Electrogravity: On a scalar field of time and electromagnetism

Electrogravity: On a scalar field of time and electromagnetism Eytan H. Suchard, Metivity Ltd, email: [email protected] Abstract It is possible to describe a universal scalar field of time but not a universal coordinate of time and to attribute its non-geodesic alignment to the electromagnetic phenomena. A very surprising outcome is that not only mass generates gravity, but also electric charge does. Charge is, however, coupled to a non-geodesic vector field and thus is not totally equivalent to inertial mass. Only the entire “Energy-Momentum” tensor has a vanishing divergence. The model can be seen as misalignment of physically accessible events in an observer spacetime and of gravity as a controlling response by volumetric contraction of the observer spacetime in the direction where events bend or accelerate to. This non geodesic acceleration is described by a generalization of the Reeb vector. Misalignment of events can be described by 1, 2, and 3 such vectors. The paper presents a term with 4 vectors but does not discuss its physical meaning. The paper also discusses particle mass ratios and the Fine Structure Constant where added or subtracted area in relation to a disk does not involve a ratio 1 24 but 1 96 due to the physical meaning of the orientation of a space foliation which is perpendicular to a time-like vector 𝛼 and due to the orientation of a plane which is perpendicular to a time-like vector 𝛼 and its Reeb vector 𝜂 where 𝛼 is mapped to a 1-Form, 𝑑𝛼 = ±𝜂^𝛼. This forgotten definition of the Reeb vector 𝜂 is not limited to contact manifolds. These two orientations mean that only one side of a 3-dimensional foliation has a physical meaning and only one side of a sub-plane of that foliation has a physical meaning then 11 1 2 2 24 = 1 1 . Another interpretation of the factor is the Bekenstein - Hawking entropy to area 96 4 4 constant. An additional coefficient describes an acceleration field strength and has a 𝜋 compelling source in mainstream physics. Other two field strength coefficients are less understood but are very intuitive, these are 95 96 and a critical value due to an imbalance equation between gravity and anti-gravity ~1.55619853719. Keywords: General Relativity, Time, Electromagnetism. Table of Contents Introduction – section 1. Electro-gravity – section 2. Term (4) Equation of gravity. Term (8) Energy density of non - geodesic acceleration. Term (13) Charge gravitational mass. Ceramic capacitors – section 3. Term (14) Term for ideal capacitor with DC baseline without AC ripple. Thrust from 1000 Pf capacitor … , assumption – section 4. Martin Tajmar experiment’s null results analysis – section 5. Particle mass rations – section 6. Term (24) – Muon / electron highly accurate mass ratio. Term (31) – W/Z mass ratio. Term (32) – W/Tau mass ratio. Term (33) – Bottom Quark pole energy / Muon. The exact inverse Fine Structure Constant – section 7. Terms (34), (35) – Maximally imbalanced gravity and anti-gravity Terms (34),(35),(36) – Tau / Muon mass ratio. Terms (34), (35), (40) – accurate Inverse Fine Structure Constant. Terms (34), (35), (41), (42) approximation equation for the inverse Fine Structure Constant. The mass hierarchy – section 8. Term (46) – An approximation of the mass hierarchy. Interesting acceleration to radius coefficients relation – Section 9. Conclusion Appendix A: Euler Lagrange minimum action equations. Terms (47) – (55). Appendix B: Proof of conservation. Terms (56) – (63). Appendix C: Generalization to more than one generalized Reeb vector. Terms (64) – (65). Appendix D: Another way to derive the Reeb vector. Terms (66) – (69). Appendix E: 95/96, the precursor of the inverse Fine Structure Constant and of the muon/electron mass ratio. Terms (70) – (79). Appendix F: The Python code for (40) and for the remark after (40) and its output References. Appendix G: Code for (81)-(86) Appendix H: Causality conservation theorem 1. Introduction – measurement of non - geodesic deviation The Result of the Geroch Splitting Theorem [1] is that a field of time can be defined. In simple geometries such as FRWL, which are Big Bang geometries, such time also has an intuitive meaning; it is a scalar field and not a coordinate of time. It is the maximal time between each event of space-time and the Big Bang as a limit, measured by a physical clock that may experience forces. Such proper time can be measured along different curves and is therefore not traceable, not geodesic under forces and cannot be a coordinate that also requires a 4-direction. The existence of a non – traceable time is not a new idea and was postulated by the philosopher R. Joseph Albo [2] in the 14th century. The approach that will be presented to make peace between General Relativity and Quantum Mechanics is not to describe Space-Time as emergent out of huge matrices and to preserve the particles approach [3], but to replace particles with events. In non-hyperbolic spacetime, a scalar field can still be defined as universal clock but will no longer be an upper limit of measurable time to an event from a Cauchy surface as an interpretation to [1]. What information can a scalar field encode, that is not already predicted by the metric tensor of space time 𝑔𝜇ν ? The answer is non - geodesic motion. The motion equations of the theory of General Relativity predict only geodesic motion. This theory is based on two assumptions, 1) The basic assumption is that matter can be described via acceleration in the gradients of scalar fields, more specifically, the electromagnetic phenomena can be described by a nonzero acceleration of the gradient of a Geroch function [1] P2 in hyperbolic space-time or PP* if P is complex. This acceleration is known as a Reeb vector field [4] in odd dimensions but can also be defined in 4 dimensions via a 1-Form 𝛼, 𝑑𝛼 = 𝜂^𝛼 where 𝜂 is the Reeb vector. Important: In odd dimensions, the Reeb field can be defined in a way that it is not the acceleration of at least one unit vector field [5]. In two dimensions, the generalized Reeb vector is not geodesic. That is an important difference that has been missed all these years. 𝑑𝛼 = 𝜂^𝛼 is the forgotten definition of a Reeb vector which is used in the definition of the Reeb Class [6] and which is not limited to contact manifolds but is also defined on Symplectic manifolds. Actions are defined for 1 Reeb field, "electromagnetic", 2 Reeb fields "electro-weak", 3 Reeb fields, "Strong" and 4 Reeb fields as a “Fifth Force” or massive gravity. A definition can be made also for 4 Reeb fields but its physical meaning is not discussed in this paper. See appendix C, (65). The motivation to use Reeb vector fields, including a complex formalism, can be seen in the paper by Yaakov Friedman [7]. To complete assumption 1, energy density is 𝑎𝜇 𝑎𝜇 8𝜋𝐾 where 𝐾 is Newton’s constant of gravity and 𝑎𝜇 describes an acceleration of a normalized vector 𝑋 = 𝑐 𝑝𝜇 √𝑃𝜆 𝑃𝜆 where 𝑝𝜇 = 𝑑𝑝 𝑑𝑥 𝜇 where 𝑝 is a scalar field, 𝑥 𝜇 are the coordinates of the spacetime manifold and c is the speed of light. In simple words, what is claimed in this paper is that starting from the field X, which is derived from a Geroch function, a physical test clock which moves along X will continue to move along X also when X is not geodesic. That is to say that 𝑎𝜇 is a field which prohibits geodesic motion. The paper will show a way to define such a field regardless of the direction of motion of the test clock in the field. 𝑋 and 𝑎𝜇 span only one two-dimensional hyperplane of spacetime. The field must be defined in 4 dimensions. If such a field is the reason for the energy of the electric field, then the components of 𝑎𝜇 must be very small, otherwise acceleration of neutral particles in a strong electromagnetic field would be easily noticeable. 2) The scalar fields quantization is 𝑃 = ∑∞ 𝑘=1 𝑃(𝑘) such that ∫Ω 𝑃(𝑘)𝑃 ∗ (𝑗)+𝑃(𝑗)𝑃 ∗ (𝑘) 𝑃(𝑘)𝑃∗ (𝑗)+𝑃(𝑗)𝑃 ∗ (𝑘) 2 √−𝑔𝑑Ω = 0 if 𝑘 ≠ 𝑗 and ∫Ω √−𝑔𝑑Ω = 1 if 𝑘 = 𝑗 where √−𝑔 is the volume element of 2 space-time, where 𝑔 is the determinant of the metric tensor. Note: The mathematical foundation of this paper is the Geroch function [1], [2], Reeb vector fields [4] for encoding trajectory curvature, symplectic geometry directly on spacetime and not on any phase space due to [7], and the idea of physically accessible events in an embedding spacetime, an idea very similar to Harland Snyder’s quantized spacetime [8] but without any assumed non-commutative relation. Challenges to the reader: The challenges to the reader are to understand Reeb vectors in their original formalism with the meaning of non-geodesic acceleration, which is not limited to contact manifolds but describes how much a gradient of a scalar field is not geodesic, to understand how two scalar fields and two Reeb vectors describe a Scarr – Friedman acceleration matrix as a field and not as a uniform acceleration as originally proposed in their paper, and to understand how such an acceleration matrix can serve as a Symplectic form that acts directly on spacetime and not on any phase space as is the usual case in mainstream physics. Another challenge, which is somewhat a quantum leap, is to understand the use of non-geodesic geometry of foliations of spacetime and its meaning as matter. The Scarr-Friedman formalism will be discussed shortly in this paper and is essential to the understanding of this paper. Most theories in mainstream physics deal with geodesic curves and not with accelerated curves, unlike this paper which speaks of both. Another challenge is to accept that lack of collaboration in solving the field equations of this paper (4), (64) requires educated guess of field strength coefficients for Leptons. It is responsible to say that 95 96 for the electron is better understood than before and that the Tau field strength coefficient is better understood too though more research and collaboration 4 would greatly benefit the paper. The muon field strength coefficient is, however, from a critical 𝜋 field value of Quantum Mechanics and not directly from the presented theory. We can describe non geodesic integral curves along a field 𝑃𝜇 ≡ 𝑑𝑃 𝑑𝑥 𝜇 for the coordinates 𝑥 𝜇 , also, 𝑃𝜇 need not be time-like in all events of space-time. We now define the square norm for real numbers as 𝑍 ≡ |𝑃𝜆 𝑃 𝜆 | and its gradient 𝑍𝜇 ≡ 𝑑𝑍 𝑑𝑥 𝜇 . We define a geometric object measure how much the field 𝑃𝜇 is not geodesic. When 𝑐𝜏 describes the evolution of the vector 𝑋 = 𝑐 formed by the field 𝑋, 1 𝑐 𝑑𝑋 𝑑𝜏 𝑝𝜇 √𝑃𝜆 𝑃 𝜆 𝑈𝜇 that will 2 along the integral curves which are must be perpendicular to 𝑋 because 𝑋𝜇 𝑋𝜇 = 𝑐 2 and then 𝑑(𝑋𝜇 𝑋 𝜇 ) 𝑐𝑑𝜏 = (𝑋̇𝜇 𝑋𝜇 + 𝑋𝜇 𝑋̇𝜇 ) = 0 which implies 𝑋𝜇 𝑋̇𝜇 = 0 since 𝑑𝜏 is a scalar. Now writing 𝑍 = 𝑃𝜆 𝑃 𝜆 we have 𝑑 𝑑𝜏 Defining: 𝑈𝜇 ≡ 𝑝𝜇 √ 𝑃𝜆 𝑃 𝜆 = 𝑍𝜇 𝑍 = 𝑑 𝑝𝜇 𝑑𝜏 √𝑍 = 𝑝̇ 𝜇 √𝑍 − 𝑝𝜇 𝑍̇ 3 2𝑧 2 = 𝑃𝜇 ;𝜈 𝑑𝑥 𝜈 √𝑍 𝜏 − 𝑝𝜇 𝑍;𝜈 𝑑𝑥 𝜈 3 2𝑧 2 𝜏 = 𝑃𝜇 ;𝜈 𝑝𝜈 √𝑍 √𝑍 − 𝑝𝜇 𝑍;𝜈 𝑝𝜈 3 2𝑧 2 √𝑍 𝑃𝜇 ;𝜈 𝑝𝜈 𝑝𝜇 𝑍;𝜈 𝑝𝜈 𝑃𝜈 ;𝜇 𝑝𝜈 𝑝𝜇 𝑍;𝜈 𝑝𝜈 𝑍𝜇 𝑍𝜈 𝑝𝜈 𝑝𝜇 − = − = − 2𝑍 2𝑧 2 𝑍 2𝑧 2 𝑍 2𝑧 2 − 𝑍𝑘 𝑃 𝑘 𝑍2 𝑃𝜇 consider, 𝑃𝜇 ,𝜈 √𝑍 𝑑 𝑃𝜇 𝑑𝑥 𝜈 √𝑍 − 𝑃𝜇 𝑍𝜈 3 2𝑍 2 𝑃𝜈 𝑍𝜇 3 2𝑍 2 − − − 𝑑 𝑃𝜈 𝑑𝑥 𝜇 √𝑍 = √𝑍 3 2𝑍 2 𝑃𝜈 ,𝜇 + 𝑃𝜇 𝑍𝜈 3 2𝑍 2 𝑃𝜈 𝑍𝜇 = = (1) (1.1) = 𝑈𝜇 𝑃𝜈 𝑈𝜈 𝑃𝜇 1 𝑍𝜇 𝑃𝜈 𝑍𝑘 𝑃𝑘 1 𝑍𝜈 𝑃𝜇 𝑍𝑘 𝑃𝑘 𝑃𝜇 𝑃𝜈 ( − 2 𝑃𝜇 )− ( − 2 𝑃𝜈 )= − 2 𝑍 √𝑍 2 √𝑍 2 √𝑍 𝑍 𝑍 2 𝑍 √𝑍 √𝑍 √𝑍 But why to use, 1 2 1 𝑍 𝜇 𝑈𝜇 = ( 𝑍 − 2 It is easy to show that 𝑈𝜇 2 𝑍𝑘 𝑃 𝑘 𝑃𝜇 𝑍2 ) and not simply, 𝑍𝜇 𝑍 ? The reason is that behaves as the acceleration of the unit vector 𝑃𝜇 √𝑍 𝑈𝜇 𝑃 𝜇 2 √𝑍 = 0. . See Appendix D for another way to derive the Reeb vector. In terms of a 4-acceleration 𝑎𝜇 , it is easy to see: 𝑈𝜇 Where 𝑐 is the speed of light. 𝑈𝜇 2 2 = 𝑑𝑐 −1 𝑋 𝜇 𝑐𝑑𝜏 = 𝑎𝜇 𝑐2 (2) is the generalization of a Reeb vector [4] to 4 dimensions. Can this 𝑎𝜇 have a simple physical meaning of accelerating any neutral mass ? There is an experimental way to find out, once we analyze the electric field in the coming sections. Defining 𝐴𝜇𝜈 ≡ 𝑈𝜇 𝑃𝜈 2 √𝑍 − 𝑈𝜈 𝑃𝜇 2 √𝑍 we get 𝐴𝜇𝜈 𝑝𝜈 √𝑍 = 𝑈𝜇 2 which means that 𝐴𝜇𝜈 is a rotation and scaling matrix, however, as a linear operator it acts only on one of two hyper-planes of spacetime. To extend 𝐴𝜇𝜈 as a rotation and scaling matrix on the entire tangent bundle 𝑇(𝑀) of the spacetime manifold 𝑀 there is a need to use a contraction of 𝐴𝜇𝜈 with an antisymmetric tensor and to sum the result with 𝐴𝜇𝜈 . This extension will be discussed. Fig 1. – The generalized Reeb vector as an acceleration vector. To describe a field that accelerates any unit vector, we need an anti-symmetric matrix of acceleration similar to the Tzvi Scarr & Yaakov Friedman’s acceleration matrix [9]. The matrix 𝐴𝜇𝜈 = 𝑈𝜇 𝑃𝜈 2 √𝑍 − 𝑈𝜈 𝑃𝜇 2 √𝑍 is insufficient for that purpose; however, it can be extended quite easily, by using the Levi-Civita alternating tensor [10], not the alternating Levi-Civita symbol, 1 We have 𝐵𝜇𝜈 = 2 𝐸𝜇𝜈𝛼𝛽 𝐴𝛼𝛽 which define an acceleration matrix in a perpendicular plane to the plane spanned by 𝑃𝜇 √𝑍 and Uμ 2 . In the complex case we define the acceleration matrix: 𝐹𝜇𝜈 = 𝐴𝜇𝜈 + 𝛾𝐵𝜇𝜈 where 𝛾𝜖𝑈(1). With a vector 𝑤 𝜈 , 𝑤 𝜈 𝑤𝜈 = 𝑐 2 , we derive its acceleration, 𝐹𝜇𝜈 𝑤𝜈 𝑐 = 𝑎𝜇(𝑤) 𝑈𝜇 𝑈𝜇 1 𝜇𝜈 𝐹 𝐹 = 4 𝜇𝜈 4 Exercise to the reader: show that the Reeb vector (3) 𝑐2 𝑈𝜇 2 of 𝑃𝜇 √|𝑍| is the same as for 𝑃𝜇 √|𝑍| 𝑒 𝑖𝜃 for 𝑖 = √−1 and a smooth scalar 𝜃. See that you understand the idea of a field of acceleration that maps 4velocity to 4-acceleration by multiplication with an anti-symmetric matrix [9], 𝐹𝜇𝜈 𝑤𝜈 𝑐 = 𝑎𝜇(𝑤) 𝑐2 . In Special Relativity, 4-velocity is perpendicular to 4-accleration and 𝑤 𝜈 𝑤𝜈 = 𝑐 2 . 𝐹𝜇𝜈 is then an Acceleration Field and it can be deconstructed into the sum of two matrices which act on two 1 perpendicular two-dimensional hyperplanes in spacetime. 𝐵𝜇𝜈 = 𝐸𝜇𝜈𝛼𝛽 𝐴𝛼𝛽 and 𝐵𝜇𝜈 = 1 − 𝐸 2 𝜇𝜈𝛼𝛽 2 𝐴𝛼𝛽 yield the same result in (3). When reduced to the three-dimensional foliation which is perpendicular to 𝑃𝜇 , If 1 − 𝐸𝜇𝜈𝛼𝛽 𝐴𝛼𝛽 mean that 𝐵𝜇𝜈 = 2 𝑈𝜇 has a divergence point, say Q, then the choice 𝐵𝜇𝜈 = 2 𝑃(2)𝜈 𝑈(2)𝜇 √|𝑍| 2 − 𝑃(2)𝜇 𝑈(2)𝜈 √|𝑍| 2 , where 𝑃(2)𝜈 𝑃𝜈 = 0, 𝑈(2)𝜈 𝑃𝜈 = 0, 𝑈(2)𝜈 𝑃(2)𝜈 = 0, and finally the real numbers case yields the following, 𝑈𝜇 ;𝜇 = ∓𝑈(2)𝜇 ;𝜇 (3.1) 𝑈𝜇 ;𝜇 + 𝑈(2)𝜇 ;𝜇 ∈ {2𝑈(2)𝜇 ;𝜇 , 0} Lagrangian generalization offer and further research offer It is possible to define a Lagrangian for two independent acceleration vectors that are related to each other by multiplication, here it is presented in a complex formalism, with a volume element √−𝑔, | 𝜁 ∗𝜆 𝜁𝜆 +𝜁 𝜆 𝜁 ∗ 𝜆 8 |= | | 1 Pk 0 𝑈(2)∗𝑘 +P∗ 0 k 2√2Z 𝑈(2)∗𝑘 𝑈 𝑘 𝑈𝑘∗ +𝑈 ∗𝑘 𝑈𝑘 8 𝑈(2)𝑘 𝑈𝑘∗ +𝑈(2)∗𝑘 𝑈𝑘 8 1 2 Pk 𝑈(2)∗𝑘 +P∗ k 𝑈(2)∗𝑘 2√2Z 𝑈(2)𝑘 𝑈𝑘∗ +𝑈(2)∗𝑘 𝑈𝑘 | √−𝑔 8 | ∗ 𝑘 ∗𝑘 𝑈(2) 𝑈(2)𝑘 +𝑈(2) 𝑈(2)𝑘 8 (3.2) The meaning of (3.2) is of a squared acceleration which is the Minkowski squared norm of a spacelike vector. In (+,-,-,-) metric convention, a negative sign has to be added, − 𝜁 ∗𝜆 𝜁𝜆 +𝜁 𝜆 𝜁 ∗ 𝜆 The following norm calculates a physical non-geodesic acceleration, √ 8 𝜁 ∗𝜆 𝜁𝜆 +𝜁 𝜆 𝜁 ∗ 𝜆 8 . . Since the 3 forces in Nature seem to be aligned with the electric field, it is reasonable to assume that 𝜁𝜆 must be either aligned or anti-aligned with 𝑈𝜆 , or in other words, 𝜁𝜆 = 𝑓(𝑥 𝜇 )𝑈𝜆 (3.2) 𝜁𝜆 + 𝜁 ∗𝜆 𝑈𝜆 + 𝑈 ∗𝜆 = 𝑓(𝑥 𝜇 ) 4 4 for some scalar function of the coordinates 𝑓(𝑥 𝜇 ). A real valued vector is then 𝑈𝜇 +𝑈 ∗ 𝜇 4 but to assume this expression is the direction of an acceleration vector, by Occam’s razor must be inferred from a variation of the Lagrangian L= involves the divergence, ( 𝑈 𝜇 +𝑈 ∗𝜇 4 𝑈𝜇 𝑈 ∗𝜇 +𝑈 ∗ 𝜇 𝑈 𝜇 8 ) ;𝜇 which implies that √−𝑔. Such a variation indeed 𝑈𝜇 +𝑈 ∗ 𝜇 4 has indeed a meaning of an acceleration of a unit vector. The zeros in (3.2) mean that the acceleration vector perpendicular to the unit vector term Pk 𝑈 ∗𝑘 +P∗ k 𝑈 ∗𝑘 2√2Z ∗𝑘 Pk 𝑈(2) +P∗ k 𝑈(2)∗𝑘 2√2Z P∗ k √𝑍 𝑍𝜇 , (2𝑍 − 𝑍𝜆 𝑃 ∗𝜆 𝑃𝜇 2𝑍 2 ) 𝑃 ∗𝜇 √𝑍 =( 𝑍𝜇 𝑃 ∗𝜇 2𝑍 − 𝑍𝜆 𝑃 ∗𝜆 2𝑍 2 𝑍) 1 √𝑍 𝑈𝑘 2 is = 0 and then the = 0. If Pk and Uk are perpendicular to P(2)k and U(2)k then of course = 0, however, this Lagrangian can define an action operator even without such an orthogonality as a prerequisite and is therefore more general. The Lagrangian above has symmetry SU(2) and is therefore offered as a generalization of this paper with properties of the “electroweak” field. To summarize the motivation of this section, saying that an energy density can be described as the negative squared norm of an acceleration of unit vectors in (+,-,-,-) metric does not mean such acceleration field can’t be a result of other Reeb fields. The description of the electric field as the simplest example is discussed later. Could acceleration vectors also explain the gravitational field? The acceleration of a test particle with velocity 𝑑2 𝑥 𝜇 𝑑𝜏2 𝑑2 𝑥 𝜇 With weak gravity: 𝑑𝑡 2 𝑑𝑥 𝜇 𝑑𝜏 in weak gravity, 𝜇 𝑑𝑥 𝛼 𝑑𝑥 𝛽 𝑑𝜏 𝑑𝜏 + Γ𝛼𝛽 𝜇 𝑑𝑥 𝛼 𝑑𝑥 𝛽 𝑑𝑡 𝑑𝑡 + Γ𝛼𝛽 =0 =0 (3.3) (3.4) 𝑔𝜇𝜈 = 𝜂𝜇𝜈 + 𝜖ℎ𝜇𝜈 𝑂( 𝑑𝑥 𝜇 𝑑𝑡 (3.5) ) = 𝑂(𝜖) (3.6) And then space terms are neglected, which reduces the equation of motion to 𝑑2 𝑥 𝜇 While 𝑑𝑥 0 𝑑𝑡 𝑑𝑡 2 = 𝑐, the speed of light, we get, 𝑑2 𝑥 𝜇 ≈0 (3.7) 1 𝜇 ≈ −𝑐 2 Γ00 = − 𝑐 2 𝜖(ℎ𝜇 0,0 + ℎ𝜇 0,0 − ℎ00 ,𝜇 ) 𝑑𝑡 2 And in a static field 𝜇 𝑑𝑥 0 𝑑𝑥 0 𝑑𝑡 𝑑𝑡 + Γ𝛼𝛽 (3.8) 2 𝜇 1 (3.9) 1 (3.10) Γ00 = − ℎ00 ,𝜇 𝑑2 𝑥 𝜇 𝑑𝑡 2 2 ≈ 𝑐 2 𝜖ℎ00 ,𝜇 2 Consider the following representation of the metric tensor, 𝑔𝜇𝜈 = 𝑃(0)𝜇 𝑃(0)𝜈 𝑍(0) − 𝑃(1)𝜇 𝑃(1)𝜈 𝑍(1) − 𝑃(2)𝜇 𝑃(2)𝜈 𝑍(2) − 𝑃(3)𝜇 𝑃(3)𝜈 (3.11) 𝑍(3) At this point no full tetradic representation is considered and 𝑃(𝑖)𝜇 𝑃(𝑗)𝜇 ≠ 0 for 𝑖 ≠ 𝑗 or 𝑃(𝑖)𝜇 𝑃(𝑗)𝜇 = 0 for 𝑖 ≠ 𝑗. Consider the weak field equation of motion while focusing on the contribution of 1 𝑃(0)𝜇 𝑃(0)𝜈 𝑍(0) so in that case it is necessary to say that we account for only of the gravity if the contribution from 4 2 all fields, 𝑃(𝑖) is equal, in that case, 𝑃(𝑖) represents time, − 1 (𝑝(0)0 )2 𝜇 , 2 𝑍(0) =− 1 2𝑝(0)0 𝑝(0)0 ,𝜇 2 √𝑍(0) ( √𝑍(0) − 𝑝(0)0 𝑍(0),𝜇 3 2𝑍 2 1 𝜇 ) ≈ Γ00 4 (3.12) 𝑝(0)0 𝑝(0)0 𝑍(0),𝜇 𝑝(0)0 ,𝜇 1 𝜇 1 2𝑝(0)0 𝑝(0)0 ,𝜇 𝑝(0)0 𝑍(0),𝜇 ( )= ( ) ≈ Γ00 − − − 3 3 4 2 √𝑍(0) √𝑍(0) √𝑍(0) √𝑍(0) 2𝑍 2 2𝑍 2 𝑝(0)0 ,𝜇 ≈ 𝑝(0)𝜇 ,0 ⟹ 𝑝(0)0 𝑝(0)0 𝑍(0),𝜇 𝑝(0)𝜇 ,0 𝑍(0),𝜇 𝑈(0)𝜇 1 𝜇 ( ) ≈ − ≈ ≈ Γ00 3 2𝑍 2 4 √𝑍(0) √𝑍(0) 2𝑍 2 The latter result is due to 𝑝(0)1 , 𝑝(0)2 , 𝑝(0)3 being neglected but not their derivatives and due to (1). The conclusion of (3.12) is that (3.11) can describe weak gravity, however the scalar fields 𝑃(0), 𝑃(1), 𝑃(2), 𝑃(3) in this case, do not represent force fields but Gauge fields. Caveat: An important caveat is that even if the complex formalism of (3.12) is used, 8 complex scalars may not be able to describe gravity. Contribution from additional fields may be needed. Caveat: Do not confuse between using generalized Reeb vectors / acceleration fields in order to describe the energy of force fields and other such fields which are used to show that unit vectors can account for gravity itself if they are not geodesic. In general, geodesic curves are not geodesic when mapped to a flat spacetime. The meaning of (3.12) was simply to show the possibility of using non-geodesic curves as the underlying field that drives gravity too and not only other force fields. This can be achieved by mapping geodesic curves to non-geodesic curves in flat spacetime. Even the complex formalism may not be sufficient: 𝑃(0)𝜇 𝑃 ∗ (0)𝜈 +𝑃 ∗ (0)𝜇 𝑃(0)𝜈 𝑔𝜇𝜈 = 2 − ∑3𝑖=1 𝑃(𝑖)𝜇 𝑃 ∗ (𝑖)𝜈 +𝑃 ∗ (𝑖)𝜇 𝑃(𝑖)𝜈 (3.13) 2 𝑃(0)𝜇 is interesting when it is not geodesic also in the curved geometry, and is not only pseudo non-geodesic simply by omission of the Christoffel symbols. Electro-gravity The action of gravity is defined as: 𝐴𝑐𝑡𝑖𝑜𝑛 = 𝑀𝑖𝑛 ∫Ω (𝑅 − 1 4ℶ 𝑈 𝑘 𝑈𝑘 ) √−𝑔 𝑑Ω The Euler Lagrange equations by the metric 𝑔𝜇𝜈 , by the scalar field of time P yield, Appendix A or [9]: 1 1 (𝑈𝜇 𝑈𝜈 − 2 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 4ℶ 𝑃𝜇 𝑃𝜈 𝑍 1 ) = 𝑅𝜇𝜈 − 2 𝑅𝑔𝜇𝜈 𝑃𝜇 𝑍𝜈 𝑃𝜈 𝜇 𝑊 ;𝜇 = (−4𝑈 ;𝑘 − 2 2 𝑈 ) ;𝜇 = 0 𝑍 𝑍 𝜇 𝑘 It is easy to prove without the right hand side that see Appendix B or [11]. (4) assumes ℶ = 1. 