Inflation model and Riemann tensor on non-associative algebra
IInflation model and Riemann tensor onnon-associative algebra
V. Yu. Dorofeev ∗ A. A. Friedmann Laboratoryfor Theoretical Physics
Abstract
In this article the reduction of a n -dimensional space to a k -dimensionalspace is considered as a reduction of N n states to N k states, where N stands for the number of single-particle states per unit of spatial length. Itturns out, this space reduction could be understood as another definitionof inflation. It is shown that the introduction of the non-associativity ofthe algebra of physical fields in a homogeneous space leads to a nonlinearequation, the solutions of which can be considered as two-stage inflation.Using the example of reduction T × R to T × R , it is shown that thereis a continuous cross-linking of the Friedmann and inflationary stages ofalgebraic inflation at times 10 − with the number of baryons 10 in theUniverse. In this paper, we construct a new gravitational constant basedon a nonassociative octonion algebra. Keyword.
Inflation, gravity, compactification.
Introduction
One of the problems of General relativity is the problem of isotropy of the spacearound us. Indeed, according to the well-established Friedmann model of theUniverse, the indefinite metric of our space-time is defined by a time dependencyin the form of ds = c dt − a ( t )( dx + dx + dx ) , (1)where a ( t ) is actually the size of our Universe, which is clearly seen in theexample of a closed Universe ds = c dt − a ( t )( dξ + sin ξ (sin θdϕ + dθ )) (2)According to Friedmann theory until the time when matter became trans-parent, i. e. since the emergence of the Universe until T dust = 380 thousandyears or 1 . · seconds, the equation of Universe’s state looked like p = ε/ ∗ e-mail: [email protected] a r X i v : . [ phy s i c s . g e n - ph ] A ug using historically preceeding approach, which does not include, for example, deSitter phase with equation of state p = − ε ). Thus, in the case of a flat topologyof the Universe, the equation for the evolution of the size of the Universe was a ( t ) = a √ t . At the moment T dust the equation of state of the Universe changedto p = 0 and the new equation of the size of the Universe became a ( t ) = a √ t or a ( t ) = (cid:26) a √ t, T born ≤ t ≤ T dust. ,a √ t , T dust ≤ t ≤ T now (3)The values of the constants a and a are found from the agreement ofmodern ideas about the age of the Universe, its current size and continuity ofthe function a ( t ). (Again, the power-law nature of the dependency on the timeof the radius of the Universe implies the absence of an event horizon, so weconsider the visible part of the Universe to coincide with its size. For the deSitter stage, the horizon arises and is determined by the inverse value of theHubble constant and, correspondingly, the size of the Universe is much largerthan its visible part.)From recent observations on PLANK, the modern age of the Universe isestimated at 13.75 billion years or T now = 4 . · seconds and accordingly thesize of the Universe is equal to a ( T now ) = 1 . · cm, where a = a ( T now ) · T − / now = 1 . · / (4 . · ) / = 2 . · . (4)comes from.Using the continuity a ( t ) at the moment of changing the equation of state a · T / dust = a · T / dust , we find a : a = a (cid:112) T dust = 2 . · · √ . · = 3 . · . (5)As a result we come to the equation of evolution of the radius of the Universein cm a ( t ) = (cid:26) . · · √ t, T born ≤ t ≤ T dust. , . · · √ t , T dust ≤ t ≤ T now (6)Substituting Planck time 10 − s as T born , we get the initial size of theUniverse is 10 − cm, and considering that the Planck length is c · T begin =3 · − cm, we find that at Planck times the universe ¡¡consisted¿¿ of 3 . · independent pieces. At the same time, according to the latest data, theanisotropy of the Universe appears only in the fourth sign of the accuracy ofthe temperature of the Universe, depending on the direction.The most widely accepted way to solve the problem of isotropy of the uni-verse is to quickly inflate. It at times 10 − s using the scalar field existing inthe early Universe in models of inflation. The history of this approach beginswith the works of Gus and is reflected in book [1].The reasons why such a scalar field existed