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A Hidden Dimension, Clifford Algebra, and Centauro Events

Carl Brannen∗ A Quality Construction, Woodinville, WA† (Dated: July 21, 2005) This paper fleshes out the arguments given in a 20 minute talk at the Phenomenology 2005 meeting at the University of Wisconsin at Madison, Wisconsin on Monday, May 2, 2005. The argument goes as follow: A hidden dimension is useful for explaining the phase velocity of quantum waves. The hidden dimension corresponds to the proper time parameter of standard relativity. This theory has been developed into a full gravitational theory, “Euclidean Relativity” by other authors. Euclidean relativity matches the results of Einstein’s gravitation theory. This article outlines a compatible theory for elementary particles. The massless Dirac equation can be generalized from an equation of matrix operators operating on vectors to an equation of matrix operators operating on matrices. This allows the Dirac equation to model four particles simultaneously. We then examine the natural quantum numbers of the gamma matrices of the Dirac equation, and generalize this result to arbitrary complexified Clifford algebras. Fitting this “spectral decomposition” to the usual elementary particles, we find that one hidden dimension is needed as was similarly needed by Euclidean relativity, and that we need a set of eight subparticles to make up the elementary fermions. These elementary particles will be called “binons”, and each comes in three possible subcolors. A calculation for the masses of the leptons will be given in a paper associated with a talk by the author at the APSNW 2005 meeting at the University of Victoria, at British Columbia, Canada on May 15, 2005. After an abbreviated introduction, this paper will concentrate on the phenomeno- logical aspects of binons, particularly as applied to the Centauro type cosmic rays, and gamma-ray bursts.

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The paper consists of three sections. The first discusses We postulate a simple potential function as the binding binons, a subparticle postulated to make up quarks and potential between binons. We discuss spin and statis- leptons, and gives some details of their theory. The sec- tics for binons as subparticles to quarks and leptons. We ond discusses various unexplained astronomical observa- briefly discuss C, P, and T symmetry from the perspec- tions mostly having to do with the unusual cosmic ray tive of binons. We generalize the Space Time Algebra of events known as Centauros. The third section connects David Hestenes[1] to a set of geometries more suited to the first two sections by showing how binons can be used the internal symmetries of particles, the Particle Internal to explain the observations. Symmetry Algebra (PISA), which includes a hidden vio- lation of Lorentz symmetry. We parameterize the various PISA geometries. We attribute the breaking of isospin symmetry to the PISA parameterization. Finally, we dis- I. A BRIEF INTRODUCTION TO BINONS cuss spin in the context of a geometrization of particle internal symmetries. This first of two sections of this paper will give a brief introduction to binons. We begin with proper time and why it is natural to postulate a hidden dimension corre- A. Proper Time as a Hidden Dimension sponding to proper time. Then we generalize the Dirac equation from an equation written in matrices and vec- tors to one written in matrices for both the operators In 1923, L. de Broglie proposed that matter possesses and the wave. We extract the natural quantum num- the same wave particle duality as light, with similar rela- bers of the gamma matrices, and give the result for a tions between frequency and energy.[2] As a consequence 1 general complexified Clifford algebra which will have a of time dilation and the relation E = ~ω, de Broglie hypercubic form. We show that the quantum numbers concluded that there must be associated with a parti- of the fundamental fermions form a cube and postulate cle travelling at speed v, a wave that travels at (phase) that the corners of the cube correspond to subparticles speed: that make up quarks and leptons that we call “binons”. p 2 We look at the quantum numbers of binons in detail. vφ = ω/k = c/ 1 − v = c/β. (1)

∗Electronic address: [email protected] 1 Some notation, for example the use of ω in preference to ν, has †URL: http://www.brannenworks.com been changed to match modern usage.

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Since this speed is greater than c [3, P23.15], he referred where k is the wave vector and ω is the frequency. In to the wave as fictitous.2 The confusion was soon cor- order to have a wave with a speed of c, we must have rected when he realized that it is only the group velocity 2 2 2 2 that can be associated with the position of the particle, c (kz + ks ) = ω . (3) and the group velocity will be less than c.3 This has remained the conventional answer to this odd The phase velocity in the +z direction is p feature of the theory for the better part of a century. For 2 2 example, in A. Messiah’s excellent introduction to quan- vφ = c/ 1 − vz /c = ω/kz, (4) tum mechanics, the details of the calculation for group Squaring the second equality gives: velocity are given, but he fails to explicitly mention that the phase velocity exceeds c.[5, CH.II, §3] The paradox is 2 2 2 2 2 c kz = ω (1 − vz /c ). (5) not eliminated in relativistic quantum mechanics, though it is rarely mentioned. For example see [6, §3.3]. Subtracting this equation from Eq. (3) gives: The presence of a hidden dimension, or multiple hidden 2 2 2 2 2 dimensions, in Kaluza-Klein theories, as well as modern c ks = ω (vz /c ). (6) string theory, suggests that an explanation for the su- perluminal phase velocities of matter waves is that the Now our assumption was that the superluminal phase wave is being considered in only 3 dimensions. A wave velocity of a de Broglie wave was due to the physical travelling with a phase velocity of c in 3 dimensions plus wave traversing a hidden dimension and possessing a true a hidden dimension, when translated into a wave in the phase velocity of c. Under this assumption, the associ- usual space, will give a phase velocity in excess of c. This ated classical particle must also have a true velocity of c. effect is commonly seen at the beach, where the velocity In order for this to occur we must have: of breakers along the shore exceeds the velocity of the c2 = v2 + v2. (7) incoming waves. See Fig. (1) where the beach lies along z s the z axis, and we use s for the coordinate in the hidden Substituting this relation into Eq. (6) puts it into the dimension. same form as Eq. (5):

2 2 2 2 2 FIG. 1: Illustration of the increase in phase velocity when c ks = ω (1 − vs /c ). (8) a hidden dimension is ignored. The true wavelength in 2 dimensions, λ, is increased to λx = λ/ cos(θ) when only the z Let us write the standard metric for special relativity dimension is considered. The phase velocity is λω/2π. with the proper time treated as a spatial dimension:

ds2 = c2 dt2 − (dx2 + dy2 + dz2). (9) ¢¢¢¢¢¢ ¢¢¢¢¢¢ ¢¢¢¢¢¢ ¢¢¢¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ 2 ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ Dividing by dt gives: ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢  2  2  2  2 ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ 2 dx dy dz ds λ c = + + + (10) s ¢¢¢ ¢¢¢ -z ¢¢¢ ¢¢¢ dt dt dt dt PP θ ¢¢¢ ¢¢¢ PPPq ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ λ ¢¢¢ ¢¢¢ Comparing with Eq. (7) shows that we must interpret ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ v as ds/dt, and therefore the hidden dimension must ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ s ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ be interpreted as corresponding to proper time. Keeping ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ s as a spatial dimension, we will use the (− + + + +) z metric for space-time. That is, gµν is the diagonal matrix with entries (−1, 1, 1, 1) or (−1, 1, 1, 1, 1) depending on With the natural assumption that the true phase ve- whether 4 or 5 dimensions are being discussed. locity of the matter wave is c, we can immediately derive By assuming that nature keeps track of proper time a physical interpretation of the hidden dimension. Let with a hidden dimension, we explain several riddles re- ψ(x, y, z, s, t) be a 5-dimensional plane wave travelling in lated to the nature of our world. First, the coincidence the +z direction: that different reference frames agree on proper time inter- vals, while disagreeing about measurements of the usual ψ(x, y, z, s, t) = ψ0(zkz + sks − ωt), (2) spatial dimensions or time, is explained by the assump- tion that Lorentz boosts do not affect the proper time dimension. Second, that stationary matter contains large amounts of energy is explained by the assumption that 2 Cette onde de vitesse plus grande que c ne peut correspondre `a stationary matter implicitly moves in a hidden dimen- un transport d’´energie;nous la consid´ereronsseulement comme une onde fictive associ´eeau mouvement du mobile.[2] sion. Third, since we suppose here that matter always 3 For an interesting derivation of gravity in flat space-time from moves at the speed of light, it becomes obvious why we the optics of de Broglie’s matter waves, see [4]. cannot arrange for it to exceed that speed in the usual 3

3 spatial dimensions. Fifth, our inability to measure the tion between the space-time symmetries and absolute phase of a quantum object can be seen to arise the other symmetries – those mixing different from our inability to measure absolute positions in the particle types (e.g. charge, flavour, color). hidden dimension. Sixth, we can attribute the ability This means that the properties and struc- of matter to interfere with itself to interference over the ture of the Poincar´egroup are of no help in hidden dimension. Arranging for all quantum waves to choosing the set of other symmetries. But travel at speed c promises to simplify quantum mechan- the most important implication is that sym- ics. Also, we can see reasons why Wick rotations have metries cannot relate particles with different been so useful in quantum mechanics. mass and spin and thus hope to describe the The use of a hidden dimension corresponding to proper full variet of particle types through symmetry time is neither unique nor original with this author, considerations was destroyed. though this exposition, as far as I know, is. J. M. C. Montanus has published numerous articles which use the The Coleman-Mandula theorem also applies to higher proper time of an object as its fourth coordinate.[7] He dimensions as the source of the above quote shows.[18] calls the result “absolute Euclidean spacetime” or AEST. The present work is distinguished from that of Montanus and similar work by others in “Euclidean relativity” such B. Generalizing the Dirac Equation as A. Gersten [8], or [9] or the “4-dimensional optics” (4DO) of J. B. Almeida [10], or the “Gauge Theory of The crowning achievement of standard quantum me- Gravity” [11–13] primarily in that this work will assume chanics is the calculation of the g-2 value for the electron that the hidden dimension is cyclic and small. We will to an accuracy of better than 10 decimal places. A the- call this version of space-time the ”proper time geome- ory claiming to be a unified field theory needs to be able try” (PTG). For other work using proper time, see [14– to reproduce this calculation. The present theory will do 16]. The above authors have mostly concentrated on this by deriving, from the principles of this theory, the showing that general relativity can be expressed in Eu- propagators and vertices used in the Feynman diagrams clidean coordinates. This paper will accept this conclu- of the standard model. sion and will concentrate on fitting elementary particles The derivation of the Dirac propagator from first prin- to this geometry. ciples will be given in a companion paper associated with When converting coordinates between classical relativ- the APSNW2005 meeting. This paper will restrict itself ity and the PTG, one simply ignores the s coordinate as to deriving a few characteristics of the subparticles. But it is infinitesimal in size. There will be a small error, but to see the necessity for these subparticles, it is useful to for a sufficiently small hidden dimension, the error will consider generalizations of the Dirac equation. be below our capabilities for measurement. Examples of The usual massive Dirac equation allows the move- calculations in special relativity on the PTG are included ment of a single spin-1/2 fermion to be modeled in a in the Appendix A. very precise manner. On the other hand, the standard Given that the above authors differ from this one pri- model [19] attributes fermion masses as arising from ex- marily in the assumption that the proper time dimen- change of Higgs bosons. Therefore it is the massless Dirac sion is cyclic and small, some defense of this position is equations that are of interest. In this sense, the various required. First, the fact that the hidden dimension is elementary fermions all share the same Dirac equation, hidden suggests that we should assume that it is small. and this makes them distinct from the various elemen- Second, having a large hidden dimension corresponding tary bosons. This suggests that it is natural to associate to proper time requires that particles be able to collide a distinct, but identical, Dirac equation with each type despite having distinct values of proper time. Third, the of elementary fermion. weak force couples only to handed particles and handed particles travel at the speed of light. This is consistent An obvious way of obtaining a multiple particle mass- with the fact that a cyclic hidden dimension as described less Dirac equation is to leave the gamma operators alone, here implies that speed in 3 dimensions is quantized, and but to replace the spinors, with matrices. Each column the lowest energy particles would travel at speed c. of the matrix making up the multiparticle wave can be This paper, which proposes to associate the internal associated with a class of elementary fermion. For ex- symmetries of particles with the geometry of space-time, ample, the first column could represent the electron, the would not be possible under an assumption of perfect second column the red down quark, the third column the Poincar´esymmetry, at least without some major mod- green down quark, etc. The massless Dirac equation for ification of quantum field theory due to the Coleman- the electron: Mandula[17] theorem: µ (γµ∂ )4×4 (ψe)4×1 = 0, (11) The implications of the theorem are far reach- ing. It implies that, at least in the domain where the gamma matrices satisfy of classical groups, and under the stated as- sumptions of the theorem, there is no connec- γµγν + γν γµ = gµν , (12) 4 can be generalized to an equation for four fermions: C. The Quantum Numbers of a Complexified Clifford Algebra µ (γµ∂ )4×4 (ψe, ψdR, ψdG, ψdB)4×4 = 0, (13) Another reason for the utility of expanding the Dirac To separate the four components of the 4×4 wave matrix equation to operate on matrices instead of vectors is that into the four different particles, one can use the projec- one can then use the gamma matrices themselves as the tion operators: 4 basis for the matrices. There are sixteen different possible multiples of gamma matrices, other than sign differences, and these sixteen products provide a basis [6, §3.4] for     4 × 4 matrices over the complex numbers. 1 0 The four gamma matrices anticommute, so one can  0   1  η = η = always reduce a product of gamma matrices to a product 00  0  11  0  that contains no matrix more than once, and one can 0 0     reorder the matrices to an appropriate standard. We 0 0 will always assume that this is done, and furthermore we  0   0  η = η = , (14) will abbreviate products of distinct gamma matrices as 22  1  33  0  0 1 γµγν = γµν . (16) where we use the notation ηjk to refer to the matrix with As we later generalize to Clifford algebras, which is nec- a one in the (i, j) location and the remainder of entries essary in order to add the hidden dimension, s, we will zero. abbreviate our notation and keep only the indices of the The four projection operators satisfy the following re- 5-vector of gamma matrices. For example, lations: γ1γ2 = γ12 = γxy = xy.c (17)

k ηjj ηkk = δj ηkk, Making this notation change will greatly ease our calcu- X lations, will more closely follow the mathematical litera- ηkk = 1ˆ. (15) ture, and will make more clear the geometric interpreta- k tion of our results. In addition, none of these projection operators can be The gamma matrices form a 4-vector so each individual written as a nontrivial sum of two other projection op- gamma matrix is associated with a particular direction in space-time. Since any 4 × 4 matrix can be written erators. This combination of attributes defines the four 5 projection operators as a set of “mutually annihilating as a sum over products of gamma matrices, this gives primitive idempotents”. us a geometric interpretation of any 4 × 4 matrix as a sum over complex multiples of products of geometric di- Of course there are other sets of primitive idempotents rections. When we extend spinors from vector form to for the 4×4 complex matrices. For example, if S is an in- matrix form, this extends the geometric interpretation of vertible matrix, then S η S−1 gives another set of prim- kk the components of 4×4 matrices to a geometric interpre- itive idempotents. But the particular set given above is tation of the components of the spinors. Of course the significant because it projects the unified wave function geometric interpretation depends on the choice of repre- Ψ = (ψ , ψ , ψ , ψ ) into distinct elementary fermions. 0 1 2 3 sentation of the gamma matrices. Thinking of the fermions in this manner promises to be Products of gamma matrices, like the matrices them- useful in that it explains how it comes to be that linear selves, square to 1 or -1, so their eigenvalues will be ±1 or combinations of fermions appear so often in the standard ±i. This is true for any of the products, but the products model. Therefore, to understand the symmetries of the that are diagonal are particularly easy to analyze, so we fermions, we should examine the symmetries of the pro- will now choose a convenient representation. When we jection operators of the gamma matrices. generalize to Clifford algebra, there will be no need to The association of the elementary particles with prim- refer to any particular representation. itive idempotents is not an original idea of this paper. Since we expect that the chiral particles are fundamen- Most famously, J. Schwinger uses the same method in tal, we will use a representation of the gamma matrices what is known as the “Schwinger measurement algebra”. that has the left and right handed electron and left and His analysis was devoted to kinematics instead of particle right handed positron separated in the components of the symmetries. For a brief introduction, see Appendix B.