1 1 (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 4ℶ 2 (4) 𝑃𝜇 𝑃𝜈 𝑍 ) ;𝜈 = 0 1 Consider 𝜌 = 𝑈 𝑘 ;𝑘 to be stationary along 𝑝𝜇 , with local coordinates such that only 𝑝0 is 2 numerically significant. We will neglect all small terms that are multiplied by 𝑈𝜇 and its 𝑝𝜇 derivatives. with 2 ( 𝑍 ) ;𝜇 ≈ 𝑝𝜇 𝑝0 2 ( 𝑍 ) ;𝜇 ≈ 2 (𝑝0𝑝 ) ,0 and 1 is due to 𝑍𝜈 𝑝𝜈 𝑍2 ≈ 0 𝑍2 𝑍𝜈 𝑝𝜈 𝑝0 (𝑝0 𝑝0 ),0 (𝑝0 𝑝0 )2 𝑍𝜈 𝑝𝜈 𝑍2 ≈ and ( 𝑍𝜈 𝑝𝜈 𝑍2 0 0 𝑝 (𝑝 𝑝0 ),0 (𝑝0 𝑝0 )2 ) ;𝜇 𝑈𝜇 ≪ 1, the first approximation is the result of 𝑝0 ≈ 2 (𝑝0 𝑝 ) ,0 the last approximation ( 0 and the fact that 𝑈𝜇 is spacelike. Then, 𝑍𝜈 𝑝𝜈 𝑍2 ) ; 𝜇 𝑈𝜇 ≪ (−4𝑈 𝑘 ;𝑘 𝑃𝜇 𝑍 −2 𝑍𝜈 𝑃 𝜈 𝑍2 𝑈𝜇 ) ;𝜇 = 0 ⟹ 2𝜌 ≈ 𝑈 𝑘 ;𝑘 (4.1) Dynamics: (4.1) implies the dynamics of the electric field of points of divergence 𝑈 𝑘 ;𝑘 ≠ 0. Theorem 1: If non-geodesic curves are prescribed to motion in material fields then zero Einstein 1 1 1 tensor implies 𝑈𝜇 = 0, i.e. 𝑅𝜇𝜈 − 𝑅𝑔𝜇𝜈 = 0 ⟹ 𝑈𝜇 = 0 i.e. geodesic motion. 2 2 Proof: 2 1 We contract both sides of (4) with 𝑈𝜇 𝑈 𝜈 so (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 2 𝑈𝜆 𝑈 𝜆 = 0 because 𝑈𝜇 𝑃𝜇 = 0 and now we contract both sides of (4) with 𝑃𝜇 𝑃𝜈 𝑍 and 1 (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 𝑃 𝜆 𝑃𝜆 𝑍 2 1 𝑃𝜇 𝑃𝜈 𝑍 1 2 words, motion must be geodesic and we are done. Remember 𝑈𝜇 2 = 𝑎𝜇 𝑐2 𝑍 𝑃𝜇 𝑃𝜈 𝑍 ) 𝑈𝜇 𝑈 𝜈 = 0 ⟹ so we have ) = − 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 = 2𝑈 𝑘 ;𝑘 = 0 because 𝑈𝜆 𝑈 𝜆 = 0 = 1 so we get 𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 2 𝑃𝜇 𝑃𝜈 𝑃𝜇 𝑃𝜈 𝑍 = 𝑈𝜇 𝑈𝜈 = 0 ⟹ 𝑈𝜇 = 0. In other as acceleration and the equation of gravity by Einstein, using the dust energy momentum tensor from General Relativity, 8𝜋𝐾 𝑐4 1 𝑇𝜇𝜈 = 𝑅𝜇𝜈 − 𝑅𝑔𝜇𝜈 (5) 2 in (-,+,+,+) convention, we will use (5) further on, to show unique gravity by electric charge. 1 𝑈 𝑘 𝑈𝑘 = 4 𝑎𝑘 𝑎𝑘 (6) 𝑐4 (6) compared to Einstein’s tensor means that the energy density in old physics terms can be seen as: 𝑎𝑘 𝑎𝑘 8𝜋𝐾ℶ = 𝐸𝑛𝑒𝑟𝑔𝑦𝐷𝑒𝑛𝑠𝑖𝑡𝑦 ⟹ 8𝜋𝐾 𝑐4 𝐸𝑛𝑒𝑟𝑔𝑦𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑎𝑘 𝑎𝑘 ℶ𝑐 4 = 1 4ℶ 𝑈 𝑘 𝑈𝑘 (7) Where ℶ = 1 relates non geodesic acceleration to geometry, direct outcomes of (7) will be shown in (13) and (43). (7) means that the energy of the classical non-covariant electric field must be hidden in a very weak acceleration field 𝑎𝑘 𝑎𝑘 1 ≅ 𝜀0 𝐸 2 (8) |𝑎|2 = 4𝜋𝐾𝜀0 ℶ𝐸 2 (9) 8𝜋𝐾ℶ 2 𝜀0 is the permittivity of vacuum, K is Newton’s constant of gravity, which means and ‖𝑎𝜇 ‖ = √4𝜋𝐾𝜀0 ℶ‖𝐸‖ (10) Indeed, a very weak acceleration if ℶ = 1. However, there is a surprise: 1 Means that 1 𝑈 𝑘 ;𝑘 = 2ℶ 𝑎 𝑘 ;𝑘 𝑐2 Now remember the term 4ℶ 1 (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 2 4𝜋𝐾 𝜌 4𝜋𝐾𝜀0 ℶ 𝜌 =√ 1 ℶ2 (−2𝑈 𝑘 ;𝑘 4ℶ 𝜀0 𝑐 2 𝑃𝜇 𝑃𝜈 𝑍 =√ 8𝜋𝐾 2ℶ 𝑍2 ≈ 1 𝑍 𝑉𝜇 𝑐 𝑐4 0 𝑃𝜇 𝑃𝜈 𝑍 ≈ 𝑉𝜇𝑉𝜈 𝑐2 = 1 8𝜋𝐾𝑐 4 (11) 2 where , in Special Relativity 𝑉𝜇 = 4𝜋𝐾ℶ 𝜌𝑐ℎ𝑎𝑟𝑔𝑒 𝑉 𝜇 𝑉 𝜈 √ ℶ2 𝜀 ∙ 8𝜋𝐾 1 ) = 𝑅𝜇𝜈 − 𝑅𝑔𝜇𝜈 where 𝜌 is charge density. ) and the relation equivalent to a normalized velocity vector 1 𝑈 𝜇 ;𝜇 𝑃 𝜇 𝑃 𝜈 ℶ𝜀0 𝑐 2 𝑃𝜇 𝑃𝜈 𝑃𝜇 is √𝑍 (𝑐,𝑣𝑥 ,𝑣𝑦 ,𝑣𝑧 ) √1−𝑣2 /𝑐 2 , so we get 4𝜋𝐾 √ ℶ𝜀 𝜌𝑐ℎ𝑎𝑟𝑔𝑒 𝑉𝜇 𝑉 𝜈 (12) 0 But that can only mean that charge density behaves like mass density except for the fact that is not geodesic and therefore for charge Q: 𝑀= 𝑄 𝑃𝜇 √𝑍 (13) √16𝜋𝐾𝜀0 ℶ Assuming ℶ = 1 where 𝜀0 is the permittivity of vacuum and K is Newton’s constant of gravity, M is a gravitational mass, from (13) ±1 𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠 is equivalent to ±𝟓. 𝟖𝟎𝟐𝟏𝟑𝟓𝟐𝟏𝟓 ∗ 𝟏𝟎𝟗 𝐊𝐠. Caveat: 𝑃𝜇 √𝑍 1 is not geodesic unless 𝑈𝜇 = 0. So 𝜌𝑐ℎ𝑎𝑟𝑔𝑒 2 𝑃𝜇 𝑃𝜈 𝑍 does not behave as inertial mass. Electric field to acceleration from far observer coordinates – the following is not the way to derive the relation between gravitational mass and charge, not only because charge is coupled to a non-geodesic bivector, however, it does serve as an indication that the results are correct. 𝑒 4𝜋𝜀0 𝑟 1 (4𝜋𝜀0 𝐾)2 = 2 𝑐2 𝑟 (13.1) Where the right-hand side stands for acceleration or the norm of the Reeb vector multiplied by the squared speed of light. ‘e’ is the charge of the electron, 𝜀0 the permittivity of vacuum and 𝐾 is the gravity constant of Newton. (13.1) is a result of (10). 𝑒 ( 𝐾 𝑐 2 4𝜋𝜀0 1 2 ) =𝑟 We will equate the right-hand side to the Schwarzschild radius of some mass, (13.2) 𝑒 From which ( 𝐾 1 2 𝑐 2 4𝜋𝜀0 ) = 1 1 2 e( 16πKε0 2𝐾𝑚 (13.3) 𝑐2 ) =m (13.4) This is a very surprising result although it is not derived from the Euler Lagrange equations but just agrees with them 100% for the choice ℶ = 1 in (13). We are now set to derive the inverse Fine Structure Constant from (13) and from a spin term. We sloppily do this by mixing ideas from General Relativity and Quantum mechanics and (13). From Quantum Mechanics, the angular momentum of the electron is, √𝑠(𝑠 + 1)ℏ 1 (13.5) where the spin number is 𝑠 = for the electron, where ℏ is the reduced Planck constant and 2 1 √𝑠(𝑠 + 1) is specific to a particle’s spin. Suppose that a positive charge with 8 (𝑈 ∗𝜇 𝑈𝜇 + 1 𝑈𝜇 𝑈 ∗𝜇 ) = 0 or in the real case 𝑈𝜇 𝑈𝜇 = 0, is spinning near the speed of light at twice the 4 Schwarzschild radius created by the charge in (13), where this radius is known as the radius of the Marginally Bound [unstable] Orbit, then by (13) this radius should be 2 where 𝑚 = 𝑒 √16𝜋𝐾𝜀0 2𝑒𝐾 √16𝜋𝜀0 𝐾𝑐2 =2 2𝐾𝑚 , 𝑐2 and e is the charge of the positron. Of course, we need to remember that 𝑃𝜇 √𝑍 is not velocity and therefore the interpretation of m is not as the familiar inertial mass, moreover, 𝑃𝜇 √𝑍 is not a geodesic vector field if 𝑈𝜇 2 is not zero. The radial metric coefficient is 1 due to Schwarzschild metric, not Kerr metric, then by (13) and remembering that the angular momentum does not mean a classical rotation, the angular momentum should be, 𝐽= 𝑒𝑐 2∗2𝑒𝐾 √16𝜋𝜀0 𝐾 √16𝜋𝜀0 𝐾𝑐 2 = 𝑒2 4𝜋𝜀0 𝑐 (13.6) and we ignore any Kerr metric because the spin effect on spacetime is not identical to the classical rotation of a black hole, otherwise positrons would dissipate their spin energy. We also assume that our field is a fundamental field to all charged particles and therefore omit the √𝑠(𝑠 + 1) which is specific to the spin number s. Now consider the ratio between 𝐽 and the spin independent coefficient ℏ, we get, 𝐽 ℏ = 𝑒2 4𝜋𝜀0 𝑐ℏ (13.7) Which is the known term for the Fine Structure Constant as an upper limit on a ratio between classical angular momentum and Quantum angular momentum. Theorem 2: If the electromagnetic energy is not zero and the charge density 𝑈 𝑘 ;𝑘 is zero in a domain D of space-time then 𝑈0 is never 0 in all events of D. Proof: 1 We write the Einstein - Grossmann equation (4) in its dual form, 𝑅𝜇𝜈 = 𝑇𝜇ν − 𝑔𝜇𝜈 𝑇𝛼𝛼 = 2 1 4ℶ 1 4ℶ 1 𝑃𝜇 𝑃𝜈 2 𝑍 (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 2 1 (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 𝑈 𝑘 ;𝑘 (𝑔𝜇𝜈 − 2 𝑃𝜇 𝑃𝜈 𝑍 𝑍 𝑃𝜇 𝑃𝜈 1 1 − 𝑔𝜇𝜈 𝑔𝑖𝑗 (𝑈𝑖 𝑈𝑗 − 𝑔𝑖𝑗 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 2 1 2 𝑃𝑖 𝑃𝑗 𝑍 − 𝑔𝜇𝜈 𝑈 𝜆 𝑈𝜆 + 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 + 𝑔𝜇𝜈 𝑈 𝑘 ;𝑘 ) = 2 1 )) = 4ℶ (𝑈𝜇 𝑈𝜈 + )). If 𝑈0 = 0 in D then there exist local coordinates such that only the 𝑃0 component of 𝑃𝜇 is not zero. We assumed 𝑈 𝑘 ;𝑘 = 0. Since 𝑈0 = 0, 𝑅00 = 0 so the electromagnetic energy is zero. On the other hand, since 𝑈𝜇 is not zero, 𝑃𝜇 cannot be geodesic and therefore 𝑃0 cannot be the only component of 𝑃𝜇 which is not zero along geodesic coordinates. Note: If there is a time-like curve 𝛾 around which 𝑈𝜇 is in relative motion in different events of every small D that contains 𝛾, then 𝑅00 is not zero in D. Note: There is one obvious peculiarity about charge generated gravity, the charge. It is dictated by a scalar field of space-time! 𝑃𝜇 √𝑍 is not the velocity of Note – physical interpretation: From (10) and (13), if 𝑎𝜇 has a simple physical interpretation as a field that accelerates any neutral mass then we have to take (13) into account as an opposite effect. The result is that a field of 1,000,000 volts over 1 mm distance will accelerate any neutral particle at 8.61 cm * sec-2 and with taking into account (13) it will be less, due to an opposite gravitational effect, see (14), will be reduced to 4.305 cm * sec-2. The quantization of P is into a sum of event wave functions and has the physical meaning of Sam Vaknin’s realization chronons [12]. The theory is easily expanded to 2 and to 3 Reeb vectors where the Lagrangian has U(1) SU(2) SU(3) symmetry if orientation is preserved, otherwise the symmetry group contains also reflections, see also an SU(4) Lagrangian, Appendix C. It is important to say that Vaknin’s approach [12] is diametrically opposed to that of Jungjai Lee and Hyun Seok Yang [3]. 2. Ceramic capacitors In this section we will examine gravitational propulsion, not an Alcubierre’s warp drive because the Alcubierre [13] extrinsic curvature condition (𝐾𝑖 𝑖 )2 − 𝐾𝑖𝑗 𝐾 𝑖𝑗 < 0 will not hold in the same geometry as in the Alcubierre warp drive bubble. However, a negative plate below and a positive plate above, will manifest weak acceleration upwards as the negative gravity will push the positive plate upwards and the negative plate will be pulled by the positive plate above it. The main problem is that due to the dielectric material, the mass of the dielectric material will not be gravitationally repelled by the negative plate. Only a small portion of the mass of the capacitor will be affected in a highly dielectric material. Overcoming the anti-alignment, see Fig 2.A., is technological challenge which cannot be achieved without a dynamic electric field, see Fig 2.B. Fig. 2.A. – Only a small portion of the mass, in purple, is affected. Fig. 2.B. – Electro-gravitational thrust engine with two capacitors and slow anti-alignment dielectric layer, which mitigates the anti-alignment by charging (right) and discharging (left) cycles. The capacitors rotate as depicted in the drawings. The ground direction is bottom. The arrows depict the direction in which the capacitors are rotated by an electric motor. Static field without dielectric anti-alignment requires about 2 * 10-4 Coulombs / cm2 in order to accelerate the dielectric layer against the gravity of the Earth. This is why with the current technology, the offered thrust engine is insufficient for a commercial flight. Measurable thrust of up to 1 Newtons is expected with voltage above 2,000,000 volts, relative dielectric constant of above 1000, dielectric polarization time of a millisecond, heavy dielectric layers with mass density similar to Ta2O5 but with a higher dielectric constant, electric motor rotation of at least 3000 RPMs and capacitor areas of about 20 x 20 cm2. Partners in this experiment are Jessica Lynne Suchard and Raviv Yatom. It is easy to see from (13) that in the classical limit near the plate, the gravitational field is mostly affected by charge density. By (13) the gravitational acceleration is 𝑎≅ 4𝜋𝐾𝑄 𝐴∗𝜀∗√16𝜋𝐾𝜀0 ℶ = 𝑉 𝑑√ℶ ∗ √𝜋𝐾𝜀0 ⟹ 𝛿𝑊𝑒𝑖𝑔ℎ𝑡 ≅ 𝑉 𝑑√ℶ ∗ 𝑀𝑑𝑖𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑔 √𝜋𝐾𝜀0 = 𝑉𝜌𝐴 𝑔√ℶ √𝜋𝐾𝜀0 (14) where K is Newton’s gravitational constant, Q is charge, A is area, 𝜌 is the dielectric layer’s density and M is its mass and 𝜀0 is the permittivity of vacuum, 𝜀 is the relative dielectric 𝐴 constant, assuming ℶ = 1, g is the Earth surface acceleration. (14) is the result of 𝑄 = 𝑉𝜀0 = 𝑑 𝑉 𝜀 𝐴 𝑑 0 𝑄 ⇔ √16𝜋𝐾𝜀0 𝑄 𝜀0 𝐴 = 4𝜋 𝑄 4𝜋𝜀0 𝐴 𝐸 where E is the classical intensity of the electric field. We saw: 𝑀 = with ℶ = 1. The gravitational acceleration by the charge is 𝑎 ≅ ℶ 4𝜋𝐾𝑀 𝐴 = 4𝜋𝐾𝑄 𝐴∗𝜀∗√16𝜋𝐾𝜀0 ℶ , if we assume an attenuation by the dielectric layer’s induced dipoles to be proportional to the attenuation of the electric field by the same induced dipoles. This assumption is problematic because the induced dipoles are the accelerated material by the gravitational dipole of the external plates, and they are in much closer proximity to local charge than to the charge on the 𝑉 external plates. 𝛿𝑊𝑒𝑖𝑔ℎ𝑡 ≅ 𝑑√ℶ ∗ model. 𝑀𝑑𝑖𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑔 √𝜋𝐾𝜀0 = 𝑉𝜌𝐴 𝑔√ℶ √𝜋𝐾𝜀0 is therefore a very optimistic Caution with (14): In reality, the charge of the induced dielectric dipoles is closer to the mass of the dipoles than the external plates. The assumptions of (14) therefore break down and the Inertial Dipole effect is much smaller. One possible technological remedy to this anti-alignment is to add an Alternating Current - AC component to the DC baseline and to disrupt the antialignment. Still, even with such a component, a feasible propulsion system may require millions of volts as a baseline. When using voltage above 2 * 511 kV, creation of electron-positron pairs is difficult to avoid (not the Schwinger limit but accelerated electrons through parasitic leakage), and the resulting gamma rays are a serious health hazard. A dynamic voltage and/or current component, renders the mathematical description of the Inertial Dipole much more difficult. The following calculations are therefore very optimistic. Suppose we have a 1000Pf ceramic capacitor and we charge it with 10000 Volts and the area of the plates is 1 cm2. The charge on the plates is then 10-5 Coulombs and its density 10-1 Coulombs per square meters. Now we want to calculate the approximate acceleration that the upper positive plate experiences due to the anti-gravity effect from the lower plate. Only a thin portion of the upper layer is affected, where the positive charge accumulates. A calculation shows: 0.48663510306 meters / sec2. Dividing 0.4866351… meters/sec2 by 9.81 meters / sec2 we get 0.049606024776763 which is less than 5 percent relative to the gravity of the Earth. If instead of a dielectric material, an insulator with relative dielectric constant 1 is used for the same charge density of 10-5 Coulombs per 1 cm2, a weight loss of the insulating slab should be measured at about 0.0496 of its weight. With a high relative dielectric constant, the affected mass could be well below 1 milligram, and it will lose 0.0496 of its weight. This renders the measurement of such an effect very hard to achieve unless the dielectric material is saturated and can no longer shield the field of the plates such as in the H4D experiment [14*]. In any other case, practically no measurable thrust is expected for an area 1cm2 with 10,000 Volts and scale resolution worse than 10-4 grams. In the case of saturation, at first the inertial dipole is expected to grow with the saturation of the dielectric material and with the amount of charge on the plates. [14*] will be discussed later. The H4D lab [14] 69 mm radius and 2mm PMMA thickness capacitor with 20,000 volts, weight loss is at least 0.0015509 grams, however the thickness of the metal plates is 1mm. It is sufficient to have a low frequency AC ripple from the DC power supply to churn the electrons on the plates such that not only a thin layer of the plates will be charged, also with an AC ripple, of typically 150 VAC for 20000 Volts DC, the induced gravitational field can no longer be considered static. Under such conditions (14) is no longer valid. 