</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref12">12</xref>] - [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>]. The RLM also utilizes the three neutrino eigenmass values to show that, due to their very small rest masses m<sub>o</sub>, neutrino masses can reach the Planck mass values at modest (up to 313 MeV) energies.</p><p>The latter implies that gravitational forces between ultrarelativistic neutrinos, at a distance d, can easily reach the value, ℏ c / d 2 , of the Strong Nuclear Force which is the strongest force for creating composite particles. Here ℏ is the Planck constant and c is the speed of light.</p><p>Consequently the RLM utilizes only two forces, i.e. gravity and electromagnetism and shows that matter is created via the rotation of neutrino triads in circular orbits with rotational speeds near the speed of light c, and corresponding Lorentz factor γ ( = ( 1 − v 2 / c 2 ) − 1 / 2 ) values up to 10<sup>10</sup>. In this way it turns out that the new mass created is ( γ − 1 ) m o c 2 . As an example, as shown in the next section, three rotating neutrinos, of rest mass 0.0437 eV/c<sup>2</sup> each, form a rotating triad with a Lorentz factor γ equal to 7.163 &#215; 10<sup>9</sup>. The mass of the composite particle formed is 3 γ m o , i.e. 939.565 MeV/c<sup>2</sup> which is the rest mass of a neutron. The importance, simplicity and effectiveness of the RLM has been analyzed and discussed recently in Research Features [<xref ref-type="bibr" rid="scirp.120214-ref20">20</xref>].</p></sec><sec id="s2"><title>2. Rotating Lepton Model of the Neutron, Muon and Pion</title><sec id="s2_1"><title>2.1. Neutron</title><p>Within the RLM approach the neutron is modeled as a rotating relativistic neutrino triad of the heaviest neutrino mass eigenstate m<sub>3</sub> (<xref ref-type="fig" rid="fig1">Figure 1</xref> and <xref ref-type="fig" rid="fig2">Figure 2</xref>). According to Special Relativity the relativistic mass, m<sub>r</sub>, of a neutrino is given by γ m o and its inertial longitudinal mass is given by</p><p>m i = γ 3 m o (1)</p><p>where m<sub>o</sub> is the rest mass of the neutrino and γ is the Lorentz factor</p><p>γ = ( 1 − v 2 / c 2 ) − 1 / 2 (2)</p><p>In instantaneous reference frames, the above Equation (1), derived initially for linear motions [<xref ref-type="bibr" rid="scirp.120214-ref10">10</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref11">11</xref>], remains valid for arbitrary motion [<xref ref-type="bibr" rid="scirp.120214-ref11">11</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref14">14</xref>]. Therefore, using the equivalence principle, the gravitational mass, m<sub>g</sub>, of all particles equals their inertial mass, m<sub>i</sub>, thus</p><p>m g = γ 3 m o (3)</p><p>Using the definition of the gravitational mass, m<sub>g</sub>, which is the mass value entering Newton’s universal gravitational law, i.e.</p><p>m g 2 = F d 2 G (4)</p><p>We obtain the following expression for the gravitational force F</p><p>F = G m o 2 γ 6 d 2 (5)</p><p>where G (=6.673 &#215; 10<sup>−</sup><sup>11</sup> m<sup>3</sup>∙kg<sup>−</sup><sup>1</sup>∙s<sup>−</sup><sup>2</sup>) denotes the usual gravitational constant. For circular motion of three m<sub>3</sub> mass neutrinos rotating along a circle of radius r, it follows that F is given by</p><p>F = G m 3 2 γ 6 3 1 / 2 r 2 (6)</p><p>and therefore the equation of motion of each rotating particle is</p><p>γ m 3 v 2 / r = G m 