at the beginning of the evolutionof the Universe are beyond the scope of these models.Nevertheless, the inflationary scenario of the early Universe today has a fairlygood theoretical basis, consistent with models of strong and weak interactions,2nd any revision of this approach should not contradict the consequences consis-tent with the modern theory of elementary particles.Thus, the main results of the Weinberg-Salam theory are related to momentshappenden after than 10 − s, so perhaps there is some freedom in creatingmodels of the early Universe before these times, although some questions arisedue to the beginning of the hadron era.Modern inflationary scenarios are based on model approaches to the evolu-tion of the early Universe. In them, geometry is described by GRG methods,and the matter that induces a geometric structure is described as a scalar field.The main inflationary scenarios in various models are obtained as self-consistentfield equations and geometries and can be divided into models of the first andsecond kind.1. The model of chaotic inflation . A primary inflaton field ϕ ( t ) [2] ispostulated, for which the Euler-Lagrange equation is obtained in curved co-ordinates as: ϕ (cid:48)(cid:48) + 3 Hϕ (cid:48) + dVdϕ = 0 (7)where H – Hubble constant. The inflaton field ϕ with potential V ( ϕ ) appearedin work [3] as external to geometry and should have existed at the beginning ofevolution.2. Starobinsky Model , which takes the Lagrangian f ( R ) [4] with an additionalterm R in addition to the renormalization of the gravitational constant, allowedus to introduce a primary inflaton field, which is already an internal field withrespect to geometry.Inflationary scenarios in type I and II models arise from the de Sitter vacuumsolution p = − ε of the Friedman Universe ρ (cid:48) a + 3( p + ε ) a a (cid:48) = 0 . (8)and ¡¡good¿¿ initial conditions.In this work, an inflation scenario is proposed, which is based on the principleof compacification of the dimensions of the original space. The proposal of thisprinciple is meant to apply it to a non-associative octonion algebra, which couldproduce insightful results. Efforts to quantize space time date back to works of Bronstein. However, itwas most successfully done by Wheeler and de Witt in their joint publicationin 1967. In this work, the authors proposed a model for quantization of theUniverse based on quantization of the space-time metric. It was shown forthe first time that taking into account the nonlinear structure of the Einsteinequations, on the basis of which the Hamiltonian of the quantum Universe wasconstructed, resulted in a loss of time. The Wheeler de Witt equation has thestructure of a quantum operator over all possible spatial Riemannian metrics in3he representation of the continuum integral [5] (cid:18) G abcd (cid:20) δδg ab (cid:21) (cid:20) δδg cd (cid:21) − √ g (3) R (cid:19) Ψ( g ab ) = 0 (9)Function of the state space of the Universe Ψ( g ab ) contains an explicit ref-erence to the metric of this space, thereby indicating the character of the earlyuniverse as a three-dimensional spatial manifold. In addition, it is obviouslypossible to proceed from the original space with a large number of dimensionsand the subsequent compactification of a part of the dimensions. In any case,the Wheeler de Witt equation points to the spatial character of quantum states.There is another important quantum approach to the Early Universe themethod of multiple interpretation of worlds in the Everett model [6]. In essence,this model goes back to Fock quantization, when the states of the Universe areinterpreted as eigenstates ξ Si of operators ˆ a, ˆ a + , defining the general state of thequantum system [7] Ψ S = (cid:88) i,j a ij ξ S i ξ S j (10)In contrast to the approach of Wheeler and de Witt, there is initially nodimension of space-time, and the states of the Early Universe are abstract,endowed with an additional space-time structure. On the other hand, in theFock approach states are characterized solely by a set of quantum variables.Take, for example, a set of various energy states and you will get the quantumenergy operator in full accordance with the Bohr interpretation of the atom.In this sense, the Aspecs trial is illustrative [8], which shows the