3. Thrust from 1000 Pf capacitor with two metallic plates and 10000 volts Assumptions: Most of the dielectric mass is not completely shielded from the plate fields and the attenuation of the influence of the external dipole on the mass within the induced dipole is by a factor 𝜀 −1 , where 𝜀 is the relative dielectric constant. If this assumption does not hold true then (14) is invalid. Such a problem may occur at least theoretically even if in total the dielectric constant is low only because of low mass density. A second assumption is that dielectric dipoles are evenly distributed within the dielectric layer. A third assumption is a low alternating current – AC component in the power supply and that the influence of the Inertial Dipole on the metal plates is negligible due to the charge concentrating on the metallic surfaces which are in contact with the dielectric material. A high AC component might disrupt electrons alignment on the plates and if the plate’s thickness is not negligible then (14) is no longer valid. Also, if the dielectric material reaches saturation and the metallic plates are thick in relation to the dielectric layer, the charge distribution on the plates can no longer be limited to the contact surfaces with the dielectric layers which also results in (14) being no longer valid. Suppose we have a high voltage ceramic capacitor of 1000Pf of Ta2O5 [15] with each plate area 1cm2 which is charged by 10,000 volts. The permittivity of vacuum is about 8.8541878128 * 10−12 Farads*meter−1. So we can calculate the distance d between the plates, 8.8541878128 * 10−12 Farads * metere−1 * 10-4 meters2 * d-1 * 25 = 10-9 Farads. That means d ~ 0.22135469532 * 10-1 mm or d ~ 0.22135469532 * 10-2 cm. Now we take into account the weight density of the Ta2O5 which is 8.2 grams perm 1cm3 volume. So we have 8.2 * 1cm * 1cm * 0.22135469532 * 10-2 cm = 0.01815108501624 grams. At 10000 volts the weight loss is of a portion of 0.04960602477676315711411588216388 of the weight of the dielectric material and the inertial dipole is attenuated by the relative dielectric constant 25 just as the electric field is. So we have 0.01815108501624 grams * 0.04960602477676315711411588216388 * 25-1 ~ ~3.60161*10-5 grams weight loss. This estimate can be much lower in a multilayered capacitor where fields cancel out or when the dielectric constant is higher and the dipoles density is not uniform. 4. Martin Tajmar experimental null results analysis Martin Tajmar [16] used a capacitor of a relative dielectric constant 4500 and a Teflon [17] capacitor with radius 50 mm and Teflon thickness d=1.5 mm and 10,000 Volts. The highly dielectric capacitor weight loss is way below the experiment scale resolution 3 * 10-4 grams due to division by 4500 of the charge which is 10-5 per 1000Pf capacitance. With a radius of 0.5cm, such a capacitor with say 6.02 grams * cm-3 density will lose about 2.077389 * 10-5 grams. Next focus is on one of the Teflon capacitors. The gravitational acceleration on the face of the Earth, about g=9.80665 meter * sec-2. By (14), the result is 7.5917876115 * 10-6 grams. This result is smaller than the resolution of 3 * 10-4 grams. The results assume ℶ = 1 in (4), (7), (13). It is important to say that unlike Martin Tajmar (sounds as Taymar), the Brazilian H4D experiment [14] used much greater capacitor areas. A significant AC ripple cannot be ruled out. 5. Particle mass ratios Motivation: solving (4) analytically is extremely hard, let alone, the more general Lagrangians that will be presented in (64) and (65) for complex Reeb vectors. One possible way to tackle this challenge is to rely on a theorem by Georges Reeb, according to which the restriction of the field to the three-dimensional foliation perpendicular to 𝑃𝜇 √𝑍 , must have a zero rotor. In other words, the field must have drains and sources, by which the divergence of the field is not zero. The result of this theorem is that as the far observer 𝑟 → 0 in source or drain of the field, particles formation is inevitable. This section will try to find a relationship between an acceleration as √|𝑎𝜇 𝑎𝜇 | = 𝜉′ 1 for some 𝜉 and the norm of the Reeb field √ |𝑈𝜇 𝑈 ∗𝜇 + 𝑈 ∗𝜇 𝑈𝜇 | = 𝜉′ 8 1 𝑟 𝑐2 𝑟 for some 𝜉′. In fact, this section considers (4) as 𝑟 → 0 as an attempt to avoid the extremely hard analytic solutions. The following section will try to reach at the Reeb field strengths of the electron, Muon and Tau Lepton. It will also try to reach the Reeb field strength for the W and Z bosons. As we shall see, for the first 3 values, the assessment is As we shall see 95 96 = 1− 1 64 + 1 192 = 95 4 , and ~1.5561985371903483965638770314399… 96 𝜋 193 192 + 63 64 − 1, which can be interpreted as the summation of two fundamental states of the field. However, to keep an open mind, other possible reasons, although less plausible, are also brought into the discussion. The only value that does not come 4 directly from this theory is . It has a compelling Quantum Mechanics source; however, other 𝜋 less plausible explanations are also considered. The last value, ~1.55619853719, is derived from maximal imbalance between gravity and anti-gravity. For the W boson, two possible field 4 4 strength coefficients are discussed and . The latter yields a higher mass for the W boson 𝜋 4 3 4 although the author tends to accept and not . 𝜋 3 In this section, equation (4) is explored in a small infinitesimal sphere, where we assume a linear relation between a far observer radius 𝑟 and acceleration 𝑎𝜇 𝑐2 = 𝑈𝜇 2 = 𝑍𝜇 − 2𝑍 𝑍 𝑘 𝑃𝑘 𝑃 𝜇 2𝑍 2 , see (1), (2). Our goal is to reduce (4) from a four-dimensional Minkowsky geometry to a three-dimensional Riemannian geometry and then to a two-dimensional Riemannian geometry of surfaces. We make the following assumption: ‖𝑎𝜇 ‖ 𝑐2 = 𝜉 𝑟𝑥 (15) Where, c is the speed of light, 𝜉 is a coefficient that depends on the field as 𝑟 → 0 and the variable 𝑥 changes with the density of the field as it passes through a two-dimensional sphere. 𝑥 is required because space-time curvature can cause such a sphere to be less than or more than 4𝜋𝑟 2. Note: A natural question due to (10) is, when does the acceleration 𝜉𝑐 2 𝑟 = 𝑒 4𝜋𝜀0 𝑟 2 √4𝜋𝜀0 𝑘 , such that 𝑒 is the charge of the electron, 𝑘 is Newton’s gravity constant, 𝜀0 is the permittivity of vacuum and 𝑐 is the speed of light? The answer is 𝑟 = 𝑒 𝜉𝑐 2 𝑘 𝑘 𝑒 √4𝜋𝜀 and for 𝜉 = 1, 𝑟 = 𝑐 2 √4𝜋𝜀 0 0 ,which by the order of the inverse of the square root of the Fine Structure Constant is smaller than the Planck length, 𝑒 𝑐2 𝑘 1 𝑒 𝑘 4𝜋𝜀0 ℏ𝑐 − √4𝜋𝜀 𝛼 2 = 𝑐 2 √4𝜋𝜀 √ 0 0 𝑒2 ℏ𝑘 =√ 𝑐3 , where ℏ is the reduced Planck constant. This calculation of course, assumes that in such a strong field, the permittivity is that of vacuum and is not affected by virtual electric fields that attenuate the electric field. It is also limited to the far observer coordinates system. We also make other assumptions as follows: 1) Assumption 1: In small radii, the energy of the gravitational field depends on the area around the source of gravity. This assumption is consistent with the paper of Ted Jacobson [18]. 2) Assumption 2: The area ratio that has a physical meaning is between a disk to which the unit vector 𝑃𝜇 √𝑍 points to and the 1 weighted Euclidean sphere ℷ ∗ 𝜋𝑟 2 so ℷ = 4. The area loss of a disk is time-like vector 𝜋 96 𝜋 𝑅𝑟 4 , where R is obtained by contracting Einstein’s tensor twice with a 24 𝑃𝜇 √𝑍 and 𝑟 is an infinitesimal radius. However, we consider 𝑅𝑟 4 . As we divide this area by Euclidean disk area, we get 1 Following are explanations to the factor . 4 𝜋 96 𝑅𝑟 4 ∗ 1 𝜋 𝑅𝑟 4 = 4 24 (𝜋𝑟 2 )−1 = 1 96 𝑅𝑟 2 . The simplest explanation: The simplest explanation is that the portion of the gravitational energy that can be used when particles decay is only from the time-like field 𝑃(0)𝜇 𝑍(0) 1 in (3.12) from which the factor comes from. This explanation is however problematic even if 𝑃(𝑖)𝜇 𝑍(𝑖) 4 are complex functions as in (3.13) because 8 scalar functions may not be sufficient to describe gravity. Blackhole thermodynamics - Bekenstein and Hawking entropy and area: see the 1 relation 𝑆𝐵𝐻 = 𝐾𝐵 4 𝐴 ℓ𝑝 2 [19] where 𝐴 is area, ℓ𝑝 is the Planck length, and 𝐾𝐵 is Boltzmann’s constant. We assume entropy is related to particles decay. Mathematically and physically compelling explanation: We return to the principles of the chronon field by Sam Vaknin [12] in which the time arrow is defined via spin and thus via orientation: There are two orientations to be considered. The first is the orientation of the foliation that is perpendicular to foliation which is perpendicular to 𝑃𝜇 √𝑍 and to 𝑈𝜇 2 𝑃𝜇 √𝑍 . The second is the plane within that . In each case only one side of a 3D 1 1 1 foliation and one side of a plane can be related to energy and ∗ = . 2 2 4 Causal triangulation explanation: A polygonal graph is a graph in which vertices on a circle relate to edges and each vertex is also connected to the center. So, for the m vertices of the polygon and one vertex of the center, the graph has m+1 vertices. We also assume m=2n for some natural number n. The graph has 2m = 4n edges, m connecting the polygon vertices, each vertex to 2 neighbors and m connecting the polygon vertices to the center. Using graph theory techniques, it is easy to see that a random walk for a large 3 1 n on such polygonal graph reaches a probability at the center and 4𝑚 at each polygon 4 vertex. The probability of moving from a vertex on the polygon to one if its two 1 1 neighbors is 3 for each neighbor and to the center 3. The probability of reaching one node 1 of the polygon from the center is . Seeing a particle as a loop with or without a center is 𝑚 beyond the scope of this paper, however, such a model under random walk reaches the 1 unique probability at the center and is worth mentioning as another approach to area 4 related to energy as 𝜋 96 𝑅𝑟 4 instead of tensor twice with a timeline vector 𝑃𝜇 √𝑍 𝜋 24 𝑅𝑟 4 , where R obtained by contracting Einstein’s and 𝑟 is an infinitesimal radius. The python code for the random walk calculations is brought here: import numpy as NP import numpy.linalg as LA print('Random walk on 24-Polygonal graph with a center.') matrix = NP.zeros((25, 25), dtype=NP.float64) a = 1/24 b = 1/3 for i in range(1, 25): matrix[0, i] = b matrix[i, 0] = a k = i + 1 if i < 24 else 1 matrix[i, k] = b k = i - 1 if i > 1 else 24 matrix[i, k] = b w, v = LA.eig(matrix) scale = v[:, 0].sum() v[:, 0] /= scale print('Eigenvector of probability:') for i in range(25): print(f'v[{i}]={v[i, 0]}') print(f'Eigenvalue {w[0]}') The output is: Random walk on 24-Polygonal graph with a center. Eigenvector of probability: v[0]=0.24999999999999997 v[1]=0.03124999999999989 ... The Causal Set interpretation (87)-(90) and its relation to the number 96 and the Fine Structure Constant cannot be ignored! Non rigid explanation: This idea is derived from a physical principle according to which a spin of a particle always either points to an observer or in the opposite direction. In this manner, the observer can only refer to the disc which is perpendicular to the spin axis and not to an entire sphere. An area ratio 𝜋𝑟 2 4𝜋𝑟 2 1 = means 0 gravity. 4 This assumption means that the delta area of a curved sphere divided by 4𝜋𝑟 2 is and not 𝛿4𝜋𝑟 2 4𝜋𝑟 2 𝛿𝜋𝑟 2 ℷ∗𝜋𝑟 2 . There could be other explanations to this assumption including a choice of 32𝜋𝐾 in (7) instead of 8𝜋𝐾 and 1 16 1 instead of in (4), however to the author’s opinion, 4 (43) does not support such other explanations. We revisit equation (4) and contract it twice with the unit vector direction 1 1 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 (𝑈 𝜇 4ℶ 2 𝑃𝜇 𝑃𝜈 𝑃𝜇 𝑃𝜈 𝑍 ) 𝑍 ∓ 𝑟2 𝑥 𝑃𝜇 √𝑍 which means a chosen time 1 = (𝑅𝜇𝜈 − 𝑅𝑔𝜇𝜈 ) 2 𝑃𝜇 𝑃𝜈 Since 𝑈𝜇 𝑃𝜇 = 0, and assuming ℶ = 1, we have around an electric charge by (15) 1 ℶ 1 (− 𝑔𝜇𝜈 2 𝑈𝜆 𝑈 𝜆 4 1 1 − 𝑈 𝑘 ;𝑘 ) = (− 2 ℶ 1 𝜉2 2 𝑟2 𝑥2 𝜉 1 ) = (𝑅𝜇𝜈 − 𝑅𝑔𝜇𝜈 ) 2 𝑍 𝑃𝜇 𝑃𝜈 𝑍 We calculated the divergence of a field of a non-geodesic acceleration from intensity (16) 𝜉 𝑟𝑥 to 0 along the distance 𝑟. The divergence 𝑈 𝑘 ;𝑘 can be either positive or negative and depends on the sign of the electric charge. We now refer to Seth Lloyd lecture [20], Fig. 3. Area gain or loss in the direction of a unit vector: As we see, to get the area loss on a disk which is perpendicular to the unit vector curvature, we need to multiply (16) by 1 𝜉2 1 𝜉 1 𝜋 𝜋 1 4 𝑟 12 2 1 𝜉2 = 𝜋 24 𝜉 𝑟 4. 𝑃𝜇 √𝑍 due to 𝜋 (− 2 𝑟2 𝑥2 ∓ 𝑟2𝑥) 24 𝑟4 = ℶ (− 2 𝑥2 ∓ 𝑥) 24 𝑟2 = 𝐴𝑟𝑒𝑎𝐿𝑜𝑠𝑠𝑂𝑓𝐴𝐷𝑖𝑠𝑘 ℶ (17) By our second assumption, the following has a physical meaning, where ℷ = 4, ℶ ∗ ℷ = 1 ∗ 4 = 4 1 𝜉2 𝜉 1 1 1 1 𝜉2 𝜉 𝜋 (− 2 𝑥2 ∓ 𝑥) 96 = ℶ∗ℷ 𝜋𝑟 2 (− 2 𝑥2 ∓ 𝑥) 24 𝑟2 = 𝐴𝑟𝑒𝑎𝐿𝑜𝑠𝑠𝑂𝑓𝐴𝐷𝑖𝑠𝑘 (18) ℷ∗𝜋𝑟 2 But x should be a ratio between an area around a charge and Euclidean area, according to assumption 2. If x is greater than 1, then by (17), the non-geodesic acceleration field density is decreased by a factor of 1 1 𝑥 . If the area ratio is smaller 1 then the non geodesic field density is increased by . So we must have the following equation: 𝑥 𝑥 = 1+ 𝐴𝑟𝑒𝑎𝐿𝑜𝑠𝑠𝑂𝑓𝐴𝐷𝑖𝑠𝑘 4𝜋𝑟2 And by (10) and (12), (18) becomes: 1 𝜉2 𝜉 1 ⟺𝑥−1= 1 𝜉2 𝜉 1 𝐴𝑟𝑒𝑎𝐿𝑜𝑠𝑠𝑂𝑓𝐴𝐷𝑖𝑠𝑘 4𝜋𝑟2 (− 2 𝑥2 ∓ 𝑥) 96 = x − 1 ⟺ 1 + (− 2 𝑥2 ∓ 𝑥) 96 = 𝑥 ⟺ 192𝑥2 ∓2𝜉𝑥−𝜉2 192 = 𝑥3 (19) The righthand side is expected to be positive around a negative charge and negative around a positive charge if we take into account the H4D experimental qualitative result [14] with imprecise balance. What are the possible values for 𝜉 if we wish to describe the electron, the Muon and the Tau lepton? It is expected that the lower value for 𝜉 and the upper value will be dictated by minimal and maximal possible values for such a field. The middle 𝜉 value which is field strength of the Muon should not be dictated by such constraints. For example, a semi classical approach to such a field can come from the understanding that a spinning Reeb field means that it is stronger in the 4 spin plane and zero in the poles. This approach dictates only one possible value of 𝜉 = . Consider the averaging of an acceleration field towards the center that depends on 𝜉 𝑐2 𝑟 π 𝑈 ∗𝜆 𝑈𝜆 +𝑈 𝜆 𝑈 ∗ 𝜆 where c is the speed of light, 8 = Then the intensity of the field with respect to the angle 𝜃 with the rotation plane should be 𝜉 𝑐2 𝑟 𝜋 𝜋 cos (𝜃). At the poles this angle is and − . In the rotation plane this angle is 0. So the 2 average field yields: 2 𝜋 2 𝑐 2 ∫02 𝜉 cos (𝜃)2𝜋𝑟cos (𝜃)𝑑𝜃 𝑟 𝜋 2 ∫02 2𝜋𝑟cos (𝜃)𝑑𝜃 = 𝑐2 𝜋 4𝜋𝑟 2 𝜉 𝑟4 4𝜋𝑟 2 Now we see that in order that the average field will be 𝑐2 𝑟 = 𝑐2 (19.1) 𝑟 then 𝜉 = 4 π which means that the field 4 strength in the rotation plane must be stronger than the factor 1 by 𝜉 = . Without other π constraints this should be field strength the defines the Muon, not as the minimal electron and the maximal Tauon in terms of mass. Of course, there are other possible explanations which will all be mentioned but this explanation is by far the simplest. We will first start with an assumption 4 𝜉 = . This assumption is also based on Ettore Majorana’s notebook [21] and on the compelling π assessment of the critical strength of the Coulomb and the Yukawa potentials [22]. It is also the well-known ratio between a star graph and a Steiner star in Euclidean spaces – star Steiner ratio in ℝ𝑑 [23]. The addition of a middle point in a ball can reduce the length of a star graph in relation to a star where the star graph is defined as straight lines between n-1 points and a single point on the sphere. And a Steiner star connects the points to the center. A physical meaning of such a ratio is that where there is a middle point, divergence of an acceleration field can be defined, where there is no such point, no such divergence can be defined. For such a case, a different value of 𝜉 should be defined. This fact is brought here as thought, not as any proof to 4 why 𝜉 = . The calculation in (19.1) is much more compelling. π Then (19) yields two solutions as follows, 192𝑥1 2 +2𝜉𝑥1 −𝜉 2 192 = 𝑥1 3 ⟹ 1 𝑥1 −1 ≅ 𝟐𝟎𝟔. 𝟕𝟓𝟏𝟑𝟑𝟗𝟖𝟖𝟓𝟎𝟐𝟐𝟎𝟐 (20) This value is surprisingly very close to the mass ratio between the Muon and the electron! 105.6583745MeV 0.5109989461MeV ≅ 206.7682826 (21) The following is an area ratio around a positive charge. The discussion about its meaning is postponed for now. 192𝑥2 2 −2𝜉𝑥2 −𝜉 2 192 = 𝑥2 3 ⟹ 1 1−𝑥2 ≅ 𝟒𝟒. 𝟔𝟑𝟗𝟓𝟓𝟎𝟏𝟕𝟓𝟗𝟔𝟒𝟎𝟏 (22) Before we continue, we need to prove another theorem which has important implications to 96 Quantum Gravity. The factor 95 is, however, not final in what will be described as Steiner Trees. Theorem 3: In Riemannian geometry, a computational model for the connection of a finite connected set of points on a sphere S2 and the center with radius r can converge in polynomial 96 time only to a minimal graph of S2 not within radius 𝑟 but within radius 𝑟 . 95 Proof: The proof of this theorem is a direct result of the complexity limit of the Minimum Steiner Tree. Finding the minimal length of such a graph is in polynomial time only above 96 95 of the minimal graph length due to [24]. As a result, to connect all the points in the sphere and its center is possible in polynomial time only for 𝑟 96 95 and we are done. The meaning of this theorem is very deep for most Quantum Gravity theories. For this specific theory, if acceleration depends on 𝑟 −1 then physically the dependence must be on 95 −1 𝑟 . 