3 2 γ 6 3 1 / 2 r 2 (7)</p><p>which in turn, yields</p><p>r = G m 3 3 1 / 2 c 2 γ 5 ( γ 2 γ 2 − 1 ) (8)</p><p>Solving Equation (8) coupled with the de Broglie quantum mechanics equation, i.e. with</p><p>γ m o v r = ℏ (9)</p><p>accounting for the number (three) of quarks in a neutron [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] and denoting by m n the neutron mass, it follows from energy conservation (i.e. from m n c 2 = 3 γ m o c 2 ) that m n = 3 γ m o and thus</p><p>r = 3 ℏ / m n c ;   m n = 3 13 / 12 ( m P l m o 2 ) 1 / 3 ;   m o = m 3 = ( m n / 3 ) 3 / 2 3 1 / 8 m P l 1 / 2 (10)</p><p>Substituting for the neutron mass m n = 939.565   MeV / c 2 and for the Planck mass m P l = 1.221 &#215; 10 28 eV / c 2 , one obtains m o = 0.0437   eV / c 2 and thus, γ ( = m n / 3 m o ) = 7.163 &#215; 10 9 . The so computed relativistic mass γ m o value is of the order of quark masses (313 MeV/c<sup>2</sup>) and the corresponding rest mass m<sub>o</sub> value (0.04372 eV/c<sup>2</sup>) is in surprisingly good agreement with the heaviest neutrino mass m<sub>3</sub>, as shown in <xref ref-type="fig" rid="fig1">Figure 1</xref>. Since this mass differs less than 2% from the mass, m<sub>3</sub>, of the neutrino produced in the W<sup>&#177;</sup> decay [<xref ref-type="bibr" rid="scirp.120214-ref18">18</xref>] it follows that the W<sup>&#177;</sup> boson comprises, similarly to neutrons and protons, the heaviest mass neutrinos ν 3 (<xref ref-type="fig" rid="fig1">Figure 1</xref>).</p></sec><sec id="s2_2"><title>2.2. Muons</title><p>The algebraic expressions for the masses m<sub>1</sub>, m<sub>2</sub> and m<sub>3</sub>, also shown and compared in <xref ref-type="fig" rid="fig1">Figure 1</xref> with the Superkamiokande measurements, are obtained by modeling the muon structure ( μ &#177; ) which is known to comprise an e<sup>&#177;</sup>, a ν &#175; e and a ν μ [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>]. Thus one considers two gravitating neutrinos, ν 1 and ν 3 , of masses m<sub>1</sub> and m<sub>3</sub>, respectively and, similarly to Equation (7), we have</p><p>γ 1 m 1 v 1 2 / r = G m 1 m 3 γ 1 3 γ 3 3 3 r 2 (11)</p><p>γ 3 m 3 v 3 2 / r = G m 1 m 3 γ 1 3 γ 3 3 3 r 2 (12)</p><p>Upon multiplying by parts, taking the square root, considering the limit v 1 , v 3 → c</p><p>and defining γ 13 = ( γ 1 γ 3 ) 1 / 2 and m 13 = ( m 1 m 3 ) 1 / 2 one obtains</p><p>γ 13 m 13 c 2 / r = G m 13 2 γ 13 6 / 4 r 2 (13)</p><p>Furthermore, by multiplying by parts the two de Broglie wavelength expressions as in Equation (9), we obtain</p><p>γ 13 m 13 c r = ℏ (14)</p><p>From Equations (13) and (14) it follows</p><p>γ 13 6 = 4 ℏ c G m 13 2 ;   γ 13 = 2 1 / 3 ( m P l m 13 ) 1 / 3 (15)</p><p>where m P l = ( ℏ c / G ) 1 / 2 . Consequently, the muon mass is computed from</p><p>m μ = γ 13 m 13 = 2 1 / 3 ( m P l m 13 2 ) 1 / 3 = ( 2 m P l m 1 m 3 ) 1 / 3 (16)</p><p>and, therefore, using the experimental muon mass, m μ = 105.66   MeV / c 2 and the m<sub>3</sub> mass eigenstate value of 0.0437 eV/c<sup>2</sup> [<xref ref-type="bibr" rid="scirp.120214-ref14">14</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref15">15</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref18">18</xref>], we obtain</p><p>m 1 = m μ 3 2 m P l m 3 = 0.001105   eV / c 2 (17)</p><p>Interestingly the same result for the muon mass obtained from Equation (16) can also be reached by considering two rotating neutrinos, each with rest mass m<sub>2</sub>. In