Relativistic particle dynamics and deformed symmetry

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1 Relativistic particle dynamics and deformed Poincare symmetry Department for Theoretical Physics, Ivan Franko Lviv National University XXXIII Max Born Symposium, Wroclaw

2 Outline Lorentz-covariant deformed algebra Deformed Poincare symmetry Classical limit Action of relativistic particle Dirac constraint analysis Free particle motion

3 Lorentz-covariant deformed algebra We study Lorentz-covariant deformed algebra proposed in [1] Quesne,Tkachuk, J. Phys. A 39 (2006) [ ˆX µ, ˆP ν ] = i [(1 β ˆP ρ ˆP ρ )η µν β ˆP µ ˆP ν ]; [ˆP µ, ˆP ν ] = 0; [ ˆX µ, ˆX ν ] = i 2β β (2β + β )β ˆP ρ ˆP ρ 1 β ˆP ρ ˆP ρ (ˆP µ ˆX ν ˆP ν ˆX µ ), (1) leading to (isotropic) nonzero minimal uncertainty in the position X min = (Dβ + β )[1 β ( P ˆ0 ) 2 ]. (2) Here g µν = g µν = diag(1, 1, 1,..., 1) being metric tensor, β and β two small nonnegative parameters. D = 3, β = 0 - Snyder algebra [2] H. S. Snyder, Phys. Rev. 71 (1947) 38.

4 Deformed Poincare symmetry Lorentz transformation In paper [1] the transformation properties under deformed Poincare algebra was considered. Namely it was shown that infinitesimal proper Lorentz transformation X µ = X µ + δx µ, δx µ = δω µ ν X ν, (3) P µ = P µ + δp µ, δp µ = δω µ ν P ν. (4) with δω µ ν = δω ν µ R (5) leaves deformed algebra (1) undeformed.

5 Deformed Poincare symmetry Lorentz transformation Generators of such transformation, satisfying the properties are given by δx µ = i 2 δωαβ [L αβ, X µ ], (6) δp µ = i 2 δωαβ [L αβ, P µ ]. (7) L αβ = (1 βp ρ P ρ ) 1 (X α P β X β P α ). (8)

6 Deformed Poincare symmetry Translation transformations It was also discussed in [1] the infinitesimal translation transformations have to be deformed to X µ P µ = X µ + δx µ, δx µ = δa µ g(p ρ P ρ )δa ν P ν P µ, (9) = P µ + δp µ, δp µ = 0, (10) with g(p ρ P ρ ) = 2β β (2β + β )βp ρ P ρ (1 βp ρ P ρ ) 2. (11)

7 Deformed Poincare symmetry Translation transformations The generators of translation are Π α = (1 βp ρ P ρ ) 1 P α (12) and the following relations are satisfied: δx µ = i δaα [Π α, X µ ], δp µ = i δaα [Π α, P µ ].

8 Deformed Poincare symmetry Algebra of generators Importantly to note that algebra of deformed generators of Lorentz and translation transformation leaves undeformed: [L αβ, L ρσ ] = i (η αρ L βσ η ασ L βρ η βρ L ασ + η βσ L αρ ), [Π α, Π β ] = 0, [L αβ, Π ρ ] = i (η αρ Π β η βρ Π α ).

9 Classical limit In classical limit 1 i [Â, ˆB] {A, B} {X µ, P ν } = [(1 βp ρ P ρ )η µν β P µ P ν ]; {P µ, P ν } = 0; {X µ, X ν } = 2β β (2β + β )βp ρ P ρ 1 βp ρ P ρ (P µ X ν P ν X µ ).(13) The requirement of Poisson bracket s possession the same properties as the quantum mechanical commutator (namely, it must be anti-symmetric, bilinear, and satisfy the Leibniz rules and the Jacobi Identity) allow us to derive the general form of our Poisson bracket for any functions of coordinates and momenta as ( F G {F, G} = X µ P ν F ) G P ν X µ {X µ, P ν } + F G X µ X ν {X µ, X ν }.(14)

