Quantum, Classical and Symmetry Aspects of Max Born Reciprocity L.M. Tomilchik B.I. Stepanov Institute of Physics, National Academy of Sciences of Belarus Minsk, 16–19 May 2017 1 / 36 Topics 1 Reciprocal Symmetry and Maximum Tension Principle 2 Maximum Force and Newton gravity 3 Extended Phase Space (QTPH) as a Basic Manifold 4 Complex Lorentz group with Real Metric as Group of Reciprocal Symmetry 5 One-Particle Quasi-Newtonian Reciprocal-Invariant Hamiltonian dynamics 6 Canonic Quantization: Dirac Oscillator as Model of Fermion with a Plank mass 2 / 36 Born Reciprocity version 1 Reciprocity Transformations (RT): xµ pµ pµ xµ ! ; ! − : qe pe pe qe 2 Lorentz and Reciprocal Invariant quadratic form: 1 1 S2 = xµx + pµp ; η = diagf1; −1; −1; −1g B 2 µ 2 µ µν qe pe 3 xµ, pµ are quantum-mechanical canonic operators: [xµ; pν ] = i~ηµν 4 Phenomenological parameter according definition pe = Mc so that qe = ~=Mc = Λc (Compton length): The mass M is a free model parameter. 5 Two dimensional constants qe[length], pe[momentum] connected by corellation: qepe = ~ (Plank constant): 3 / 36 Relativistic Oscillator Equation (ROE) Selfreciprocal-Invariant Quantum mechanical Equation: @2 − + ξµξ Ψ(ξ) = λ Ψ(ξ); µ µ B @ξ @ξµ µ µ where ξ = x =Λc. The (ROE) symmetry ≈ U(3; 1). There are solution corresponding linear discrete mass spectrum. 4 / 36 Maximum Tension Principle (MTP) The space-time and momentum variables are to be just the same dimentions of a quantity. It is necessary to have the universal constant with dimention: momentum/length, or equivalently: mass/time, energy/time, momentum/time. In reality: MTP [Gibbons, 2002] Gibbons Limit: dp c4 Maximum force Fmax = = = FG; dt max 4GN dE c5 Maximum power Pmax = = = PG: dt max 4GN [momentum]=[length] – Universal constant 3 pe c æ0 = = [LMT, 1974, 2003] qe 4GN −1 −2 It is evident: æ0 = c Fmax = c Pmax. All these values are essentially classic (does not contain Plank constant ~). 5 / 36 Born Reciprocity and Gibbons Limit Let the parameter a to be some fixed (nonzero!) value of action S. We replace initial Born’s relation qepe = ~ by πqepe = a without beforehand action discretness supposition. We demand the nonzero area Sfix = πqepe of the phase plane only. The second suggestion is: 3 pe c æ0 = = (universal constant). qe 4GN 6 / 36 Now the parameters qe, pe are defined separately r a raæ qe = ; pe = πæ π Evidently: the phenomenological constants Ee = pec [Energy] −1 and te = c qe [time] satisfy the conditions 5 Ee 2 c Eete = πa; = c æ = te 4GN Under Born’s parametrization 2MGN pe = Mc; qe = = rg(M) – gravitational radius. c2 General mass-action correlation ac M 2 = : 4πGN It is valid as in classical as in quantum case. Under supposition when amin = h we receive s 2 hc 1 2 ~c M = = MP l;MP l = – Plank mass. 4πGN 2 GN 7 / 36 Maximum Force