96 As a caveat, 96 95 is not believed by the author to be an absolute limit to the hardness of the Steiner Tree problem. Before continuing, a much more compelling explanation for the choice of 𝜉 = be brought. Right now, different options are described. 95 96 for the electron’s field strength will By the principle of parsimony, the electron field strength should be explainable by ground state roots of (19). Consider 𝜉 = 𝑥1 and 𝜉 = 𝑥2 in the following area ratio equations. 1 𝜉1 𝜉1 = 𝑥1 ∧ ( 2 2 𝑥1 2 𝜉2 = 𝑥2 ∧ ( 1 𝜉2 𝜉 1 + 𝑥1 ) 96 = 𝑥1 − 1 ⟹ 𝑥1 = 2 2 𝑥2 2 1 𝜉 − 𝑥2 ) 2 1 96 𝛿𝑥1 + 𝛿𝑥2 = 193 192 and 𝜉2 = 63 64 95 192 = 𝑥2 − 1 ⟹ 𝑥2 = Adding these two delta area ratios yields Definition: 𝜉1 = 193 1 192 + −1 64 63 64 =− 𝜉 = 1 + 𝛿𝑥1 + 𝛿𝑥2 = 95 96 ⇔ 𝛿𝑥1 = 𝑥1 − 1 = ⇔ 𝛿𝑥2 = 𝑥2 − 1 = 1 (22.1) −1 (22.2) 192 64 1 (22.3) 96 will be called Stability Field Strengths and 𝜉 = (22.4) 95 96 is called Joint Stability Field Strength. 𝜉 = 96 is the first candidate for the electron field strength that will be used in the Muon/electron mass ratio assessment. It is not difficult to see that for the choice of 95 𝜉 = 96, also see motivation in Appendix E, (74), (75), (79), and the surprising relation between the Fine Structure Constant and exponential perturbations of 𝜉 = polynomials yield, 95 2 1 (96) (− 2 𝑎2 95 + 95 1 ) 96 = 𝑎 − 1 ⟹ 𝑎 96 95 2 192𝑏2 −2 𝑏−( ) 96 192 96 = 𝑏3 𝑎𝑛𝑑 95 2 95 192𝑎2 +2 𝑎−( ) 96 1 (𝑎−1)(1−𝑏) 96 192 3 95 96 = 𝑎 𝑎𝑛𝑑 (− in (81)-(86), the following 95 2 1 (96) 2 2 𝑏 − ≅ 𝟏𝟐𝟐𝟎𝟐. 𝟖𝟖𝟖𝟕𝟒𝟎𝟔𝟔𝟒𝟔𝟕𝟕𝟐𝟒 95 1 ) 96 = 𝑏 − 1 ⟹ 𝑏 96 (23) (𝑎 − 1)(1 − 𝑏) answers the question of what happens when the test particle is neutral. To better understand the above expression, it is best to contract the acceleration matrix 𝐴𝛼𝛽 (3), [9] with the Levi-Civita tensor (not symbol), 𝐸𝜇𝜈𝛼𝛽 but with a possible orientation change from 𝐵𝜇𝜈 = 1 2 𝐸𝜇𝜈𝛼𝛽 𝐴𝛼𝛽 . This description is of a second plane in which the divergence of a Reeb-like acceleration vector can be of an opposite sign, perpendicular to both 𝑈𝜇 2 and to 𝑃𝜇 √𝑍 ̃𝜇 𝑈 2 1 = 𝐸𝜇𝜈𝛼𝛽 𝐴𝛼𝛽 𝑉𝜐 where 𝑉𝜐 is a unit vector 2 . One field is then of a positive charge and one of a negative charge which is the explanation for the term (𝑎 − 1)(1 − 𝑏) where 𝑎 denotes the area addition ratio around a negative charge and b is the area loss ratio around a positive charge. One would expect to see √(𝑎 − 1)(1 − 𝑏) however, roots will be discussed regarding spin 1 mass ratios. Roots of such a value also have a meaning, see appendix C, (64). Combining (20) and (23), the following holds: (𝑥1 − 1)𝟏𝟎𝟓. 𝟔𝟓𝟖𝟑𝟕𝟒𝟓𝟓𝑀𝑒𝑉 ≅ 𝟎. 𝟓𝟏𝟎𝟗𝟗𝟖𝟗𝟒𝟔𝟏𝑀𝑒𝑉 1 + (𝑎 − 1)(1 − 𝑏) 1 1 1 1 (− (1 − )2 𝑎−2 + (1 − ) 𝑎−1 ) = 𝑎 96 2 96 96 1 1 1 2 −2 1 1 + (− (1 − ) 𝑏 − (1 − ) 𝑏 −1 ) = 𝑏 2 96 96 96 2 4 1 4 1 1 + (− ( ) 𝑐 −2 + 𝑐 −1 ) = 𝑐 π 2 π 96 𝑀𝑢𝑜𝑛𝑀𝑎𝑠𝑠 ∗ (𝑐 − 1) = 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑀𝑎𝑠𝑠 + 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑀𝑎𝑠𝑠 ∗ (𝑎 − 1)(1 − 𝑏) By (23) the ratio is ~206.76828270441461654627346433699131011962890625 1+ (24) Where 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑀𝑎𝑠𝑠 ∗ (𝑎 − 1)(1 − 𝑏) = ~𝟒𝟏. 𝟖𝟕𝟓 𝒆𝑽/𝑐 2 looks like a new particle or resonance. Corroboration requires to detect excess in cosmic ~20.9 𝒆𝑉/𝑐 2 photons. Verification of this theory by Muon decays can be done by observing rare excess of 20.9376221059304 eV photons. With electron energy 0.5109989500 MeV the Muon energy is ~105.658375355 MeV. We only needed a small correction to the 2014 Muon energy from 105.6583745 MeV to 105.65837455 MeV with electron energy 0.5109989461055 MeV to arrive at the energy ratio and therefore mass ratio of the Muon and the electron. Is that a mere coincidence? The extremely small ratio error and the choices of 𝜉 = − 1 64 1 192 + and 1 1 192 𝜋 and 𝜉 = 95 96 193 highly disfavors a mere coincidence. It is 1 193 2 −2 193 ) 𝑎 + ( ) 𝑎−1 ) = 𝑎 has a biggest root 𝑎 = 2 192 192 192 1 1 63 2 −2 63 63 1 + (− ( ) 𝑏 − ( ) 𝑏 −1 ) = 𝑏 has a biggest root 𝑏 = = 1 − . The 96 2 64 64 64 64 1 important to notice that 1 + 1+ 1 4 =− (− ( = delta is a delta of energy ratios between the two stable states with field strength 96 193 coefficients 𝜉 = 96 192 and 𝜉 = shows that our choice of 𝜉 = 63 64 95 96 and roots 𝑎 = 193 192 and 𝑏 = 63 . 1 plus this delta yields 64 95 , which 96 was not at random but is the result of the summation of negative and positive area ratios for which the field strengths are equal to the biggest roots. 95 Let us define the electron field strength as 𝜉 = 96 and consider a perturbation of this value and its close link to the Fine Structure Constant. Recall (23), 1+ 1+ 1 96 1 (− (1 − 2 1 2 −2 ) 𝑎 96 + (1 − 1 96 ) 𝑎−1 ) = 𝑎 (24.1) 1 1 1 1 (− (1 − )2 𝑏 −2 − (1 − ) 𝑏 −1 ) = 𝑏 2 96 96 96 Consider a little more accurate result than the one used before in (23): 1 1 (𝑎−1) (1−𝑏) ≅ 12202.8887406646790623199 Now consider the following perturbation on 𝜉 = 95 96 =1− 1 , and raise this 96 (24.2) 95 96 to the power 1 + 𝛼 where 𝛼 is the Fine Structure Constant - FSC. We will take the assessment from (40) of the inverse FSC, about 137.0359990368270075578… and not the larger assessment 95 137.0359992990990 from the remark after (41). Our new field strength will be 𝜉′ = ( )1+𝛼 , 96 which is an exponential perturbation with the help of the Fine Structure Constant. We want to calculate the new value of the neutral area ratio of addition and subtraction in two acceleration planes, one positive and one negative, as expected from the electron Neutrino because it is electrically neutral and see what we get. Before that, please view the following illustration, in reality, it is a 4-dimensional model with two perpendicular planes. The neutral charge is of one positive two-dimensional plane and one negative two-dimensional plane, both defined by two acceleration matrices and two generalizations of two Reeb vectors to 4 dimensions. Fig. 4.: this is an over-simplification of two Reeb fields in two Symplectic Lagrangian planes: 95 And now we have for 𝜉′ = ( )1+𝛼 , 96 1 1 (𝑎′−1) (1−𝑏′) And the following shows up: (1 − ′ (𝑎′ −1)(1−𝑏 ) (𝑎−1)(1−𝑏) ) ≅ 12204.188931677483196836 1 2 − 1 (24.2) ≅ 96.8837368186132295022617 (24.3) which is remarkably close to 𝛼 −1 2−2 ≅ 96.899084185613574504714052, from (40). The fact that perturbations of the field strength 95 96 yield the Fine Structure Constant is easy to see in other cases other than (81)-(86) and (24.3). Using a simple datasheet without the accuracy of 95 1+𝛽 the Python math libraries, consider 𝜉 ′ = ( ) 96 95 where 𝛽 ≈ 1.00370694 for which ( 1 1 95 96 − 𝜉 ′ )( )−1 ≅ 25762.75334−1 then (23) yields (𝑎′−1) (1−𝑏′) ≅ 12203.54919. Now consider the 96 following function 𝜉 ′′ = 1 + ln ( 𝜉 ′ ) for which (23) yields ln ( ln((𝑎′′ −1)(1−𝑏 ′′ )) ln((𝑎′ −1)(1−𝑏′ )) 1 1 (𝑎′′−1) (1−𝑏′′) − 1)2 ≈ 137.0359991 ≅ 12204.49567, (24.4) In other cases, the inverse Fine Structure Constant can emerge from trigonometric perturbations of a higher field strength. In both cases, fractional powers of roots are involved. This is not surprising if we consider that the Fine Structure Constant must be related to electromagnetic waves, and these should be a result of perturbations of the field strength of elementary particles such as the electron or even of the Tau lepton as an upper limit of an allowed leptonic field strength. Although (24) is not a rigid mathematical proof of the mass ratios between the Muon and the electron, and although only 𝜉 = 4 𝜋 is a well understood field strength, not directly from this paper, one can argue that the result in (24) is too accurate to be ignored, especially if (24.3), (24.4) and (81)-(86) in “Appendix E” are taken into account. 𝜉 = 2 as a field strength is a critical value much higher than the highest field strength for leptons which is offered in this paper, simply because for a negative charge, the gravitational field vanishes. 192𝑥 2 +2∗2∗𝑥+22 192 3 = 𝑥 with a stable root x=1, 𝑥(𝑛 + 1) = ( 1 192𝑥(𝑛)2 +2∗2∗𝑥(𝑛)−22 3 ) . But then 192 1 𝑥−1 as an added area portion around the negative charge is undefined but with a left limit 0. So, asking whether a logarithmic scale that starts at 2 has a physical meaning is legitimate. We choose our scale to be: 95∗96 95∗96 {2 95 ln (2) 𝑛−1 ln (2) ,2 95∗96−1 95∗96 ,…,2 1 95∗96 1 95∗96 }, now consider 𝑦 = ((2 − 1) ∗ 96) −1 ≅137.050820617 and 1 (𝑛−1)∗𝑛 ≅ 137.05602888445 and it is easy to show that as n grows, ((2 . It is easy to see a nice result, 𝑦−137.0359990368270075578 137.0359990368270075578 −1 − 1) ∗ 𝑛) ≈ ≅ 96.1546032−2 so the relative error to one of the assessment of the inverse Fine Structure Constant, see (40), is nearly expressible as a power of 96. This is one good reason to search for a relation that involves 2 and powers of 96 or of 95*96 as the mathematical term that will yield the Fine Structure Constant, however, such a term should appear out of a perturbation of a field strength because the Fine Structure Constant defines the Quantum electric strength, but which field strength? Important: A leading idea is that the Fine Structure Constant should be related to perturbations of a maximal allowed field strength for leptons, i.e., the Tau lepton field strength. Any perturbation exceeding this limit must be dissipated as waves. The exploration which is performed here is not out of analytic solutions to (4) or a complex version of (4) or to the further-on mentioned (64), which may take many years to yield fully analytic solutions. It is a “reverse engineering” of Nature by assessment of (4) and field strengths in an infinitesimal limit. It will require more discussion to reach more comprehensible terms for the inverse Fine Structure Constant. Another clue to where the Fine Structure constant comes from is the following: Consider a search for the number 96 and to keep the idea simple and related to the roots of the third order Gravity and Anti-gravity area ratio polynomials. π4 Consider the following known equation: 96 = ∑ ∞ 1 4 𝑘=0 (2𝑘+1) which results from Parseval’s identity when developing the Fourier series of the function 𝑓(𝑥) = |𝑥| in (−𝜋, 𝜋). Notice that π4 the fourth root of 96 is relation to 1 is. π 1 964 π 1 964 −1≅ ≅ 1.00364948118 … . We can see what the error of this value in 1 274.01155134419542… = 1 2∗137.00577567209771179617192026613 . We may therefore search for an expression in which twice the inverse Fine Structure Constant appears. If we choose the value from (40), and not the higher value 137.0359992990990 after (41) we get the following error assessment (1 − 137.00577567209771179617192026613 ) 137.0359990368270075578038813546 2 ≅ (95.227180726406028880040436362512)2 The residual error is related to a number between 95 and 96 and the factor 2 appears again. Although it is not any mathematical proof, it is still difficult to ignore such a lead in the search for where the inverse Fine Structure comes from. It is worthy of mentioning that getting the exact Fine Structure Constant in (81)-(86) requires a very small addition 𝜉 = 95 96 + 𝜀 for some small 𝜀. Then (24) would require the mass of the Muon to be slightly higher than 105.65837455 MeV, about 105.658375 MeV. The following Python code was used to reach the result in (24), import numpy as np x1 = 1 third = 1 / 3 f = 4 / np.pi # Ettore Majorana's ring of a disk, potential factor. f2 = f * f # Iterate to most stable root. for i in range(2000): x1 = np.power((192 * x1 * x1 + 2 * x1 * f - f2) / 192, third) a = 1/(x1 - 1) # Negative charge. print('Xi = 4/Pi, a = %.48f' % a) x3 = 1 x4 = 1 f = 95 / 96 f2 = f * f # Iterate to most stable roots. for i in range(2000): x3 = np.power((192 * x3 * x3 + 2 * x3 * f - f2) / 192, third) x4 = np.power((192 * x4 * x4 - 2 * x4 * f - f2) / 192, third) c = 1/(x3 - 1) # Negative charge. d = 1/(1 - x4) # Positive charge. print('Xi = 95/96, c = %.48f, d = %.48f' % (c, d)) print('Xi = 95/96, c * d = %.48f' % (c * d)) print('Approximated mass ratio between the Muon and the electron %.48f' % (a * (1 + (x3-1)*(1-x4)))) 1 Few words about 𝜉 = 1 − 96 . What is so special about 𝜉 = 1 − 1 ideal area loss ratio and an ideal area addition ratio + 1. biggest root of 1 + root of 1 + 1 96 1 1 96 (− (1 − 2 1 (− (1 + 2 1 2 −2 ) 𝑥 64 𝑃𝜇 2 )2 𝑥 −2 − (1 + − (1 − 𝑈𝜇 𝑈 𝜇 How about null Reeb vectors √𝑍 1 192 2 1 ) 𝑥 −1 ). 64 1 192 − 1 1 96 64 ? It is twice the average of an =− 1 96 where 𝑥 = 1 + ) 𝑥 −1 ) = 𝑥 and 𝑥 = 1 − 192 1 64 1 192 is the is the biggest = 0. It is not difficult to see that in this case, the unit vector should be space-like at least in the near vicinity of the test particle as 𝑟 → 0 and 𝑈𝜇 may not be all 0 at the center of a sphere but can be a null vector. With 𝜉 = case: 1+ and 1+ From (25) 𝑐1 = From (26) 𝑏1 = 1 𝑐1 −1 4 1+(1+ 2 1 95 1 𝑐1 96 we have in this ≅ 76.38530, 1 1−𝑐2 ≅72.98648402, −1 1 1−𝑐2 4 ≅ 74.3845968, ≅70.98571137, (25) =𝑏 962 (26) 1 1 2 ) 6𝜋 1+(1− 2 ≅ 1.010204037 … , 𝑏2 = With 𝜉 = , (28) is a bit different: 3 95𝑏−1 ≅ 1.0130915 … , 𝑐2 = 1 95 2 ) 1+(1+ 96∗24 2 95 (± π 𝑐 −1 ) = 1 ± 24π = 𝑐 96 (± 96 𝑏 −1 ) = 1 ± 96 1 1 2 ) 6𝜋 and 𝜉 = 𝑐 −1 4 1 4 𝜋 ≅ 0.986556 … and 1 95 2 1+(1− ) 96∗24 1 2 √(𝑐1 −1)(1−𝑐2 ) 1 √(𝑐1 −1)(1−𝑐2 ) = 95 96 and ≅ 75.3783115, ≅71.9791462, (27) (28) (28.1) Important: Where does this 𝜉 = come from? The reader is advised to check that the average 3 4 distance between two points on the Euclidean ring is 𝜉 = . The average distance between two 𝜋 4 points on the Euclidean sphere is 𝜉 = and is left as an exercise to the reader. We may say that 3 4 4 𝜉 = 𝜋 means a geometric ring field strength and 𝜉 = 3 is a geometric sphere field strength. If we take into account particle decay through Bosons with two different field strengths, 𝜉 = 4 4 𝜋 if the Muon is involved and 𝜉 = in other cases, then there is a new interaction that is not covered by 3 the W Boson alone! 1 ≅ 98.00042535, 𝑏1 −1 We now look at: 1 1 = 96, 1−𝑏2 √(𝑏1 −1)(1−𝑏2 ) √(𝑐1 −1)(1−𝑐2 ) √(𝑏1 −1)(1−𝑏2 ) ≅ 96.99505572 (29) ≅ 1.134361808−2 (30) Roots are attributed in this case to spin 1 or 2. It is easy to see that also: (1 + (𝑐1 − 1)(1 − 𝑐2 )) ( ≈ (𝑐1 −1)(1−𝑐2 ) 1/4 ) (𝑏1 −1)(1−𝑏2 ) 91.1876 𝐺𝑒𝑉 80.3725 𝐺𝑒𝑉 ≅ 1.134561453 (31) Which is remarkably close to the ratio between the energy of the Z boson and the energy of the W boson and for W Boson of 80.3725 𝐺𝑒𝑉 the relative error of this ratio is about 1/1528961.689. For where the idea of 4th roots came from, please refer to Appendix C, (65). 4 Another research direction is to use the inverted value of 𝜉 = , i.e., 𝜉 = 𝜋 𝜋 4 in the negative and positive charge area ratio equations as in (24). That yields two new maximal roots 𝑎1 2 + 1 1 𝜋 2 𝑏1 2 + (− ( ) + 𝑏1 ) = 𝑏1 3 and 𝑏2 2 + 96 2 𝜋 (− ( ) + 𝑎1 ) = 𝑎1 3 and 𝑎2 2 + 96 2 1 4 4 1 4 2 (30), we have, 4 𝜋 𝜋 √(𝑏1 −1)(1−𝑏2 ) √(𝑎1 −1)(1−𝑎2 ) 1 96 1 𝜋 2 𝜋 (− ( ) − 𝑎2 ) = 𝑎2 3 along with the older ones 2 1 96 4 4 1 4 2 𝜋 201.6240447 ∗ 86.46523917 ≅√ 4 (− ( ) − 𝑏2 ) = 𝑏2 3 . Quite like the ratio in 2 206.7513399 ∗ 44.63955018 𝜋 ≅1.374383282 which is close to the following mass ratio between a Higgs Boson of 125.3267 GeV and a Z Boson of 91.1876 GeV which yields, 1.37438314, close to 1.374383282. It is interesting though not sufficiently accurate to draw any conclusion at this stage. The idea behind using charge equations without null Reeb vectors is because the Higgs boson is supposedly responsible for non-zero mass. From (31) and using s instead of c, √(𝑠1 − 1)(1 − 𝑠2 ) ≅ 75.3783115 …−1 and 91.1876 GeV * √(𝑏1 −1)(1−𝑏2 ) √(𝑎1 −1)(1−𝑎2 ) ∗ (1 + (𝑠1 − 1)(1 − 𝑠2 )) ≅ 125.3487702 GeV. A similar (1 + (𝑠1 − 1)(1 − 𝑠2 )) value was used in (29) as (1 + (𝑐1 − 1)(1 − 𝑐2 )). If the reasoning here is correct, the Higgs boson interacts as an electric dipole. Returning to (22) 1 1−𝑥2 ≅ 𝟒𝟒. 𝟔𝟑𝟗𝟓𝟓𝟎𝟏𝟕𝟓𝟗𝟔𝟒𝟎𝟏 and written as 1 1−𝑐 , 80372.88 MeV (1−𝑐) 1+√(𝑐1 −1)(1−𝑐2 ) 4 ≈ 𝟏𝟕𝟕𝟔. 𝟗𝟏 𝑴𝒆𝑽 (32) ≈ 𝟏𝟕𝟕𝟔. 𝟗𝟏 𝑴𝒆𝑽 (32.1) With 𝜉 = as in (28.1), (32) gets the same result for a higher value of the W Boson mass, 3 80422.57 MeV (1−𝑐) 1+√(𝑐1 −1)(1−𝑐2 ) The root, √(𝑐1 − 1)(1 − 𝑐2 ) can be better understood as a result of taking the root of a determinant of a Gram matrix of two Reeb vectors in Appendix C or is related to spin 1. The value 1776.91 MeV will be discussed in (36) with a reference. A very surprising relation between Quarks and Leptons with the same 1 1−𝑐 ≅ 44.63955017596401 as in (22) is the relation between the pole energy of the Bottom/Beauty Quark [25], [26] and the anti-Muon, this time we take the Muon value that yields in (24) along with the denominator of (23), the exact mass ratio between the Muon and the electron 105.65837455 𝑀𝑒𝑉 instead of the 2014 value 105.6583745 𝑀𝑒𝑉, 105.65837455 𝑀𝑒𝑉 (1 + √(𝑐1 − 1)(1 − 𝑐2 )) (1 − 𝑐)(1 + (𝑎 − 1)(1 − 𝑏)) = 44.63955017596401 ∗ 105.65837455 𝑀𝑒𝑉 ∗ (1 + 75.378311502572868277860789009693−1 ) ∗ (1 + 12202.88874066467724−1 )−1 ≅ 4,778.7223164425585113299 MeV ≈ 𝟒. 