quantumstate is a state in a certain state space. The peculiarity of such a space of statesis that, being spatially unrelated, they nevertheless form a single state that feelschanges in each of its parts even when it is separated in coordinate space. Thisfact has now been experimentally confirmed and has already been applied tothe model of quantum signal shifting.Due to existence of this kind of analogy, we can further apply Everett’s ap-proach to the Early Universe as a space of Fock states. That would allow usto consider the Early Universe in the spirit of the original n -dimensional space.This space in the Early Universe can only be the space of Bose particles as fields.This is indicated by the Weinberg-Salam model, according to which Fermi par-ticles are formed due to interaction with the Higgs fields [9] in its minimumpotential energy. Since in the Early Universe all energy is localized in a smallphysical space, there can be no minimums of energy. However, this is also in-dicated by the Landau model of a quantum liquid, which is where the analogywith the Higgs fields came from. In the Landau model, the bound states of aquantum liquid are destroyed when the temperature increases. Current temper-ature does not allow the formation of superconductors at normal temperature[10].The proposed approach to the Early Universe, according to which differentdimensions of space-time are possible, turns out to be fruitful in another impor-tant way. Thus we can view the reduction of spatial dimensions as an inflation4f space-time. This idea emerges if you try to imagine cubes of the same lin-ear size along one of the axes, but with a different number of cells that havethe same length for each of their dimensions. In this case, a decrease in thedimension of the space immediately leads to an increase in the effective lengthat which n -dimensional cubes are placed for the same number of k -dimensionalcubes as cells of possible spatial states of Bose particles in a new space with asmaller number of dimensions.It is assumed that the velocity at each coordinate in n -dimensional spaceis stored, but new spatial states are formed in k -dimensional space. Fig. 1illustrates the change in the effective length of the space on the example of thetransition from a two-dimensional space to a one-dimensional one. In this case,both the OX axis and the OY axis propagate a wave with the same speed c . Itis noteworthy that Fig. 1 describes the process of inflation. During this period,no new states are formed – so the number of states is constant. Let’s assume that at time T there is one state in the space R n that occupiesthe volume V = L n , L = cT = 1 · T . And allowt for each moment of time T for each dimensions of space R n , the number of states increase by one, that is,at time 2 T , the number of states is equal to 2 n , and the volume V ( t ) occupiedby these states at time t = 2 T is equal to V ( t = 2 T ) = (2 L ) n = 2 n L n , andso on.Let there be the maximum possible number of states in the space R n perone dimension R , equal to N . Then the maximum possible number of states inthe rn space is N n . Let’s look at the moment when the space R n is filled withall possible valid states, the states are reduced to the space R k ⊂ R n ( k < n ).We define the principle of reduction from the state of the space R n to thespace R k . Since R k ⊂ R n ( k < n ), the original principle of evolution-the emer-gence of new states must remain the same, that is, let one state appear in theone-dimensional space of space R n for a unit of time T .Then ( N + 1) n states can appear in an n dimensional space per unit oftime. Accordingly, the additive is equal to ( N + 1) n − N n . If N >> N k ∼ nN n − .Let this number of states of an n -dimensional space pass into states of a k -dimensional space.At the moment of time, N T in k -dimensional space became n k = N k states.In the next moment, N T + T of them will be n k = N k + ∆ N k ∼ N k + nN n − .Thus, if during a single period T in the space R n the number of states perdimension increased by one, then in the space R k the number of states willincrease by∆ n ∼ k (cid:113) N k + nN ( n − − N = N (cid:18) k (cid:113) N k + nN n − − k − (cid:19) . (11)5et n > k + 1, then ∆ n ∼ N (cid:18) k (cid:113) nN