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Quantum, Classical and Symmetry Aspects of Max Born Reciprocity

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Quantum, Classical and Symmetry Aspects of Max Born Reciprocity 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.
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