𝟕𝟖 𝐆𝐞𝐕 Which is equivalent to: 𝑃𝑜𝑙𝑒𝐸𝑛𝑒𝑟𝑔𝑦𝑂𝑓𝐵𝑜𝑡𝑡𝑜𝑚𝑄𝑢𝑎𝑟𝑘∗(1−𝑐) (1+√(𝑐1 −1)(1−𝑐2 )) = 𝑀𝑢𝑜𝑛𝐸𝑛𝑒𝑟𝑔𝑦 (1+(𝑎−1)(1−𝑏)) (33) In which the root in the left denominator is attributed to spin 1. New physics? Looks like it! A.M. Badalian’s prediction 4,778 MeV [26] is too close to 4,778.72 MeV to be ignored. The outcome of the Muon being the electro-gravitational energy of the pole energy of the Bottom Quark is as follows: a) Lepton universality should be broken in decays of anti-Bottom Quark that involve Muons. b) High energy p-p collisions can no longer be considered for the calculation of the W Boson mass. Before we proceed, it is worthy of mentioning the following Simon Plouffe identity [27]: consider the functions, 𝑆𝑛 (𝑟) = ∑∞ 𝑘=1 1 𝑘 𝑛 𝑒 𝜋𝑟𝑘 −1 then there is a well-known relation between 𝜋 and 96, 𝜋 = 72𝑆1 (1) − 96𝑆1 (2) + 24𝑆1 (4). Notice that the sum of the positive coefficients 72 + 24 = 96 and the negative coefficient is -96. While this identity is not a direct relation between 𝜉 = and 𝜉 = 95 96 =1− 1 , it does show an example of how 𝜋 and 96 can be related to each other 96 through Zeta functions in a simple and straight forward manner. A deeper and a very surprising relation will be seen in a note after (40). 4 𝜋 6. The exact inverse Fine Structure constant – critical imbalance between gravity and anti-gravity The following endeavor originated in the search for a field strength coefficient near 𝜋 2 for quite a simple reason. If a motion in a small circle is with the constant velocity c, then after half a circle the velocity will be -c. The difference c-(-c) is 2c and the time between the two velocity measurements is acceleration 2 𝑐2 𝜋 𝑟 𝜋𝑟 𝑐 𝜋𝑟 so 2𝑐( )−1 = 𝑐 2 𝑐2 𝜋 𝑟 while the acceleration of the motion is 𝑐2 𝑟 . The inferred can be interpreted only when the velocity can take one of two values c or -c, or in other words when velocity itself is quantized. The correction in this situation is by a factor and 𝜋 2 𝑐2 2𝜋 𝑟 = 𝑐2 𝑟 . Given a radius r and an upper speed limit c, the correction coefficient considered as a possible upper field strength coefficient. The way a coefficient near 𝜋 2 𝜋 2 𝜋 2 should be was found will be discussed along with its relation to the inverse Fine Structure Constant. The fine structure constant is surprisingly reached through the mass ratio between the Tau lepton and the Muon and an interesting perturbation of the field strength of the Tau lepton that will be found in this section. Recommended reading for this section is Appendix E, (70) - (79). Note: The more advanced parts of this section require basic knowledge of electrical engineering and especially a good understanding of the trivial subject of Dissipation Factor and Loss Tangent and especially of Power Factor [28]. Note: Why dissipation factor? The reason is that any perturbation of the Reeb field, which behaves as acceleration, above a maximal allowed limit, must be emitted and in mainstream physics, the electromagnetic field is dissipated as photons. The denominator 1 + √(𝑐1 − 1)(1 − 𝑐2 ) in (32), (33) and (1 + (𝑎 − 1)(1 − 𝑏)) in (24) can be used together to yield a nice result that seems to be more than just a mathematical coincidence. Consider the following imbalance equation as in (23) of negative and positive charge: 1 1 −2 1 + (− 𝜉 2 𝑔1 + 𝜉𝑔1 −1 ) = 𝑔1 96 2 1 2 −2 1 1 + (− 𝜉 𝑔2 − 𝜉𝑔2 −1 ) = 𝑔2 2 96 1 1 Such that (𝑔1 − 1)−2 = 2 (1 − 𝑔2 )−1 (34) With biggest roots 𝑔1 ≅1.003629541 and 𝑔2 ≅0.969877163. 𝑔1 means an area portion ~𝟐𝟕𝟓. 𝟓𝟏𝟔𝟗𝟑−𝟏 is added around a negative charge and ~𝟑𝟑. 𝟏𝟗𝟕𝟒𝟎−𝟏 of the area is subtracted around a positive charge, which reflects a possibly maximal allowed gravitational imbalance between negative and positive charge. A calculation that uses an electronic datasheet, yields, 1 𝜉 ≅ 𝟏. 𝟓𝟓𝟔𝟏𝟗𝟖𝟓𝟑𝟕𝟏𝟗𝟎𝟑𝟒𝟖𝟒, (𝑔1 − 1)−2 ≅ 𝟏𝟔. 𝟓𝟗𝟖𝟕𝟎𝟐𝟎𝟑 (35) which is close to the known mass ratio between the Tauon and the Muon, ≅16.817 where 𝜉 denotes a maximal allowed coefficient. Multiplying this value by 1 + √(𝑐1 − 1)(1 − 𝑐2 ) from (32), (33) and dividing by (1 + (𝑎 − 1)(1 − 𝑏)) from (24) yields, 𝑀𝑢𝑜𝑛 105.6583745 𝑀𝑒𝑉 (1+(𝑎−1)(1−𝑏)) ≅ √𝑔1 −1 𝑇𝑎𝑢𝑜𝑛 1776.9127923826 𝑀𝑒𝑉 (36) (1+√(𝑐1 −1)(1−𝑐2 )) Which is ≅16.81752914. So, this calculation predicts a Tauon energy of about 1776.9127923826 MeV which agrees with [29]. Please note the remark after (28.1) for a possible additional W Boson. We now need to check the consistency of (36) with (32) as a test to this theory. We take 1 1−𝑥2 from (22) and 1 √(𝑐1 −1)(1−𝑐2 ) from (28) and check the following: 1776.91279322344…MeV∗(1+√(𝑐1 −1)(1−𝑐2 )) 1−𝑥2 ≅ 80372.8876666694𝑀𝑒𝑉 (36.1) Which is consistent with (32) but less with (32.1) of a higher W Boson energy as the approximation of the W Boson’s energy with 𝜉 = 4 𝜋 Boson energy is a bit higher. (35) is strikingly related to (20) and (22). and 4 2 𝜋 4 𝜋 192𝑦1 2 −2( )𝑦2 −( ) 192 4 and a null Reeb vector. For 𝜉 = the W 3 4 4 2 192𝑦1 +2( )𝑦1 −( ) 𝜋 𝜋 = 𝑦2 3 in the following way: 2 192 = 𝑦1 3 Assessing the following yields, − 1 1 log (𝑦1 ) log (𝑦2 ) and on the other hand from (35), 1 1 (𝑔1 −1) (1−𝑔2 ) ≅ 9147.571874743285661679692566 ≅ 9146.446148044115034281276166 The relative error in these two values in relation to (𝑔 1 1 1 −1) (1−𝑔2 ) (36.2) (36.3) is Relative error ≅ 8124.926018710571952397003770−1 . Please note that for a small d the following holds. 1 log (𝑑+1) and also 1 𝑑 ≈− 1 1 𝑑 ≈ . This relation alludes to a possible exponential relation between log (1−𝑑) the roots of (20), (22) and the roots of (35) but before we actually check an exponential perturbation on the field strength 𝜉 ≅1.5561985371903483965638770314399 from (35) we notice the following for the same field strength coefficient of (35): 2 cos(𝜉) ≅ 2 cos(1.5561985371903484) ≅ 137.011909869, (37) tan−1(952 962 (1 − 𝑔2 )+4 ) ≅1.5561948778250207190765973767615 (38) remarkably approximate 𝜉 ≅1.5561985371903484 from (34), (35). 𝐸𝑟𝑟𝑜𝑟 = 𝜉−(952 962 (1−𝑔2 )4 ) 𝜉 ≅ 425,263.60132816790517958824157133−1 (39) In terms of electrical engineering Dissipation Factor and Loss Tangent, we can write, 𝐷𝐹 = 952 962 (1−𝑔2 )−4 ≈ tan (𝜉) where the numerator is known as the Resistive Power Loss and the denominator as the Reactive Power Oscillation. It is expected that an oscillating charge will generate oscillation in area due gravity changes, however, it is not expected that the area portion that is lost due to gravity will appear as the power of 4. This is a very rare property that connects between trigonometry and the electro-gravity polynomials (34). We can get from this relation two insights, the first is that if (37) is not a mathematical coincidence, then the inverse Fine Structure constant should come out of a trigonometric function and a numbers relation. The second is that 952 962 (1 − 𝑔2 )4 should be part of this equation. We may think that perhaps scaling of the value of 𝜉 in a rational way, will yield the exact inverse Fine Structure Constant. So we want to find some d such that 2 1 𝑑 cos(1.5561985371903484∗(1+ )) will yield the constant we are looking for. We will soon find such d, 𝑑 ≅ 606400.8 that complies with [30] and we get, 2 cos(1.5561985371903484∗(1+ 1 )) 606400.8 ≅ 137.0359990462475253. The motivation for this endeavor is taken from electrical engineering [28] where the cosine term means a ratio between delivered power and actually measured power in motors and other electric devices. In our case, we are interested in the ratio between radiation's energy and the energy it delivers upon interaction. Until now, d is not very interesting because we could not find d out of any new theory. Well, not 1 very accurate. First, (1 − 𝑔2 )−4 ≅ 607276.5368006824282929 ≈ 606400.8 and 2 2 cos(1.5561985371903484∗(1+2(1−𝑔2 )4 )) ≅ 137.0359643018112763 If we test the following values for 𝑑 ≅ 606400.8 we get: 954 𝑑 ≅ 134.3181357940161.., 140.0635619214.. and the geometric average of these two values is ( 1 954 964 2 𝑑 𝑑 ) = 952 962 𝑑 964 𝑑 ≅ ≅ 137.1607689. It is not difficult to see the following: As a result of the conclusions of (38), (39), the exact inverse Fine Structure Constant was found by the following, although some aspects of the following calculation are not resolved yet. We put together (20), (22), (34), (35), (37), (35) 𝜉 ≅ 1.5561985371903484 1 2 (1 − 𝑔2 )−4 ≅ 607276.5368006824282929, and from 1 1 4 2 −2 4 −1 1 1 + (− ( ) 𝑎 + 𝑎 ) = 𝑎 ⟹ ≅ 206.75133988502202 π 96 2 π 𝑎−1 1 4 2 −2 4 −1 1 1 ≅ 44.63955017596401 1 + (− ( ) 𝑏 − 𝑏 ) = 𝑏 ⟹ 2 π 1−𝑏 π 96 1 1 𝑑 = ( (1 − 𝑔2 )−4 )1+(𝑎−1)(1−𝑏) ≅ 𝟔𝟎𝟔𝟒𝟎𝟏. 𝟎𝟑𝟕𝟐 ≈ 𝟔𝟎𝟔𝟒𝟎𝟎. 𝟖 2 2 1 𝑑 cos(1.5561985371903484∗(1+ )) ≅ 𝟏𝟑𝟕. 𝟎𝟑𝟓𝟗𝟗𝟗𝟎𝟑𝟔𝟖𝟐𝟕𝟎𝟎𝟕𝟔 ≈ 𝟏𝟑𝟕. 𝟎𝟑𝟓𝟗𝟗𝟗𝟎𝟑𝟕 (40) 1 1.5561985371903483965638770314399 ∗ (1 + ) exceeds the maximal allowed value of the 𝑑 field strength 𝜉 = 1.5561985371903483965638770314399 and therefore must account for emission of what we know in mainstream physics as photons. 1 Note: 𝑝 = ((𝑎 − 1)(1 − 𝑏))− 2 ≅ 96.0691772148863 is a very special number in the following property that bridges between area ratios and powers as follows, denote 𝑠 = 1 2 (1 − 𝑔2 )−4 then 𝑠 1 ) 1+(𝑎−1)(1−𝑏) ( ≈ 𝑠(2 − 1 ) or written as numbers 606401.0372 ~ 962 (𝑎−1)(1−𝑏) 1 606401.0194 with a relative error of about 34,109,836.56-1. An exact equality, 𝑠 1+𝑝−2 = 𝑝2 𝑠 (2 − 962 ) ≅ 606401.0371, follows from replacing 𝑝 = ~96.0691772148863 with p = ~96.06917582 with a relative error in 96.0691772148863 = ((𝑎 − 1)(1 − 𝑏))−1/2 of ~1.45953 * 10-8. If the reader still thinks (40) is a fluke of chance, then this note does not agree with such a hypothesis. Also note that p comes from (20), (22) which resulted in (24). See Python code and it’s more exact output in Appendix F. Another result is by finding the variable s where a and b are given in (40): 952 ∗962 ( 𝑠 1+(𝑎−1)(1−𝑏) ) 1+(𝑎−1)(1−𝑏) 952 ∗ 962 ) ( 𝑠 = 2 1 )) cos (𝜉(1+ 1 ( ) 1+(𝑎−1)(1−𝑏) 𝑠 ⟹ ≅ 𝟏𝟑𝟕. 𝟎𝟑𝟓𝟗𝟗𝟗𝟎𝟑𝟔𝟒𝟐𝟖𝟖𝟕𝟔𝟐𝟓𝟐 137.0359992990990, 𝑠 ≅607280.4243559269234538 and 𝑠1/(1+(96∗95) 606394.43614689458627253770 With 𝑎 = 𝑔1, 𝑏 = 𝑔2 from (34), (𝑎 − 1)(1 − 𝑏) yields in (41), 𝑠 ≅ 607279.477540519786998629570007 and ( 𝟏𝟑𝟕. 𝟎𝟑𝟓𝟗𝟗𝟗𝟐𝟑𝟒𝟗𝟓𝟖𝟒𝟔𝟕𝟐𝟒𝟏𝟎𝟖𝟓𝟔𝟎𝟎 (41) 952 ∗962 𝑠 1+(𝑎−1)(1−𝑏) ) −1 ) ≅ ≅ The latter choice needs to be better explained as follows: Consider the Airy function Bi, Airy functions Ai and Bi are very popular in Quantum Mechanics [31], 1 ∞ 𝑡3 𝑡3 𝐵𝑖(𝑥) = ∫ [exp (− + 𝑥𝑡) + sin ( + 𝑥𝑡)] 𝑑𝑡 3 3 𝜋 0 The reason Bi(x) is tested here is because it maps the highest 𝜉 value from (34) near 2 and along with Ai(x) it is used to describe the wave function of a particle in a triangular potential well. Table 1. 𝜉 Electron, Muon, 4 𝜋 Airy function 𝐵𝑖(𝜉) 95 96 Tauon, 1.556198537190348396563877 0314399… 2 arccos ( ) 137.0359990368270078. . ) 2 arccos ( ) 137.0359992349584672. . 1.1977758259 63505749636 1.5153444515 1380815816 1.9895372119 25930116713 1.9895424786 76644522794 1.9895424787 19955590708 We can see a surprising possible relation between 1 2 − 𝐵𝑖(𝜉) 1.2465343632 91960163318 2.0633210599 2280961894 95.576818809 731802443136 4884 95.624954430 323135605547 95.624954826 365255405784 1 , see (19), a, b ((a−1)(1−b)) are the largest roots, a>1, b<1. 110.46668611244152202743862 2899353504180908203125 96.069177214886309457142488 099634647369384765625 95.637054262686888250755146 14582061767578125 1 1 , 2−𝐵𝑖(𝜉) ((a−1)(1−b)) and the Fine Structure Constant. This relation is one of the motivations to try the value 1 + (a − 1)(1 − b) in (41) where a = g1 and b = g 2 in (34). Another idea is to solve the following equation where s is given by (34) and p is a variable: 1 𝑠 = ( (1 − 𝑔2 )−4 ) ≅ 607276.536800682428292930 2 952 ∗962 ( 1+1/(𝑝∗𝑝) 952 ∗ 962 ) ( 𝑠 𝑠 1+1/(𝑝∗𝑝) ) = 2 1 )) cos (𝜉(1+ 1 ) ( 𝑠 1+1/(𝑝∗𝑝) ⟹ (42) ≅ 𝟏𝟑𝟕. 𝟎𝟑𝟓𝟗𝟗𝟗𝟎𝟑𝟓𝟕𝟒𝟕𝟏𝟖𝟏𝟓𝟓𝟏, 𝑝 ≅ 𝟗𝟔. 𝟎𝟕𝟎𝟔𝟔𝟔𝟔𝟕𝟎𝟑𝟎𝟓𝟖𝟒𝟎𝟐𝟖𝟓 For comparison, if we set p=96 in the right-hand side of (42) we get the value 137.035999086935760260530515. Combining (41) and (42) we find a numerical attractor at 1 (42) with 𝑠 ≅ 𝟔𝟎𝟕𝟐𝟕𝟔. 𝟓𝟑𝟔𝟖𝟎𝟎𝟔𝟖𝟐𝟒𝟐𝟖𝟐𝟗𝟐𝟗𝟑𝟎𝟏𝟐𝟔𝟐 ≅ ( (1 − 𝑔2 )−4 ), 𝑠 2 ( 1 ) 1+1/(𝑝∗𝑝) ≅ 𝟔𝟎𝟔𝟒𝟎𝟏. 𝟎𝟔𝟒𝟐𝟗𝟔𝟖𝟏𝟐𝟔𝟑𝟑𝟗𝟖𝟑𝟕𝟗𝟏, 𝜉 ≅ 𝟏. 𝟓𝟓𝟔𝟏𝟗𝟖𝟓𝟑𝟕𝟏𝟗𝟎𝟑𝟒𝟖𝟒 from (35). Before we close this discussion, it is nice to mention another relation (1 − ln ((1 + 1 137.035999035747181551 ) 137.035999035747181551 ))−1 ≅ 275.4045237287 ≈ 275.51693 ≅ 1 𝑍 (𝑔1 − 1)−1 in (43.10). That is not a total surprise because (1 − ln ((1 + ) ))−1 ≈ 2𝑧 for big z. 𝑍 Reverse engineering Nature – Looking for simple but not random relations In this section a much less significant result than (24), (40), remark after (40), (41), (42), will be considered as an interesting course of research. This time, an approximation of the inverse Fine Structure Constant will not be as nearly as accurate and will not be a result of exponential perturbations of a Reeb field strength. The search for meaningful field strength coefficients for the electron, Muon and Tau lepton 95 4 reached the following 𝜉 ∈ { , , ~1.5561985371903483965638770314399} 96 𝜋 But these field strength coefficients did not appear out of solutions to equation (4). In fact, there has been no collaboration with mainstream physics to reach such solutions and especially to the complex form of (4). The analytic solutions of such an equation make take decades and without collaboration on solving the Lagrangians in (4), (64), (65), other approaches are required in order to convince the reader that the choices of field strength coefficients are not a mere mathematical pareidolia. The assessment of the mass ratio between the Muon and the electron in (24) is already with a sufficiently small error to trigger interest, especially when considering the 4 simplicity of (24) and that the choice of came out of an existing theory [22]. (40), the remark 𝜋 after (40), (41) and (42) are also strong indicators that this research is on the right path. It will be wrong not to mention other findings which are straight forward from the method which had been presented in (16), (17), (18), (19) and the first interesting result (20). In this method, the Reeb 𝑃𝜇 vector term was collapsed with the non-geodesic or accelerated time direction 1 1 the contraction (𝑈𝜇 𝑈𝜈 − 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 4 With acceleration field 𝜉𝑐 2 𝑟 2 , where 𝜉 = 4 𝜋 𝑃𝜇 𝑃𝜈 𝑃 𝜇 𝑃 𝜈 𝑍 ) 𝑍 1 𝜉2 2 𝑥2 𝜉 ∓ ) 𝜋 𝑥 24 and we saw that resulted in (20), (22). denotes the field strength and 𝑥 is the adjustment factor of the acceleration field because of area loss, we used the term (− (− √|𝑍| 1 𝜉2 2 𝑟 2𝑥2 ∓ 𝜉 ) 𝜋 𝑟 2 𝑥 24 𝑟4 = 𝑟 2 to express area loss due to a gravitational field at small 𝑟 in the far observer coordinates. Now it is time to look at area loss in a direction perpendicular to the direction of time, namely the momentum direction in spacetime, or as expressed through a bivector derived from a unit vector, consider 2𝑈 ∗𝜇 𝑈 ∗𝜈 complex bivector 𝑈 ∗𝜆𝑈 1 𝜆 +𝑈 𝑈𝜇𝑈𝜈 𝑈 𝜆 𝑈𝜆 𝜆𝑈 ∗ 𝜆 1 and for the sake simplicity, the contraction is not with a . From 𝑈𝜇 𝑃𝜇 = 0, it is easy to see the following, (𝑈𝜇 𝑈𝜈 − 2 𝑔𝜇𝜈 𝑈𝜆 𝑈 𝜆 − 2𝑈 𝑘 ;𝑘 4 𝑃𝜇 𝑃𝜈 𝑍 ) 𝑈𝜇 𝑈𝜈 𝑈𝜆 𝑈𝜆 1 = 𝑈𝜆 𝑈 𝜆 = 8 1 𝜉2 2 𝑟2 𝑥2 (42.1) The latter is to achieve a reduction of the curvature calculation from Lorentzian to Riemannian geometry. 𝑈 𝜇𝑈𝜈 𝑈𝜇𝑈𝜈 Caveat: Notice that using 𝑈 𝜆𝑈 and not |𝑈 𝜆 𝜆 𝑈𝜆 is done here in order to achieve 𝑔𝜇𝜈 | 𝑈𝜇 𝑈𝜈 𝑈𝜆 𝑈𝜆 = +1 as expected from a unit vector in (+,-,-,-) metric convention. The reader may criticize this choice of a bivector because Reeb vectors in this paper are space-like and not time-like because they represent non-geodesic acceleration as a result of misaligned events in an observer spacetime object. Multiplying by 𝜋 24 𝑟 4 due to [20], see lecture of Seth Lloyd, and dividing by 4 times the area of an Euclidean disk, due to assumption 2 after the note after (15), yields, From which 1 𝜉2 192 𝑥3 2 = 1 𝜉2 𝜋 2 𝑟 2 𝑥3 2 24 𝑟4 192𝑥3 2 −𝜉 2 192 1 4𝜋𝑟 2 = 𝑥3 − 1 (42.2) = 𝑥3 3 (42.3) Which is an iterative equation that converges to the most stable root, a technique that had been used in all previous third order polynomial equations. Solving for 𝜉 = 4 𝜋 as in (20), (22), yields, (𝑥3 − 1)−1 ≅ 120.410611116112391982824192382395267486572265625 (42.4) (𝑥1 − 1)−1 ≅ 206.751339885022019871030352078378200531005859375 (1 − 𝑥2 )−1 ≅ 44.63955017596401120272275875322520732879638671875 And the following calculation yields an interesting result, 1 ln ((𝑥1 − 1)−1 (1 − 𝑥2 )−1 (𝑥3 − 1)−1 )2 2−2 1 ≅ ln(1111304.0650477090384811162948608398437)2 2−2 ≅ 137.0341023246677139013627311214804649353 (42.5) Which is a surprisingly simple and unexpected approximation of the inverse Fine Structure Constant. The relative error of (42.5) in relation to the result in (40) is about 72249.23316-1 which is not even closely significant as (40), (41), (42) or the remark after (40) and yet, if this result joins other approximations of the inverse Fine Structure Constant in this paper, it is not wise to ignore (42.5). In (24.3), (40), (41), (42), (81)-(86), the inverse Fine Structure Constant comes out of exponential field perturbations as in (24.3), (40), (41), (42) or as exponential 4 functions of 2 or with coefficients (95*96)-1 or 95 and 96 as seen in (81)-(86). Notice that both 𝜋 (24.3) and the last result, involve the square root of 2. Hypergeometric tests - Dr. Sam Vaknin’s suggestion from 2013 A suggestion from Dr. Sam Vaknin regarding the possible solutions of the equations of the Geometric Chronon Field Theory was that they are related to Hypergeometric functions [32]. His idea was lately checked regarding the stable roots of third order polynomials of gravity and antigravity, area ratio loss and gain, see (22.1) and (22.2). The stable field strength coefficients were defined as 𝜉 = 193 192 = 1+ 1 192 for negative charge and 𝜉 = charge. The summation of the two deltas + 1+ 1 192 − 1 64 =1+ 1 192 − 3 192 = 1− 1 96 = 1 192 95 − 1 64 63 64 =1− 1 64 = 1− 3 192 for positive to 1 yields the field strength coefficient 𝜉 = . The question is what do these values 96 1 192 and 3 192 teach us about any possible grand theory of particle physics? 