n − − k (cid:19) . (12)Thus, the rate of increase of the one-dimensional space in R k is equal to(12), which can be called the rate of reduction of the space R n in R k .If N is large, i. e. N >> n , then ∆ n k << N n (remember that n > k + 1),so all the states of the k -dimensional space will not be filled in one period. Since k -dimensional space is a subspace of n -dimensional space, the number of statesin n -dimensional space is preserved.Let’s find out the number of states that are formed in k -dimensional space inthe next unit of time, counting the rate of formation of states in n -dimensionalspace as the same: n k = N k + nN n − + nN n − . (13)Thus, for each time period, the number of new states in the R k space remainsunchanged. N n / ( nN n − ) = N /n. (14)The total number of such steps is equal to N n / ( nN n − ) = N /n. (15)And the total time to fill the space R k with all states will be T N /n .The volume V of the space R k at the moment of its full filling is equal to V end = ( n end L ) k = L k N n , and the linear size of the space is equal to L end = LN n/k The situation described in this model can be commented on as in figure 2.Figure 2 shows the starting situation, when all energy is concentrated in asingle volume L = cT , equal to the entire volume of the Universe at the initialtime T . Then, during T , the universe, expanding in a flat n-dimensional space,occupies the volume L n . At this point, the expansion in the entire n -dimensionalspace stops, since the minimum energy value has been reached. After this, theprocess of inflation begins, as shown in fig. 2 by an arc. Thus fig. 2 shows theone-stage process of evolution of the Universe. Consider the reduction of the space R n +1 at n = 7 to the space R k +1 -a physicalfour-dimensional space, when k = 3 - as cosmological inflation, and we assumethat R n +1 = T ⊗ R n and R k +1 = T ⊗ R k .Let’s consider the next stages of the evolution of the early Universe.1. The evolution of the Universe occurs when a single Bose state appearsat the moment T = 1 / ( N k ). Since the total energy of all states is equal to E all = N k (14), and the energy of one state is equal to N k , then N = N k and L = c · T . 6. Inflation occurs when all single states of the Universe are constructed inthe space R n : L b = c · T EUIbegin = c · N T = c · N / T .3. The size of the Universe before the first inflation is L b = c · T EUIbegin = c · N T = c · N / T .4. During inflation, for equal periods of time, the physical size of the Universeincreases by the same amount, so there is a linear law of growth of the size ofthe Universe.5. The duration of the first inflation is equal to T EUItime = N T /n = N / T /n .6. The first stage of inflation ends at time T EUIend = ( n + 1) N / T /n .7. The Linear size of the Universe at the end of the first inflationary periodof evolution is L = L EUI = L N / = L N .8. The Number of states of particles in the Universe at the end of the firstperiod of inflation becomes equal to N , and in each state there is one particle.After inflation, a new situation arises. The space size has increased and anew solution ( ?? ) appears with N = 1 in (14). The total number of states afterthe first stage of evolution is N = N . Since the particles are free, this space isagain M , and the number of states becomes 2 n now after T = N T = N / T .9. Repeating the previous arguments, we get that in the space R all thestates of N will be filled at the moment T EUIIbegin = N · T = N / T andthe second non-inflationary period of the evolution of the Universe will be over.10. By the end of the second non-inflationary period, the size of the Universewill increase to L EUII = c · T EUIIbegin = c · N / T .11. At the time T EUIIbegin , the second inflationary period, which will con-tinue in T EUIItime = N T /n = N / T /n .12. The linear size of the Universe to the end of the second inflation periodwill begin of evolution increased to L EUII = N L .13. The second period of inflation, will end at T EUIIend = ( n + 1) N / T /n . a ( T EUIIend ) = L EUII = N L = N L = cN T (16)or 87 a (cid:112) T N / = cN T (17)jnrelf N − / = 6449 c a T − (18)be fulfilled. a ( T EUIIend ) = L EUII = N L = N L = cN T (19)or 87 a (cid:112) T N / = cN T (20)then N − / = 6449 c a T − (21)7ubstituting T = 10 − s, c = 3 · cm/s and a from (16), we get N = 0 . · (22)For N = 10 , we find that inflation II ended in 8 . · − s, and the size ofthe Universe was 6 . · cm.From 1. - 13. it is not difficult to find that at the initial moment of T =10 − s, the linear size of the Universe is L = 3 · − cm, and the totalnumber of states with an energy of E = 10 k in each was N = 10 . At themoment of T EUIbegin = 1 . · − s, when the linear size of the Universe becomes4 . · − cm, the first inflation will occur with a duration of 0 . · − s, and thelinear size of the Universe will increase to 3 · − cm. The second inflation willstart at T EUIIbegin = 7 . · − s, last 10 − s, and end at T EUIIend = 8 . · − s, while the size of the Universe will increase from 3 · − cm to 6 . · cmduring the second stage.The linear size of the Universe and the beginning of the Friedman stage ofthe evolution of the Universe were well consistent with the classical model. Butthere is a serious contradiction to the proposed model with the total number ofparticles, if we consider the model N = 10 as the number of baryons, whilethe energy E = 10 k , which follows from the model, agrees very well withthe Planck mass. The situation can be corrected if we do not try to link theFriedman and algebraic stages of inflation, immediately putting N = 10 -anumber that reflects the current idea of the number of baryons in the Universe.In this case, the algebraic stage of evolution ends in T EUIIend = 3 . · − s. The size of the Friedman universe at this moment is 4 . · cm, and thealgebraic Universe is 4 . · cm. Reducing the initial time of 10 − to 10 − s allows you to stitch the Universe with a size of 1 . · cm at the end of theal-Hebraic inflation in 3 . · − s.From (18) we get N T = 6449 c a (23) We set the task to find the field equation in the early Universe.Let the state space of the early Universe be described by some algebra, whichwe define on some linear space. We define the dual space as a coordinate space.The space dual to the coordinate space is defined as a vector space. If we takethe physical space T × R as the coordinate space, we can, for example, use itsspinor representation [12].[ physical world ] −→ V (vector space) V −→ G = U (1) × SU (2) , V −→ V ∗ , V ∗ −→ V ∗∗ ∀ v ∈ V : v = v σ + v σ + v σ + v σ = (cid:18) v − v g + iv v − iv v + v (cid:19)
8s a result, we get: V ∗ – dual coordinate covector space of Minkowski and ∀ x ∈ V ∗ : x µ = x µ σ + x µ σ + x µ σ + x µ σ or x = cT, x = X, x = Y, x = Z, x = (cid:18) cT − Z X + iYX − iY cT + Z (cid:19) (24)As a result, we get a model that basically coincides with the usual quantumfield theory (the implementation of the corresponding Penrose program has anumber of important results and is currently developing).It is known that a group with a layer - by-layer transition can be realized ona V ∗∗ - vector space, A (cid:48) µ = g ( x ) A µ ( x ) g − ( x ) + ∂ µ g ( x ) g − ( x ) (25)where g ( x ) is a group element U (1) × SU (2) that matches vector fields on thealgebra u (1) × su (2) at different group points with the same coordinate x ∈ R .We research the vector field A µ ( x ) on the algebra u (1) × su (2) by [11]. Todo this, consider the isometry of P e (cid:72) l A µ dx µ = I + (cid:73) l A µ dx µ + P (cid:90) (cid:90) x >x A µ ( x ) dx µ A ν ( x ) dx ν + o ( l ) == I + (cid:73) ∂ Ω A µ dx µ + 12 (cid:73) Ω F µν dx µ ∧ dx ν + o ( l ) (26)For further research, we will write out the corresponding coordinate repre-sentations of the above equations. F µν = ∂ µ A ν − ∂ ν A µ − [ A µ , A ν ] (27)In the case of commutative fields on the group U (1), the tensor (27) has theform of a coordinate function F µν = ∂ µ A ν − ∂ ν A µ . On the group SU (2), thetensor F µν is written in layers: F aµν = ∂ µ A aν − ∂ ν A aµ − ([ A µ , A ν ]) a .The covariant derivative ∇ is formed by raising the index µ using the metrictensor g µν . Metric tensor g µν appears as a transition to arbitrary coordinates inthe original flat Minkowski space R on layer a , so the index of layer a remainsupper, and the coordinate index µ rises from the bottom to the top. As a result ∇ µ F µν = ∂ µ F µν − [ A µ , F µν ] = 0 (28)or ∂ µ ∂ µ A ν − ∂ µ ∂ ν A µ − [ A µ , ∂ µ A ν − ∂ ν A µ − [ A µ , A ν ]] − ∂ µ [ A µ , A ν ] = 0 (29)As follows