5 Since the gamma matrices anticommute, one can substitute com- 4 Mathematicians frequently refer to projection operators as mutation for multiplication in forming this group, provided one “idempotents”. includes a factor of 0.5, for example see [6, §3.4] 5 usual spinor. In a later paper we will derive the massive simultaneously diagonal, a topic we will pursue by first Dirac propagator from the massless chiral propagators looking at the form of the gamma matrices that we have used here, but for now, we will simply treat the handed chosen. particles and antiparticles as distinct. The Weyl or chiral With the diagonal gamma products in this form, it’s representation [6, §3.2] satisfies our requirements: particularly easy to solve for the ηkk primitive idempo-     tents in terms of gamma matrices. The results are: 0 0 1 0 0 0 0 1  0 0 0 1   0 0 1 0  ˆ tˆ= xˆ = η00 = (1 − ztb + ixyc − ixyztd )/4  1 0 0 0   0 −1 0 0  ˆ b d 0 1 0 0 −1 0 0 0 η01 = (1 + zt − ixyc − ixyzt)/4     ˆ b d 0 0 0 −i 0 0 1 0 η02 = (1 + zt + ixyc + ixyzt)/4  0 0 i 0   0 0 0 −1  ˆ b d yˆ = zˆ = . (18) η03 = (1 − zt − ixyc + ixyzt)/4. (20)  0 i 0 0   −1 0 0 0  −i 0 0 0 0 1 0 0 We can factor ixyztd = ixyc ztb and put these into the more suggestive result: The diagonal group elements are as follows:     ˆ b ˆ 1 0 0 0 −1 0 0 0 η00 = (1 − zt)(1 + ixyc)/4  0 1 0 0   0 −1 0 0  ˆ b ˆ 1ˆ = ixyztd = η01 = (1 + zt)(1 − ixyc)/4  0 0 1 0   0 0 1 0  η = (1ˆ + ztb )(1ˆ + ixyc)/4 0 0 0 1 0 0 0 1 02     η = (1ˆ − ztb )(1ˆ − ixyc)/4 (21) −1 0 0 0 1 0 0 0 03  0 1 0 0   0 −1 0 0  ztb = ixyc = (19) This can be continued for the remaining η , but the  0 0 1 0   0 0 1 0  nm primitive idempotents are enough for this paper. 0 0 0 −1 0 0 0 −1 The four primitive idempotents ηkk, have the following Note that the above four elements are linearly indepen- eigenvalues with respect to the diagonal gamma matrix dent and thus give a basis for the diagonal 4×4 matrices. products: Also note that we have included a factor of i so that these matrices each square to 1.ˆ When we generalize to Clifford 1ˆ ixyc ixyztd ztb algebra, we will use the “complexified” Clifford algebra η00 1 1 −1 −1 so this can always be done. η11 1 −1 −1 1 (22) It is also possible to do a similar analysis using real η22 1 1 1 1 Clifford algebra. The idempotent structure of a Clifford η33 1 −1 1 −1 algebra is referred, in the mathematical literature, to the “spectral basis” or “spectral decomposition” [20, pp201- Of course 1ˆ is a trivial operator, and ztb is simply the 2]. The real Clifford algebras are more complicated than product of ixyc and ixyztd . Thus we can use the eigen- the complexified ones because one can always multiply values of ixyc and ixyztd alone to define the ηkk primitive the basis vectors of a complexified Clifford algebra by i idempotents as shown in Fig. (2). Note that our choice of 6 and so the signature does not matter much . For simplic- these two is somewhat arbitrary, we could instead have ity, this paper will restrict itself to the complexified case chosen ixyc and ztb or ztb and ixyztd . with only one signature, and for familiarity we will use a Fig. (2) shows that the quantum numbers of the geometry as similar as possible to the usual Minkowski. gamma matrices form a square. This is a general at- When one is working with spinors, the effect of a choice tribute of Clifford algebras, namely that their quantum of gamma matrix representation is to choose the interpre- numbers form hypercubes[23, §17.5], also see [22]. tation of the four components of the spinor. For exam- The presence of hypercubes in the eigenvalue spectrum ple, one could choose the gamma matrices so that the of Clifford algebras, along with the fact that Clifford alge- four components of a spinor correspond to electrons and bras are the natural generalization of the Dirac gamma positrons with x-component of spin ±1/2. There are lim- matrices, suggests that we should look for hypercubic itations as to which gamma matrices can be chosen to be symmetry in the elementary particles, a subject we will now turn to.

6 For the case of real Clifford algebras, an interesting analysis of the signature issue is given by [21]. To see an analysis of the D. The Fermion Cube fermions for the general Clifford algebra case, both real and com- plexified, in various dimensions and signatures, see [22], where the signature and geometry choices of this paper are included The fermions in a family can be designated by their as the second to the last line in the table showing symmetries SU(2) and U(1) symmetry quantum numbers t3 and t0, towards the end of section II. or alternatively, by their electric charge Q and “neutral 6

from square to cube form.7 FIG. 2: The natural primitive idempotents of 4 × 4 matrices plotted according to their eigenvalues with respect to the ztb and ixyc operators.

6 rη00 rη22 FIG. 4: The fermion cube. The ν is not shown for clarity. +1 R λixy 1.0 r 6 "be¯R t0 " b " b b " u b 0.5 "r ∗R br e¯ bν¯L " L 0 b b ¯ " r b d∗R " r m l d∗L b " u∗L QkQ 3 0.0 b"r Q ν¯R r u¯∗L r d¯∗L r rη11 rη33 d −1 -0.5 r e ∗R r ν n ? bL " L b r " u¯∗R λ b " ixyzt- b r " -1.0 b"eR −1 0 +1 - −0.5 0.0 0.5 t3 charge” (or “weak charge”) Q0.[24, Table 6.2] The values 0 for Q and Q are related to t3 and t0 by

Q = t3 + t0, In Fig. (4), note that our designation of the handedness 0 Q = t3 cot(θw) − t0 tan(θw), (23) of the antiparticles is reversed from the usual. The stan- dard notation assigns handedness of the antiparticles in 2 where θw is the Weinberg angle. We will use sin (θw) = such a way that a spinor that corresponds to spin 1/2 for 1/4. A table of the usual quantum numbers for fermions a particles gives spin −1/2 for the antiparticle. The stan- is shown in Fig. (3). Values for antiparticles are the neg- dard model definition is compatible with the hole model atives of the values shown. of positrons: “If the missing electron had positive Jz, its absence has negative Jz”.[6, Ch. 3.5] We will be mod- FIG. 3: Table of standard model fermion quantum numbers. eling particles and antiparticles on an equal footing (no √ 0 Dirac sea and no holes), so we will use the same definition t3 t0 QQ 3/2 of handedness for both types.8 eR 0 -1 -1 1/2 eL -1/2 -1/2 -1 -1/2 Both of the right handed neutrinos, νR andν ¯R, end νL 1/2 -1/2 0 1 ν 0 0 0 0 up at the origin, and we have to choose which goes with R the visible part of the cube and which is hidden. Since d∗R 0 -1/3 -1/3 1/6 the rest of the top part of the cube (i.e. {ν¯ , e¯ , e¯ }) are d∗L -1/2 1/6 -1/3 -5/6 L R L u∗L 1/2 1/6 2/3 2/3 all antiparticles, we will place theν ¯R with them. The n u∗R 0 2/3 2/3 -1/3 vector therefore runs in the direction from theν ¯R towards the eR, the l runs towards thee ¯L, and the m runs towards theν ¯ . In Fig. (3), the alert reader will notice that we are L using nonstandard notation for the electron and positron. Fig. (4) shows that the leptons do have a cubic struc- We will specify all antiparticles by putting a bar on top ture and can therefore be interpreted as primitive idem- and will suppress extraneous indications of the particle’s potents of a Clifford algebra. According to the illustrated electric charge. choice of n, l, and m, and using the order |n, l, mi, the When the fermions are plotted according to their Q and Q0 quantum numbers, the cubic nature of their sym- metry is apparent. We will define the cube according to the n, l, and m vectors as shown in Fig. (4). The fact that 7 In a complexified Clifford algebra, the number of quantum num- the symmetry comes in cube form, rather than square, as bers increases by one for every two dimensions added. we would be led to expect from the quantum numbers of 8 Note that Fig. (4) suggests that the way that the standard model the usual gamma matrices, indicates that our earlier ex- distinguishes between particle and antiparticle is arbitrary. In pectation that there is a hidden dimension corresponding our model, for example, the up quark is a mixture of binons from a particle (neutrino) and binons from an antiparticle, (positron). to proper time, s is vindicated. The hidden dimension Rather than the usual definition, a better definition of “antiparti- expands the number of gamma matrices by one. This cle” is relational: the antiparticle of a binon is the binon opposite adds a single quantum number, changing their structure it in the fermion cube. 7 assignments for the eight leptons are as follows: only that two binons swap their positions. Thus the rela- tive strength of the strong and weak force is likely due to ν¯R ≡ | − −−i = |000i = |0i the difference in the complexity of the binon interactions ν¯L ≡ | − −+i = |001i = |1i that comprise them.

e¯L ≡ | − +−i = |010i = |2i e¯ ≡ | − ++i = |011i = |3i R E. Binon Quantum Numbers eR ≡ | + −−i = |100i = |4i eL ≡ | + −+i = |101i = |5i As noted above, we must add a dimension in order to νL ≡ | + +−i = |110i = |6i get a cube form for the binons. It is possible to add extra ν ≡ | + ++i = |111i = |7i (24) Clifford algebra vector basis elements without associating R them with a hidden dimension, for example [28, 29], but The multiplicity of the quarks, along with their interme- this seems somewhat contrived, and we will instead only diate positions between the leptons, suggests that each add one vector basis element to correspond to the hidden of the fermions is a composite particle composed of three dimension corresponding to proper time. Adding this subparticles, with the three subparticles possessing three extra vector to the basis set for the gamma matrices is different colors. Since the quantum numbers for a Clif- impossible with 4x4 matrices, so we now pass to the more ford algebra can be described with binary numbers, we general Clifford algebra description. will call these subparticles “binons”. See Fig. (5) for the Earlier, we concluded that space-time is a 5 dimen- composition making up a column of leptons and quarks. sional manifold with 4 spatial dimensions and one time The column shown is the right most edge of the cube of dimension. One of the space dimensions, for which we Fig. (4). will use the coordinate s, corresponds to proper time and is small and cyclic. We will use t for time, and x, y, and z for the usual spatial dimensions. The movement of FIG. 5: A column of quarks and leptons shown as bound particles is defined by the usual metric: states of three binons each. dt2 = dx2 + dy2 + dz2 + ds2. (25) r 1 6 e¯R |e¯1Ri|e¯2Ri|e¯3Ri By the principles of Geometric Algebra[1, 30], there is Q a naturally associated Clifford algebra that is generated |e¯1Ri|e¯2Ri|ν3Ri (u3R) by vectors corresponding to these five coordinates, with a u r r r 2/3 ∗R |e¯1Ri|ν2Ri|e¯3Ri (u2R) signature of (−++++). With five basis vectors, there are |ν i|e¯ i|e¯ i (u ) 1R 2R 3R 1R then 32 different products that form the “canonical basis” for the Clifford algebra. These 32 products correspond

|ν1Ri|ν2Ri|e¯3Ri (d¯3R) to the 16 products of gamma matrices discussed earlier. ¯ r r r 1/3 d∗R |ν1Ri|e¯2Ri|ν3Ri (d¯2R) Since we are using a complexified Clifford algebra, we ¯ |e¯1Ri|ν2Ri|ν3Ri (d1R) can always multiply canonical basis elements by i so as to arrange for their squares to be 1. Since physical oper- r ators need real eigenvalues, this will be convenient. With 0 νR |ν i|ν i|ν i 1R 2R 3R this modification, the 32 canonical basis elements are as follows:

A similar scheme for subparticles making up quarks 1ˆ ixyc xtb ixzsd ystc iyzstd and leptons is that of Harari[25], which differs from xˆ ixzc ytb iyzsd zstc ixyzst\ the present mostly in that this scheme treats the chiral yˆ ixsc ztb xytd xyzs[ fermions as the subparticles, rather than the fermions (26) zˆ iyzc stb xztc ixyztd themselves. More complicated schemes abound, for ex- c d ample, [26]. sˆ iysb ixyzd xst ixyst ˆ c d In distinction from previous attempts at describing it izsb ixysd yzt ixzst quark / lepton subparticles, we will provide a binding po- ˆ tential and give geometric derivations of the various par- Of the 32 canonical basis elements, Eq. (26) two, 1 and ticle symmetries. For an analysis of the requirements of ixyzst\ commute with everything. Since xyzst\ squares to color in Clifford algebras, especially noting that complex- −1 and commutes with the whole algebra, it’s natural to ification is required, but not assuming that the quarks think of it as i, and most workers in Geometric algebra and leptons are composed of subparticles, see [27]. do just that. A weak force interaction, for example the conversion of Assigning xyzst\ = i makes the interpretation of com- an electron to a neutrino, can be seen, in Fig. (5) to re- plex numbers geometric and this sort of thing common quire the modification of all three binons comprising the in work in the Geometric algebra.[31] Two exceptions are electron. A strong force interaction, by contrast, requires Almeida’s insertion of SU(3) × SU(2) × U(1) symmetry 8 into complexified Geometric algebra[32] and Avramidi’s applying it to ψ gives: [33] generalization of the Dirac operator as a square root ˆ of the D’Alembert operator to Riemannian geometry. (t∂t − x∂ˆ x − y∂ˆ y − z∂ˆ z − s∂ˆ s)ψ+z(t − z) = 0 (tˆ+z ˆ)ψ+z = 0 Since this paper deals with both SU(3)×SU(2)×U(1) (1ˆ + ztb )ψ = 0, (27) symmetry and uses a modification of the derivation of +z the Dirac equation to derive symmetry breaking, it is not where the last equation is obtained by multiplying the surprising that this paper uses a complexified geometric previous by −tˆ. A general solution to Eq. (27) is: algebra. While it is true that with the inclusion of the 1 hidden dimension, the unit psuedoscalar, xyzst\ squares ψ+z = (1ˆ − ztb )η, (28) to −1 and commutes with everything in the algebra, we 2 will distinguish between xyzst\ and i. However, it is clear where η is an arbitrary Clifford algebraic constant, and that the two elements are difficult to distinguish and that the 1/2 is included in order to make the 1ˆ−ztb term into an this can be confusing, especially, as will be seen later in idempotent. It is evident that the operator corresponding this paper, in the area of symmetry breaking. to movement in the +z direction is given by ztb , with eigenvalues of ±1 corresponding to movement in the ∓z The 31 nontrivial canonical basis elements all square directions. to 1, and possess quantum numbers of ±1. Any sub- Since our binons are to be interpreted as particles that set of them that commutes, and are not related triv- move in various directions, we will choose ztb as a quan- ially by multiplication, therefore provides a natural set tum number. More generally, for a particle travelling in of operators for the multiplicative quantum numbers of thev ˆ = vxxˆ + vyyˆ + vzzˆ direction, the appropriate oper- binons. Additive quantum numbers, in the physics lit- ator will bev ˆtˆ. erature, carry eigenvalues, at least for the fundamental We still have two more quantum numbers to pick. We particles, of ±1/2 so the natural operators for those quan- would like the remaining quantum numbers to commute tum numbers will be canonical basis elements divided by not only with ztb , but to also commute with other possible 2. choices such asv ˆtˆ. Otherwise more than just the one unavoidable quantum number will lose validity when we The distinction between additive and multiplicative consider different directions of travel. This restriction quantum numbers need not concern us here. The only greatly reduces the problem of defining the remaining additive quantum number for a single binon that will quantum numbers. In order to commute with xtb , ytb and possibly concern is is spin, and since we can simultane- ztb , a canonical basis element will have to either avoid ously find eigenvectors of spin with particle type, we can, x, y, z and t, entirely or include them all. Eliminating for the purpose of distinguishing individual binons, treat unity, there are only three choices: spin as a multiplicative quantum number. In fact, this turns out to be necessary for simplification. ixyztd sˆ ixyzst\ (29)

With these basis elements, we can now begin the task We can take as our commuting operators ztb and any two of associating idempotents with the binons. In a sense, of the above three. this is the Clifford algebra generalization of the choice Since these operators all commute, their various prod- of representation that we made for the gamma matri- ucts will all possess simultaneous good quantum num- ces, in that how we choose to write the idempotents, and bers. There are a total of eight such products: therefore choose quantum numbers, corresponds to which 1ˆ s ˆ ixyztd ixyzst\ gamma matrices were chosen to be diagonal. Since we (30) ztb zstc ixyc ixys.d have an extra dimension, instead of choosing two gamma matrices diagonal, we now may choose three canonical We can count the number of ways of assigning these to basis elements (from the list in Eq. (26) ), and these ele- the n, l and m quantum numbers of Fig. (4). There are ments will define the quantum numbers that distinguish 7 choices for n and then 6 choices for l. For m, there the binons. How we choose quantum numbers among are five left, but one of these is the product of n and the canonical basis elements is somewhat arbitrary. See l, so there are 4 choices. Thus there are 168 different AppendixC for a more general treatment. assignments. Once we have determined a correct solution for n, l One of the most important characteristics of a wave and m, we can use the resulting quantum numbers to is the direction it is travelling in. We will want one of define the primitive idempotents associated with the bi- our quantum numbers to define the direction of travel in nons. For example, if n l and m are the quantum num- geometric terms. To derive this form, let ψ(x, y, z, s, t) = bers associated with the ztb ,s ˆ and xyztd operators, then ψ+z(t−z) be a plane wave solution, of the massless Dirac the eight binon idempotents are given by the generaliza- equation in 5 dimensions, travelling in the +z direction. tion of Eq. (20): Note that ψ+z takes values among the Clifford algebra. Writing the Dirac equation in explicit geometric form and (1 ± ztb )(1 ± sˆ)(1 ± ixyztd )/8, (31) 9

iδnχ where each choice of the three ± defines one of the eight where βnχ = αnχ e . This substitution is convenient idempotents. for evaluating the integral of the binon potential over s:

When we write down the binon binding potential we Z 2π will find that energy will be minimized if certain idem- 2 (ψ1(s) + ψ2(s) + ψ3(s)) ds potents are paired up with others. The patterns involved 0 Z will greatly reduce the number of choices. The choices 2π is is is
2 will be further reduced when we consider the manner in = R(β1χe + β2χe + β3χe ) ds 0 which the electron vs neutrino symmetry can be broken, Z 2π
but for now, let us move on to the binon binding poten- 1 is ∗ ∗ ∗ −is 2 = (β1χ+β2χ+β3χ)e +(β1χ+β2χ+β3χ)e ds tial. 4 0 Z 2π ∗ ∗ ∗ = 0.5(β1χ + β2χ + β3χ)(β1χ + β2χ + β3χ) ds 0 F. The Binding Potential 2 = π|β1χ + β2χ + β3χ| . (36)

It should be remembered that we complexified the Clif- We will absorb the factor of π into V0 and use the com- ford algebra only under the interpretation that the com- plexified idempotents with coefficients βnχ. plex phase would correspond to a rotation in the hidden Thus we see that complexifying the Clifford algebra dimension. The point here is that our use of complex has the effect of simplifying our potential energy calcula- numbers is for the sake of convenience in calculation only. tion by allowing us to use complex numbers to emulate Underlying this theory is a real valued geometric algebra rotations in the hidden dimension. We replace the inte- in the tradition of Hestenes’ work.[34] In our definition of gral over 4 spatial dimensions in Eq. (32) with an integral the binon binding potential, we will therefore temporarily over the usual 3 spatial dimensions of the complex wave return to a real valued Clifford algebra. functions: Let three binons be given the real valued wave func- Z 2 tions ψ1(r), ψ2(r), and ψ3(r), where r = (x, y, z, s) de- V = V0 |ψ1 + ψ2 + ψ3| dx dy dz, (37) fines a position in space. The definition of the potential that we will use will be the simplest function possible: For the chiral fermions, we will assume that the wave Z functions are the lowest energy possible, so that as far V = V |ψ + ψ + ψ |2 dx dy dz ds, (32) as dependence on position goes, each of the ψn is an 0 1 2 3 identical Gaussian. By making this assumption, we need not consider the spatial parts of their wave functions. 2 where V0 is a positive real constant, and | | is defined Obviously we can minimize this potential by arranging as the usual norm for a (real) Clifford algebra: for ψ1 +ψ2 +ψ3 = 0 everywhere. That is, we could assign the exact same idempotent to each of the three binons, X 2 X 2 2 and arrange their phases so that the sum cancelled. This |φ| = αχχˆ = αχ , (33) would be in violation of the spirit of the Pauli exclusion principle, so it is clear that in order to obtain realistic whereχ ˆ runs over the 32 canonical basis elements, and solutions, we must require that the combined wave func- α =< φ > are real numbers. For example if χ χ tion of a group of binons be antisymmetrical under the φ =x ˆ + 5ˆy − 10xys,d then exchange of two binons.[35, CH. XIV] 2 2 2 2 The three binons making up a single chiral lepton can |φ| = 1 + 5 + (−10) = 126. (34) only be distinguished by arranging for them to have dif- fering directions. On the other hand, the chiral lepton For χ a canonical basis element, let the χ component of they make up has a preferred direction, that is, the di- ψn be given by the positive (real) value αnχ with a phase rection of its spin and / or its direction of travel. The rotation in the hidden dimension s given by δnχ, where component of velocity of the binon in the direction of δnχ is given by the overall rotation of the nth binon, plus travel of the chiral fermion has to be c, so the binon a rotation that depends on χ. That is, if < ψn >χ is pos- must be travelling faster. Therefore we will assume that itive real, positive imaginary, negative real or negative the binons are tachyonic particles.9 imaginary, we arrange for αnχ to be positive real by set- For a comment on tachyons and quantum mechan- ting δnχ equal to the overall rotation plus, respectively, ics, in particular how Lorentz symmetry is altered, see 0, π/2, π or 3π/2. Then the real value of the χ component of ψn as a function of s is given by: 9 For a phenomenological approach to the situation of different < ψn >χ (s) = αnχ cos(s + δnχ) particles possessing different maximum velocity (effective speeds is+iδnχ of light), in the context of high energy cosmic rays see [36]. Also = R(αnχ e ), see [37] which proposes that tachyonic neutrinos are responsible is = R(βnχe ), (35) for the knees in the cosmic ray spectrum. 10

θB. The three binons are arranged equidistantly around FIG. 6: Direction vectors for the three binons making up a the cone. See Fig. (6). fermion. The fermion vector is marked e. The binons are marked R, G and B. Triangle RGB is equilateral with e at Having the binons travel faster than c is compatible with the assumptions of the version of general relativity its center. \R0e = \G0e = \B0e = θB . associated with this theory. Euclidean relativity defines a coordinate system similar to that of Newton, and in such a preferred rest frame, there is no causality problem with travelling faster than c. G B For other examples of tachyonic particles and preferred ¨ 30 30 ¨¨ frames, see [46]. An entertaining review of tachyons is  30 ¨¨ [47]. Also see [48, 49]. M. Consoli and E. Costanzo’s[50] ¨ 120 0r¨ - e reopening of the historical conclusions of the Michelson- H 120 HH Morley experiments is also of interest. HH A perhaps less radical alternative to the assumption of H 30 tachyonic binons would be to suppose that they carry HH HH R fractional spin. While fractional spin, and the case H of spin-1/6 in particular, has been explored in string theories[51], the match with experimental observation is better with tachyonic binons, and the theory is easier to derive from standard quantum mechanical principles. For now, let us note that for normal energies and un- [38, 39]. For an extension of quantum field theory to exotic matter the tachyonic nature of binons is hidden tachyons, see [40]. For a paper on Lorentz symmetry vi- by their being bound into composite particles, and the olations at high energies in general, see [41]. Having neu- next subsection will discuss further the question of the trinos travel at other than c could be a part of neutrino relationship between binons and fermions. mixing according to [42]. For an extensive and recent re- view of Lorentz violation in theory and experiment, see [43] which has 202 references. G. Spin and Statistics This paper does not postulate a violation of Lorentz symmetry for normal matter. What it does instead is The operators {ixˆxyz,d iyˆxyz,d izˆxyzd} form a represen- to postulate that subparticles flagrantly violate Lorentz tation of SU(2). That is, their multiplication and addi- symmetry, but that their bound states achieve exact (to tion rules are identical to the Pauli spin matrices. Since within experimental error) Lorentz symmetry. Conse- izˆxyzd = ixyc is included in the list of good quantum num- quently, the absence of experimental detection of Lorentz bers for a particle moving in the +z direction Eq. (30), symmetry violation in normal matter is not of concern it is true that a binon travelling in the +z direction does here.10 possess, individually, a good quantum number that cor- This paper’s approach to tachyons is different from the responds to spin in the z direction. But the three binons usual unstable or false vacuum seen in QFT.[45] We will making up a fermion cannot point in the same direc- assume that the true speed of waves is faster than c, and tion as the fermion, and the binons individually are not that chiral fermions and photons are composite particles spin-1/2 particles. This is a subtle point that is easy to that are restricted to travelling at speed c. The chiral overlook and bears comparison with the usual situation. fermions then combine to produce massive fermions that The spinor that represents a particle of spin-1/2 has move at speeds less than c. There are thus two levels of two complex degrees of freedom. An assembly of three combination, each associated with a reduction in speed. such particles has a total of eight degrees of freedom. An In order to allow color to be an unbroken symmetry, elementary exercise in quantum mechanics is to convert it is natural to expect that the three binons travel at the between the representations of the spinors of the indi- same angle with respect to the fermion. Let that angle vidual particles and the spinor of the combination using be θB. Then the relative speeds of the particles is given Clebsch-Gordon coefficients. The Clebsch-Gordon series by: can be written as:

0 cB = c/ cos(θB). (38) 2 × 2 × 2 = 4 + 2 + 2 1 1 1 3 1 1 0 (39) | 2 i × | 2 i × | 2 i = | 2 i + | 2 i + | 2 i The three binons travel on a cone centered on the fermion velocity / spin vector. The opening angle of the cone is On both sides of the equation there are eight complex degrees of freedom, two in each of the spin-1/2 spinors, and four in the spin-3/2 spinor. The spinor multiplica- tion forms a tensor product. The usual Clebsch-Gordon 10 For an interesting solution to the problem of combining relativity coefficients give the conversion with respect to a partic- and quantum mechanics, see [44]. ular spin axis, for example Sz. That is, when each of the 11   3c2 − 1 1 − 3c2 1 + 3c2 six spinors in the above are written in their Sz basis, the rˆgˆ = B + √ B (ˆr+ˆg) + √ B ˆb xyz,d Clebsch-Gordon coefficients give the conversion from one 2 2 3c 2 3c  B  B side to the other. 3c2 − 1 √ 1 − 3c2 = B + 3 B uˆ + c ˆb xyz,d The three binons making up a fermion have only one 2 2 B complex degree of freedom each. Rather than spin-1/2,   3c2 − 1 1 − 3c2 1 + 3c2 fermions, as far as degrees of freedom go, the binons gˆrˆ = B − √ B (ˆr+ˆg) + √ B ˆb xyz,d 2 2 3c 2 3c are non commutative spin-0 fermions. Consequently, the  B  B Clebsch-Gordon series for three binons combining to form 3c2 − 1 √ 1 − 3c2 = B − 3 B uˆ + c ˆb xyz,d (43) a chiral fermion is trivial: 2 2 B

1 × 1 × 1 = 1 , (40) |0i × |0i × |0i = |0i 9c (1−c2 ) 1−3c2 rˆgˆˆb = + B √ B xyzd + B (ˆr − gˆ − ˆb), 2 3 2 where the three particles on the left are written in three 2 2 9cB(1−cB) 1−3cB different bases, that is, they are written as idempotents gˆrˆˆb = − √ xyzd + (ˆg − rˆ − ˆb). (44) 2 3 2 according to their direction of travel. The binon idempotents can be written as being made One can rotater ˆ,g ˆ and ˆb to get the multiplication rules d up of a nondirectional part, for example, (1 + ixyzt) (1 − forg ˆˆb, etc. sˆ), and a part which, under spatial rotations, acts as the To show how binon idempotents combine into quarks scalar unit plus a vector or psuedo-vector, for example, and leptons, let ι , ι and ι be the nondirectional idem- b r g b (1+zt) or (1+ixyc. We will split our binon idempotents in potents associated with three binons, and define the di- this manner because it simplifies multiplication between rectional idempotents as (1+a∗ ∗ˆ i xyzd)/2, where a∗ takes idempotents oriented in distinct directions. In particular, the values ±1 for ∗ = r, g, b. Thus the three binons will any non directional part of an idempotent commutes with take spin quantum numbers of a , a and a in the di- all the other idempotents. It is only the directional parts r g b rectionsr ˆ,g ˆ and ˆb, respectively. Then the full binon that fail to commute. values11 with complex phases are: All the binons oriented in a particular direction com- mute. That is, the product of two such idempotents is iδr ηr = 0.5(1 + ar rˆ i xyzd )ιr e , zero if they are distinct and unchanged if they are iden- η = 0.5(1 + a gˆ i xyzd )ι eiδg , tical. This implies that an anticommutation of any set of g g g ˆ iδb such binons will give zero. However, binons travelling in ηb = 0.5(1 + ab b i xyzd )ιb e . (45) different directions may neither commute nor mutually annihilate so the antisymmetrization operation can give Note that the vectorsr ˆ,g ˆ, ˆb andu ˆ commute with ixyzd, a nonzero result. and while these vectors do not commute with all the In order to simplify calculations involving binon bound nondirectional idempotents, their products with ixyzd do states, we will, when convenient, use a non orthogonal co- so commute. ordinate system oriented in the directions of the binons. The nondirectional idempotents, ι∗ in Eq. (45) com- Letu ˆ be the spin orientation of the chiral fermion. We mute with all the directional idempotents. Consequently, wantr ˆ,g ˆ and ˆb to be three unit vectors evenly spaced if any two of the nondirectional idempotents are differ- ent, then, since these idempotents are self annihilating, around the cone with opening angle θB oriented in the uˆ direction. For example, ifu ˆ is oriented in the +z di- the result of computing the combined wave function by rection, then a possible set of orientations for the binons antisymmetrization will be zero. Therefore, let ιr = ιg is: = ιb = ι. On the other hand, since the directional idempotents are not aligned, they cannot self annihilate even if they uˆ ≡ (0, 0, 1) have different eigenvalues. Consequently, ar, ag and ab rˆ ≡ (s , 0, c ) B √B may be different. There are eight cases where ar = ag = gˆ ≡ (−sB/2, sB 3/4, cB) ab and these are the eight chiral leptons. The mixtures √ occur along one axis of the fermion cube, which cor- ˆb ≡ (−s /2, −s 3/4, c ) (41) B B B responds to the one eigenvalue that must be taken to be directional. This is the verticaln ˆ axis as shown in where cB and sB are cos(θB) and sin(θB), respectively. Fig. (4). Thus the particles that mix ar, ag and ab are the Some useful relations between these unit vectors in- clude:

11 uˆ2 =r ˆ2 =g ˆ2 = ˆb2 = 1, The inclusion of a nonzero phase δ∗ prevents these binon val- ues from being idempotents but we will undoubtedly call them ˆ rˆ +g ˆ + b = 3cBu,ˆ (42) idempotents anyway. 12 quarks, and the simple rule that “three binons can bind This is an eigenvector of spin in theu ˆ direction (i.e. an together only if their antisymmetrization is not equal to eigenvector of Su = iuˆxyz/d 2), with eigenvalue of ±1/2 if zero” gives the standard model chiral fermions.

Since we are using complex phases to represent rota- cB = ±1/3. (50) tions in the hidden dimension, we must multiply our bi- non idempotents by phases appropriate to the three bi- 12 We will use cB = 1/3 for the remainder of this paper. nons that make up a fermion. The phases turn out to be −1 Accordingly, we will take the value of θB to be cos (1/3) important in the calculation of the masses of the three ≈ 70.528 degrees. For this particular angle, the results of generations of leptons, which is a subject to be covered in the anticommutation program is 0.5ι times Eq. (49) or: the next paper by the author. Since the complex phases that represent rotations in the s dimension are scalars, p they will factor out of any product of binons such as an η = −i 16/3 (1 + iuˆxyzd) ι. (51) antisymmetrization, and we will therefore ignore them for the remainder of this paper. Setting ar = ag = ab = −1 changes the relative signs of According to the antisymmetrization postulate, the terms with odd numbers of µ∗, but leaves unchanged the combined wave function for the binons (after removing terms that are even. This complements the imaginary iδ scalar factors such as e r ) will be given by: term in Eq. (48), and results in an eigenvector for Su with eigenvalue −1/2. In addition, the overall sign is η = ηrηgηb + ηgηbηr + ηbηrηg complemented.

− ηrηbηg − ηgηrηb − ηbηgηr. (46) Before antisymmetrization, the three binons broke the radial symmetry around theu ˆ direction. That is, while In the above, the value of the right hand side will depend the particle as a whole was oriented in theu ˆ direction, on θB, which we will now calculate. For simplicity, we the individual binons were oriented on a cone centered will postpone the quark calculation till later and will set in that direction. This gave the collection a geometric ar = ag = ab = 1. content more complicated than simply theu ˆ direction. First, we will pay attention only to the directional As a result of the antisymmetrization, the wave function parts of ηr, ηg and ηb and we will leave off the 0.5. We can corresponding to the bound state, Eq. (51), has no ori- do this because the remaining terms will all simply give entation but that ofu ˆ. Thus the fermions, despite being a factor of 0.5ι. It will be useful to write the directional composed of three particles, possess only a single axis of parts in this form: orientation. One might expect that cos(θ ) = 1/3 if one were to 2ι = 1 + µ = 1 + a rˆ i xyz,d B r r r suppose that the binons each carry spin 1/2, but at an 2ιg = 1 + µg = 1 + aggˆ i xyz,d angle so as to attribute each binon as contributing 1/6 ˆ 2ιb = 1 + µb = 1 + abb i xyz,d (47) to the total fermion spin. This paper associates the elementary particles with the where µr, µg and µb are the spin operators that give +1/2 idempotents rather than the nilpotents of the Clifford for the given binon. The required sum, η, is 0.5ι times: algebra. This is somewhat contrary to expectations in that the Grassmann algebra used for fermions in QFT η = 1+µr +µg +µb +µrµg +µrµb +µgµb +µrµgµb consists of nilpotents. However, the antisymmetrization +1+µr +µg +µb +µgµr +µbµr +µgµb +µgµbµr assumption is sufficient to arrange for products of iden- 13 +1+µr +µg +µb +µrµg +µbµr +µbµg +µbµrµg tical fermions to give zero.

−1−µr −µg −µb −µrµg −µrµb −µbµg −µrµbµg

−1−µr −µg −µb −µgµr −µrµb −µgµb −µgµrµb 12 −1−µr −µg −µb −µgµr −µbµr −µbµg −µbµgµr, Other values for θB give a result of the form κ(1 + βixyc) where = [µ , µ ] + [µ , µ ]µ + cyclic β 6= 1. It may be possible to interpret these as eigenvectors r g r g b of a spin operator more general than ones constructed from = [ˆr, gˆ] + (ˆrgˆˆb − gˆrˆˆb)ixyzd + cyclic. the Clifford algegbra. In doing so, one would bring back the usual quantum distinction between operators and eigenvectors. (48) However, the author’s next paper will show that the choice of cos(θB ) = 1/3 also leads to the correct formula for the lepton Using Eq. (42) through Eq. (44): masses. 13 √   If you use the idempotent/nilpotent structure of a Clifford al- = 3 (1 − 3c2 )ˆu + 2c ˆb xyzd gebra to separate a nonlinear wave equation into linear (Dirac) B B equations and nonlinear interactions, then you find that, in gen- √ +3i 3c (1 − c2 )xyzd2 + (1 − 3c2 )i(ˆr − gˆ)xyzd + cyclic eral, the linear equations for the propagation of the idempotents √ B B √ B and nilpotents have a different speed. Idempotents cannot bind = −9i 3c (1 − c2 ) + 3 3(1 − c2 )ˆuxyzd together when sharing the same direction, but nilpotents, par- B B  B  √ ticularly after symmetry breaking, may be able to. However, 2 1 the speed of such bound nilpotents may not be equal to the free = −9i 3cB(1 − cB) 1 + iuˆxyzd . (49) 3cB speed of the idempotents. 13