1,3 and 192 with 192 in the denominator should hint us about such a theory. As we saw in (40), (41), (42) a key number in the calculation of the positive perturbation over 𝜉 was 𝑠 = 0.5/(1 − 𝑔2 )4 ≅ 607276.536800682428292930126190185546875, see (42). Can this number be a result of combinatorial mixing by the Gauss hypergeometric function 2F1? The question is if 2f1 (a, b, c, z) = ∑ { ∞ 𝑘=0 (𝑎)𝑛 (b)𝑛 𝑧 𝑛 (𝑐)𝑛 𝑛! , such that (𝑞)𝑛 = 1 𝑛=0 } can yield such a number in a meaningful way. | 𝑞(𝑞 + 1)(𝑞 + 2) … (𝑞 + 𝑛 − 1) 𝑛 ≥ 1 If Dr. Sam Vaknin was right, we may be able to find a meaningful z that solves 2f1 (−3,1,192, z) = 1 − 2(1 − 𝑔2 )4 (42.6) This is exactly what was done numerically. The result was very surprising, and it is very unlikely that it is a fluke of chance: 𝑧≅ 2 (137.0362714026169470571403508074)2 (42.7) The relative error of 137.0362714026169470571403508074 from the assessment 137.0359990368270075578… in (40) is about 137.0359990368270075578−137.0359990368270075578… 137.0359990368270075578… ≅ 503132.1997830774052999913692−1 Also, quite near the higher value 137.0359992990990… after (41). It is quite compelling to say that Dr. Sam Vaknin was right already back then in 2013. There is even stronger evidence in his favor. Consider a second order perturbation on the hypergeometric coefficients -3, 1: 2f1 (−3 ∗ and 2f1 (−3 63 193 2 )≅ ,1∗ , 192, (137.0359990368270075578 … )2 64 192 (42.8) 63 2 193 ) , 1 ∗ , 192, (137.0359990368270075578 … )2 64 192 (42.9) 1− ∗ ≅1− 1 607299.792079592822119593620300292968750 1 607299.806042340234853327274322509765625 Comparing the right hand side denominator to 1 1 2 (1−𝑔2 )4 ≅ 607276.536800682428292930126190185546875 from the remark before (40) and from (42), the results in (42.8) and (42.9) are very interesting although not within the ranges of (40)(42) with a highest value of 137.0359992990990… . Instead of have 192 63 64 if we consider all the powers of − = ∑∞ 𝑘=0 ( 191 1 ) 192 𝑘 1 64 we have 64 65 1 𝑘 1 = ∑∞ 𝑘=0 (− ) and with + 192 we 64 Consider a second order perturbation on the hypergeometric coefficients -3, 1: 2f1 (−3 and ∗ 192 2 64 )≅ ,1∗ , 192, (137.0359990368270075578 … )2 191 65 (42.8.1) 192 64 2 ) , 1 ∗ , 192, (137.0359990368270075578 … )2 191 65 (42.9.1) 1− 2f1 (−3 ∗ ≅1− 1 607135.055724701262079179286956787109375 1 607135.069557101931422948837280273437500 Here is the code in Python for (42.6) and (42.7): import numpy as NP from scipy.special import hyp2f1 as SCIPY_SPECIAL_hyp2f1 a = 137.035999036827007557803881354629993438720703 q = 607276.536800682428292930126190185546875 #s = NP.power(q * 2, 0.25) s = NP.sqrt(NP.sqrt(q * 2)) s = NP.sqrt(s * s * s * 0.25) print(f's={s:.42f}') # Was a numerical analysis output: w = 137.0362714026169470571403508074 u = 1/(w/a - 1) print(f'u={u:.42f}') r = SCIPY_SPECIAL_hyp2f1(-3, 1, 192, 2/(w ** 2)) r = 1/(1-r) r /= 607276.536800682428292930126190185546875 r = 1/(1-r) r /= s print(f'r={r:.42f}') r = SCIPY_SPECIAL_hyp2f1(-3, 1, 192, 2/137.035999036827007557803881354629993438720703 ** 2) r = 1/(1-r) r /= 607276.536800682428292930126190185546875 r = 1/(1-r) print(f'r={r:.42f}') Here is the code in Python for (42.8), (42.9) import numpy as NP from scipy.special import hyp2f1 as SCIPY_SPECIAL_hyp2f1 Xi = 1.556198537190348396563877031439915299415588378906 r1 = \ SCIPY_SPECIAL_hyp2f1(-3 * 63/64, 1 * 193/192, 192, 2/137.035999036827007557803881354629993438720703 ** 2) r1 = 1/(1-r1) print(f'1/(1-r1)={r1:.33f} compared to' f' 607276.536800682428292930126190185546875') inverse_alpha1 = 2/NP.cos(Xi*(1+1/NP.power(r1, 1/(1+1/(95*96))))) print(f'Inverse alpha(r1) {inverse_alpha1:.33f}') r2 = \ SCIPY_SPECIAL_hyp2f1(-3 * 193/192, 1 * 63/64, 192, 2/137.035999036827007557803881354629993438720703 ** 2) r2 = 1/(1-r2) print(f'1/(1-r2)={r2:.33f} compared to' f' 607276.536800682428292930126190185546875') inverse_alpha2 = 2/NP.cos(Xi*(1+1/NP.power(r2, 1/(1+1/(95*96))))) print(f'Inverse alpha(r2) {inverse_alpha2:.33f}') r = r1 / r2 r = 1/(1-r) print(f'1/(1-r1/r2)={r:.33f}') r3 = \ SCIPY_SPECIAL_hyp2f1(-3 * 64/65, 1 * 192/191, 192, 2/137.035999036827007557803881354629993438720703 ** 2) r3 = 1/(1-r3) print(f'1/(1-r3={r3:.33f} compared to' f' 607276.536800682428292930126190185546875') inverse_alpha3 = 2/NP.cos(Xi*(1+1/NP.power(r3, 1/(1+1/(95*96))))) print(f'Inverse alpha(r3) {inverse_alpha3:.33f}') r4 = \ SCIPY_SPECIAL_hyp2f1(-3 * 192/191, 1 * 64/65, 192, 2/137.035999036827007557803881354629993438720703 ** 2) r4 = 1/(1-r4) print(f'1/(1-r4)={r4:.33f} compared to' f' 607276.536800682428292930126190185546875') inverse_alpha4 = 2/NP.cos(Xi*(1+1/NP.power(r4, 1/(1+1/(95*96))))) print(f'Inverse alpha(r4) {2/NP.cos(Xi*(1+1/NP.power(r4,1/(1+1/(96*96))))):.33f}') r = r3 / r4 r = 1/(1-r) print(f'1/(1-r3/r4)={r:.33f}') print(f'(1/Alpha1+1/Alpha3)/2: ' f'{(inverse_alpha1+inverse_alpha3)/2:.33f}') print(f'(1/Alpha2+1/Alpha4)/2: ' f'{(inverse_alpha2+inverse_alpha4)/2:.33f}') 7. The mass hierarchy ℏ𝑐 By (13) and considering the Planck mass √ and the Fine structure constant Alpha: 𝐾 ℏ𝑐 𝑒2 √ ∗ 𝐾 4𝜋𝜀 0 ℏ𝑐 = 2𝑒 2√4𝜋𝐾𝜀0 = 2𝑒 √16𝜋𝐾𝜀0 = 𝑃𝑙𝑎𝑛𝑐𝑘𝑀𝑎𝑠𝑠 ∗ √𝐴𝑙𝑝ℎ𝑎 (43) So, multiplication of the Plank mass by the square root of the Fine Structure Constant yields twice the electro-gravitational mass of a charge e! If we take 𝜉 ≅ 1.5561985371903484 from (35) to be the maximal allowed field coefficient of an electric charge, then the field around a single charge as a normalized quantity is obtained as 11 1 𝑃𝑙𝑎𝑛𝑐𝑘𝑀𝑎𝑠𝑠 ∗ √𝐴𝑙𝑝ℎ𝑎 = 𝜉 𝜉2 𝑒 √16𝜋𝐾𝜀0 (44) Now we recall from (24) the following root a around a negative charge: 1+ 1 96 1 95 95 (− ( )2 𝑎−2 + ( ) 𝑎−1 ) = 𝑎 ≅ 1 + 192.0463944−1 2 96 96 We take from (24), (40), (𝑎 − 1)(1 − 𝑏) ≅ 11 𝑃𝑙𝑎𝑛𝑐𝑘𝑀𝑎𝑠𝑠∗√𝐴𝑙𝑝ℎ𝑎 𝜉2 ( 𝑀𝑒 1 206.75133988502202 ∗ 44.63955017596401 (𝑎−1)(1−𝑏) ) (45) and calculate ≅ 1 + 192.04864774452−1 (46) Where 𝑀𝑒 ≅ 0.5109989461 𝑀𝑒𝑉, e is the electron’s charge 1.602176634×10-19 Coulombs, K is Newton’s constant of gravity 6.674×10−11 m3⋅kg-1⋅s-2, Planck mass 1.22091×1022 MeV, from (40) 𝐴𝑙𝑝ℎ𝑎 ≅ 137.0359990368270076−1 . The relative error between (46) and (45) is 192.04864774452−192.0463944 ≅ 85,227.266539382−1 . 192.0463944 that 𝜉 = 95 96 We are also led to the following conclusion is the Reeb field strength coefficient of the electron field, 𝜉 = 4 𝜋 is the Muon field strength and from the solution to (35) 𝜉 = 1.5561985371903484 … is the field strength coefficient of the Tau lepton. Of course, a lot of work has to be done to achieve exact analytic solutions to (4) and as we shall see also to (64), because only 𝜉 = 4 𝜋 has a compelling source [22]. Serendipity is part of physics and mathematical rigor must follow. 8. Interesting acceleration to radius coefficients relation – the field strength coefficients Consider the coefficients 95 96 4 , from (23) , from (24) and 𝜉 = 𝜋 1.556198537190348396563877031439915299415588 from (34), (35). Note the following table Table 2. 𝜉 of Electron, Muon, Tau 95 96 4 𝜋 4 −1 𝜉( ) 𝜋 7 0.7772169325287248897~ 1= 9 9 9 1.5561985371903483965638770314399 1.22223547299109529~11 9 4 −1 𝜉∙9∙( ) 𝜋 6.994952392758524 9 11.0001192569 The numbers in the left column suggest that the field strength coefficients are related to natural numbers. Models of natural numbers occur for example in Heisenberg’s XXZ model of spin 4 chains. The normalization factor for L=4 and ∆= 1 in such a spin chain is remarkably close to , 𝜋 and remarkably close to 1.556198537190348… with L=11 and ∆= 1. The normalization factor is close to 95 when L=13 and ∆= 0. Here the model is not of any spin chain and spin chains are 96 only brought as an example of how natural numbers can be related to numbers such as this model’s field strength coefficients. The reader can check that 7 * 9 * 11 = 693 is the integer floor of ( 693.634239847 , see the note after (40) with 2 − 1 1 962 (𝑎−1)(1−𝑏) − 1)−1 ≅ . We have yet to show more 962 (𝑎−1)(1−𝑏) compelling evidence the choice of 𝜉 is not by chance. Some readers will remain skeptical no matter what evidence is brought in this paper. This section is not meant for such readers but for readers who agree that serendipity is important for new discoveries in physics. The coefficient 4 𝜋 from (22), (80) is well understood [22], however, 95 from (23), (79), (86) and 96 1.5561985371903483965638770314399 from the solution to (35) are not well understood. Evidence, except from the previous table and the note after (40), can be found if we look at the polynomial term that means loss or addition of area in relation to 4 times the area of a disk. The factor 4 was thoroughly discussed before (16) and led to the number 96 from 𝑅(3)∗𝑟 2 96 = where R(3) is obtained by double contraction of the Einstein tensor with a direction of time. We return to (18), (− 1 1 𝜋 ∗ ∗𝑅(3)∗𝑟 4 4 24 𝜋𝑟 2 1 1 𝜉2 2 𝑥2 𝜉 ∓ ) 𝑥 1 96 = 𝛿𝐴𝑟𝑒𝑎 4𝜋𝑟2 and consider the following polynomials 1 (− 2 𝜉2 ∓ 𝜉) 96 of the field strength coefficient 𝜉. Like before in (23), we consider the terms 𝑎−1 and 1 1−𝑏 from the biggest and stable roots a, b. Not too surprisingly, these terms are approximated 1 1 −1 1 1 −1 by 𝛼 = 𝑝1(𝜉) = ((− 𝜉 2 + 𝜉) ) and 𝛽 = 𝑝2(𝜉) = ((− 𝜉 2 − 𝜉) ) 2 2 96 96 which only depend on the field strength coefficients. Consider the following relative error terms 𝑅𝑎𝑡𝑖𝑜𝐴 = ( 1)−1 and 𝑅𝑎𝑡𝑖𝑜𝐵 = ( 1−𝑏 𝛽 − 1)−1or as an output of a python code: Field strength coefficient analysis: Xi=0.9895833333333334, p1=192.02083559413998, p2=64.89902805794975 Xi=0.9895833333333334, 1/(a-1)=192.04639436012951, 1/(1-b)=63.54135822920768 Xi=0.9895833333333334, RatioA=-7513.91496909199486, RatioB=46.80177527998884 ----------------------------------------------------Xi=1.2732395447351628, p1=207.49126659259227, p2=46.06948110927548 Xi=1.2732395447351628, 1/(a-1)=206.75133988502202, 1/(1-b)=44.63955017596401 Xi=1.2732395447351628, RatioA=279.42137750905704, RatioB=31.21797643232097 ----------------------------------------------------- 𝑎−1 𝛼 − Xi=1.5561985371903484, p1=278.00172875202145, p2=34.69366870085835 Xi=1.5561985371903484, 1/(a-1)=275.51690891864394, 1/(1-b)=33.19740405023536 Xi=1.5561985371903484, RatioA=110.88003452715024, RatioB=22.18685313217340 ----------------------------------------------------- This output shows proximities between functions of the field strength coefficient, 𝜉 or Xi in the Python output. The proximities are p1 of the next 𝜉 to RatioA and p2 of the next 𝜉 to RatioB. The first value ~-7513.91496909199486 is in red as an exception because it is not matched to the 4 value of p1 for the next field strength coefficient 𝜉 = . Following is the code in Python that was used for the last calculations, 𝜋 import numpy as NP def function_p(p_x): return (-0.5 * p_x * p_x + p_x)/96, -(-0.5 * p_x * p_x - p_x)/96 def function_cubic_viete(a, b, c, d): # If all roots are real. # Viete's formula when all roots are real. b2 = NP.longdouble(b * b) b3 = NP.longdouble(b2 * b) a2 = NP.longdouble(a * a) a3 = a2 * a p = (3 * a * c - b2) / (3 * a2) q = (2 * b3 - 9 * a * b * c + 27 * a2 * d) / (27 * a3) offset = b / (3 * a) t1 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) \ * (3 * q) / (2 * p)) / 3) t2 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) * \ (3 * q) / (2 * p)) / 3 - NP.pi / 3) t3 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) * \ (3 * q) / (2 * p)) / 3 - 2 * NP.pi / 3) x1 = t1 - offset x2 = t2 - offset x3 = t3 - offset return (x1, x2, x3) ma_list = [95/96, 4/NP.pi, 1.5561985371903484] print('Field strength coefficient analysis:') for ma_x in ma_list: ma_tuple = function_p(ma_x) ma_a,_,_ = function_cubic_viete(1, -1, -ma_x / 96, (ma_x * ma_x) / 192) ma_b,_,_ = function_cubic_viete(1, -1, ma_x / 96, (ma_x * ma_x) / 192) print('Xi={}, p1={:.14f}, p2={:.14f}'.format(ma_x, 1/ma_tuple[0], 1/ma_tuple[1])) print('Xi={}, 1/(a-1)={:.14f}, 1/(1-b)={:.14f}'.format(ma_x, 1/(ma_a-1), 1/(1ma_b))) ma_a = (ma_a - 1) / ma_tuple[0] ma_b = (1 - ma_b) / ma_tuple[1] ma_a = 1 / (ma_a - 1) ma_b = 1 / (ma_b - 1) print('Xi={}, RatioA={:.14f}, RatioB={:.14f}'.format(ma_x, ma_a, ma_b)) print('-----------------------------------------------------') 95 We now return to the field which is smaller than 1, namely to 𝜉 = 96. It is easy to see that if we pick 𝜉1 = 193 192 =1+ equation 𝑥1 2 + 1 96 1 192 and 𝜉2 = 𝜉1 𝑥1 − 1 2 63 64 = 1− 1 64 we get rational roots for the following anti-gravity 𝜉 = 𝑥1 3 and gravity equation 𝑥2 2 − 192 1 which 𝑥1 = 𝜉1 and 𝑥2 = 𝜉2 , interestingly 1 + 𝜉2 −𝜉1 2 = 95 96 and 𝜉1 −𝜉2 2 1 96 = 𝜉2 𝑥2 − 1 . 96 2 1 𝜉 = 𝑥2 3 for 192 2 Some nice relation between the roots of gravity and anti-gravity of area ratio polynomials with field strength coefficients 𝜉 = 95 96 and 𝜉 ≅ 1.5561985371903483965638770314399 as in (35) is considered. We saw that for 𝜉 ≅ 1.5561985371903483965638770314399 the following holds: 2(1−𝑥2 ) 1 (𝑥1 −1)2 = 1. There is another relation not less illuminating, of a simple datasheet we can see that for 𝜉 = 95 96 we get 1 4(1−𝑥2 )2 𝑥1 −1 1.5561985371903483965638770314399 as in (35), we get 1 4(1−𝑥2 )2 𝑥1 −1 . With low accuracy ≅ 96.36912199 and for 𝜉 ≅ 1 4(1−𝑥2 )2 𝑥1 −1 ≅ 191.2741085 which is almost 192=2*96. Multiplying these two values together we have 96.36912199 … ∗ 191.2741085 … ≅ 18432.9179 ≈ 18432 = 2 ∗ 962 , and we can see ( 96.00239033. 18432.9179 1 2 )2 ≅ Conclusion The presented model predicts gravity not only by mass but also by electric charge. It offers a technological breakthrough by generating inertial dipoles and it offers mass ratios between particles that are not accessible through the Standard Model. (33) and (65) can only be interpreted as the existence of a fifth force of Nature with symmetry SU(4), or by (3.12) is related to gravity, while (24) results in a new neutrally charged particle of energy ~41.8752442118608 eV. The muon field strength coefficient is different than the electron’s and Tauon field strength, which implies different physics. (33) indicates a deep relation between leptons and hadrons and especially between the Muon and the Bottom Quark. As for the theoretic approach that this paper took, not any Gauge fields are a blessing. There was a big expectation from Albert Einstein that the Palatini action, which is identical to EinsteinHilbert action, would be a great insight into Quantum Gravity, especially since spinor equations require tetrads because they are limited to an orthogonal reference frame. However, this paper took a very different approach, to leave the metric tensor as is and instead of using tetrads or Ashtekar variables, to consider the metric as of a reference manifold, like coordinates but as an entire geometric reference object, not as a physically accessible object. Then in this framework. the idea was that time must be the engine of the model and that acceleration of that time in the sense of a generalized Reeb field - not limited to contact manifolds - will describe the possibility of non-geodesic curves and will predict the electric force. In (64) it becomes an electro-weakstrong action, using indeed 5 fields, but unlike tetrads, time is a meaningful Geroch function while the other fields are Gauge fields. There is a redundancy in the system because this time can be accounted for by 3 vectors just as Ashtekar variables. This redundancy is cancelled out in action (64), instead of using an ADM formalism or Ashtekar variables, and orthogonality is no longer needed, which renders the spin connections redundant. 4 out of the 5 scalar fields describe an additional geometric information to the metric as foliations. The same theory can be written with tetrads and generalized Reeb vectors of these tetrad fields, but the Einstein-Hilbert action will be the same. On the other hand, action (64) in this case, does add geometric information as non-geodesic alignment of curves and thus of forces. It is a far simpler approach than that of Abhay Ashtekar and it reaches new results. Adding a summation constraint to the action of (64), e.g. that each chronon probability sums to 1, keeps the same action but then PP* is replaced by an event function and the integration of PP* becomes 1. That requires the only constant in the theory except for the speed of light to be with the units of 𝐿𝑒𝑛𝑔𝑡ℎ−2 . Appendix A: Euler Lagrange minimum action equations  We assume  = 8 (from the previously discussed term, − a  a / 8K as an energy density). Z = N 2 = P P  and U  = Z Z Pk P 1 − k 2  and L = U k U k Z Z 4 R = Ricci curvature. 8   Min Action = Min   R − L  − g d =     1   Min   R − U k U k . − g d s.t.  = 8  4   (47) The variation of the Ricci scalar is well known. It uses the Platini identity and Stokes theorem to calculate the variation of the Ricci curvature and reaches the Einstein tensor [33], as follows, R = R g   and  ( R − g ) = ( R  − 1 Rg   )g   2 1 4  −g =− 1 g  g   − g 2 by which 𝜕𝑔𝜇ν − 𝑑 𝑑𝑥 𝑚 𝜕 𝜕𝑔𝜇𝜈 , 𝑚 𝑑2 + 𝑑𝑥 𝑚 𝑑𝑥 𝑠 𝜕 𝜕𝑔𝜇𝜈 ,𝑚 ,𝑠 infer which will be later added to the variation of ((𝑅 − 𝑈𝑗 𝑈𝑗 )) √−𝑔 by g   . The following Euler Lagrange equations have to hold, 𝜕 we 1 ((𝑅 − 𝑈𝑗 𝑈𝑗 )√−𝑔) = 0, 4 𝜕 𝜕 𝜕 1 𝑑 𝑑2 ((𝑅 − 𝑈𝑗 𝑈𝑗 )√−𝑔) = 0 − 𝑚 + 𝑚 𝑠 𝜕𝑃 𝑑𝑥 𝜕𝑃,𝑚 𝑑𝑥 𝑑𝑥 𝜕𝑃,𝑚 ,𝑠 4 Uk Uk = Z Z  Z2 − (Z s P s ) 2 which we obtain from the minimum Euler Lagrange equation because Z3 Z  P  Z k P k P P  − = 0 . In order to calculate the minimum action Euler-Lagrange equations, Z Z2  s 2 we will separately treat the Lagrangians, L = Z  Z2 and L = (Z s P3 ) to derive the Euler Lagrange Z Z  s 2 equations of the Lagrangian L = Z  Z2 − (Z s P3 ) = U U  . The Euler Lagrange operator of the Ricci Z Z U P = scalar (  d2   d . − + μν m μν m s μν g dx (g ,m ) dx dx (g ,m , s ) The reader may skip the following equations up to equation (53). Equations (53), (54) and (55) are however crucial. Note: the relation 𝐿= (−2 ( 𝐿= 𝑍,𝑠 𝑃 𝑠 𝑍3 𝑍 𝜆 𝑍𝜆 𝑍2 (𝑃𝜆 𝑍 𝜆 ) 𝑍3 2 𝑑 𝑑𝑥 𝜈 𝜆 √|𝑔| is used in the next equations. √|𝑔| = Γ𝜆𝜈 𝑠. 