and Modified Newton Gravity 4 mM c 2mGN 2MGN 1 (NG): f = G ≡ N 2 2 2 2 r 4GN c c r 2 rgRg r0 p = FG ≡ FG (r 0; r0 = rgRg) r2 r2 > r2 max (ModNG): f mod = F 0 (r r ); f mod = F N G 2 > 0 N G r r=r0 The corresponding gravitational potential Energy 2 mod r0 U = −FG gr r possess a minimum under r = r0 4 2 c 2GN p c p U mod(min) = −F r = − mM = − mM gr G 0 2 4GN c 2 We have dealing (instead of point-like masses) with a pair spacely extended simple massive gravitating objects like the elastic balls or drops with a high surface tension. 8 / 36 r0 M m Rg rg p 2 2 1 2 Erest(M; m) = Mc + mc − mMc 2 9 / 36 The whole energy such a “binary” system is as follows: 1p E(M; m) = Mc2 + mc2 − mMc2 2 1r µ = (M + m)c2 1 − ; 2 M + m where µ = mM=(m + M). Mc2; mc2; (M + m)c2 – rest energy 1 1p ∆E = pµ(M + m)c2 = mMc2 – energy corresponding 2 2 1 p the “mass defect” ∆m = 2 mM 10 / 36 “Gravitational fusion” picture [eliminated energy] ∆E λ = = [rest energy] (M + m)c2 p p mM 1 δ = = ; 2(M + m) 2 1 + δ where m Mmin δ = ; δ0 6 δ 6 1; δ0 = M Mmax ∆Eeliminated 1 Maximum λmax = = Erest 4 under δ = 1 (m = M) Such a “fusion” of two equal rest mass M lead to mass = 3M=2, the Energy ∆E = Mc2=2 eliminate (3 : 1). 11 / 36 Extended Phase Space as a Basic Manifold Space of states (QTP) and Energy (H) (sec [J. Synge, 1962]) in general: 2 + 2N dimensions. The metric is N z }| { ηAB = diagf1; −1; −1;:::; −1g (X; Y ) – pair of conjugated pseudoeuclidean vectors XA, YA, A; B = 0; 1;:::;N (N + 1 dimention). The Poisson brackets for '(X; Y ), f(X; Y ), N X @' @f @' @f f'; fg = − A;B @Xk @Y @Y k @X k=0 k k so that fXA;YAgA;B = ηAB 12 / 36 Dynamic System in (QT; P H) is defined, when the Energy Surface( 2N + 1 dimention) Ω(X; Y ) = 0 is defined. Solving this equation on Y0 we obtain Energy equation Y0 = F (Xk;Yl;X0) k; l = 1; 2;:::;N The corresponding Hamilton function of the System is by def: H ≡ Y0 = H (Xk;Yl;Y0) We shall consider (2 + 2 · 3) – dimensional version (QT; P H) A; B ! µ, ν = 0; 1; 2; 3; ηµν = diagf1; −1; −1; −1g; Xµ, Xµ are real 4-dimensional pseudoeuclidean SO(3; 1) – vectors. 13 / 36 Complexification of (QT; P H) Let us introduce the complex 4-vector Zµ according definition: µ µ µ Z = X + iY ; ηµν = diagf1; −1; −1; −1g and define the real norm µ ∗ µ µ∗ Z Zµ = Z ηµν Z 2 2 2 2 = jZ0j − jZ1j − jZ2j − jZ3j = inv 14 / 36 Complex Lorentz Group with a real metric (Barut Group [Barut 1964]) The group of 4 × 4 complex matrices Λ of transformations Z0 = ΛZ; Z = X + iY satisfying the condition ΛηΛy = η; (y – hermitien conjugation) Notice: the diagonal matrices ΛR and Λ¯ R 2 (BG) ΛR = −iI4 = −idiagf1; 1; 1; 1g and Λ¯ R = −iηµν = −idiagf1; −1; −1; −1g Metric invariant (no positive-defined) µ ∗ 2 2 2 2 Z Zµ = jZ0j − jZ1j − jZ2j − jZ3j = inv 2 2 Three possibilities: jZj ? 