H. Quarks We will compute the transition probabilities for binons using the traditional quantum mechanical principle that The previous section derived the combined wave func- the probabilities are proportional to the squared magni- tion for the leptons which are given by the situation tude of the inner product of the two states being trans- formed between. where ar = ag = ab. The alternative, for example, ar = ag = −ab corresponds to the quarks, and we will This rule for transition probabilities will allow a bi- now discuss these. non of one color to transform to a different color. This Our calculations were made under the assumption that means that a colorless particle, such as an electron, has a we could commute the nondirectional idempotent shared nonzero probability of turning into a colored particle. Of by the three binons (i.e. ι) with the directional parts of course it will immediately turn right back for energetic the idempotents. Written in this manner, the three idem- considerations. potents must share the same nondirectional part. This The act of transforming from one state to another is means that there can only be one directional contribution similar to the act of wave function collapse in that it hap- to the idempotent, for example, 0.5(1 + arrˆxyzd). pens at a point in space. While color is not conserved, While we are writing the idempotents with only one position will be, and in particular, the position in the directional part, it is still the case that the idempotents hidden dimension will be conserved. Since we are mod- are eigenvalues of the usual full set of 4 directional op- eling the position in the hidden dimension with complex erators and 4 nondirectional operators listed, for thez ˆ phases, this means that when a binon transforms from orientation, in Eq. (30). In particular, ixyztd is a good one state to another, it will preserve its phase. nondirectional quantum number and therefore must be Were it not for the issue of complex phases, we could shared between three binons making up the quark. That model the sequence of transformations of a binon with a is ixyztd is either one of the two quantum numbers that Markov chain. The addition of complex phases implies make up ι, for example ι = (1 − ixyztd )(1 +s ˆ), or it is the that an appropriate transition matrix will have proba- product of the two operators as in ι = (1−ixyzst\)(1+s ˆ). bilities multiplied by complex phases. Since there are In either case ixyztd is a good quantum number for the three colors of binons, the complex transition probability three binons and they share the same value. matrix will be of size 3 × 3. Since ixyztd = (iuˆxyzd)(ˆutˆ), the fact that ixyztd is a For a solution to the problem of propagating in such a shared quantum number implies that if two binons mak- way that a particle is stable, we will look for the eigenvec- ing up the quark have opposite quantum numbers for tors of the transition probability matrix. There will be iuˆxyzd, then then must also have opposite quantum num- three such eigenvectors and they will correspond to the bers foru ˆtˆ. In words, if two binons in a quark have three generations of fermions. Their associated eigenval- ues will give their respective masses. In our next paper we different spin directions, (in the sense that ar = −ag) then they must also have different velocity directions (in will derive some surprising relationships between fermion the sense thatg ˆ must be reversed from the direction it masses from these principles. Since color is not conserved by binon interactions, a would point relative to ar for the lepton case). Referring back to Eq. (47), it is clear that if we negate colored high energy particle, upon collisions with normal matter, will have a branching ratio to transform into a rˆ at the same time as we negate ar, there will be no change to the calculation for η. Consequently, the value combination of particles that are colorless. This effec- tively confines color. of cb = ±1/3 will give spin−1/2 quarks as well as leptons. Another way of putting this is to note that when we The fact that binons group together in 3s to form nor- compute the combined wave functions η, the difference mal matter, along with pair production of binons and between quarks and leptons is lost. the ability of binons to transform amongst themselves, suggest that a single free binon, upon colliding with nor- mal matter, will first combine with two other binons to I. Lepton Generations and Masses form a fermion. The other binons involved will combine similarly. This subsection is a brief sketch of the author’s next paper, but without the mathematical details. It is in- tended as a brief introduction for the purpose of justi- J. The Particle Internal Symmetry Algebra fying certain possible behavior of binons as observed in very high energy cosmic rays. An electron is composed The binon model of the elementary particles is based of two chiral halves, eR and eL. As in the old zitterbe- on an assumption that the internal symmetries of the wegung model, these two move at speed c and transform particles can be given geometric meaning in terms of ba- from one to the other. sis vectors that have very specific meaning in the usual In the present theory, the eL and eR are each composite world. Only the proper time vector,s ˆ, would be entirely particles composed of three binons each. In order for an unfamiliar to a geometry student of ancient Egypt. eR to transform to an eL, each of its three binons must In this primitive framework, the complexities of the transform from one sort to the other. standard model of quantum mechanics are difficult to 14 envisage. Under our assumptions, we expect to find the symmetric manner, but leave each spinor still satisfying symmetries of our apparently Lorentz symmetric world a standard Dirac equation. Since we are treating mass repeated in the structure of the elementary particle in- as a particle interaction, we need only do this operation teractions, but the elementary particle interactions are on the massless Dirac equation: not nearly so symmetric as Lorentz symmetries. Weak hypercharge has a U(1) symmetry, which can be (t∂ˆ t − (ˆx∂x +y∂ ˆ y +z∂ ˆ z +s∂ ˆ s))ψ = 0. (52) represented by a single complex number. Weak isospin has SU(2) symmetry and is represented by a spinor. The This paper has already assumed a version of gen- Weinberg angle, θW , determines the relative strength of eral relativity that suggests a preferred, Euclidean, rest the weak and electromagnetic forces in terms of weak frame. It has further assumed binons that travel at speed hypercharge and weak isospin. Thus these two symme- 3c, an assumption that demands a preferred rest frame. tries, weak hypercharge and weak isospin, are intimately It should not shock the reader that we will now promote related in the structure of the standard model. the speed of light from a scalar to a Clifford algebraic Any spin−1/2 fermion has a spin axis, a 3−vector that constant. defines a direction that its spin always gives +~/2 when While doing this, we will want to leave the Clifford measured along. This suggests that a geometric repre- algebraic structure unaltered. Suppose that cα satisfies sentation of spin should be accomplished by an object the following commutation relations 14: that can be thought of as a vector, (or axial vector). Ac- ˆ ˆ cording to the theory of relativity, space and time are in- tcα = cαt, −1 timately related, so whenever we have a spatial 3−vector, xcˆ α = cα x,ˆ it must be part of a 4−vector with the missing component −1 ycˆ α = c y,ˆ corresponding to time. α zcˆ = c−1z,ˆ That there is, in the standard model, no 4 − vector α α −1 whose spatial components give the spin axis of a particle scˆ α = cα s.ˆ (53) suggests that the internal symmetries of the particles are not Lorentz symmetric. If the internal symmetries were Under these restrictions, it is easy to verify that the five Lorentz symmetric, then we would expect a very partic- elements (note that they are no longer vectors): ular relationship between spin and its time component. ˆ Later in this paper we will devote a subsection to deriv- t, cαx,ˆ cαy,ˆ cαz,ˆ cαs,ˆ (54) ing, from geometric principles, that the time component of spin is helicity. satisfy the same relations as the usual basis vectors of a Clifford algebra (or the gamma matrices of the Dirac The relationship between weak hypercharge, which is equation). For example, a U(1) symmetry, and weak isospin, which is an SU(2) symmetry also suggests a violation of Lorentz symmetry (c xˆ)2 = c xcˆ xˆ (55) in the internal states of particles. Here, neither of the α α α −1 2 symmetries is apparently connected to space-time, but = cαcα xˆ in a geometric theory based on a generalization of the = 1 Dirac equation, it is inevitable that weak hypercharge −1 (cαxˆ)(cαyˆ) = cαc xˆyˆ (56) will be associated in some way with time, while weak α = −c c−1yˆxˆ isospin will be associated in some way with space. α α For both the spin / isospin 4−vector, and the weak hy- = −(cαyˆ)(cαxˆ) percharge / weak isospin relationship, we need to look for (cαxˆ)tˆ = −cαtˆxˆ (57) ways to violate Lorentz symmetry in the internal symme- = −tˆ(cαxˆ). tries of our theory. But at the same time, we must retain (58) Lorentz symmetry in terms of matching the results of the standard model. Therefore, the asymmetric Dirac equation: A typical calculation for quantum mechanics is given in terms of Feynman diagrams. These diagrams consist of (t∂ˆ t − cα(ˆx∂x +y∂ ˆ y +z∂ ˆ z +s∂ ˆ s))ψ = 0, (59) propagators and vertices. The propagators for fermions arrange for the particles to obey the Dirac equation. If, when we break Lorentz symmetry, we can do it in such a way as to leave the Dirac equation unmodified, then all 14 The commutation relations, in addition to allowing a modifica- we need do, to obtain the same results as the standard tion of the Clifford algebra canonical basis vectors in a way that model, is derive the vertex values. preserves the Clifford algebra structure, also happen to be pre- If we were still working with a Dirac equation that only cisely the modifications that can be made to the Dirac equation spatial components so as to retain it as a square root of the un- dealt with a single spinor, we would be stuck. But since modified Klein-Gordon equation. This is how the author first our Dirac equation has been generalized, we can consider discovered these relations as well as the parameterization shown modifications to it that mix the spinors in a non Lorentz in the next subsection.[52] 15 can in no way be distinguished from the usual symmetric If an operator has an even number of spatial compo- one Eq. (52). In terms of the mathematics of the algebras nents, then it will be unmodified in the conversion from they define, they are both the same Clifford algebra, and GA to PISA. Otherwise, it will be multiplied on the left are as indistinguishable as different representations of the by cα. For example: gamma matrices. 0.5ˆ −0.5 ˆ While our use of the asymmetric Dirac equation is no cα tα = t 0.5 −0.5 change from the usual Dirac equation in terms of Clifford cα xαˆ = cαxˆ algebra, it does mark a departure from the Dirac equation 0.5 b −0.5 b cα xtα = cαxt of Hestenes’ geometric algebra. What we have done is to 0.5 −0.5 modify the definition of the spatial vectors so as to give cα xyαc = xyc 0.5d −0.5 d them what will turn out to be the form of mixed vectors. cα xytα = xyt At the same time, we will persist in thinking of these 0.5 −0.5 cα xyzαd = cαxyzd (65) vectors in terms of the tangent vectors of the space-time manifold. We will call the resulting algebra for space- In the next subsection, we will derive the solutions to the time, the Particle Internal Symmetry Algebra, or PISA. cα commutation relations, parameterize these solutions Up to now, all our work has been with the Geometric in a convenient way, and establish some useful relations. algebra. We need some tools to allow conversion between solutions to Geometric algebra and PISA wave equations. Such a tool will allow us to make difficult PISA compu- K. PISA Parameterization tations using easy Geometric algebra techniques. Let ψS be a solution to the usual Dirac equation Write cα as a sum over canonical basis elements: Eq. (52). Assuming that c has a square root and that α ˆ \ the square root satisfies the same commutation relations cα = c11 + cxxˆ + ... + cxyzstxyzst, (66) given in Eq. (53), we can calculate as follows: where cχ is a complex number. According to Eq. (53), c must commute with tˆ and with any even product of (t∂ˆ t − (ˆx∂x +y∂ ˆ y +z∂ ˆ z +s∂ ˆ s))ψ = 0, α 0.5 −0.5 0.5 spatial vectors. This requirement forces all but four of c (t∂ˆ t − (ˆx∂x +y∂ ˆ y +z∂ ˆ z +s∂ ˆ s))c c ψ = 0, α α α the cχ coefficients to be zero. The remaining coefficients ˆ 0.5 0.5 0.5 (t∂t − cα cα (ˆx∂x +y∂ ˆ y +z∂ ˆ z +s∂ ˆ s))(cα ψ) = 0, are: 0.5 −0.5 (t∂ˆ t − cα(ˆx∂x +y∂ ˆ y +z∂ ˆ z +s∂ ˆ s))(c ψc ) = 0(60). α α cα = c11ˆ + cttˆ+ cpxyzs[ + ccxyzst,\ (67)

Therefore, if ψS is a solution of the usual Dirac equation, where we have shortened the names of three of the co- then a solution to the PISA Dirac equation is given by: efficients. Two of these are identical to the four good nondirectional quantum numbers, but the other two are ψ = c0.5ψ c−0.5. (61) A α S α not. 2 ˆ If ψ is written as a primitive idempotent, for example: From Eq. (53) we have that (cαxˆ) = 1. Putting S Eq. (67) in and comparing terms gives four equations:

ψS = (1 + e1)(1 + e2)(1 + e3)/8, (62) 1ˆ (c1c1 − ctct + cpcp − cccc) = 1ˆ, then ψA will also be a primitive idempotent, but with tˆ(c1ct − ctc1 + cpcc − cccp) = 0, modified operators: xyzs[ (c1cp − ctcc − cpc1 + ccct) = 0, 0.5 −0.5 0.5 −0.5 0.5 −0.5 ψS = (1 + cα e1cα )(1 + cα e2cα )(1 + cα e3cα )/8. xyzst\ (c1cc − ctcp − cpct + ccc1) = 0. (68) (63) The modified primitive idempotent will operated on by These reduce to give the equations relating the coeffi- the modified operator, will possess the same quantum cients as: numbers that the unmodified idempotent had when op- c2 − c2 + c2 − c2 = 1, erated on by the unmodified operator. For example: 1 t p c c1cc − ctcp = 0, (69) e1(1 − e1) = −(1 − e1), 0.5 −0.5 0.5 −0.5 0.5 −0.5 the other two equations giving 0 = 0. Since we have cα e1cα (1 − cα e1cα ) = −(1 − cα e1cα )(64) two equations in four unknowns, the solution set can be In general, making the change from Geometric algebra parameterized with two free parameters. to PISA will not modify any relation that is based on Let cα be given as follows: multiplication or addition of Clifford algebraic numbers. cα(αt, αs) = exp(iαttˆ) exp(αsxyzs[). (70) However, our definition of the inner product used an ab- solute value, and therefore will be modified. Eventually We now show that cα(αt, αs) so defined satisfies the com- we will associate probabilities with the inner product, so mutation relations and is therefore the parameterized so- these probabilities will change. lution to Eq. (69) 16

Since tˆ and xyzs[ commute, we have that: To derive the time component of spin, we simply ex- tend the 3−vector (ˆx, y,ˆ zˆ to a 4−vector, (t,ˆ x,ˆ y,ˆ zˆ. Thus ˆ ˆ exp(iαtt + αsxyzs[) = exp(iαtt) exp(αsxyzs[), (71) the time component of spin is simply: and therefore, St = ±tiˆ xyzd = ±ixyztd (77) exp(−iαttˆ− αsxyzs[) = 1/ exp(iαttˆ+ αsxyzs[), (72) The time component of spin is ±ixyztd , where the sign is ˆ ˆ Since t commutes with iαtt + αsxyzs[, we have that somewhat arbitrary. We will take the positive sign. This ˆ ˆ cα(αt, αs) commutes with t. Since iαtt + αsxyzs[ anti- can be written as ixyc ztb , which is the product of the commutes withx ˆ, we have: operator for spin in the z direction, and the operator for exp(iα tˆ+ α xyzs[)ˆx =x ˆ exp(−iα tˆ− α xyzs[). velocity in the z direction. This is therefore the helicity t s t s operator. This completes the verification that cα(αt, αs) satisfies An analysis of the bound states of binons will provide the commutation relations Eq. (53). insight into such diverse subjects as the mass relations Summing up, a set of useful equations involving of the leptons and the generalized Cabibbo angles. The cα(αt, αs) are: brief introduction included here is enough to establish n the general properties of binons as is needed for the ob- cα(αt, αs) = cα(αtn, αsn), (73) servational section of this paper. −1 cα(αt, αs)u ˆ =u ˆ cα (αt, αs), (74) cα(αt, αs) tˆ = tˆ cα(αt, αs), (75) II. HIGH ENERGY COSMIC RAYS & C. whereu ˆ is any spatial vector. This section will include a brief description of the facets of high energy cosmic rays and related astronomical oddi- L. The Time Component of the Spin Vector ties that may be explained by binons. In addition to high energy cosmic rays in general, we will cover the myste- In the standard model, spin is a vector operator, and rious Centauro and Chiron events, gamma ray bursts, it remains so in this theory: galactic jets and the GZK limit.