𝑡. 𝑍 = 𝑃𝜇 𝑃𝜇 𝑎𝑛𝑑 𝑍𝑠 ≡ 𝑍,𝑠 = 𝑑𝑍 𝑑𝑥 𝑠 𝜕(𝐿√−𝑔 𝑑 𝜕(𝐿√−𝑔 ) − 𝑑𝑥 𝑚 𝜕𝑔𝜇𝜈 ,𝑚 𝜕𝑔𝜇𝜈 𝑠 𝑍,𝑠 𝑃 𝑍,𝑠 𝑃 𝑠 𝑚 = (−2 ( 3 𝑃𝜇 𝑃𝜈 𝑃 ) ;𝑚 + 2 ( 3 ) (Γiμm 𝑃𝑖 𝑃𝜈 𝑃𝑚 + Γiνm 𝑃𝜇 𝑃𝑖 𝑃𝑚 ) 𝑍 𝑍 𝑠 𝑍,𝑠 𝑃 𝑠 𝑍,𝑠 𝑃 + 2 ( 3 ) (𝑃𝜇 𝑃𝜈 );𝑚 𝑃𝑚 − 2 ( 3 ) (Γiμm 𝑃𝑖 𝑃𝜈 𝑃𝑚 + Γiνm 𝑃𝜇 𝑃𝑖 𝑃𝑚 ) 𝑍 𝑍 𝑠 𝑠 2 𝑍,𝑠 𝑃 (𝑍,𝑠 𝑃 ) 1 (𝑍,𝑠 𝑃 𝑠 )2 + 2 ( 3 ) 𝑍𝜇 𝑃𝜈 − 3 𝑃 𝑃 − 𝑔𝜇𝜈 ) √−𝑔 = 𝜇 𝜈 𝑍 𝑍4 2 𝑍3 𝑃𝑘 ) ;𝑘 𝑃𝜇 𝑃𝜈 − 2 (𝑍,𝑠 𝑃𝑠 )2 𝑃𝜇 𝑃𝜈 𝑍3 𝑍 − (𝑍,𝑠 𝑃𝑠 )2 𝑃𝜇 𝑃𝜈 𝑍3 𝑍 + 2( 𝑠. 𝑡. 𝑍 = 𝑃𝜇 𝑃𝜇 , 𝑠. 𝑡. 𝑍 = 𝑃𝜇 𝑃𝜇 𝑎𝑛𝑑 𝑍𝑠 ≡ 𝑍,𝑠 = 𝜕(𝐿 √−𝑔 2 𝜕𝑔𝜇𝜈 − 𝑑 𝜕(𝐿√−𝑔 ) 𝑑𝑥 𝑚 𝜕𝑔𝜇𝜈 ,𝑚 (Γiμm Pi Pν Zm +Γiνm 𝑃𝑖 𝑃𝜇 𝑍 𝑚 ) 𝑍2 = (−2 ( + 𝑍𝜇 𝑍𝜈 𝑍2 𝑍 𝑚 𝑃𝜇 𝑃𝜈 −2 𝑍2 𝑍𝑠 𝑍 𝑠 𝑍3 We subtract (48) from (49) 𝑍 = 𝑃𝜇 𝑃𝜇 , 𝑠. 𝑡. 𝑍 = 𝑃𝜇 𝑃𝜇 𝑎𝑛𝑑 𝑍𝑠 ≡ 𝑍,𝑠 = 𝑑𝑍 𝑑𝑥 𝑠 2 𝑍𝑠 𝑍 𝑠 𝑍3 𝑑𝑍 𝑍𝜆 𝑍 1 (𝑍,𝑠 𝑃 𝑠 )2 2 𝑍3 (Γiμm Pi Pν Zm +Γiνm 𝑃𝑖 𝑃𝜇 𝑍 𝑚 ) 1 𝑍𝑚 𝑍 𝑚 2 𝑃𝜇 𝑃𝜈 − , 𝑈𝜆 = 𝑍3 ) 𝑍𝜇 𝑃𝜈 − 𝑑𝑥 𝑠 ) ;𝑚 + 2 𝑃𝜇 𝑃𝜈 − 𝑍,𝑠 𝑃 𝑠 − 𝑍2 𝑍2 𝑍2 𝑍𝑘 𝑃 𝑘 𝑃 𝜆 𝑍2 +2 𝑔𝜇𝜈 ) √−𝑔 = (−2 ( 1 𝑍𝑚 𝑍 𝑚 2 𝑔𝜇𝜈 ) √−𝑔 (48) 𝑔𝜇𝜈 + 𝑍𝜇 𝑍𝜈 𝑍2 𝑍𝑚 𝑍2 ) √−𝑔 , 𝐿 = 𝑈 𝑘 𝑈𝑘 = 𝑍𝜆 𝑍 𝜆 𝑍2 (𝑃𝜇 𝑃𝜈 );𝑚 𝑍 𝑚 𝑍2 − ) ;𝑚 𝑃𝜇 𝑃𝜈 − − (49) (𝑍𝑘 𝑃 𝑘 )2 𝑍3 1 (𝑍𝑚 𝑃 𝑚 )2 2 𝑍3 𝜕(𝐿 √−𝑔 𝜕𝑔𝜇𝜈 𝑔𝜇𝜈 + ((+2 ( 𝑑 − 𝑑𝑥 𝑚 𝜕(𝐿√−𝑔 ) 𝜕𝑔𝜇𝜈 , (𝑍𝑚 𝑃 𝑚 )2 𝑃𝜇 𝑃𝜈 𝑍3 𝑍𝑚 𝑃 𝑚 𝑍3 1 𝑍𝑘 𝑍 𝑘 2 𝑍2 𝑍 = (+2 ( 𝑚 𝑍𝑚 𝑍𝑚 𝑃 𝑚 𝑍3 𝑃𝑘 ) ;𝑘 𝑃𝜇 𝑃𝜈 + 2 + (−2 ( 𝑍 2 ) ;𝑚 𝑃𝜇 𝑃𝜈 − 2 𝑍𝑚 𝑃𝑘 ) ;𝑘 − 2 ( 𝑍 2 ) ;𝑚 ) 𝑃𝜇 𝑃𝜈 + 2 𝑔𝜇𝜈 + 𝑍𝑚 𝑍𝜇 𝑍𝜈 𝑍2 − 2( 2 ( 𝑍 2 ) ;𝑚 ) 𝑃𝜇 𝑃𝜈 + 2 1 𝑍𝑠 𝑃 𝑠 𝑍3 ) 𝑍𝜇 𝑃𝜈 + 2 (𝑃 𝜆 𝑍𝜆 ) 𝑃𝜇 𝑃𝜈 𝑍3 (𝑈𝜇 𝑈𝜈 − 𝑈 𝜆 𝑈𝜆 𝑔𝜇𝜈 − 2𝑈 𝑘 ;𝑘 2 𝑃𝜇 𝑃𝜈 𝑧 𝑧 𝑍𝜆 𝑍 𝜆 𝑃𝜇 𝑃𝜈 𝑍2 2 𝑍 (𝑃 𝜆 𝑍𝜆 ) 𝑃𝜇 𝑃𝜈 𝑍3 2 𝑧 (𝑃 𝜆 𝑍𝜆 ) 𝑃𝜇 𝑃𝜈 −2 𝑍3 𝑍 𝑍 𝜆 𝑍𝜆 𝑃𝜇 𝑃𝜈 𝑍2 𝑧 (𝑍𝑚 𝑃 𝑚 )2 𝑃𝜇 𝑃𝜈 − 𝑍3 1 𝑍𝜆 𝑍 𝜆 2 −2 𝑍2 𝑍 𝑔𝜇𝜈 + 𝑍 𝜆 𝑍𝜆 𝑃𝜇 𝑃𝜈 𝑍2 −2 𝑍 + 𝑍𝑚 𝑃 𝑚 𝑍3 𝑍𝜇 𝑍𝜈 𝑍2 )) √−𝑔 = 1 (𝑃 𝜆 𝑍𝜆 ) 2 ) √−𝑔 = ((+2 ( 1 𝑍𝜇 𝑃𝜈 + 𝑍3 𝑍𝑚 𝑃 𝑚 𝑍3 2 𝑔𝜇𝜈 − 𝑃 𝑘 ) ;𝑘 − + 𝑈𝜇 𝑈𝜈 − 𝑈 𝜆 𝑈𝜆 𝑔𝜇𝜈 ) √−𝑔 = 2 ) √−𝑔 (50) (Z s Ps ) 2 s.t. Z = P λ Pλ and Z m = (P λ Pλ ),m Z3 (L − g ) d (L − g ) = − Pμ ,ν Pμ dx ν L=   (Z P s ) (Z P s ) μ ν μ  − 4( s 3 Γi ν Pi Pν + P P );ν + 4 s 3 Z Z   s   (Z P ) (Z s P s ) μ μ + 4 Γi k Pi Pk +  − g = P ;ν P ν − 4 s 3 3 Z Z     ZmPmZ μ (Z m P m ) 2 μ  + 2  − 6 P Z3 Z4   ( − 4( Z PmZ μ (Z P m ) 2 μ (Z s P s )P ν −6 m 4 P ) −g );ν P μ + 2 m 3 3 Z Z Z (51) s Z Zs s.t. Z = P λ Pλ and Z m = (P λ Pλ ),m Z2 (L − g ) d (L − g ) = − Pμ ,ν Pμ dx ν L=   PμZ ν 4 μ  − 4( );ν + 2 Γ i k P i Z k +  2 Z Z     4 4 μ μ ν i k  + 2 P ;ν Z − 2 Γ i k P Z +  − g = Z   Z m   ZmZ μ P −g  − 4 3 Z   ( − 4( ZmZ m Zν − 4 )P μ ); ν Z3 Z2 s We subtracted the Euler Lagrange operators of (Z P3s ) Z 2 −g (52) − g in (48) from the Euler Lagrange λ operators of Z Z2 λ − g in (49) and got (50) and we will subtract (51) from (52) to get two tensor Z equations of gravity, these will be (53), and (55). Assuming  = 8 , where the metric variation equations (47), (48), (49) and (50) yield Z Z Pk P 1 − k 2  , L = U iU i and Z = P k Pk Z 4 Z  m m   (P P ), m P Z  + 2(( P k ); k −2( 2 ); m ) P P +  3 Z Z    Z  Z  P P ( P  Z  ) 2 P P 8 1  = + 2 2 − +  4 Z Z Z3 Z2    1 k   + U  U  − U kU g   2   P P 1 1 8 1   ) = R   − Rg   (U  U  − U kU k g   − 2U k ; k  4 Z 2 2 s.t. R = R   g   Z = N 2 = P P  , U  = s.t. Rk R  = ( j k ), p −( p k ), j +  p   jk P j P P   −  p j k  p (53) is the Ricci tensor and R  − 1 Rg   is the Einstein tensor [33]. In general, by (4) and  = 8 , 2 (53) can be written in (−1, +1, +1, +1) metric convention, so 𝑍 = |𝑃𝜇 𝑃𝜇 | as, P P 1 1 1 (U U − U kU k g   − 2U k ; k ) = R  − Rg   Z 4 2 2 Charge-less field: The term − 2U k ;k − 2((U k ;k + U *k ;k ) / 2) 4( A  ;  P * Z + A *  ;  Pμ Pν Z P Z in (54) can be generalized to: Z ( Pμ P *ν + P * μ Pν ) / 2 (54) and can be zero under the following condition: ) = U U * +U * U   U k ;k + U *k ;k = 0 Note: The complimentary matrix 𝐵𝜇𝜈 = 1 √2 𝐸𝜇𝜈𝛼𝛽 𝐴𝛼𝛽 , see few lines before (3), can be transformed to a real matrix due to the SU(2) x U(1) degrees of freedom and also be imaginary. From (51), (52) we have, d (  − d   dx We recall, W  = (  − d  P dx P  )(U k U k dx P ,  )(U k U k P , − g) − g ) = W  ; −g =0 W = ( Z P s ) P (Z P m ) 2  Z Zm Z PmZ  Z ); −4 m 3 )P  + 4( s 3 ); P  − 2 m 3 +6 m 4 P = 2  Z Z Z Z Z Z Zm Z − 4( 2 ); P  − 4 m 3 P  + Z Z s  (Z P ) P (Z P m ) 2  + 4( s 3 ); P  + 4 m 4 P Z Z Z Pm Z  ZmPmP −2 m 2 ( − )= Z Z Z2 U kU k  Z Pm Uk − 4(( );k + )P − 2 m 2 U  = 0 Z Z Z ( −4(  (Z m P m )  P  W  ; =  U  − 4U ; Z − 2 Z2    ;  = 0  (55) Appendix B: Proof of conservation Theorem: Conservation law of the real Reeb vector. From the vanishing of the divergence of Einstein tensor and (54), we have to prove the following in (−1, +1, +1, +1) metric convention: PP 1 1 U U − U kU k g   − 2U k ; k   Z 4 2  1 ; = G  ;  = ( R  − Rg   );  = 0 2  (56) Proof: From the zero variation by the scalar time field (55)  (Z m P m )  P   W ;  =  − 4U ; U −2 Z Z2    (Z m P m )   P     = 2 ; ; U − U      Z Z2     P  P  − 2U k ; k Z   (Z m P m )    Z2 U  Now let t  ZmPm  (Z P m )  ;  =  m 2 U    Z  Z ;  P − U k ; k  Z   ;  = 0    ;    (57) (58)   P   ;  P −  2U k ; k P ; =  Z    (59) Z t Z t  t    k = ( 2 ); U  P + 2 U  ;  P − U k ;k =  2 U ;  P − U ;k Z Z Z Z  And we have  Z t − U  ;  U  + ( 2 ); U  P Z This is because −U ν = − Zν Z + t Z P ν ⟹ −U μ ;μ 2 twice the real numbered Reeb vector. So, (−2U k ; k Zν Z + t Z2 U μ ;μ P ν = −U μ ;μ U ν . Note that −U ν is minus P  P t );  = −U  ;  U  + ( 2 );  U  P Z Z (60) Returning to the theorem we have to prove and using equation (60), we have to show,    1 P  P   ; = U U − U kU k g   − 2U k ; k 2 Z   1 U  ;  U  + U  U  ;  − (U k ;  U s + U kU s ;  ) g ks g   − 2 t U  ;  U  + ( 2 );  U  P = Z 1 t U  U  ;  − (U sU s ); +( 2 );  U  P = 0 2 Z (61) Notice that 1 U  U  ;  − U sU s ; = 2 t t    Zk U  ( );  −( 2 );  Pk − ( 2 ) Pk ;   g k − Z Z  Z  t t  Z  U s  ( s ); k −( 2 ); k Ps − ( 2 ) Ps ; k  g k = Z Z Z   t − U  ( 2 );  P Z (62) Since − ( t 2 ); k PsU s = 0 because the Reeb vector is perpendicular to the foliation kernel Z Pλ , √Z Pk Uk √Z 2 = 0. Equation (62) is also a result of ln(Z),k ;μ U μ g kν = ln(Z),s ;k U s g kν and of Pk ;μ U μ g kν = Ps ;k U s g kν . 1 t t t U U  ;  − (U sU s ); +( 2 );  U  P = −U  ( 2 );  P + ( 2 );  U  P = 0 2 Z Z Z (63) and we are done. Appendix C: Generalization to more than one generalized Reeb vector Considering (3.2), (3.3) and given the previous fields 𝑈(2)𝜇 2 , 𝑆𝜇 𝑊𝜇 𝑇𝜇 2 , 2 , 2 , Pk √Z and 𝑈𝜇 2 and additional Reeb vector fields The following Lagrangian can be defined with the determinant of the metric g: 1 0 𝐿=| 𝑈 𝑘 𝑈𝑘∗ +𝑈 ∗𝑘 𝑈𝑘 | √−g + 0 | | | | 1 8 Pk 𝑈(2)∗𝑘 +P∗ k 𝑈(2)∗𝑘 0 0 Pk 𝑈(2)∗𝑘 +P∗ k 𝑈(2)∗𝑘 2√2Z 2√2Z 𝑈(2)𝑘 𝑈𝑘∗ +𝑈(2)∗𝑘 𝑈𝑘 𝑈 𝑘 𝑈𝑘∗ +𝑈 ∗𝑘 𝑈𝑘 8 𝑈(2)𝑘 𝑈𝑘∗ +𝑈(2)∗𝑘 𝑈𝑘 8 𝑈(2)𝑘 𝑈(2)∗𝑘 +𝑈(2)∗𝑘 𝑈(2)𝑘 8 ∗ 𝑆𝜇 𝑝𝜇 𝑆 ∗𝜇 +𝑝𝜇 ∗ 𝑊𝜇 𝑝𝜇 𝑊 ∗𝜇 +𝑝𝜇 ∗ 𝑇𝜇 𝑝𝜇 𝑇 ∗𝜇 +𝑝𝜇 2√2Z ∗ 𝑊𝜇 𝑝𝜇 𝑊 ∗𝜇 +𝑝𝜇 8 𝑊𝜇 𝑆 ∗𝜇 +𝑊𝜇∗ 𝑆 𝜇 8 𝑊𝜇 𝑊 ∗𝜇 +𝑊𝜇∗ 𝑊 𝜇 8 𝑊𝜇 𝑇 ∗𝜇 +𝑊𝜇∗ 𝑇 𝜇 2√2Z 8 8 8 1 ∗ 𝑆𝜇 𝑝𝜇 𝑆 ∗𝜇 +𝑝𝜇 2√2Z ∗ 𝑇𝜇 𝑝𝜇 𝑇 ∗𝜇 +𝑝𝜇 2√2Z ∗ 𝑆𝜇 𝑆𝜇 𝑆 ∗𝜇 +𝑆𝜇 8 𝑇𝜇 𝑆 ∗𝜇 +𝑇𝜇∗ 𝑆 𝜇 2√2Z ∗ 𝑊𝜇 𝑆𝜇 𝑊 ∗𝜇 +𝑆𝜇 8 𝑇𝜇 𝑊 ∗𝜇 +𝑇𝜇∗ 𝑊 𝜇 2√2Z ∗ 𝑇𝜇 𝑆𝜇 𝑇 ∗𝜇 +𝑆𝜇 8 𝑇𝜇 𝑇 ∗𝜇 +𝑇𝜇∗ 𝑇 𝜇 | | 8 1 3 1 2 | √−g + | √−g (64) The last term of (64) has SU(3) * reflections symmetry, however, when considering the spacelike foliation which is perpendicular to 𝑝𝜇 √𝑍 , extremal solutions, not saddle variations, have a physical meaning of rotating fields. See Fig. 5. There could be better action operators than (64), after all, (64) is no more than a research offer although it has its own logic which is not fully explained in this paper. Uniform gravity and forces formalism, challenges, and an open problem: Apparently the first and the last additives of (64), 2x2 and 4x4 matrices imply that the Einstein-Hilbert action can be written in a tetradic formulation with 4 scalar functions where: 𝑒 𝐽𝜇 = Caveat: 𝑝𝐽 𝜇 √𝑍(𝐽) 𝑝𝐽 𝜇 √𝑍(𝐽) = 𝑝(𝐽),𝜇 √𝑍(𝐽) and 𝜂𝐾𝐽 𝑒𝐾𝜇 𝑒𝐽𝜈 = 𝑒 𝐽𝜇 𝑒𝐽𝜈 = 𝑔𝜇𝜈 or 1 2 (𝑒 ∗𝐽 𝜇 𝑒𝐽𝜈 + 𝑒 𝐽𝜇 𝑒 ∗𝐽𝜈 ) = 𝑔𝜇𝜈 and 𝐾 ≠ 𝐽 ⇒ 𝜂𝐾𝐽 = 0, 𝜂00 = −1, 𝜂11 = 𝜂22 = 𝜂33 = 1 may not be related to any force fields but only to gravity. Please refer to the remark after (3.13). (64.1) Clarification: Notice that P(0)P*(0) is a Geroch function. Adopting Sam Vaknin’s methods [12] in a scalar function formalism, P(0)P*(0), can be a probability density of a single event in 4volume, instead of time. The good news is that 𝑃(0), 𝑃(1), 𝑃(2), 𝑃(3) yield 4 gradients 𝑃(0)𝜇 , 𝑃(1)𝜇 , 𝑃(2)𝜇 , 𝑃(3)𝜇 and 4 non geodesic accelerations as generalized Reeb vectors -not limited to contact manifolds, unlike the usual Reeb vectors- 𝑈𝜇 𝑆𝜇 𝑊𝜇 𝑇𝜇 2 , 2 , 2 , 2 as explained in (64). The much less good news is that 4 complex functions 𝑃(0), 𝑃(1), 𝑃(2), 𝑃(3) are 8 real functions which mean that the 2 degrees of freedom in the 10 independent General Relativity equations are gone (without considering the 4 vanishing divergence equations). This problem can be partially mitigated by multiplication of the Tetradic metric tensor by a scalar function 𝜙, however, such a 1 solution to a unified formalism of forces and gravity is awkward, 𝜙𝜙 ∗ (𝑒 ∗𝐽 𝜇 𝑒𝐽𝜈 + 𝑒 𝐽𝜇 𝑒 ∗𝐽𝜈 ) = 𝑔𝜇𝜈 is still incomplete. The problem is the following condition: 𝑃∗𝜇 (0)𝑃𝜈 (0)+𝑃𝜇 (0)𝑃∗𝜈 (0) 2√𝑍(0) − ∑3𝑖=1 1 𝑃∗𝜇 (𝑖)𝑃𝜈 (𝑖)+𝑃𝜇 (𝑖)𝑃∗𝜈 (𝑖) 2√𝑍(𝑖) 2 = 𝑔𝜇𝜈 and from the vanishing of the ordinary covariant derivative: (𝑒 ∗𝐽 𝜇 𝑒𝐽𝜈 + 𝑒 𝐽𝜇 𝑒 ∗𝐽𝜈 );𝜆 = 𝑔𝜇𝜈 ;𝜆 = 0. This is not a good place to be begin 2 with. For the Riemann tensor in the real case, we have: 𝑅 𝐽 𝐾𝐿𝑀 = 𝑒 𝐽 ((∇𝐿 ∇𝑀 − ∇𝑀 ∇𝐿 − [𝑒𝐿 , 𝑒𝑀 ])𝑒𝐾 ) and 𝑅 = 𝑅 𝐽𝑀𝐽𝑀 and the volume element is √|𝐷𝑒𝑡𝑒𝑚𝑖𝑛𝑎𝑛𝑡(𝑒 𝐽𝜇 𝑒𝐽𝜈 )| with Greek letters denoting the ordinary Riemann indices. There are, however, some caveats. 𝑝 𝐽=0𝜇=1 = 𝑝0 2 = 𝑝0 3 = 0 means that P(0)P ∗ (0) must be a Geroch time function [1]. The rest of the tetrads need not be perpendicular to each other, 𝑒 𝐽𝜇 𝑒 ∗𝐾𝜇 + 𝑒 ∗𝐽𝜇 𝑒 𝐾𝜇 ≠ 0 but using spin connections they may be formulated in such a way that 𝑒 𝐽𝜇 𝑒 ∗𝐾𝜇 + 𝑒 ∗𝐽𝜇 𝑒 𝐾𝜇 = 0. Fig. 5. The Aryeh Aldema’s offer of a Relative curvature action and it’s meaning Let 𝑝𝜇 (0) √|𝑍(0)| = 𝑝𝜇 √|𝑍| and 3 other scalar fields are defined 𝑝𝜇 (1) , 𝑝𝜇 (2) , 𝑝𝜇 (3) √|𝑍(1)| √|𝑍(2)| √|𝑍(3)| with real numbers. Taking the Gaussian curvature but without demanding that the embedding manifold will be flat, consider 𝐾 √−𝑔 ≡ 𝑝 (0) 𝑝𝜇 (1) 𝑝𝑗(0) 𝑝𝜈 (2) 𝑝𝑘(0) 𝑝𝜁 (3) 𝑝𝑠(0) 𝑖𝑗𝑘𝑠 ( 𝑖 );𝜇 ( );𝜈 ( );𝜁 𝜖 √|𝑍(0)| √|𝑍(1)| √|𝑍(0)| √|𝑍(2)| √|𝑍(0)| √|𝑍(3)| √|𝑍(0)| 𝑝𝑎 (0) 𝑝𝑏 (1) 𝑝𝑐 (2) 𝑝𝑑 (3) 𝑎𝑏𝑐𝑑 𝜖 √|𝑍(0)|√|𝑍(1)|√|𝑍(2)|√|𝑍(3)| √−𝑔 (64.2) where 𝜖 𝑖𝑗𝑘𝑠 is the Levi-Civita alternating symbols. It is not difficult to prove that this definition does not depend on the choice of and independent of 𝑝𝑠 (0) 𝑝𝜇 (1) , . Obviously, ( √|𝑍(0)| 𝑝𝜇 (2) , 𝑝𝜇 (3) √|𝑍(1)| √|𝑍(2)| √|𝑍(3)| 𝑝𝑠 (0) √|𝑍(0)| as long as they are mutually independent ) ;𝑖 𝑝𝑖 (0) = 𝑝𝑠 ;𝑖 (0) √|𝑍(0)| 𝑝𝑖 (0) − 1 𝑝𝑠 (0) 2 |𝑍(0)|32 𝑍𝑖 (0)𝑝𝑖 (0) = 0 so there is no need to worry about 𝑝𝜇 (1), 𝑝𝜇 (2), 𝑝𝜇 (3) being within the foliation. A full proof replaces 𝜖 𝑖𝑗𝑘𝑠 with the Levi-Civita tensor at first 𝐸 𝑖𝑗𝑘𝑠 = 𝑠𝑔𝑛(𝑔) immediately see that 𝐾≡ 𝜖 𝑖𝑗𝑘𝑠 √−𝑔 𝑝 (0) 𝑝𝜇 (1) 𝑝𝑗(0) 𝑝𝜈 (2) 𝑝𝑘(0) 𝑝𝜁 (3) 𝑝𝑠 (0) 𝑖𝑗𝑘𝑠 ( 𝑖 );𝜇 ( );𝜈 ( );𝜁 𝐸 √|𝑍(0)| √|𝑍(1)| √|𝑍(0)| √|𝑍(2)| √|𝑍(0)| √|𝑍(3)| √|𝑍(0)| 𝑝𝑎 (0) 𝑝𝑏(1) 𝑝𝑐 (2) 𝑝𝑑 (3) 𝑎𝑏𝑐𝑑 𝐸 √|𝑍(0)|√|𝑍(1)|√|𝑍(2)|√|𝑍(3)| =− 𝜖 𝑖𝑗𝑘𝑠 √−𝑔 so we (64.3) Which means that K is the quotient of two scalar functions and is therefore a scalar function. Without loss of generality, (64.4) 𝑝𝑗 (0) 𝑝 (0) 𝑝𝜇 (1) + 𝑟𝑝𝜇 (2) 𝑝𝜈 (2) 𝑝 (0) 𝑝𝜁 (3) 𝑝𝑠 (0) 𝑖𝑗𝑘𝑠 ( 𝑖 ) ;𝜇 ( ( ) ;𝜈 ( 𝑘 ) ;𝜁 𝐸 √‖𝑝𝜇 (1) + 𝑟𝑝𝜇 (2)‖ √|𝑍(0)| √|𝑍(0)| √|𝑍(2)| √|𝑍(0)| √|𝑍(3)| √|𝑍(0)| 𝑝𝑎 (0) 𝑝𝑏 (1) + 𝑟𝑝𝑏 (2) 𝑝𝑐 (2) 𝑝𝑑 (3) 𝑎𝑏𝑐𝑑 𝐸 √|𝑍(0)| √‖𝑝𝑏 (1) + 𝑟𝑝𝑏 (2)‖ √|𝑍(2)| √|𝑍(3)| 𝑝𝑗 (0) 𝑝𝜇 (1) 𝑝𝜈 (2) 𝑝 (0) 𝑝𝜁 (3) 𝑝𝑠 (0) 𝑖𝑗𝑘𝑠 𝑝 (0) ) ;𝜇 ( ( ) ;𝜈 ( 𝑘 ) ;𝜁 𝐸 ( 𝑖 √‖𝑝𝜇 (1) + 𝑟𝑝𝜇 (2)‖ √|𝑍(0)| √|𝑍(2)| √|𝑍(0)| √|𝑍(3)| √|𝑍(0)| √|𝑍(0)| = 𝑝𝑎 (0) 𝑝𝑏 (1) 𝑝𝑐 (2) 𝑝𝑑 (3) 𝑎𝑏𝑐𝑑 𝐸 √|𝑍(0)| √‖𝑝𝑏 (1) + 𝑟𝑝𝑏 (2)‖ √|𝑍(2)| √|𝑍(3)| 𝑝𝑗 (0) 𝑝 (0) 𝑝𝜇 (1) 𝑝𝜈 (2) 𝑝 (0) 𝑝𝜁 (3) 𝑝𝑠 (0) 𝑖𝑗𝑘𝑠 ( 𝑖 ) ;𝜇 ( ( ) ;𝜈 ( 𝑘 ) ;𝜁 𝐸 √‖𝑝𝜇 (1)‖ √|𝑍(0)| √|𝑍(0)| √|𝑍(2)| √|𝑍(0)| √|𝑍(3)| √|𝑍(0)| = 𝑝𝑎 (0) 𝑝𝑏 (1) 𝑝𝑐 (2) 𝑝𝑑 (3) 𝑎𝑏𝑐𝑑 𝐸 √|𝑍(0)| √‖𝑝𝑏 (1)‖ √|𝑍(2)| √|𝑍(3)| This is because the Levi-Civita symbols are alternating so if the same vector occurs twice with two different indices, this component vanishes. For example, in the numerator ( 𝑝𝑖 (0) √|𝑍(0)| ) ;𝜇 √‖𝑝 𝑟𝑝𝜇 (2) is annihilated with ( 𝜇 (1)+𝑟𝑝𝜇 (2)‖ 𝑝𝑗 (0) √|𝑍(0)| ) ;𝜈 𝑝𝜈 (2) √|𝑍(2)| when these are multiplied by the Levi-Civita tensor. So we find that K does not depend on the choice of 𝑝𝜇 (1), 𝑝𝜇 (2), 𝑝𝜇 (3) and is therefore intrinsic to the foliation perpendicular to 𝑝𝜇 (0). Our argument was about the importance of the “relative curvature”. My opinion was clear, that the need for 𝑝𝜇 (1), 𝑝𝜇 (2), 𝑝𝜇 (3) is only if they can reveal a value which is not anticipated by the geometry of the 3D foliation because the Reeb vector 𝑈𝜇 2 already contains geometric information about the foliation which is perpendicular to the vector 𝑝𝜇 = 𝑝𝜇 (0). This information is of how the fields 𝑝𝜇 (1), 𝑝𝜇 (2), 𝑝𝜇 (3) are misaligned and do not make geodesic curves. This was the reason behind (64). Aryeh’s input clarified the importance of (64). Unfortunately, Aryeh Aldema, who was my colleague passed in Dec/30/2022. The vorticity action of 4 Reeb vectors Possibly a fifth force of Nature or by (3.12) and mapping curves to a flat spacetime, massive gravity is described by the following SU(4) symmetry Lagrangian of 4 Reeb vectors: ℸ𝜇 2 ℵ𝜇 ℶ𝜇 ℷ𝜇 , with Hebrew letters Alef, Beit, Gimmel, Dalet, ℵ𝜇 ℵ∗𝜇 +ℵ∗𝜇 ℵ𝜇 ℵ𝜇 ℶ∗𝜇 +ℵ∗𝜇 ℶ𝜇 ℵ𝜇 ℷ∗𝜇 +ℵ∗𝜇 ℷ𝜇 8 ℵ𝜇 ℷ∗𝜇 +ℵ∗𝜇 ℷ𝜇 8 ℷ𝜇 ℶ∗𝜇 +ℷ∗𝜇 ℶ𝜇 8 ℷ𝜇 ℷ∗𝜇 +ℷ∗𝜇 ℷ𝜇 8 8 ℶ𝜇 ℶ∗𝜇 +ℶ∗𝜇 ℶ𝜇 | ℵ𝜇 ℶ∗𝜇 +ℵ∗𝜇 ℶ𝜇 | 8 ℵ𝜇 ℸ∗𝜇 +ℵ∗𝜇 ℸ𝜇 8 8 ℶ𝜇 ℸ∗𝜇 +ℶ∗𝜇 ℸ𝜇 | 8 ℷ𝜇 ℸ∗𝜇 +ℷ∗𝜇 ℸ𝜇 | 8 ℸ𝜇 ℸ∗𝜇 +ℸ∗𝜇 ℸ𝜇 8 ℸ𝜇 ℷ∗𝜇 +ℸ∗𝜇 ℷ𝜇 8 , 2 , 2, 1 ℵ𝜇 ℸ∗𝜇 +ℵ∗𝜇 ℸ𝜇 4 8 ℶ𝜇 ℷ∗𝜇 +ℶ∗𝜇 ℷ𝜇 8 ℸ𝜇 ℶ∗𝜇 +ℸ∗𝜇 ℶ𝜇 2 8 √−g (65) 8 The determinant of two Reeb vectors can help to understand the roots in (30), (31), (32), and (33). It describes accelerations in two perpendicular planes. Three Reeb vectors describe accelerations in the foliation perpendicular to Pμ . It is not clear whether (65) is related to (3.12) in which case it does not represent a new field but a massive gravitational field. Appendix D: Another way to derive the Reeb vector We may now write the Lie derivative [34] of 𝑃 ∗𝑚 𝑃𝑖 𝐿𝑖𝑒 ( , √𝑍 √𝑍 )= 𝑃 ∗𝑚 √𝑍 𝑃 𝑃𝑖 √𝑍 ( 𝑖 ) ,𝑚 + ( √𝑍 with respect to the vector field 𝑃 ∗𝑚 √𝑍 ) ,𝑖 𝑃𝑚 √𝑍 In which the second term is positive because the differentiated The first term becomes, 𝑃 ∗𝑚 √𝑍 The second term is, ( 𝑃 ∗𝑚 𝑃 ( 𝑖 ) ,𝑚 = 𝑃𝑚 √𝑍 = 𝑃 ∗𝑚 𝑃𝑖 , ) √𝑍 √𝑍 = √𝑍 ) ,𝑖 √𝑍 𝑃 ∗𝑚 𝑃𝑖 ,𝑚 𝑍 𝑃 ∗𝑚 ,𝑖 𝑃𝑚 𝑍 − − 𝑃 ∗𝑚 𝑃𝑖 𝑍𝑚 √𝑍 2𝑍 3/2 𝑃 ∗𝑚 𝑃𝑚 𝑍𝑖 2𝑍 2 = which (66) becomes 𝑍𝑖 𝑍 𝑍 − 2𝑍𝑖 − 𝑃 ∗𝑚 𝑍𝑚 𝑃𝑖 2𝑍 2 𝑃 ∗𝑚 𝑃𝑖 ,𝑚 𝑍 𝑃 ∗𝑚 ,𝑖 𝑃𝑚 We add (67) and (68) to get (66) and notice that 𝐿𝑖𝑒 ( = 𝑍 − − 𝑃𝑖 √𝑍 = 2𝑍𝑖 − 𝑃 ∗𝑚 𝑍𝑚 𝑃𝑖 2𝑍 2 , (66) vector has a low index. (67) 2𝑍 2 𝑍 2𝑍 𝑃 ∗𝑚 𝑃𝑖 ,𝑚 𝑃 ∗𝑚 ,𝑖 𝑃𝑚 𝑍 √𝑍 𝑃 ∗𝑚 𝑍𝑚 𝑃𝑖 𝑍𝑖 + 𝑃 ∗𝑚 𝑍 = 𝑈𝑖 2 (68) = 𝑃 ∗𝑚 𝑃𝑚 ,𝑖 𝑍 + 𝑃 ∗𝑚 ,𝑖 𝑃𝑚 𝑍 = 𝑍𝑖 𝑍 from (69) Appendix E: 95/96, the precursor of the inverse Fine Structure Constant and of the muon/electron mass ratio Results (24), (36), (40), (41), (42), were not reached immediately. There was one finding that was a total serendipity that later lead to these results. The observation was the following, given a scaling factor 1+d of area addition with d=1 as a maximal value, 1+d = 2. More precisely (1 + 𝛼)95 < 2 ∧ (1 + 𝛼)96 > 2 (70) 1 And ℵ = (296 − 1)−1 ≅ 137.999325615 (71) 1 ℶ = (295 − 1)−1 ≅ 136.5566369 (72) √ℵℶ ≅ 137.27608605 (73) And the geometric average is: Which is close to the result from (40), 137.0359990368270076. An immediate observation is ℵ=( And ℶ= 95 2−296 −1 95 296 ) (74) 96 295 −2 ( 2 )−1 Where we expressed a power which is close to 1, namely 𝜉 = (75) 95 and 𝜉 −1 = 96 96 95 . as such, 𝜉 was nominated as polynomial coefficient because it was in the range between 0 and 2, unlike 𝜉 = which has a geometric interpretation thanks to Ettore Majorana, 𝜉 = algebraic meaning. 95 96 seems to have an 4 𝜋 We continue with a rather surprising relation 1 (295∗96 − 1)−1 ≅ 13,156.87877924 And it is quite easy to notice the following: (76) 1 96(1+96−2 ) 1 (295∗96 − 1)−1 ≅ 137.03595126474 (77) which is very close to the inverse Fine Structure Constant. Actually, if we replace the factor 1 96(1+96−2 ) by is when n=96 1 𝑛(1+𝑛−2 ) for some integer n, the closest result to the inverse Fine Structure Constant In fact 1 (295∗96 −1)−1 137.0359990368270076 See (40). The factor 1 95∗96 ≅ 96.010383196499723 ≅ 96(1 + 96.1546032−2 ) (78) can be seen as 1 95∗96 = 95 96 + 96 95 −2 (79) The factor 95 ∗ 96 found expression in (41), (42) and is the final missing piece in the puzzle. It is the bridge between trigonometry and electro-gravitational polynomials (35) which resulted in: 1 𝜉 ≅ 1.556198537190348396563877031439915299415588378906 and (1 − 𝑔2 )−4 ≅ 2 607276.5368006824282929301262, provided here with more accuracy if required for further research. 4 In (78) plugging in from (24) instead of 2 and dividing by 2 ∗ 137.03599903682700762 𝜋 instead of by 137.0359990368270076 we get another indication of a deep theoretical relation, 4 𝜋 1 (( )95∗96 −1)−1 2∗137.03599903682700762 ≅ 1 + (2 ∗ 95.974269533437)−1 (80) We now explore another approach, exponential perturbation of the field strength coefficient 95 . 