from [11], only the physical part of (cid:73) A µ dx µ = (cid:73) ( A || µ + A ⊥ µ ) dx µ = (cid:73) A || µ dx µ (30)9an be left due to transformations (25).Therefore, we consider A aµ ( x ) = A || µ ( x ) = n aµ A ( x ) , x = k x − (cid:126)k(cid:126)x (31)with a normalized ( n aµ ) = 1 vector in M space. As a result, we come to theequation for transverse vector fields k µ A µ = 0 ∂ µ ∂ µ A ( x ) = 0 (32)Note that in (32), in accordance with [11], the Greek index µ numbers thereal numbers of the dx µ coordinates in R . the Group index carries a field, forexample, in the case of the group SU (2), we get F µν = F aµν σ a . In our case, thisis not the case for the simple reason that the coordinate space is constructed asdual to the vector space, so it turns out to be from the same algebra. In a sense,this is not new-Penrose’s twistor approach just means an attempt to constructa physics in which the coordinate space is an element of the SU (2) algebra.When finding the tensor F µν in (30), the commutativity and associativity ofthe field A µ and the coordinates dx µ are used, so they must be output in a newway.Mapping a space element with a basis e leads to a component record ofthe vector A = A µ e µ = A aµ Σ a e µ and, respectively, the covector x = x µ e µ = x µ ( b ) e µ Σ b .Therefore A · dx = A µ dx µ = A aµ dx µ ( b ) Σ a Σ b (33)Note that in (33) A aµ x µ ( b ) are real numbers, and Σ a Σ b are the basis elementsof the algebra.When passing from point x to point x (cid:48) = x + ∆ x , the vector V changes dueto the change of the coordinate and the element of the algebra δV ( x ) = V ,µ ∆ x µ − V A µ ( x )∆ x µ (34)Therefore, as shown in [11], we could come to (26). However, it is necessaryto take into account the non-associativity of the algebra. To do this, we calculate (cid:72) A µ dx µ to the second-order infinitesimal accuracy using (33) and taking thevector components of A aµ Σ a as a vector V in (34). (cid:73) A µ dx µ = A µ ( x )∆ x µ + A (cid:48) µ ( x + ∆ x )∆ x µ − A (cid:48)(cid:48) µ ( x + ∆ x )∆ x µ − A µ ( x )∆ x µ =+( A µ ( x ) − ( A µ ( x + ∆ x ) − A µ ( x ) A ν ( x )∆ x ν ))∆ x µ −− ( A µ ( x ) − ( A µ ( x + ∆ x ) − A µ ( x ) A ν ( x )∆ x ν ))∆ x µ == (( ∂ ν A µ + A µ A ν )∆ x ν ))∆ x µ − (( ∂ ν A µ + A µ A ν )∆ x ν ))∆ x µ == ( ∂ ν A µ + A µ ( x ) A ν ))(∆ x ν ∆ x µ − ∆ x ν ∆ x µ )++((( ∂ ν A µ + A µ A ν )∆ x ν ))∆ x µ − ( ∂ ν A µ + A µ A ν )(∆ x ν ∆ x µ )) − ((( ∂ ν A µ + A µ A ν )∆ x ν ))∆ x µ − ( ∂ ν A µ + A µ A ν (∆ x ν ∆ x µ ) == 12 ( ∂ ν A µ − ∂ µ A ν − [ A ν , A µ ])(∆ x ν ∆ x µ − ∆ x ν ∆ x µ ) + N A (35)
N A = ((( ∂ ν A µ + A µ A ν )∆ x ν )∆ x µ − ( ∂ ν A µ + A µ A ν )(∆ x ν ∆ x µ )) −− ((( ∂ ν A µ + A µ A ν )∆ x ν )∆ x µ − ( ∂ ν A µ + A µ A ν )(∆ x ν ∆ x µ )) (36)The first term in (36) corresponds to the usual stress tensor and has the form F µν ( dx µ ∧ dx ν ) of a non-Abelian associative field. Note that the parenthesesindicate the order of operations on the one hand and the two dual structureson the other hand. The term N A makes additional corrections due to the non-associativity of the vector and covector spaces. We separate the dual spatialpart from the vector fields in the term
N A by writing the expression in the base.
N A = 12 [((( ∂ ν A aµ − ∂ µ A aν )(∆ x ν ( b )1 ∆ x µ ( c )2 − ∆ x ν ( b )2 ∆ x µ ( c )1 ) · ((Σ a Σ b )Σ c − Σ a (Σ b Σ c )) + ( A aµ A bν − A aν A bµ )(∆ x ( c ) ν ∆ x ( d ) µ − ∆ x ( c ) ν ∆ x ( d ) µ ) · (((Σ a Σ b )Σ c )Σ d − (Σ a Σ b )(Σ c Σ d ))] == ( ε adck ( ∂ ν A aµ − ∂ µ A aν ) + ε abl ε lcdk ( A kµ A cν − A kν A cµ )) ·· (∆ x ( c ) ν ∆ x ( d ) µ − ∆ x ( c ) ν ∆ x ( d ) µ )Σ k where ε abcd – associator of basic elements:(Σ a Σ b )Σ c − Σ a (Σ b Σ c ) = 2 ε abcd Σ d in which there is no summation by repeating indexes. We will look for a solution in the space T ⊗ R n with metric ds = c dt − dx − . . . − dx n = dx − dx − . . . − dx n , (37)for the vector field A µ ( x ) of the algebra su ( n + 1) in the form [13] ∇ µ F µν = ∂ µ F µν − [ A µ , F µν ] = 0 (38)or ∂ µ ∂ µ A ν − ∂ µ ∂ ν A µ − [ A µ , ∂ µ A ν − ∂ ν A µ − [ A µ , A ν ] − ∂ µ [ A µ , A ν ] = 0 (39)We will search for wave homogeneous transverse solutions of the form A aµ = n aµ A ( x ) , n aµ n bµ = δ ab , n aµ k µ = 0 , x = k x − (cid:126)k(cid:126)x .11n the case of associative