S = (Sx,Sy,Sz) = (ˆx, y,ˆ zˆ)ixyz.d (76) The special theory of relativity uses 4−vectors instead of A. Extreme High Energy Cosmic Rays treating space and time separately. It therefore is natural to ask what the time component of the spin vector is. While the composition and behavior of most cosmic Similarly, in this geometric theory, time is treated on rays are well understood, extremely high energy cosmic an equal footing with the spatial directions, so given a rays exhibit behavior which is difficult to understand. spatial vector, it is possible to find a more or less unique There are several excellent reviews of the problem. For an time component associated with it. extensive and recent review of high energy cosmic rays, In the special theory of relativity, a Lorentz boost see [56]. For a review concentrating on potentials for new mixes the space and time components of a 4−vector. physics, see [57]. However, a Lorentz boost on a (Pauli) spinor, has only The most interesting data for high energy cosmic rays the effect on the spinor of rotating the vector associated comes form film or emulsion based ground targets. This with the spinor. The same effect happens with the four data is frequently ignored in reviews of high energy cos- component bispinors of the Dirac theory.[6, §3.3] From mic rays, such as the two mentioned above, as they are of the principles of relativity, one would expect that the considerably smaller energies, but there is much more de- three spatial components of the vector associated with tailed information for these events, and they are strange. the spinor would be mixed with the time component of For an in depth review with plenty of data on these events the spinor.[53] see [58], which the reader is urged to obtain. For a brief In theories where there is a preferred reference frame, review of Centauros from the point of view of accelerator which includes the “Lorentz Ether Theory” that Ein- physics, see [59]. We will include here a brief description stein’s relativity replaced [54] this discrepancy in the use of these unusual events. of spinors in the standard model is not a surprise. In- Anomalous events on emulsion and film have now been stead, the Lorentz symmetry is assumed to be acciden- unexplained for three decades. From a recent review ar- tal15, so no assumptions about the internal symmetries ticle: of the particles can be made from it. The most striking, unexpected phenomena observed in emulsion chamber experiments are Centauro events with an exceptionally 15 For an interesting note on how Lorentz symmetry can appear small number of photons, events with par- naturally in a physical system without relativity, see [55] ticles or groups of particles being aligned 17

along a straight line, halo events character- Because of the thickness of the atmosphere, most po- ized by an unusually large area of darkness in tential Centauros will begin at high altitude, and conse- the X-ray film, and deeply penetrating cas- quently will be classified as something other than “Cen- cades. Whether these phenomena are related tauro” when their tracks are observed. In fact, Chirons to fluctuations and the measurement tech- are observed at a rate of about 0.1 per square meter per nique of emulsion chamber experiments or year, while Centauros are about 10 to 100 times less fre- signs of new physics is controversially debated quent, at the Chacaltaya altitude.[58] for more than 30 years.[60] The debris from a Chiron is spread around by the time it reaches the target. Typically, there will be a number of The term “Centauro” refers to a cosmic ray that ex- “mini-clusters”. Each mini-cluster is thought to be the hibits an anomalous ratio of charged to neutral pion pro- particle shower that results from a collision by the pri- duction. This is a violation of isospin symmetry. In the mary particle. The mini-clusters are frequently aligned, standard model only the weak force violates isospin sym- and they indicate a high level of transverse momentum, metry, so one would expect that Centauro events are col- pT . The fact that mini-clusters tend to be aligned sug- lisions mediated by the exchange of a weak vector boson. gests that all the miniclusters are produced in a single However, Centauros exhibit cross sections that are even collision. larger than strong force cross sections for protons. The cosmic ray emulsion experiments that are the most Theoreticians have come up with plenty of possi- extensive in surface area are the ones that are most likely ble explanations for the odd behavior of high energy to observe Centauros. Unfortunately, the more extensive cosmic rays. Causes include correlations in produc- experiments typically have less target depth and conse- tion of pions[61, 62], “disordered chiral condensates”[63] quently can give less information about how the Centauro or “quark gluon plasma”[58], highly energetic gold- primary particles interact with matter. Since Centauros stone bosons[64], strangelets[65], or evaporating black are known to be deeply penetrating particles, these ex- holes[66]. periments can do little to measure the true total energy The π±/π0 ratio is measured by relying on the fact that of a Centauro. π0s quickly decay into photons. Thus the particle shower Very thick lead targets with many layers of emulsion initiated by a π0 consists, at least at first, of gamma- are most suited at measuring cross sections, energy and rays, electrons and positrons. These “electromagnetic” momenta of high energy cosmic rays, but since these tar- particles are quickly absorbed in relatively thin layers of gets are small in surface area, they are unlikely to see matter. On the other hand, a π± produces a shower of any low altitude, and therefore clean, Centauro showers. hadrons which are much more penetrative. Instead, what they typically see are mixed showers of The electromagnetic part of a shower is detected by particles from high altitude primary collisions. the uppermost layer of the cosmic ray target, while its As a result of these material limitations, our data on deeper layers detect the hadron rich π± portion. high energy cosmic rays is split between fairly good en- For a typical cosmic ray event, the size of the elec- ergy, momentum and cross section data about the parti- tromagnetic shower and the hadron shower are approxi- cles resulting from high altitude collisions, for which we ± 0 mately equal. It was the absence of an electromagnetic can know little about the π /π ratios, and low altitude shower in the presence of a very strong hadron shower collisions with good ratios but poor energy and cross sec- that caught the attention of the Brazil-Japan cosmic tion data. And these cosmic rays are only a fraction of ray collaboration at Chacaltaya.[67, 68] They named the the energy of the largest cosmic rays seen, which are de- event “CentauroI” in allusion to the mismatching of the tected at long distance by their air showers rather than top and bottom parts of a Centaur. by emulsion or film. In addition to the Centauros, other high energy cos- mic ray events appear to be balanced in their hadronic / electromagnetic components, but contain particles with B. Anomalous Transition Curves larger than expected penetrability. In addition, these events tend to be aligned on the target to an extent With each collision with an atom of atmosphere or tar- greater than that predicted by QCD. These events are get, the cosmic ray particle produces a shower of debris. called “Chirons”. With time, or equivalently, with distance, the shower If a Centauro event were to occur at high altitude, grows until its energy is converted to particles that do the particle showers produced would degrade by the time not interact much. The resulting curve, called a “tran- they penetrated the remainder of the atmosphere, with sition curve”, gives the size of the shower as a function the hadrons producing electromagnetic showers and the of its depth. Transition curves for a single shower look electromagnetic particles producing hardonic showers. It more or less like an upside down parabola, with the max- would therefore be difficult or impossible to distinguish imum of the parabola corresponding to the depth where the ratio of neutral to charged pions. It is generally be- the shower reached its maximum energy deposition. lieved that such a Centauro event would be classified as After developing the film in a cosmic ray experiment, a Chiron. the experimenters measure the transition curves by mea- 18 suring the amount of exposure to the film or by counting the number of particle tracks in emulsion. As a primary FIG. 8: Sketch of a Chiron type transition curve anomaly in which the primary particle disappears inside the detector. In particle travels through the detector, each collision it has this case, the particle has 6 collisions in the first half of the with a target particle produces new particle showers, each detector. with its own transition curve. The result is a series of bumps in the energy deposition graph, with one bump corresponding to approximately one collision. Chirons produce odd transition curves in that they, as well as some of the collision products that produce, some- times penetrate quite deeply. In addition, their transition ln(n) curves sometimes fail to decrease much with depth. A transition curve for a typical Chiron that fully penetrates a deep lead chamber is sketched in Fig. (7). 20 40 60 80 100 120 FIG. 7: Sketch of a transition curve showing Chiron type t(c.u) anomalies in deep lead cosmic X-ray film. This is a sketch only, for real data see [58]. The horizontal axis is depth, in cascade units. The vertical axis is the log of the number of tracks. In this case, the primary particle exits through are the results of high altitude collisions of the cosmic ray the bottom of the detector after 11 collisions. There are two anomalies in this sort of transition curve. First, the parti- primary, they are expected, in the standard model, to be cle is extremely penetrative. Second, the distance between hadrons generally smaller than protons. That the colli- collisions is too short for a typical hadron to be the primary sion mean free path for the mini-clusters is less than that particle. of the proton is therefore particularly unexpected.[58]

C. High Transverse Momementum ln(n) The transverse momentum of mini-clusters in Chiron events is computed using the formula:

20 40 60 80 100 120 pT = E r / h, (78) t(c.u) where E is the energy of the mini-cluster, r is the mea- sured offset of the mini-cluster from the center of the Since ultra high energy cosmic rays are not generally event, typically a few mm, and h is the height at which seen emerging from the earth, it is clear that Chirons the collision occured, typically a kilometer. For most Ch- must eventually be absorbed by normal matter. Every iron events, h is estimated by assuming a figure for the now and then, this is observed in deep lead chambers. magnitude of p . For a few, the height can be estimated For example, see PB73 Block8 S127-100 shown in Fig- T by triangulation of the particle paths. ure 2.16 of [58]. In such a case, the primary particle will cease colliding somewhere above the bottom of the detec- Since the mini-clusters are themselves made up of tor. A sketch of a typical case is shown in Fig. (8). About hadrons, the same equation can be applied to the hadrons half of all mini-clusters exhibit unusually high penetra- of the mini-cluster itself. When one does this, however, tive ability.[58] the evidence from the mini-clusters suggests much lower The high rate at which Chiron fragments collide can be values of h, and therefore much higher values of pT than illustrated by comparing the geometric mean free path in QCD can account for.[59] The value of E r between mini- the target with the collision mean free path for the cosmic clusters is about 300 times as large as the value inside a ray particles. The Chirons are able to travel only about given mini-cluster.[58] 1/2 as far as the geometric mean free path would indi- The low transverse momentum of the debris in a mini- cate. This is shown by examining the distribution of the cluster is what we expect from an extremely relativistic position of the first shower in a Chiron mini-cluster.[58, air shower. What is anomalous is the very high trans- Fig. 2.13] verse momentum in the primary collision. The conflict The geometric mean free path that the mini-cluster suggests that the primary collision (or collisions) were data gives are compared with that of typical low energy not governed by Lorentz symmetry, as we will propose cosmic rays, which are mostly protons. If mini-clusters later in this paper. 19

D. Alignment and Multiplicity time duration of a Centauro, as compared to more usual cosmic ray showers as will be later discussed. The mini-clusters of Chiron events have a tendency If there were astronomical sources of Centauros, we to be aligned to a degree in excess of that predicted by would expect to see particle showers when these sources QCD. For an extensive discussion of the alignment issue interact with matter. Such a particle shower would be of Centauro events, with simulations of what would be recognizable because it would be, like the Centauro air expected, see [69]. For a discussion of alignment in a par- showers themselves, extremely energetic and very well ticularly clean high altitude Centauro, see [70]. Also see collimated. [71]. Since the alignment occurs between mini-clusters, it In fact, astronomers do observe highly relativistic is clear that alignment is part of the problem of excessive “jets”. Some of these jets are extragalactic and are asso- transverse momentum. ciated with quasars or the centers of other galaxies.[72, If the primary for a Chiron were a heavy nucleus, as 73] A variet of quasar known as a “blazar” exhibits ex- would explain the high cross section and penetrability, tremely fast variation in intensity. The cause of the vari- we would expect to see a large number of fragments in ation, as well as the source of jets in general, is still mys- the initial collision at high altitude. Each of these frag- terious though the relation with “active galactic nuclei” ments, which would be called “jets” in accelerator lan- that are responsible for quasars is now clear.[74–76] guage, would produce its own mini-cluster. The number In addition to the nuclei of galaxies, including our of mini-clusters, or “multiplicity”, therefore, suggests an own, there are also small relativistic jets present around obvious way of deducing the complexity of the primary our own galaxy. These are associated with black holes, cosmic ray. neutron stars or x-ray binaries. These sources, be- If the primary for a Chiron were even as large as a cause of their similarity to quasars, are sometimes called proton, then we would expect to see a fairly large number “microquasars”.[77] Microquasars are also associated of jets. It would be natural for these to be aligned, as with gamma-rays.[78] jets tend to be coplanar. However, about half of Chirons Relativistic jets, when observed from small angles with include only a single mini-cluster, which is sometimes respect to their direction of propagation, will sometimes called a “uni-cluster”. This suggests that these Chirons appear superluminal. This easily explained effect is from must be simple particles. the geometry only, and has nothing to do with the tachy- Occassional Chirons are observed with extremely high onic nature of binons. numbers of mini-clusters, but these are believed to be the result of air degradation of a parent collision with 4 or 5 particles. For the standard model of QCD, the problem F. Gamma-Ray Bursts here is that the low multiplicity of some Chiron events is incompatible with the high transverse momentum and If an astronomical jet is caused by Centauros, the re- alignment of the high multiplicity events. sulting particle shower will include a lot of charged par- Mini-clusters that have passed through a sufficient ticles that will be diverted by magnetic fields. However, amount of atomsphere appear to degrade into a “halo”, there will also be gamma-rays, and some of these may which describes a region of the film where a large amount have energies low enough to survive the journey to the of energy is deposited more or less evenly over a wide earth still pointing back towards their source. region. Halos are difficult to analyze in terms of total Every now and then, very large numbers of high energy energy deposited, but they appear to be hadron rich.[58] gamma-rays hit the earth, all from the same direction. The presence of halos gives clear evidence that the low These are called gamma-ray bursts (GRB). Gamma-ray transverse momentum of mini-clusters is due to the ex- bursts are sometimes followed by delayed bursts from treme speed of the particle generating the mini-cluster the same direction. A delay as long as 77 minutes has shower, rather than some effect which collimates the de- been observed, and this is difficult for models to explain, bris of an air shower. mostly due to the fact that the delayed burst is not ther- mal, though there are some efforts.[79] In addition to delayed bursts, GRB are also afflicted E. Jets, Quasars, & c. with precursor activity that is difficult to explain. Pre- cursors have been found up to 200s before the main burst The other half of the Centauro question is where they (the longest time searched), and possessing non-thermal, come from. The film based cosmic ray targets, given softer spectrums than the main burst: the rotation of the earth, give only limited indications of what direction these cosmic rays arrive from, beyond We have shown that a sizable fraction of “up”. A useful experiment would be to combine a film bright GRBs are characterised by weak but based cosmic ray target with a cosmic ray detector that is significant precursor activity. These precur- sensitive only to high energy cosmic rays, and that would sors have a delay time from the main GRB give an indication of the time of arrival of a Centauro. which is surprisingly long, especially if com- In addition, the experiment would give an idea of the pared to the variability time scale of the burst 20

itself. The precursor emission, contrary to model predictions, is characterised by a non- FIG. 9: The “ankle” of the cosmic ray flux.[92] Flux values are given by log(F lux ∗ E3) in units of eV 2m−2s−1/sr. Energy thermal spectrum, which indicates that rela- is given by log(E) in units of eV . The ankle begins at around tivistic electrons are present in the precursor E = 1019eV , and is shown as a thin line due to its uncertainty. emission region and that this region is opti- cally thin.[80]