this case, similarly to Equation (7), we have</p><p>γ 2 m 2 v 2 / r = G m 2 2 γ 2 6 4 r 2 ;   γ 2 m 2 v r = ℏ (18)</p><p>resulting to</p><p>4 ℏ c = G m 2 2 γ 2 6 ;   γ 2 = 2 1 / 3 ( ℏ c G m 2 2 ) 1 / 6 (19)</p><p>and, therefore, to</p><p>m μ = γ 2 m 2 = ( 2 m P l m 2 2 ) 1 / 3 = 105.66   MeV / c 2 (20)</p><p>Then, from Equations (16) and (20) it follows that</p><p>m 2 = ( m 1 m 3 ) 1 / 2 = 0.00695   eV / c 2 (21)</p><p>The above equation suggests that two neutrinos, of masses m<sub>1</sub> and m<sub>3</sub> each, can hybridize to form two neutrinos with equal masses m 2 = ( m 3 m 1 ) 1 / 2 . The occurence of neutrino hybridization can be attributed to the need of synchronization when two neutrinos of different initial masses are caught on the same circular orbit in the process of forming a bound rotational state [<xref ref-type="bibr" rid="scirp.120214-ref14">14</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref18">18</xref>]. The phenomenon of hybridization is quite common in chemistry [<xref ref-type="bibr" rid="scirp.120214-ref21">21</xref>]. The phenomenon of neutrino hybridization may be related to the very important phenomenon of neutrino oscillations [<xref ref-type="bibr" rid="scirp.120214-ref4">4</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref5">5</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref6">6</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref7">7</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref8">8</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref22">22</xref>].</p></sec><sec id="s2_3"><title>2.3. Pions</title><p>Interestingly, the same m<sub>1</sub> mass value obtain in Equation (17) can be computed by modeling the structure of the pion, which is a meson, comprising (<xref ref-type="fig" rid="fig2">Figure 2</xref>) three rotating neutrinos [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>] which, as shown by the RLM, form a rotating triad. The analytical computation of the pion mass is shown below.</p><p>We consider three rotating neutrinos on a circle of radius r, two of which have the mass m<sub>1</sub> and the third with a mass m<sub>3</sub> as shown in <xref ref-type="fig" rid="fig2">Figure 2</xref>(c) and <xref ref-type="fig" rid="fig3">Figure 3</xref>.</p><p>The gravitational forces between the three rotating particles are shown in <xref ref-type="fig" rid="fig3">Figure 3</xref>, i.e.</p><p>F 11 = G m 1 2 γ 1 6 4 r 2 sin 2 φ ;   F 13 = F 23 = G m 1 m 3 γ 1 3 γ 3 3 r 2 sin 2 φ sin 2 ( φ / 2 ) (22)</p><p>and thus</p><p>F 11 F 13 = m 1 γ 1 3 4 m 3 γ 3 3 sin 2 ( φ / 2 ) (23)</p><p>From <xref ref-type="fig" rid="fig3">Figure 3</xref> it also follows</p><p>α = 2 r sin φ (24)</p><p>α sin φ = β sin ( 90 − φ / 2 ) (25)</p><p>Thus,</p><p>β = α cos ( φ / 2 ) sin φ = α 2 sin ( φ / 2 ) (26)</p><p>and from Equation (24)</p><p>β = r sin φ sin ( φ / 2 ) (27)</p><p>Then, the tangential force at B in obtained from</p><p>F 11 ⋅ sin ( 90 − φ ) = F 13 sin ( φ / 2 ) (28)</p><p>or</p><p>F 11 ⋅ cos φ = F 13 sin ( φ / 2 ) (29)</p><p>and, therefore,</p><p>F 11 F 13 = sin ( φ / 2 ) cos φ = m 1 γ 1 3 4 m 3 γ 3 3 sin 2 ( φ / 2 ) (30)</p><p>giving</p><p>4 sin 3 ( φ / 2 ) cos φ = m 1 γ 1 3 m 3 γ 3 3 (31)</p><p>In view of the de Broglie condition</p><p>γ 1 m 1 c r = γ 3 m 3 c r = ℏ (32)</p><p>we have</p><p>γ 1 m 1 = γ 3 m 3 (33)</p><p>and since</p><p>m 3 / m 1 = 0.0437 1.174 &#215; 10 − 3 = 37.22 (34)</p><p>it follows that</p><p>4 sin 3 ( φ / 2 ) cos φ = 37.22 (35)</p><p>which, by trial and error, gives</p><p>φ = 89.95 ∘ (36)</p><p>Considering the equations of motion of the three particles and assuming v 1 ≈ c , v 3 ≈ c , it follows that</p><p>ℏ c = γ 3 m 3 c 2 r = 2 G m 1 m 3 γ 1 3 γ 3 3 4 cos ( φ / 2 ) (37)</p><p>and also</p><p>( ℏ c ) 2 = ( γ 1 m 1 c 2 r ) 2 = [ G m 1 2 γ 1 6 4 sin φ + G m 1 m 3 γ 1 3 γ 3 3 4 cos ( φ / 2 ) ] 2 (38)</p><p>After multiplying by parts, accounting for the fact that</p><p>γ 1 m 1 = γ 3 m 3 = ℏ r c (39)</p><p>and defining</p><p>x = m 1 m 3 ;   y = γ 1 γ 3 (40)</p><p>we obtain</p><p>γ 1 2 γ 3 3 m 1 2 m 3 r 3 c 6 = ( ℏ c ) 3 G 3 m 3 6 γ 3 18 = ( m P l m 3 ) 6 γ 3 − 18 = 2 4 3 [ x 5 y 15 sin 2 φ cos ( φ / 2 ) + 2 x 4 y 12 cos 2 ( φ / 2 ) sin φ + x 3 y 9 cos 3 ( φ / 2 ) ] (41)</p><p>Accounting for the condition x y = 1 , it follows from Equation (41) that</p><p>[ ℏ c G m 3 2 γ 3 6 ] 3 = 1 32 [ y 10 sin 2 φ cos ( φ / 2 ) + 2 y 8 cos 2 ( φ / 2 ) sin φ + y 6 cos 3 ( φ / 2 ) ] (42)</p><p>and after substitution for φ = 89.95 ∘ , we obtain</p><p>[ ℏ c G m 3 2 γ 3 6 ] 3 = 2.37 &#215; 10 14 ;   ℏ c G m 3 2 γ 3 6 = 6.19 &#215; 10 4 (43)</p><p>Thus, it turns out that</p><p>γ 3 = [ ( m P l m 3 ) 2 1 6.19 &#215; 10 4 ] 1 / 6 = 0.159 ( m P l m 3 ) 1 / 3 (44)</p><p>and, therefore,</p><p>m π = 3 γ 3 m 3 = 3 &#215; 0.159 ( m P l m 3 2 ) 1 / 3 = 0.159 m n (45)</p><p>Consequently, it was computed that</p><p>m π = 136.6   MeV / c 2 (46)</p><p>in good agreement with m π ∘ = 134.977   MeV / c 2 which is the experimental value [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>] and with the mass value of 137.82 MeV/c<sup>2</sup> computed via the use of equilateral triangular geometry and hybridization between m<sub>1</sub> and m<sub>3</sub> neutrinos leading to m<sub>2</sub> neutrinos [<xref ref-type="bibr" rid="scirp.120214-ref16">16</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>].</p></sec></sec><sec id="s3"><title>3. Kaon Decay Products</title><p>In this section we use the RLM to investigate the structure and mass of the Kaons.</p><p>Kaon is a meson, that comes in two types: The charged Kaon (K<sup>+</sup> or K<sup>−</sup>), which has a mass of 493.677 MeV/c<sup>2</sup>, and the neutral Kaon (K<sup>o</sup>), which has a mass of 497.648 MeV/c<sup>2</sup>.</p><p>Out of these two types, only the charged one is known to decay into a charged pion and a neutral pion. The charge of the pion depends on the charge of the Kaon that decays.</p><p>Therefore, for the charged Kaon structure, one can consider six of these hybridized neutrinos (two trios, one staggered onto the other), all rotating around a common axis, as shown in <xref ref-type="fig" rid="fig4">Figure 4</xref> and <xref ref-type="fig" rid="fig5">Figure 5</xref>. Moreover, for the charged</p><p>Kaon (K<sup>+</sup> or K<sup>−</sup>) there is a small charged particle (e<sup>+</sup> or e<sup>−</sup>) at the center of this configuration, not shown in <xref ref-type="fig" rid="fig4">Figure 4</xref>, which dictates the Kaon charge. This e<sup>&#177;</sup> particle does not rotate and, thus, it contributes little by the electron or positron rest mass of (~0.511 MeV/c<sup>2</sup>) to the Kaon mass.