algebra, there is only a homogeneous solution ofthe form ∂ µ ∂ µ A ν = 0.But for nonassociative fields, a new term appears and the equation takes theform ∂ µ ∂ µ A ( x ) + A ( x ) = 0 (40)Let the conditions A (cid:48)(cid:48) ( x ) + A ( x ) = 0 (41)and A ( x = 0) = 1 , A (cid:48) ( x = 0) = 0 (42)or A ( x = 0) = 0 , A (cid:48) ( x = 0) = 1 / √ . (43)be set at the point (cid:126)x = 0.Equation (41) is converted to the form ∂ ( 12 ( ∂ A ( x )) + 14 A ( x ) − C ) = 0 . (44)The solution (44) for C = 1 / √ sn ( x | / √ cots ( x ) and sitn ( x ), respectively. Both functions cots ( x ) and sitn ( x ) are periodic with aperiod T sn = √ π Γ(5 / / ∼ . > π .In Fig. 4, the functions cots ( x ) and sitn ( x ) are combined with the trigono-metric functions cos( k x ) and sin( k x ), so it is almost certain that cots ( x ) =cos( k x ) and sitn ( x ) = sin( k x ) where k = 2 π/T sn ∼ . L of a scalar field a of the form A ( x ) L = 12 ∂ µ A ( x ) ∂ µ A ( x ) − A ( x ) (45)which corresponds to the Hamiltonian density H = ( ∂ A ) − L as H = 12 ∂ A ( x ) ∂ A ( x ) + 12 ( ∇ A ( x )) + 14 A . (46)As noted above, solutions (41 – 43) are close to trigonometric functions A tr ( x ) of the form cos( k x ) and sin( k x ), so replace (45) with L = 12 ∂ µ A tr ( x ) ∂ µ A tr ( x ) − m A tr ( x ) (47)and H = 12 ∂ A tr ( x ) ∂ A tr ( x ) + 12 ( ∇ A tr ( x )) + m A tr ( x ) . (48)the constant C in (48) based on (44) can be interpreted as the energy densityof the scalar field. 12he Lagrangian (47) can be quantized by assuming A tr as an operatorˆ A tr ( x ) = (cid:88) k e ikx ˆ a ( k ) + e − ikx ˆ a + ( k ) (49)for which the quantization condition is met in the pulse representation[ a ( k ) , a + ( k (cid:48) )] = δ kk (cid:48) . (50)For C = N/ √
2, the solution (41) is the function A ( x ) = N sitn ( N x ) (51)that is, the solution is a single mode, which, in accordance with Fig. 4, can becompared with single-mode harmonic oscillations with the total energy E = N k . (52)This is a solution to the main mode of the form A ( x ).Since the first instant of the Universe corresponds to the minimum time,which cannot be less than the time of natural vibrations of a physical particleand is equal to T = ( N k ) − , we have obtained that at the very first momentthe universe is concentrated in one state. This state contains N k ∼ N particles. The energy of each such particle is equal to N k . Thus, the totalenergy of the Universe in the first instant of the Universe is equal to (14).On the other hand, the original equation is the wave equation, so the solutionis the wave. Therefore, the physical linear size of the Universe at the firstmoment is equal to L = cT .In the next few moments, the size of the Universe will increase. However, byvirtue of the law of conservation (44), the solution (51) is preserved. The sizeof the Universe will increase, so the density of the number of particles per unitlength in the space R n will fall to a certain number N – when there is exactlyone particle in the space R in the same state. Previously, this linear space size R was calculated as L , and the corresponding evolutionary period was calledthe first stage.The first inflation phase is a transition from the solution of (51) the firstevolutionary phase the second evolutionary phase A ( x ) = sitn ( x ) (53)which at the point x = 0 is mapped to harmonic new massive scalar particle,by analogy with the first step A tr ( x ) = a k sin( k x ) , m = k (54)Note that now the energy is not N k , but just k .If the first scalar particle is understood as the Planck Higgs ( m = N k ),then the second scalar particle is an ordinary Higgs boson with ( m = k ) thesame evolutionary logic of the Early Universe.13e can assume that by analogy with the Higgs boson, Planck bosons ofHiggs particles form Planck electrons, Planck protons and Planck neutrons. Inthe future, Planck particles can form Planck atoms, which are a good candidatefor the dark matter of the Universe. In this regard, it is interesting to studythe possibility of the formation of Planck stars in our Universe and their furtherevolution.The second evolutionary stage of the Universe ends at the moment whenthere is only one particle with mass k in each state of space R . At this point,the second inflationary stage begins until there