25 G. The GZK Limit n

In the 1960s, soon after the discovery of the cosmic mi- crowave background, Greisen, Kuzmin and Zatsepin real- ized that protons and neutrons would encounter the cos- 24 mic microwave background as high energy gamma rays, 15 16 17 18 19 20 would scatter off it with production of pions, and would E consequently slow down. This should result in a cutoff in the spectrum of cosmological cosmic rays at just be- low 1020eV. For protons of energy 1020 to 1021eV, the attentuation length is of order a few Mpc.[81] III. SIGNATURES OF FREE BINONS Because of the GZK limit, there are proposals that high energy cosmic rays are generated in the cosmologi- We expect that free binons would be extremely rare, cal neighborhood[82] or are from the creation of charged but it is not unlikely that some have already been ob- black holes[83]. These proposals fail to give any hint as served. This section compares what we would expect of to why the Centauro events are anomalous. binons to the various astronomical puzzles reviewed in A similar argument to the GZK limit has been put the previous section. forward in [84] to argue that the speed of light is very close to the maximum speed that a material body can achieve. These arguments do not apply to a subparticle A. Elastic Binon Collisions and Mini-Clusters that may not possess an electric charge. For an article on Cherenkov radiation emitted by superluminal particles, Since binons are supposed to underly both the weak see [85]. and strong forces, it is natural that they interact more Gonzalez-Mestres argues that evidence for extreme strongly than the strong force, but still violate isospin tachyons, (i.e. ci  c) might be seen in cosmic rays[86] symmetry. This combination is an attribute that specif- and that this could be an explantion for the evasion of ically defines the Centauro events. the GZK limit.[87] See [88] for a discussion of small dif- Consider a binon travelling in the +z direction that ferences in particle speeds allowing the GZK cutoff to be collides with a electron or quark. Since we are here con- broken. See [89] for a general discussion of Lorentz sym- sidering the possibility that binons are responsible for metry violation in the context of extreme high energy Centauro events, and we suspect that the primary par- cosmic rays. ticles responsible for these events do not interact much It has been suggested that ultra high energy cosmic with photons, we will assume that the binon is one of the rays could be associated with quasars, provided that one neutrino type binons. avoids the GZK limit by assuming a “messenger” particle The collision will involve 4 binons. There is some that does not interact with photons.[90, 91] Given the chance that there will be an elastic collision, in the sense evidence from emulsion and film targets, this assumption that the result will give one free binon and another quark is natural. The classical Centauro produced very few or fermion. photons in its debris, so one would expect that it would Since the collision products of such an elastic collision have a very low cross section for collisions with photons. travel at different speeds (that is, the binon is tachyonic Cosmic rays are always more numerous at lower ener- while the normal matter is not), the binon will separate gies. If one draws a curve showing the number of cosmic from the normal matter particle shower. When the binon rays seen at energies from the lowest to the highest, the later produces a normal particle shower, the two showers curve drops off at about E−3. Accordingly, it is custom- will arrive in reverse order. That is, the first normal ary to plot frequency multiplied by energy cubed. The particle shower will arrive last. approximate shape of the curve is shown in Fig. (9). The The separation of the normal particle shower from the presence of the ankle suggests that the particles respon- binon will have observable consequences. First, the par- sible for Centauro events either evade the GZK cutoff or ticle shower will be spread out in time. Second, the are produced fairly locally, but whether or not the ankle movement of the tachyonic particle does not obey Lorentz exists is still being debated.[93] symmetry, and so must be analyzed in the preferred rest 21 frame. the normal particle showers. These should be rare, and If, in the preferred rest frame, the target has motion in fact, there are few examples of Chirons or Centauros perpendicular to the binon, then the particle showers for where there is an estimate of the collision height based the later collisions of the binon will arrive at different on angular data. Note that the original triangulation times, and therefore at different positions on the target. data[58, 67] for the CentauroI event was retracted[68] in For example, if the binon has two elastic collisions, 2004. and then is converted into normal matter, there will be a The faulty CentauroI triangulation of 50±15m resulted total of three separate particle showers produced. In the in pT of about 0.35GeV/c. This value was large, but preferred rest frame, relativistic effects will cause these seemed reasonable, and was used to estimate the heights three showers to be closely collimated and they will travel of other Centauro and Chiron events.[58] Consequently, down the same path. one is likely to see many estimates of heights of Cen- However, if the target is moving in the preferred refer- tauro events in the earlier literature that should now be ence frame, the three showers will hit the target at three ignored. different points. Furthermore, the three points where the showers hit will tend to be aligned. Therefore, we In 1977, a balloon borne emulsion chamber was lucky will expect to have aligned showers despite each shower enough to be hit by a Chiron close enough that it could being extremely relativistic. The effect will be a series be triangulated. This event was reanalyzed in 2001.[70] of aligned particle showers, just as is seen in the Chiron The result was that the apparent point from which the events. We will assume that each elastic binon collision collision emanated was about 120±40m above the cham- produces a new mini-cluster. ber. The target was 100 nuclear emulsion layers, with a From the alignment data for Chirons, along with an lead calorimeter. Estimated transverse momentum was estimate of the speed of the earth against the preferred 23±7 GeV/c. The alignment parameter for the 4 highest reference frame, we can estimate the distance between energy particles was .9878, where 1 is perfect alignment. elastic binon collisions, as well as the attentuation length The spread in high energy particles was about 1.4cm.[70, for binons converting into normal matter in the high at- Fig. 6] mosphere. The figures of 4cm and 120m allow us to reanalyze this Not very many physicists believe in a preferred refer- event by attributing the apparent pT to a preferred refer- ence frame, so estimates of the earth’s movement through ence frame effect. Over a distance of 120±40m, the time it have not received as much attention as one might de- of flight for a c and a 3c particle will be 400 ± 130ns and sire. Arbitrarily, we will use Consoli’s figure of about 130 ± 44ns, respectively. The difference in arrival time 200km/s.[50] Ignoring the orbital motion of the earth, will be 270 ± 90ns, so a separation of 1.4cm indicates a the average perpendicular motion of the earth in the preferred reference frame velocity of 39 to 77km/s. This preferred reference frame will be about π/4 of this or is low, but comparable with the average perpendicular 160km/s. preferred reference frame calculation of 150km/s. Unfor- According to [69], the average radius of aligned 4-core tunately, the exposure time was 160 hours, so accurate events is 1.8cm, which gives a total diameter of 3.6cm, or information about the direction of the track, relative to 1.2cm for the collision length. At 150km/s, this gives a the various purported preferred reference frames, is not time delay between particle showers of an average of 80ns. available. −1 Assuming that θB = cos (1/3), this gives a distance between collisions of about 50m. After disregarding the CentauroI triangulation, there are two other triangulation heights listed in [58]. Cen- A distance of 50m seems small, except when you con- tauroVII which gave a height of 2000 to 3000m with a sider the possibility that the binons that are impinging p comparable to CentauroI. ChironI gave a height of on the earth are not necessarily the same as the ones re- T 330 ± 30m and p of 1.42GeV/c. The triangulation data sulting from the “elastic” collisions discussed here. Since T is: Centauros are definitely low altitude collisions, and Chi- rons are still fairly clean, the incoming binon may be one with a particularly low cross section for collisions with Event Height(m) pT (GeV/c) matter. CentauroVII 2500 ± 500 0.35 (79) In this model, the apparent transverse momentum of ChironI 330 ± 30 1.42 most of the Chiron events is caused not by angular de- Balloon1977 120 ± 40 23 ± 7 viation at the collision, but instead by error induced by movement of the target with respect to the preferred ref- erence frame. Despite the high transverse momentum, This is about what we expect for elastic binon collisions we expect the particle showers to be parallel, and, in- with respect to a preferred reference frame. The low deed, highly parallel, widely separated, particle showers triangulation data is generally more accurate and leads is a characteristic of Centauros and Chirons. to higher estimates. We do not expect a binon, which If the binon manages to survive to the target, then should have a very large cross section, to survive a 2500m an angle will be produced between the binon track and passage through the atmosphere. 22

B. High Transverse Momentum binon travels a distance of l meters at a speed of 3c, the duration of the resulting air shower, at the core, will be extended by Disregarding preferred reference frame effects, the pT values for binon interactions should be higher than for 1 1 2 l standard quantum mechanics. Of the interactions in τshower = l( − ) = (80) standard quantum mechanics, the only coupling between c 3c 3c left and right handed particles consists of the mechanism seconds. In addition, if there is a distance of l between that gives mass. consecutive collisions, there may be a corresponding time In the standard model, the various spin-1 forces apply interval between consecutive peaks in the intensity of the to the leptons and quarks as a whole. This follows over shower. to the binon model in that the vector bosons are coupled The spread in time signature will be proportional to to the bound states of the binons. These forces, since the distance between the binon’s first collision and when they couple to bound states, use the spin of the bound it converts to regular matter. From the previous section, state rather than the details of the individual binons. this should be around 100 meters. Accordingly, the air The usual forces can only couple between same handed shower will be extended by about 200ns. states. This time is barely within the ability of high energy Binon theory treats mass as another force, one that cosmic ray experiments to register. For example, the binds together two chiral fermions into a single particle. HiRes experiment uses an analog to digital converter with Mass, as a force, is therefore unique in that it must treat a 100ns sampling period.[93] The Pierre Auger cosmic ray the binons individually. Seen in this way, mass is a more observatory has a better chance with a sampling rate of fundamental force than the various forces mediated by 40MHz for their Cherenkov detectors, and 100MHz for vector bosons. their air fluorescence telescopes.[98] The AGASA exper- The high transverse momentum of mass treated as a iment uses photomultiplier tubes with the light intensity force is hinted at in the zitterbewegung model of the encoded as the logarithm of the pulse duration. The pulse electron.[94, Eq. 58] In this model, the electron is shown corresponding to a single particle is 10 microseconds, so to travel at speed ±c as these are the only eigenvalues it is quite impossible for them to detect the duration of of the velocity operator in the Dirac equation. Thus a a air shower.[99] stationary electron alternately moves back and forth at While a typical shower might only be extended by speed c, with its handedness reversing at each turning 200ns, there will be some showers that are extended much point. This results in an electron with an average posi- longer. Since the AGASA encodes their light intensities tion that is stationary, and with a consistent spin oriented by pulse duration, this could be the cause of their cosmic in the direction of the movements. Thus the mass force ray energy flux measurements being higher, and extend- is not only capable of completely reversing longitudinal ing to higher energies as compared[100, Fig. 8] to the momentum, but must do so constantly. other cosmic ray observatories. When this model is applied to binons, the fact that In cosmic rays, the time of arrival of hadrons is more binons are travelling off axis relative to the electron im- precise than the arrival times of the electromagnetic par- plies that they intrinsically possess extremely high pT , ticles, as the hadrons travel paths that are more straight. and reactions involving them, like the mass force itself, In showers that arrive at a shallower angle, the additional should be expected to have high pT as well. distance travelled through the atmosphere has the effect of stripping off the electromagnetic part of the shower. This leaves a tighter time signature. The signature time C. Tachyonic Precursors of Cosmic Ray Showers is probably short enough to detect a broadening due to a tachyonic primary particle with a survival distance long If the primary particles in some cosmic rays are enough to generate the Centauro spreads.[98, Fig. 6] tachyons, it is only natural to expect that this will have an observable effect on how the shower evolves in time. It turns out that the effect will be quite small and difficult D. Inelastic Binon Collisions to observe. Several attempts to explicitly find tachyons in cosmic If we assume that extreme separation between mini- rays have met with success[95, 96] or failure[97]. From clusters is mostly due to preferred reference frame ef- the point of view of binons, a search of this sort will be fects, the statistics for mini-cluster formation will give rather difficult. Instead of surviving to the ground as a us information about the branching ratio for elastic bi- tachyonic particle, an individual binon will likely convert non collisions. After the binon is converted to quarks into quarks and leptons at high altitude. and leptons, we expect that the leptons will cascade in Instead of a tachyonic precursor, at low altitude the the manner predicted by the standard model, while the time signature of a binon is likely to be a highly energetic quarks will continue to exhibit anomalous behavior. air shower that has a duration somewhat longer than As opposed to binons, free quarks are understood by usual. If, between its first collision and its final one, the the standard model and are not likely to violate isospin 23 symmetry, so we can expect that the showers produced by the same time, visible radiation and light will be gener- free quarks will have both electromagnetic and hadronic ated by the binons who were not so lucky. portions. The extremely high energy of free quarks The difference in speed between the binons and the should give them high penetrative power. Both these associated particle shower debris is quite large. Conse- effects are seen in mini-clusters. quently, even if we have very good data on the direction The very large potential energy of free quarks should of the source of the binon, and even if the binon is un- give them a larger than normal cross section, and exactly charged and so is undeflected by magnetic fields, it may that is observed in mini-clusters. In addition, since color be difficult or impossible to match the binon up with a refers to an angle around the direction of propagation, we visible source. This is due to several effects. First, the can expect free quarks to violate cylindrical symmetry, black hole could have begun radiating so recently that no and therefore to have relatively high pT collisions. This radiation of speed c could have reached us. Second, mo- could be the explanation for the approximate 0.5GeV/c tion of the black hole could cause its apparent position collisions seen. at the time of arrival of the two signals to be displaced. Under the binon model, quarks are composed of a Finally, even if the black hole radiates continuously mixture of electron and neutrino binons. The electron and is not moving, our own proper motion with respect and neutrino binons share the nondirectional parts of to the preferred reference frame will cause the apparent their idempotents but are distinguished by the directional directions of the black hole, as determined by radiation parts. Therefore there is a coupling between these pairs with speed c and its direction as determined by radiation of electron and neutrino binons. with speed 3c radiation to diverge. This effect will be Eventually, for energetic reasons, a free quark will take comparable in angle to about v/c = 2.0 × 105/3 × 108 advantage of this coupling to convert to color-free matter. = 0.0007 radians. From that point on, the shower will lose its anomalous This effect, that is, that the binons from a black hole character and the standard model will take over. could reach us before normal matter and energy, suggests that when we use ancient photons to look for sources of high energy cosmic rays, we should consider the possibil- E. Black Holes and Binons ity that the photons are providing an obsolete descrip- tion. Thus we should not be too surprised to see binons Since binons travel at 3c, they can penetrate the event emitted from a direction in which a black hole has not horizon of a black hole. Since the conditions inside a yet appeared. Instead, we should look for evidence that black hole are expected to be extreme, and the only par- a black hole is likely to form there “in the future”. ticles that can escape the black hole are tachyons, it is natural to suppose that black holes are binon factories. Black holes are expected to form accretion disks. Bi- F. Gamma-ray Bursts and Binons nons emitted into the accretion disk will scatter and downconvert to normal matter. They will contribute to Since gamma-ray bursts have been associated with the heating of the accretion disk, but will not be observed black holes and jets, there is likely a connection to bi- at a distance. nons. That is, gamma rays may be the remains of particle The same transverse momentum effects that are seen showers produced by binons created by black holes. For in Centauro events, as well as heating in general, can be a paper that proposes that tachyons may cause gamma expected to knock normal matter out of the plane of the ray bursts and high energy cosmic rays, see [40]. We accretion disk. This will have the effect of thickening the imagine that a sudden influx of matter into a black hole, accretion disk near the singularity. Only the regions near or perhaps the initial formation of the black hole in a su- the axis of rotation of the black hole may be left open for pernovae, will create a burst of binons that then interact binons to escape, and the thickening of the accretion disk with nearby matter to create particle showers that we may leave the angles available for binons to escape to be eventually observe as gamma ray bursts. small. Later, when the binons impact matter at some As mentioned in the previous subsection, the difference distance to the black hole, they will produce collimated in propagation delay between photons and binons, over particle showers that will appear as jets to distant ob- interstellar distances, will eliminate any observable time servers. correlation between binon cosmic rays and the associ- We can expect that the region open for binons to es- ated gamma ray bursts. However, if the particle showers cape will not always be exactly aligned with the black are created over an extended region near the black hole, hole spin axis, and so the surviving particle showers will there will be a corresponding spread in arrival time of change in direction and opening angle. Since the occlu- the debris from the associated showers. sion may happen near the throat of the black hole, we First, as was noted in Subsec. (III C) particle showers can expect to see changes in jet strength and direction induced by single particles in the atmosphere naturally at rates suitable to the size of the black hole. spread in time, quite apart from any tachyonic effects, on Any binons that escape the region of the black hole un- the order of 1 to 10 microseconds. Gamma ray bursts, converted to normal matter will continue travelling. At by comparison, have typical durations around 10 to 100 24 seconds, around 107 times longer. Perhaps the density of holes. Since the black hole event horizon is fundamentally the media in which binons produce gamma ray bursts is a velocity barrier, rather than a momentum barrier, it about 10−7 that of the atmosphere. could be that the proper way of classifying the binons Gamma ray bursts are often accompanied by nonther- that escape a black-hole is in their velocity eigenstates mal precursors 10 to 200 seconds ahead of the main burst rather than their momentum eigenstates. with about 1% of the energy.[80] If a small percentage of Massless particles in velocity eigenstates do not carry the binons emitted at the black hole were to penetrate specific momenta or energy. Instead, if one carries out the region where most of the gamma rays were produced, an energy measurement of such a particle, one obtains a and then were to encounter matter in a region around 6 set of possible values according to a probability density. to 60×109 meters away, the result would be a precursor Since the energy deposited by a binon in standard mat- with the appropriate delay. ter will largely depend on the number of collisions that The effect of jets emitted by a black hole would be the binon survives before converting to standard mat- to sweep matter away from the axis of the black hole. ter, we can model the energy content of a binon velocity This could clear the way for later binon bursts to post- eigenstate by assuming a model based on a Poisson pro- pone showering until they reached the end of the empty cess. region. This mechanism could produce delayed gamma If we assume that each elastic collision deposits an en- ray bursts that are created from the difference in travel ergy E0, and that the binon survives unaltered with prob- time of radiation created in the main shower media and ability p, then the spectrum of energies deposited by the radiation created at the inner edge of the accretion disk. binon will follow the probability density:

n 1/E0 nE0 ρ(nE0) = p = (p ) . (81) G. Velocity Eigenstates of Binons 3 Replacing nE0 with E, multiplying by E and converting It is customary, in the standard model of QM, to an- to logarithms gives a form compatible with Fig. (9): alyze particle interactions in the momentum representa- tion. In the momentum representation, energy and mo- 3 log(E/E0) log(E ρ) = log(ρ0) + 3 log(E/E0) − 10 . (82) mentum are naturally conserved, but particle positions are not. The position representation, while losing energy For small values of E, the second term dominates, and and momentum conservation, is a natural representation will give a fairly steep positive slope of 3. For large val- for a theory that must take into account positions in the ues of E, the last term dominates and will give an even hidden dimension. steeper drop off. If one assumes a Lorentz symmetric world, then it is At this time, even the simplest fact about high en- natural that a model of one photon at some specific en- ergy cosmic rays, namely their energy spectrum, is ergy would imply the existence of a continuum of models uncertain.[100] Fitting the above curve to the data re- of photons at various other energies. quires knowledge of both the energy flux of cosmic rays A scheme, such as the present one, that assumes a pre- and their composition, so we will refrain from attempting ferred reference frame does not have this natural method a fit of the above curve to the data. of assigning various energies to a single massless parti- cle type. Since there is a preferred reference frame, an electron travelling at one speed can be distinguished, in principle, from an electron travelling at some other speed. H. Binon / Tachyon Cosmology Similarly, a photon or binon travelling with one energy level can be distinguished from the same particle travel- Since tachyons would dominate the early universe, they ling at energy. This leads to a question of how it comes would allow parts of the universe that could not be in about that two identical massless particle can possess dif- communication by light limited particles to come to equi- ferent energies and still be identical. librium. This result is a natural inflation that is useful A solution is to assign a “bare” propagator that is a in cosmology.[101–104] Other writers suggest tachyons as velocity eigenstate to represent a particle, and then to an explanation for dark matter or energy.[40, 105, 106] obtain effective propagators for the various possible mo- menta and energy through a resummation. One then obtains the quantum mechanics of the standard particles as an effective field theory from a very simple but unusual APPENDIX A: CLASSICAL SPECIAL underlying field theory. RELATIVITY CALCULATIONS IN PTG These bare propagators do not correspond to any spe- cific energy or momenta. This is a natural extension of In order to make clear how classical calculations in the the zitterbewegung model of the electron to massless par- PTG can match those of special relativity, this appendix ticles composed of chiral fermions. provides detailed calculations for time dilation and length The application to astrophysics has to do with black contraction using both techniques. 25

1. Time Dilation flight will indicate (when multiplied by c) twice the length of the rod. Problem: A spaceship travels 3 light years away form So let the rod begin at position (0, 0, 0, 0) through earth, at a speed of 0.6c, and then returns at the same (6m, 0, 0, 0), and set the velocity vector for the rod to speed. What is the proper time experienced on the Earth be (0.923, 0, 0, 0.384) so that it moves in the +x direc- during the voyage, and what is the proper time experi- tion. The light signal starts at (0, 0, 0, 0) and proceeds enced on the spaceship? with a velocity vector of (1, 0, 0, 0) until it meets with Special Relativity Solution: The voyage requires 3/0.6 the other end of the bar at time t1. The light direction = 5 years each way for a total of 10 years. This is the is then reversed, and it travels with velocity (−1, 0, 0, 0) proper time experienced on the Earth. The spaceship until it meets up with the trailing end of the bar at time experiences a time dilation of (1 − 0.62)0.5 = 0.8, so the t2. The length of the bar, in the reference frame of the proper time experienced on the spaceship is 10 × 0.8 = 8 bar, is then 1/2 the proper time experienced by the trail- years. ing end of the bar from 0 to t2. The equations for t1 and PTG Solution: The spaceship starts at the point t2 are therefore:

(x, y, z, s) = (0, 0, 0, 0). Align the x axis with the direc- (0, 0, 0, 0) + (1, 0, 0, 0)t1 tion of travel. The velocity of the spaceship on the outgo- = (6, 0, 0, 0) + (0.923, 0, 0, 0.384)t , (A2) ing voyage is therefore given by the vector (0.6, 0, 0, 0.8). 1 The 0.8 value is required to make the speed of the space- (1, 0, 0, 0)t1 + (−1, 0, 0, 0)(t2 − t1) ship work out in total be 1. The spaceship’s position as = (0, 0, 0, 0) + (0.923, 0, 0, 0.384)t2. (A3) a function of the global time t is therefore: Since our world does not distinguish between the hidden s coordinate, the equalities need only be established for (0, 0, 0, 0) + (0.6, 0, 0, 0.8)t (A1) 1 the first three coordinates. The solution is t1 = 78 meters, and t2 = 81.12 meters. Setting this equal to (3, 0, 0, s1) gives t1, the global coor- dinate time for the arrival of the spaceship at its desti- The proper time experienced on the trailing edge of the nation, and t is therefore 5 years. Note that the value rod is, by time dilation, 0.384 of t2, which gives 31.2. Half 1 of this is the proper length of the bar, which is the same of s1 is unspecified, as the total length of the hidden dimension is negligible as compared to the many light as the value given by special relativity. Therefore, both years of travel. Since the proper time component of the theories show the Lorentz contraction of the bar to be the velocity of the spaceship is 0.8, the total elapsed proper same. Since the two theories are the same in both time time on the outgoing voyage of the spaceship is therefore dilation and Lorentz contraction, any dynamical problem 0.8 × 5 = 4 years. Similarly, the return trip uses a ve- can be converted between them, and the results of stan- locity of (−0.6, 0, 0, 0.8) and results in a coordinate time dard relativity translate directly. Thus the problem can passage of 5 years and a proper time for the spaceship of be worked with the simpler methods of SR with only a another 4 years. The result is, of course, identical to the philosophical difference. Special Relativity result. APPENDIX B: SCHWINGER MEASUREMENT ALGEBRA 2. Lorentz Contraction Following Julian Schwinger’s 1955 lectures onquantum A rod flies lengthwise through a laboratory with a kinematics[107, Chap. 1.1], but specializing to the case speed of 0.923c. The lab measures the length of the rod of the electron fermion family, let A denote a set of char- as 6 meters. How long is the rod measured in a coordi- acteristics that distinguish the 32 particles in a family F nate system moving with the rod? of elementary fermions. We will define the QCD colors Special Relativity Solution: The Lorentz contraction as {1, 2, 3}, and use the following designations for such a 2 −0.5 factor is (1 − 0.923 ) = 2.6, so the proper length of family: the rod is 6m × 2.6 = 15.6 meters. In any given coordinate system, the constancy of the F = {eL, eR, e¯L, e¯R, νL, νR, ν¯L, ν¯R, speed of light provides a technique for measuring length. u1L, u1R, u¯1L, u¯1R, u2L, u2R, u¯2L, u¯2R, Accordingly, the rod can be measured in its own frame ¯ ¯ u3L, u3R, u¯3L, u¯3R, d1L, d1R, d1L, d1R, of reference by calculating the time required for light to d , d , d¯ , d¯ , d , d , d¯ , d¯ } travel the length of the rod. Since proper time is a prop- 2L 2R 2L 2R 3L 3R 3L 3R erty of individual particles, rather than dimensional ob- Let a1 be an elementary particle in F. Let M(a1) sym- jects such as rods, the length of the rod will have to be bolize the selective measurement that accepts particles of measured by computing the time required for the light type a1, and rejects all others. One can imagine some sort to travel down the rod, be reflected at the end, and then of Stern-Gerlach apparatus, though since quarks are ap- travel back to the point of origin on the rod. The proper parently permanently bound it will have to be an imagi- time experienced by the end point of the rod during this nary apparatus. We can define addition of measurements 26 to be the less selective measurement that accepts parti- where η is any Clifford algebra constant that doesn’t an- cles of any of the included types: nihilate (1 − xytd). Sixteen values for η that give linearly independent eigenvectors include any nonzero term in the M(a1) + M(a2) = M(a1 + a2). (B1) product (1 +x ˆ +y ˆ + tˆ)(1 +z ˆ)(1 +s ˆ). Two successive measurements can be represented by mul- Clifford algebra elements of the form (1±e)/2 are idem- tiplication of the measurement symbols. Because of the potent, if e is any element that gives 1 when squared. If physical interpretations of the symbols, addition is as- e1, e2 and e3 mutually commute and square to 1, then sociative and commutative, while multiplication is at least associative. One and zero represent the trivial measurements that accept all or no particles. Clearly, ι = (1 ± e1)(1 ± e2)(1 ± e3)/8, (C2) 0 + M(a1) = M(a1), 1M(a1) = M(a1)1 = M(a1), and where the ± are to be taken independently, are eight 1 1 0M(a ) = M(a )0 = 0, so the set of measurements form idempotents. If e1, e2, and e3 are canonical basis ele- an algebra. The “elementary” measurements associated ments and generate a group of size 8 under multiplica- with these 32 fermions satisfy the following equations: tion, then these eight idempotents are distinct, and in the context of the Clifford algebra used here, are mutu- M(a1)M(a1) = M(a1), (B2) ally annihilating primitive idempotents.[109] An example of a set of e that satisfy these requirements is e1 = ztb , e = ixyc, e =s ˆ. M(a1)M(a2) = 0, a1 6= a2, (B3) 2 3 Out of respect for the contributions of Pertti Lounesto to the mathematical understanding of spinors, we will n=32 X refer to a set of e that satisfy these requirements: M(an) = 1 (B4) n n=1

Schwinger goes on to analyze incompatible measure- enem = emen, ments, such as spin in two different directions, but these en ∈ Canonical basis, simple results are enough to establish the similarity to 2 en = 1, the approach of the present paper. 3 This paper specializes the (complex) Schwinger mea- {en}n=1 generates a group of order 8, (C3) surement algebra to a complexified Clifford algebra. For a paper that generalizes the complex Schwinger mea- surement algebra from a complex algebra to a quater- as a set of “Lounesto group generators”. The group of nionic measurement algebra, see [108]. In yet another size 8 will then be the “Lounesto group”. application of Schwinger’s methods, H. M. Fried and Since the Lounesto group generators commute, so do Y. Gabellini used the Schwinger action principle to de- their various products. Therefore Eq. (C2) gives prim- rive a quantum field theory generalization appropriate to itive idempotents that possess good quantum numbers tachyons.[40] with respect to all eight Lounesto group elements. Seven of the Lounesto group elements are nontrivial, the trivial one is 1. APPENDIX C: THE LOUNESTO GROUP The choice of generators among the Lounesto group Other than 1,ˆ the remaining 31 canonical basis ele- is somewhat arbitrary. So long as we pick three group ments have eigenvalues of −1 as well as +1. The multi- elements that are distinct and nontrivial, and we avoid plicity of each is sixteen, and the eigenvectors are easy picking a set that multiplies to unity, our three elements to write down. For example, the eigenvectors of xytd with will generate the other five. Translated back into physics, eigenvalue −1 are given by: this means that we need only keep track of three quantum numbers for each of the primitive idempotents. The rest xytd(1 − xytd)η = −(1 − xytd)η, (C1) are related by multiplication.

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