</p><p>The decay products of the K<sup>o</sup> and K &#175; o (mass 497.65 MeV/c<sup>2</sup>) are π π ( K o o = 8.95 &#215; 10 − 11     s ) and π e ν e , π μ ν μ and π π π ( K i o = 5.11 &#215; 10 − 8     s ) [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>]. These decay products, i.e. the muon and the pion, have recently been shown [<xref ref-type="bibr" rid="scirp.120214-ref16">16</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>] to contain exclusively the above intermediate mass neutrinos (i.e. with mass m<sub>2</sub> (≈6.95 &#215; 10<sup>−</sup><sup>3</sup> eV/c<sup>2</sup>) in the normal hierarchy) [<xref ref-type="bibr" rid="scirp.120214-ref14">14</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref15">15</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref16">16</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>].</p><p>From the principal hadronic decay ( π + π o ) of the K<sup>+</sup>, since the pion comprises three hybridized neutrinos [<xref ref-type="bibr" rid="scirp.120214-ref16">16</xref>], one may conclude that the Kaons comprise six ν μ e hybridized neutrinos. This is also confirmed by the leptonic Kaon decay [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>]. According to the RLM [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>], it is possible to make an estimate of the order of magnitude of the K mass via the expression</p><p>m K ≈ 6 ( m P l m 2 2 ) 1 / 3 (47)</p><p>Using m P l = 1.221 &#215; 10 28 eV / c 2 and Equation (21) one obtains m K ≈ 503.3   MeV / c 2 and this value without any detailed modeling differs already, only approximately 1% and 1.5% from the experimental values [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>] of 497.65 MeV/c<sup>2</sup> and 493.57 MeV/c<sup>2</sup>, respectively, of the neutral Kaon K<sup>o</sup> and of the charged Kaon K<sup>+</sup>.</p></sec><sec id="s4"><title>4. RLM Model for the Kaon Structure and Mass Computation</title><p>The simplest geometric model for accommodating six rotating neutrinos placed on two equilateral triangles is a normal triangular octahedron (<xref ref-type="fig" rid="fig4">Figure 4</xref> and <xref ref-type="fig" rid="fig5">Figure 5</xref>). We therefore hypothesize that the K structure comprises six neutrinos of mass m<sub>2</sub> arranged at the six vertices of a rotating normal triangular octahedron (<xref ref-type="fig" rid="fig4">Figure 4</xref> and <xref ref-type="fig" rid="fig5">Figure 5</xref>). This structure corresponds to one of the five platonic solids and belongs to the O<sub>h</sub> symmetry point group [<xref ref-type="bibr" rid="scirp.120214-ref21">21</xref>]. Therefore it comprises three C4 rotational axes (<xref ref-type="fig" rid="fig4">Figure 4</xref>, left) and four C3 axes (<xref ref-type="fig" rid="fig4">Figure 4</xref>, right).</p><p>Consequently, any of the six neutrinos can be rotating on a total of seven different axes. The energy equipartition theorem suggests that all these axial rotations contribute equally to the energy of each Kaon particle.</p><p>The total kinetic energy of the six neutrinos with rest mass m<sub>2</sub> each, corresponds to the rest energy, E, of the composite particle and is given by</p><p>E = 6 γ &#175; m 2 c 2 (48)</p><p>where γ &#175; denotes a mean Lorentz factor value accounting for the rotation of all six neutrinos.</p><p>Due to the Kaon particle symmetry, the Lorentz factor γ of each rotating particle can be computed by considering the projection of each force on the seven planes of rotation (<xref ref-type="fig" rid="fig5">Figure 5</xref>).