0, jZj = 0 – isotropic case 15 / 36 SU(3; 1) – subgroup structure (QTPH) SOv 3, 1 QT = fr; ctg = xµ H L Lk PH = p; E=c = pµ β~ = ~v=c C v C SO 3 U U * SU 3 center 4 kl lk (PTQH) H L H L H L µ PT = p;F0t = Π Mk µ QH = fr; E=F0g = Ξ SO f 3, 1 β~f = f=F~ 0 H L F0 = Fmax k; l = 1; 2; 3 16 / 36 2 Reciprocal Invariant SB as BG metric invariant µ ∗ Z Zµ By def: Zµ = Xµ + iYµ 1 1 1 1 Xµ = xµ = fx0; xkg;Yµ = pµ = fp0; pkg: qe qe pe pe Then: 3 ! 3 ! 1 X 1 X S2 = x2 − x2 + p2 − p2 = B q2 0 k p2 0 k e k=1 e k=1 2 2 3 2 2 3 x0 p0 X xk pk 2 X 2 µ ∗ = + − + = jZ0j − jZkj = Z Z q2 p2 q2 p2 µ e e k=1 e e k=1 Let xµ, pµ to be 4-vectors under SOv(3; 1)-transformations 17 / 36 Notice! There exist alternative: Z¯µ = X¯µ + iY¯µ where def x0 pk def p0 xk X¯µ = Ξµ = ; ; Y¯µ = Πµ = ; : qe pe pe qe It is evident: 2 3 2 2 3 2 2 µ µ x0 X pk p0 X xk jZ¯µj = Ξ Ξµ + Π Πµ = − + − = q2 p2 p2 q2 e k=1 e e k=1 e 1 1 = xµx + pµp = jZ j2: 2 µ 2 µ µ qe pe The natural supposition: Ξµ and Πµ are 4-vectors under SOf (3; 1)-transformations 18 / 36 2 Selfreciprocal Case: SB = 0 2 µ ∗ Quadratic form SB = Z Zµ is not positive defined µ ∗ µ ∗ Three possibilities Z Zµ = 0, Z Zµ ? 0. Especial interesting is: 1 1 S2 = pµp + xµx = 0 B 2 µ 2 µ pe qe Two possibilities 1 µ µ p pµ = x xµ = 0 (light cone, massless particles, standart clock synchronisation) 2 µ 2 µ 2 x xµ = −qe ; p pµ = pe; 2meGN qe = rg(me) = ; pe = mec: c2 19 / 36 BG: some kinematic outcome (1) Reciprocal-Invariant time TR 3 −1 pe c zµ = xµ + iæ0 pµ; æ0 = = qe 4GN def −1q µ ∗ dTR = c z zµ q −1 2 2 −2 2 2 = c dx0 − dr + æ0 (dp0 − dp ) q = dt 1 − β2 − β2 + β2β2 cos2 α v f v f r 2 = dt 1 − β2 1 − β2 − β × β : v f f v where r_ p_ c4 d β = ; β = ;F = ; (·) = ; v f G c FG 4GN dt t – is the laboratory time (comoving observer time). α – is the angle between r_ and p_ . 20 / 36 (2) In the comoving RF : q dT (β = 0) = dt 1 − β2 R v v There must exist an extra time retardation. (3) cos α = ±1; (β × β ) = 0 f v q q dT k = dt 1 − β2 1 − β2 R v f (4) cos α = 0; (β × β )2 = β2 · β2 f v f v q dT ? = dt 1 − β2 − β2 6= dT k R v f R p (5) cos α = ± 2=2, k ? dTR = dTR (6) The upper limit of jr_j and jp_ j in the expression for dTR: p p jr_jmax = c= 2; jp_ jmax = FG= 2 21 / 36 Complex Lorentz Boost Transformations Let us represent a BG-transformations using the following parametrization: ΓΓ · β ! ∗ y Λ(β) = ∗ β·β ; Λ(β)ηΛ (β) = η Γ · β I3 + jβj2 (Γ − 1) where β = β + iβ = v=c + iF =F , v F G v – is a const velocity 4 F – is a const force, FG = c =4GN – max force −1 ∗ Γ = detΛ (β);I3 = diagf1; 1; 1g; β·β – 3×3 tensor-diada: Q0 + iP 0 Q + iP Z0 = Λ(β)Z : 0 0 = Λ(β) 0 0 Q0 + iP 0 Q + iP 22 / 36 The simplest case β = fβ ; 0; 0g, β = fβ ; 0; 0g v v F F 1 β Λ(β) = Γ(β) ; β = β + iβ ; β∗ 1 v F 1 v F Γ(β) = q ; βv = ; βF = ; 1 − jβj2 c FG F – is the constant force uniformly accelerated (noninertial) FR.