96 This approach was not further investigated due to numerical stability issues, but the author finds it quite interesting. The field strength coefficient coefficients 95 4 95 96 that appears in (23) is the lowest among 3 , , 1.5561985371903484 … . At first this fact was an incentive to search for a 96 𝜋 relation between the fine structure constant and perturbations around the value We return to (23): 95 96 95 96 192𝑎2 +2 𝑎−( )2 192 And to the multiplication in (23) = 𝑎3 𝑎𝑛𝑑 1 (𝑎−1)(1−𝑏) 95 96 95 96 192𝑏 2 −2 𝑏−( )2 192 = 𝑏3 ≅ 12202.88874066467724. 95 . 96 (81) We look at the following exponential 𝑛−1 𝑛−1 95 2 95 192𝑐 2 +2( ) 𝑛 𝑐−( ) 𝑛 96 96 192 𝑛−1 𝑛 3 95 perturbation of the coefficient 96, = 𝑐 𝑎𝑛𝑑 𝑛−1 𝑛−1 95 2 95 192𝑑2 −2( ) 𝑛 𝑑−( ) 𝑛 96 96 192 = 𝑑3 (82) And we check how relatively close is (𝑐 − 1)(1 − 𝑑) to (𝑎 − 1)(1 − 𝑏). The calculation is: (𝑐−1)(1−𝑑) 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟 = (𝑎−1)(1−𝑏) − 1 The strange fact is that (83) 2 (𝑐−1)(1−𝑑) 𝛼 −1 = ((𝑎−1)(1−𝑏) − 1) approximates the inverse fine structure constant. Not as good as (40), 𝑛 (41), (42) but good enough to trigger interest. The last term can be written as in (40) 𝛼 −1 = 2 cos (𝜂) for 𝜂 ≡ 𝑐𝑜𝑠 −1 (2𝛼). It turns out that 𝛼 −1 is maximal or locally maximal at 𝑛 = 964 − 805 or if 𝑛 is allowed to take real values, 𝑛 ≅ 964 − 805.9334 Putting the terms together: 𝛼 −1 ≅ 137.0158482935 95 96 95 96 192𝑎2 +2 𝑎−( )2 192 𝑛−1 (84) 3 = 𝑎 𝑎𝑛𝑑 (85) 95 96 95 96 192𝑏 2 −2 𝑏−( )2 192 = 𝑏3 (86) 𝑛−1 95 𝑛 95 𝑛−1 95 2𝑛−1 95 𝑛 192𝑑 2 − 2 ( ) 𝑐 − ( )2 𝑛 𝑑 − ( ) 𝑛 192𝑐 + 2 ( ) 96 96 96 96 = 𝑐 3 𝑎𝑛𝑑 = 𝑑3 192 192 2 2 (𝑐−1)(1−𝑑) max ((𝑎−1)(1−𝑏) − 1) ≅ 137.015848292861875279413652606308460235595703, 𝑛 𝑛 𝑛 ≅ 964 − 805.933 See appendix G for the code in Python for (81)-(86). Consider the same type of perturbation of 4 the field strength 𝜉 = , 𝜋 4 4 192𝑎2 +2 𝑎−( )2 𝑛−1 𝜋 192 𝜋 = 𝑎3 𝑎𝑛𝑑 4 4 192𝑏 2 −2 𝑏−( )2 𝜋 192 𝜋 = 𝑏3 𝑛−1 4 𝑛 4 𝑛−1 4 𝑛−1 4 𝑛 192𝑐 2 + 2 (𝜋) 𝑐 − (𝜋 )2 𝑛 𝑑 − (𝜋 )2 𝑛 192𝑑 2 − 2 (𝜋) = 𝑐 3 𝑎𝑛𝑑 = 𝑑3 192 192 (86.1) 1 2 (𝑐 − 1)(1 − 𝑑) max ( − 1) ≅ 136.42 𝑛 𝑛 (𝑎 − 1)(1 − 𝑏) Which is close to the square root of the inverse Fine Structure Constant with 𝑛 ≅ 964 − 140631.4697265625. In both cases, numerical stability issues in (86) and (86.1) made it very difficult to check how close such exponential perturbations of the field strength coefficient can be to the inverse Fine Structure Constant trough the error in the polynomial roots. Numerical stability does exist up to 𝑛 = 963 . Before we proceed, consider the following, 𝜉 = 1 4 4 1 ( )1+151.06357822765725984 which is approximately 𝜋 (1 + 624.85524), then it is easy to check that 𝜋 192𝑏 2 − 2𝜉𝑏 − 𝜉 2 192𝑎2 + 2𝜉𝑎 − 𝜉 2 3 = 𝑎 𝑎𝑛𝑑 = 𝑏3 192 192 2 2 2 2 2 4 2𝑛−1 2 2 192𝑐 + 2 𝑐 − ( ) 192𝑑 − 2 𝑑 − ( ) 𝜋) 𝑛 𝜉 𝜉 𝜉 𝜉 3 = 𝑐 𝑎𝑛𝑑 = 𝑑3 ⇒ 192 192 2 (𝑐−1)(1−𝑑) 1 (𝑎−1)(1−𝑏) 2 ≅1 (86.2) 4 This result is expected from 𝜉 = 22 = 𝜉 but not from a field strength so close to 𝜋. It is easy to see that from 𝜉 = 1.25 to 𝜉 = 1.5, (86.2) is very close to 1 within %1 but not as close as when 4 1 1 2 𝜉 = ( )1+151.06357822765725984 or when trivially 𝜉 = 22 = . 𝜋 𝜉 The Fine Structure Constant as a result of Poisson Distribution of events within radius r: We proceed with the methods we have discussed until now. Consider the following expression, 𝑓(𝑥) = 𝑥𝑒 −𝑥 (87) which is the Poison distribution for one event and with 𝜆 = 𝑥. Consider the following perturbation equations in two variables in x around 1. 1 1 𝜂 = 𝑓 (1 − ) = 𝑓 (1 + ) 𝑎 (88) 𝑏 With the following condition for a wide range of 𝜂 > 10000, 1 1 −1 𝛼 −2 = (− ln(𝜂) − 1)−1 and 2( + 𝑎) 𝑏 1 ≅ 𝛼 −1 2−2 Then the system of equations (88), (89) approximates the Fine Structure Constant with the following approximated solution: (89) 𝑎 ≅ 97.2332790992 (90) 𝑏 ≅ 96.56660927693 𝛼 −2 = (− ln(𝜂) − 1)−1 ≅ 18778.86503 1 1 −1 2( + 𝑎) 𝑏 1 = 𝛼 −1 2−2 ≅ 96.89879752 With 𝛼 −1 ≈ 137.03559363 These estimates can be greatly improved with better numerical precision than that of an Excel datasheet, however, this paper does not deal with the Causal Set interpretation of the presented theory and chooses to focus on other subjects. Also, (90) depends on the choice of 𝜂. The Causal Set interpretations can be written as Probability(n=k) = 𝜉 𝑘 𝑒 −𝜉 𝑘! where k is the number of events within a sphere of some small radius r and n is the number of events if this number has the Poisson distribution. Appendix F: The Python code for (40) and for the remark after (40) and its output import numpy as NP def function_cubic_viete(a, b, c, d): # If all roots are real. # Viete's formula when all roots are real. b2 = NP.longdouble(b * b) b3 = NP.longdouble(b2 * b) a2 = NP.longdouble(a * a) a3 = a2 * a p = (3 * a * c - b2) / (3 * a2) q = (2 * b3 - 9 * a * b * c + 27 * a2 * d) / (27 * a3) offset = b / (3 * a) t1 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) \ * (3 * q) / (2 * p)) / 3) t2 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) * \ (3 * q) / (2 * p)) / 3 NP.pi / 3) t3 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) * \ (3 * q) / (2 * p)) / 3 2 * NP.pi / 3) x1 = t1 - offset x2 = t2 - offset x3 = t3 - offset return (x1, x2, x3) def function_fsc_polynomials(): # If all roots are real. fp_f, fp_a, fp_b = 1, 1, 1 fp_start, fp_end = 1.556, NP.pi / 2 for i in range(2000): # Get the biggest roots. These are the closest to 1. # One is above 1 and one is below 1. fp_f = (fp_start + fp_end) * 0.5 fp_a, _, _ = function_cubic_viete(1, -1, -fp_f / 96, (fp_f * fp_f) / 192) fp_b, _, _ = function_cubic_viete(1, -1, fp_f / 96, (fp_f * fp_f) / 192) fp_result_middle = 1/NP.sqrt(fp_a-1) - 0.5/(1-fp_b) if fp_result_middle >= 0: fp_end = fp_f else: fp_start = fp_f fp_s = 1/(1 - fp_b) fp_s *= fp_s fp_s *= fp_s * 0.5 fp_xi = fp_f print('1/(x1-1): %.42lf\n1/(1-x2): %.42lf' %(1/(fp_a-1), 1/(1-fp_b))) print('Xi: %.42lf\ns=0.5/(1-x2)^4: %.42lf' %(fp_f, fp_s)) fp_f = 4 / NP.pi # Get the biggest roots. These are the closest to 1. # One is above 1 and one is below 1. fp_a, _, _ = function_cubic_viete(1, -1, -fp_f / 96, (fp_f * fp_f) / 192) fp_b, _, _ = function_cubic_viete(1, -1, fp_f / 96, (fp_f * fp_f) / 192) fp_mul = (fp_a - 1) * (1 - fp_b) fp_inv_fsc = 2 / NP.cos( fp_xi * (1 + 1/NP.power(fp_s,1/(1+fp_mul)))) print('Inv FSC: %.42lf' %(fp_inv_fsc)) fp_p2 = fp_mul fp_start, fp_end = fp_mul, fp_mul + 0.00001 for i in range(2000): # Get the biggest roots. These are the closest to 1. # One is above 1 and one is below 1. fp_f = (fp_start + fp_end) * 0.5 fp_result_middle = \ fp_s * (2 - 1/(96*96*fp_f)) - NP.power(fp_s, 1/(1+fp_f)) if fp_result_middle >= 0: fp_end = fp_f else: fp_start = fp_f fp_p = 1/NP.sqrt(fp_mul) fp_miracle_p = 1/NP.sqrt(fp_f) fp_relative_p_error = fp_p / (fp_p - fp_miracle_p) print('P: %.42lf\nMiracle P: %.48lf\nRelative error in P: %.48lf^-1' % (fp_p, fp_miracle_p, fp_relative_p_error)) function_fsc_polynomials() ''' Output when run from PyCharm and Python 3.6: 1/(x1-1): 275.516908918643935066938865929841995239257812 1/(1-x2): 33.197404050235356010034593055024743080139160 Xi: 1.556198537190348396563877031439915299415588 s=0.5/(1-x2)^4: 607276.536800682428292930126190185546875000000000 Inv FSC: 137.035999036827007557803881354629993438720703 P: 96.069177214886295246287772897630929946899414 Miracle P: 96.069175812725177365791751071810722351074218750000 Relative error in P: 68515077.183215767145156860351562500000000000000000000000^-1 ''' Appendix G: The Python code for (81)-(86) import numpy as NP def function_cubic_viete(a, b, c, d): # If all roots are real. # Viete's formula when all roots are real. b2 = NP.longdouble(b * b) b3 = NP.longdouble(b2 * b) a2 = NP.longdouble(a * a) a3 = a2 * a p = (3 * a * c - b2) / (3 * a2) q = (2 * b3 - 9 * a * b * c + 27 * a2 * d) / (27 * a3) offset = b / (3 * a) t1 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) \ * (3 * q) / (2 * p)) / 3) t2 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) * \ (3 * q) / (2 * p)) / 3 NP.pi / 3) t3 = 2 * NP.sqrt(-p / 3) * NP.cos(NP.arccos(NP.sqrt(-3 / p) * \ (3 * q) / (2 * p)) / 3 2 * NP.pi / 3) x1 = t1 - offset x2 = t2 - offset x3 = t3 - offset return (x1, x2, x3) def function_f_polynomials(fp_n=96*96*96*96): # If all roots are real. fp_f = 95/96 fp_a, _, _ = function_cubic_viete(1, -1, -fp_f / 96, (fp_f * fp_f) / 192) fp_b, _, _ = function_cubic_viete(1, -1, fp_f / 96, (fp_f * fp_f) / 192) fp_mul1 = (fp_a - 1)*(1 - fp_b) fp_f = NP.power(fp_f, (fp_n-1)/fp_n) fp_a, _, _ = function_cubic_viete(1, -1, -fp_f / 96, (fp_f * fp_f) / 192) fp_b, _, _ = function_cubic_viete(1, -1, fp_f / 96, (fp_f * fp_f) / 192) fp_mul2 = (fp_a - 1)*(1 - fp_b) fp_combine = 2/(fp_n *(fp_mul2/fp_mul1-1)) #print('%.42lf' %fp_combine) return fp_combine def main(): ma_best_val = 0 ma_best_m = 0 #function_f_polynomials(96 * 96 * 96 * 96 - 1) #function_f_polynomials(96 * 96 * 96 * 96) #function_f_polynomials(96 * 96 * 96 * 96 + 1) print('Coarse search:') for i in range(-1000, 1000): ma_r = function_f_polynomials(96 * 96 * 96 * 96 - i) if ma_best_val < ma_r: ma_best_val = ma_r ma_best_m = i print('Best value %.42lf' %ma_best_val) print('Best m = 96^4-%d' % ma_best_m) print('Fine search:') ma_best_val = 0.0 ma_best_m = 0.0 for i in range(8050000-10000, 8050000+10000): ma_d = i/10000 ma_r = function_f_polynomials(96 * 96 * 96 * 96 - ma_d) if ma_best_val < ma_r: Fappen ma_best_val = ma_r ma_best_m = ma_d print('Best value %.42lf' %ma_best_val) print('Best m = 96^4-%.42lf' % ma_best_m) ''' Coarse search: Best value 137.015846787740116496934206224977970123291016 Best m = 96^4-805 Fine search: Best value 137.015848292861875279413652606308460235595703 Best m = 96^4-805.932999999999992724042385816574096679687500 ''' if __name__ == '__main__': main() Appendix H – Causality conservation theorem Theorem: If p is real, any monotone function 𝑓(𝑝), called causality function will yield the same Reeb vector. The reader is advised to check the case when p is an imaginary function. Then the Reeb vector is defined as Proof: 𝑢𝜈 2 𝑧 𝑧 = 2𝑧𝜈 − 2𝑧𝑘2 𝑝 ∗𝑘 𝑝𝜈 . We will use capital letters for 𝑃 = 𝑓(𝑝) and as in previous pages, 𝑧 = 𝑝𝜆 𝑝 𝜆 and here 𝑍 = 𝑃𝜆 𝑃 𝜆 . 𝑃 = 𝑓(𝑝) 𝑃𝜇 = 𝑓′(𝑝)𝑝𝜇 𝑍 = 𝑓 ′ (𝑝)𝑝𝜇 𝑓 ′ (𝑝)𝑝𝜇 = 𝑓 ′ (𝑝)2 𝑧 𝑍𝜈 2𝑓 ′ (𝑝)𝑓′′(𝑝)𝑝𝜈 𝑧 𝑓 ′ (𝑝)2 𝑧𝜈 2𝑓′′(𝑝)𝑝𝜈 𝑧𝜈 = + ′ 2 = + 𝑍 𝑓 ′ (𝑝)2 𝑧 𝑓 (𝑝) 𝑧 𝑓 ′ (𝑝) 𝑧 𝑈𝜈 = 𝑈𝜈 = 2𝑓 ′′ (𝑝)𝑝𝑘 𝑧𝑘 𝑓′(𝑝)𝑝𝑘 𝑓′(𝑝)𝑝𝜈 2𝑓′′(𝑝)𝑝𝜈 𝑧𝜈 + − ( + ) 𝑓 ′ (𝑝) 𝑧 𝑧 𝑓 ′ (𝑝)2 𝑧 𝑓 ′ (𝑝) 2𝑓′′(𝑝)𝑝𝜈 2𝑓 ′′ (𝑝)𝑝𝜈 𝑧𝜈 𝑧𝑘 𝑘 𝑧𝜈 𝑧𝑘 − + − 2 𝑝 𝑝𝜈 = − 2 𝑝𝑘 𝑝𝜈 = 𝑢𝜈 ′ ′ 𝑓 (𝑝) 𝑓 (𝑝) 𝑧 𝑧 𝑧 𝑧 𝑈𝜈 2 = 𝑢𝜈 2 (91) References [1] R. Geroch, Domain of dependence, J. Math. Phys. 11 (1970) 437–449, https://doi.org/10.1063/1.1665157 [2] J. Albo, Book of Principles (Sefer Ha-ikarim), “Immeasurable time – Maamar 18”, “measurable time by movement”. (Circa 1380-1444, unknown), The Jewish publication Society of America (1946), ASIN: B001EBBSIC, Chapter 2, Chapter 13. [3] Jungjai Lee, Hyun Seok Yang, "Quantized Kahler Geometry and Quantum Gravity", April 2018 Journal Korean Physical Society 72(12), DOI: 10.3938/jkps.72.1421 [4] G. Reeb. Sur certaines propri´t´s topologiques des vari´et´es feuillet´ees. Actualit´e Sci. Indust. 1183, Hermann, Paris (1952). [5] Tilman Becker, "GEODESIC AND CONFORMALLY REEB VECTOR FIELDS ON FLAT 3-MANIFOLDS", arXiv:2207.03274v3, Theorem 2 on page 3 boils down to the vector field Y, which is not accelerated and to X as a geodesic field, which is very different from 2 dimensions, where X cannot be geodesic. [6] STEVEN HURDER, "DYNAMICS AND THE GODBILLON-VEY CLASSES: A HISTORY AND SURVEY", September 5, 2000, Page 2 - Reeb Class, http://homepages.math.uic.edu/~hurder/papers/54manuscript.pdf [7] Yaakov Friedman, “A physically meaningful relativistic description of the spin state of an electron”, June 2021, DOI: 10.13140/RG.2.2.10573.15842 [8] Hartland S. Snyder , "Quantized Space-Time", 1 January 1947, Phys. Rev. volume 71, pages 38-41, Published: Americam Physical Society. doi: 10.1103/PhysRev.71.38 [9] Yaakov Friedman, Tzvi Scarr, "Uniform Acceleration in General Relativity", 8 Feb 2016, arXiv:1602.03067v1 [10] D. Lovelock and H. Rund, “The Numerical Relative Tensors”, Tensors, Differential Forms and Variational Principles, 4.2, Dover Publications Inc. Mineola, N.Y. , ISBN 0-486-65840-6, p. 113, 2.18, p. 114, 2.30. [11] Eytan H. Suchard, Electro-gravity via geometric chronon Field, arXiv:1806.05244v16, 5/April/2021 [12] S. Vaknin, “Time Asymmetry Re-visited”, LC Classification: Microfilm 85/871 (Q). [microform], Library of Congress.LC Control Number: 85133690, Thesis (Ph. D.)--Pacific Western University, 1982, c1984, Ann Arbor, MI : University Microfilms International, 1984. [13] M. Alcubierre, “The warp drive: hyper-fast travel within general relativity”, Class.Quant.Grav.11:L73-L77, 1994, DOI: 10.1088/0264-9381/11/5/001. [14] Omar Rodrigues Alves, Elio Battista Porcelli, Victor S. Filho, "Experimental Verification of Anomalous Forces on Shielded Symmetrical Capacitors", March 2020, Applied Physics Research 12(2), DOI: 10.5539/apr.v12n2p33 [15] C.Chaneliere, J.L.Autran, R.A.B.Devine,B.Balland, "Tantalum pentoxide (Ta2O5) thin films for advanced dielectric applications" Materials Science and Engineering, Volume 22, Issue 6, 25 May 1998, Pages 269-322 DOI: https://doi.org/10.1016/S0927-796X(97)00023-5 [16] M. Tajmar and T. Schreiber, "Put Strong Limits on All Proposed Theories so far Assessing Electrostatic Propulsion: Does a Charged High-Voltage Capacitor Produce Thrust?" May 2020 Journal of Electrostatics 107:103477, DOI:10.1016/j.elstat.2020.103477 [17] Paul Ehrlich, "Dielectric Properties of Teflon from Room Temperature to 314 0 C and fron Frequencies of 103 to 105 c/s1", Journal of Research of the National Bureau of Standards, Vol. 51, No. 4, October 1953, Research Paper 2449. Page 186, 4. Results and Discussion. [18] Ted Jacobson, Thermodynamics of Spacetime: “The Einstein Equation of State”, Physical Review Letters. 75, 1260 - Published 14 August 1995, DOI:https://doi.org/10.1103/PhysRevLett.75.1260 [19] Hawking, S. W (1975). "Particle creation by black holes". Communications in Mathematical Physics. 43 (3): 199–220, doi:10.1007/BF02345020 [20] Seth Lloyd - “Deriving general relativity from quantum measurement”, Institute For Quantum Computing - IQC, public lecture loaded to YouTube on 16/August/2013, http://www.youtube.com/watch?v=t9zcBKoFrME [21] Ettore Majorana: "Notes on Theoretical Physics", Edited By Salvatore Esposito, Ettore Majorana Jr, Alwyn van der Merwe and Erasmo Recami, Springer-Science+Business Media, B.V., 2003, ISBN 978-90-481-6435-6 ISBN 978-94-017-0107-5 (eBook), DOI: 10.1007/978-94-017-0107-5, 19. ENERGY OF A UNIFORM CIRCULAR DISTRIBUTION OF ELECTRIC OR MAGNETIC CHARGES, (1.183), Pages 33, 34. [22] "Techniques for solving bound state problems", M. van Iersel, C.F.M. van der Burgh, and B.L.G. Bakker, "5.2 The collapse of wave functions", page 14 one line below (38), see the discussion about estimation of the critical value of the strength of the Yukawa potential in the relativistic case, october 25, 2018, arXiv:hep-ph/0010243v1 [23] Adrian Dumitrescu, Csaba D.Tóth, Guangwu Xu "On stars and Steiner stars", Volume 6, Issue 3, August 2009, Pages 324-332, https://doi.org/10.1016/j.disopt.2009.04.003 [24] Miroslav Chlebík, Janka Chlebíková, "The Steiner tree problem on graphs: Inapproximability results", theoretical Computer Science 406 (2008) 207–214, doi:10.1016/j.tcs.2008.06.046, See: “It is NP-hard to approximate the Steiner Tree problem within a factor 1.01063 (>96/95)”. [25] Thomas Johannes Georg Rauh, "Precise top and bottom quark masses from pair production near threshold in -e +e collisions", doctorate dissertation for Technische Universitat Munchen, 2016. [26] A.M. Badalian (Moscow, ITEP), A.I. Veselov (Moscow, ITEP), B.L.G. Bakker (Vrije U., Amsterdam) "The Pole and heavy quark masses in the Hamiltonian approach", e-Print: hep-ph/0311010 [hep-ph], DOI: 10.1134/1.1777292, Published in: Phys.Atom.Nucl. 67 (2004), 1367-1377, Yad.Fiz. 67 (2004), 1392-1402 [27] "Formulas for Odd Zeta Values and Powers of Pi", Journal of Integer Sequences, Vol. 14 (2011), Article 11.2.5, Page 2, idetity, identities by Simon Plouffe. [28] Mutmainnah et al, "Power factor correction of the industrial electrical system during large induction motor starting using ETAP power station", see page 2, "3. Power Factor Correction", 2020 IOP Conf. Ser.: Mater. Sci. Eng. 885 012009 [29] M. Ablikim et. al. "Precision Measurement of the Mass of the τ Lepton", 13 July 2014, https://arxiv.org/abs/1405.1076, DOI: 10.1103/PhysRevD.90.012001 [30] Chenghui Yu, Weicheng Zhong, Brian Estey, Joyce Kwan, Richard H. Parker (UC, Berkeley), Holger Müller, "Atom‐Interferometry Measurement of the Fine Structure Constant", (UC, Berkeley & LBL, Berkeley) Annalen Phys. 531 (2019) no.5, 1800346, (2019-05-01), DOI: 10.1002/andp.201800346 https://onlinelibrary.wiley.com/doi/full/10.1002/andp.201800346 [31] M Belloni and R W Robinett, “Constraints on Airy function zeros from quantum-mechanical sum rules”, arXiv:1007.1623v1, 9/July/2010. [32] Kazuhiko Aomoto, Michitake Kita, Theory of Hypergeometric Functions, Sringer Monographs in Mathematics, ISSN 1439-7382, ISBN 978-4-431-53912-4 e-ISBN 978-4-431-53938-4, DOI: 10.1007/978-4-431-53912-4, Springer Tokyo, Dordrecht, Heidelberg, London, New York. [33] D. Lovelock and H. Rund, Tensors, Differential Forms and Variational Principles, Dover Publications Inc. Mineola, N.Y., 1989 ,ISBN 0-486-65840-6, p. 262, 3.27 is the Einstein tensor. [34] D. Lovelock and H. Rund, Tensors, Differential Forms and Variational Principles, Dover Publications Inc. Mineola, N.Y., 1989, ISBN 0-486-65840-6, p. 121-126, 4.4 The Lie Derivative.