</p><p>Consequently, in order to find first the speeds v 4 and v 3 as well as the corresponding Lorentz factor values γ 3 and γ 4 , we compute, from <xref ref-type="fig" rid="fig5">Figure 5</xref>(a),</p><p>F 3 = G m 2 2 r 2 [ 1 3 + 1 3 3 + 1 3 6 ] (49)</p><p>where F<sub>3</sub> is the centripetal force exerted on each of the rotating particles in the C4 mode by all the other five particles. Then, similarly as in Equation (52) below, we obtain</p><p>γ 3 m 2 c 2 r = G m 2 2 γ 3 6 r 2 [ 1 3 3 + 1 3 6 + 1 3 ] (50)</p><p>We also have from <xref ref-type="fig" rid="fig5">Figure 5</xref>(b) that</p><p>F 4 = G m 2 2 r 2 [ 2 2 3 + 1 6 ] (51)</p><p>where F<sub>4</sub> is the centripetal force exerted on each of rotating particles in the 3 mode by all the other five particles.</p><p>Accounting for the fact that F = γ m v 2 / r ≈ γ m c 2 / r we obtain for each particle the equation</p><p>γ 4 m 2 c 2 6 r 2 = G m 2 2 γ 4 6 r 2 [ 2 2 3 + 1 6 ] (52)</p><p>Using the De Broglie angular momenta quantization equation (i.e. the De Broglie equation) (see <xref ref-type="fig" rid="fig5">Figure 5</xref>(a) &amp; <xref ref-type="fig" rid="fig5">Figure 5</xref>(b)), we obtain</p><p>r = ℏ γ 3 m 2 c (53)</p><p>6 2 r = ℏ γ 4 m 2 c (54)</p><p>and, thus, Equations (52) and (50), yield respectively</p><p>ℏ c G m 2 2 = γ 3 6 [ 1 3 3 + 1 3 6 + 1 3 ] (55)</p><p>and</p><p>ℏ c G m 2 2 = γ 4 6 [ 2 + 1 4 ] (56)</p><p>Accounting for the fact that ℏ c / G = m P l 2 , we establish the following expressions for the two Lorentz factors</p><p>γ 3 = ( m P l m 2 ) 1 / 3 1 [ 1 3 3 + 1 3 6 + 1 3 ] 1 / 6 (57)</p><p>γ 4 = ( m P l m 2 ) 1 / 3 1 [ 2 + 1 4 ] 1 / 6 (58)</p><p>On using the energy equipartition principle and recalling that there are four 3p axes and three 4p axes, the total energy per particle E is given by the expression</p><p>E = [ 4 γ 3 m 2 c 2 + 3 γ 4 m 2 c 2 ] / 7 (59)</p><p>where the values of γ 3 and γ 4 are computed from Equations (57) and (58) as</p><p>γ 3 = 1.0166 ( m P l m 2 ) 1 / 3 (60)</p><p>γ 4 = 0.9186 ( m P l m 2 ) 1 / 3 (61)</p><p>Then, it turns out that</p><p>γ &#175; = [ 4 7 γ 3 + 3 7 γ 4 ] ( m P l m 2 ) 1 / 3 = 0.9746 ( m P l m 2 ) 1 / 3 (62)</p><p>and, thus, accounting for the six particles in the Kaon, it follows that</p><p>m K = 6 γ &#175; m 2 = 0.9746 &#215; 503 &#215; 10 6 eV / c 2 = 490.5   MeV / c 2 (63)</p><p>which differs less than 0.7% from the experimental value of 493.677 MeV/c<sup>2</sup> [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>].</p><p>Consequently, the final formula for the Kaon mass is</p><p>m K = 6 { 4 7 1 [ 1 3 3 + 1 3 6 + 1 3 ] 1 / 6 + 3 7 1 [ 2 + 1 4 ] 1 / 6 } ( m P l m 2 2 ) 1 / 3 = 490.5   MeV / c 2 (64)</p><p>If we add the mass of the central e<sup>+</sup>, then the computed m<sub>K</sub> value attains the value m K = 491.0   MeV / c 2 and the deviations from the experimental K<sup>&#177;</sup> value is 0.55%, while from the experimental K<sup>o</sup> value is 1.4%.</p></sec><sec id="s5"><title>5. Conclusions</title><p>The present study shows that the Rotating Lepton Model (RLM) which combines the de Broglie wavelength equation, which has been the basis of quantum mechanics, with special relativity and with Newton’s Universal gravitational Law, provides, without any adjustable parameters, a very good fit to the masses of Pions and of Kaons. Kaons are found to be three-dimensional rotating structures comprising hybridized rotating relativistic neutrinos with a mass corresponding to the m<sub>2</sub> neutrino mass of the normal hierarchy. So far the RLM has been used to model only one 3D particle structure, i.e. that of the Higgs boson [<xref ref-type="bibr" rid="scirp.120214-ref19">19</xref>].