is exactly one particle in eachstate of the space R . In the future, the four-dimensional space-time manifoldevolves, and the metric becomes Riemannian. Back to (35). Let’s write a nonassociative quadratic part over a vector field. ε abl ε lcdk ( A aµ A bν − A aν A bµ ) · (∆ x ( c ) ν ∆ x ( d ) µ − ∆ x ( c ) ν ∆ x ( d ) µ )Σ k (55)By construction, we consider the nonassociative part corresponding to theCalley octave algebra, so we write non-zero values of the fully antisymmetricassociator ε lcdk and also the fully antisymmetric structural constants ε abc foroctonions ε = ε = ε = ε = ε = ε = ε = 1 , (56) ε = ε = ε = ε = ε = ε = ε = 1 (57)It follows from (56–57) that the associator is different from zero only if atleast two of its elements belong to the algebra of the higher fields (the higherfields have indexes from 4 to 7). If all indices belong to older fields, and theindexes of the structural constants in (56) must belong to the younger fields(lower field have an index from 1 to 3), which should not be, as younger fieldsmust be vector representations of the boson algebra su (2). Hence the index of l in (55) must belong to the younger field, but then the index k must be for theyounger fieldsIt was previously shown that there is a homogeneous solution for higherfields on a nonassociative algebra in the form of Jacobi functions (40), so wecan assume that ∆ x ( c ) νi = sitn ( x )∆˜ x ( c ) νi , i = 1 , x ( c ) νi it is already a constant value, and sitn ( x ) is a Jacobi function.Enter a new value R κλµν ( k )Σ k = ε ( κ +4)( λ +4) l ε l ( ν +4)( µ +4) k ( A κ +4 µ A λ +4 ν − A κ +4 ν A λ +4 µ )Σ k (59)Since the isotropic solution is considered, the introduced tensor is the samefor all indices k and is equal to just R κλµν = R κλµν ( k ).14rom the form (59), we get the properties R κλµν : R κλµν = − R λκµν = − R κλνµ (60)Omitting indexes using the metric tensor g µν of some space-time R , fromthe duality of vector and coordinate variables for an isotropic solution, we canadditionally assume that R κλµν = R µνκλ (61)Thus, a tensor with the properties of the Riemann tensor appears on anonassociative algebra. As you can see, the second term in (36), which containsderivatives of vector fields, does not have the necessary properties.Thus, the nonassociative term of the Wigner loop (35) on a nonassociativealgebra can be rewritten as (cid:73) A µ dx µ = sitn ( x ) R κλµν ( k ))∆ S νµ Σ k + . . . (62)where all indexes take values from 0 to 3.Since the loop rushes to the point, we enter the notation1˜ T sn (cid:90) ˜ T sn sitn ( N x ) dx = 1 N T sn (cid:90) T sn sitn ( x ) dx = 12 κ (63)where κ is the gravitational constant.Thus, the Riemann tensor is obtained on a four-dimensional space-time( µ, ν, κ, λ = 0 , , ,
3) with the desired gravitational constant, and thus thejustification for the Friedman stage appears (see Fig. 3) after T EUIIe earlyuniverse.
The solution for a vector field of the form (40) in an isotropic space was obtainednot only by introducing non-associativity of the interaction, but also by using theidea of Penrose [12] to represent coordinates on an external algebra. However,it turned out that associative algebra did not give a nonlinear interaction in ahomogeneous space. But the idea of the duality of vector and co-vector fields,according to which the algebra of field representation can also coincide, hasbecome fruitful.Solution (10) shows that in the early Universe, in addition to free masslesswave solutions, there is a solution that can be interpreted as a mass solution. Itturns out that two types of solutions are clearly distinguished as solutions withtwo masses: one solution can be understood as Planck particles and the secondsolution-hadron particles. Formally, intermediate states cannot be excludedand this issue requires further research. Planck’s massive solutions occur attimes of 10 − seconds. If these solutions are considered as Planck-type Higgsbosons, then formally such solutions can form part of a massive Universe ofapproximately equal mass to the visible part. This part of the Universe couldrepresent dark matter particles. 15 Acknowledgements
I am grateful to RFBR for financial support under grant No. 18-02-00461 ”Ro-tating black holes as the sources of particles with high energy”.