</p><p>The neutrinos of the Kaon structure result from the hybridization of the m<sub>1</sub> and m<sub>3</sub> flavor type neutrinos, and are the same ones used already to compute the masses of muons [<xref ref-type="bibr" rid="scirp.120214-ref16">16</xref>] and pions [<xref ref-type="bibr" rid="scirp.120214-ref16">16</xref>]. It is worth noting that, similarly to the cases of the proton and the neutron masses, (938.272 MeV/c<sup>2</sup> and 939.565 MeV/c<sup>2</sup> respectively), which comprise only mass m<sub>3</sub> neutrinos and where the neutral composite particle (n) is heavier, here also the neutral composite particle (K<sup>o</sup>) has higher mass than the charged one (K<sup>&#177;</sup>). This has been attributed to the fact that electrostatic forces (between quarks [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>], or resulting from charge-induced dipole interactions, similar to van der Waals forces in chemistry, [<xref ref-type="bibr" rid="scirp.120214-ref21">21</xref>], enhance the total attraction and thus lower the required γ value for the gravitational confinement of the rotating neutrinos [<xref ref-type="bibr" rid="scirp.120214-ref14">14</xref>].</p><p>In conclusion, it should be pointed out that various Yukawa-type potentials [<xref ref-type="bibr" rid="scirp.120214-ref4">4</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref23">23</xref>] have been proposed to model the rapid decay of strong nuclear force at a rate much faster than the inverse square low (~1/r<sup>2</sup>) of Coulombic interactions for separation distances at the femtometer range. This is done by solving a “homogeneous” Helmholtz equation involving an additional phenomenological coefficient. An alternative nonlocal gradient intreaction force has been introduced more recently [<xref ref-type="bibr" rid="scirp.120214-ref24">24</xref>] leading to an “in homogeneous” Helmholtz equation, the solution of which also involves an adjustable phenomenological coefficient, which conveniently serves to effectively interpret non-Coulombic internuclear interactions. The RLM free-parameter approach can be directly used to properly determine the values of the aformentioned phenomenological coefficients.</p><p>The present study also confirms the conclusion [<xref ref-type="bibr" rid="scirp.120214-ref17">17</xref>] that neutrinos, electrons and positrons are apparently the only undividable elementary particles, as evidenced by examining the decay products of all the Tables of Elementary particles [<xref ref-type="bibr" rid="scirp.120214-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.120214-ref3">3</xref>]. Consequently these elementary particles are the equivalent of atoms in chemistry.</p></sec><sec id="s6"><title>Acknowledgements</title><p>This research was co-financed by the Greek State and the European Union (European Social Fund—ESF) through the Operational Programme Human Resources Development, Education and Lifelong Learning in the context of the project Reinforcement of Postdoctoral Researchers—2<sup>nd</sup> Cycle (MIS-5033021), implemented by the State Scholarships Foundation (IKY).</p></sec><sec id="s7"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s8"><title>Cite this paper</title><p>Vayenas, C.G., Tsousis, D., Grigoriou, D., Parisis, K. and Aifantis, E.C. (2022) Rotating Lepton Model of Pions and Kaons: Mechanics at fm Distances. 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