10 Action of relativistic particle Action of free relativistic particle of mass m is S = dτ[ ṗ µ x µ λ(p µ p µ m 2 c 2 )] (15) Let us build action which corresponds to free relativistic particle in deformed space (1). We suggest the action in the form S = dτ[ Π µ X µ λ(p µ P µ m 2 c 2 )], (16) and demand it to be invariant under deformed Poincare symmetry. Using (9), (10) the variation of the action yields δs = dτ Π µ δx µ = dτ Π µ (δa µ + g(p ρ P ρ )δa ν P ν P µ ) = dτ[ d g(p ρ P ρ ) dτ (Πµ δa µ ) δ{ 1 + g(p ρ P ρ )P ρ P ρ (X µ P µ )( Π ν P ν )}]. (17)

11 Action of relativistic particle It is obviously that to preserve the invariance of the action under the deformed Poincare transformation one have to add to Lagrange function the term in braces. We obtain the expression for the action S = dτ[ Π µ X µ + g(p ρ P ρ ) 1 + g(p ρ P ρ )P ρ P ρ (X µ P µ )(Ṗ ν P ν ) λ(p µ P µ m 2 c 2 )] And substituting the expresion for Π µ from ( 12), we finally write S = dτ [ Ṗµ X µ 1 βp ρ P ρ + f (P ρ P ρ )(X µ P µ )( Π ν P ν ) λ(p µ P µ m 2 c 2 ) ] (18) with β f (P ρ P ρ ) = (1 βp ρ P ρ )(1 (β + β )P ρ P ρ ). (19)

12 Action of relativistic particle It is interesting to note that the obtained expression of the action can be rewritten by S = dτ[ ṗ µ x µ λ(p µ P µ m 2 c 2 )], (20) with X, P and x, p connected by X µ = (1 βp ρ p ρ )x µ β (p ρ x ρ )p µ, P µ = p µ. (21) Quantities x, p satisfy usual Poisson brackets.

13 Dirac constraint analysis We interpret X and P in (18) as the configuration variables in an extended space. The canonical momentum conjugate to X, P and λ are Π X µ = L Ẋ µ = 0, Π P µ = L Ṗ µ = X µ 1 βp ρ P ρ + f (P ρp ρ )(X α P α )P µ, (22) Π λ µ = L = 0. λ µ The primary constraints which correspond to (22) are Φ = Π λ µ 0, Φ 1,µ = Π X µ 0, Φ 2,µ = Π P µ + X µ 1 βp ρ P ρ f (P ρp ρ )(X µ P µ )P µ 0.

14 Dirac constraint analys The Poisson algebra of the constraint is given by {Φ, Φ} = {Φ, Φ 1,µ } = {Φ, Φ 2,µ } = 0, {Φ 1,µ, Φ 1,ν } = 0, η µν {Φ 1,µ, Φ 2,ν } = 1 βp ρ P ρ f (P ρp ρ )P µ P µ, ( {Φ 2,µ, Φ 2,µ } = f (P ρ P ρ ) + 2β (1 βp ρ P ρ ) 2 ) (P µ X ν P ν X µ ). The second class constraint sector is next eliminated by using Dirac brackets. The constraint matrix is ( ) {Φ1,µ, Φ 1,ν } {Φ 1,µ, Φ 2,ν } Λ µν =. (23) {Φ 2,µ, Φ 1,ν } {Φ 2,µ, Φ 2,ν }

15 Dirac constraint analys Dirac bracket we find from {f, g} DB = {f, g} {f, Φ i,µ }Λ µν ij {Φ j,ν,, g} (24) Here matrix Λ µν is an inverse matrix to Λ µν such that Λ µν ij Λ jk,νρ = δ µ ik,ρ (25) and is given by Λ µν = 2β β (2β+β )βpρp ρ 1 βp ρp (P µ X ν P ν X µ ) (1 βp ρp ρ )η µν + β P µ P ν ] ρ (1 βp ρp ρ )η µν β P µ P ν 0.

16 Dirac constraint analys For our model the Dirac brackets among the configuration space variables are {X µ, P ν } DB = [(1 βp ρ P ρ )η µν β P µ P ν ]; {P µ, P ν } DB = 0; {X µ, X ν } DB = 2β β (2β + β )βp ρ P ρ 1 βp ρ P ρ (P µ X ν P ν X µ ). In the case of the сonsidered algebra leading to Snyder s one, our results coincide with those obtained in [3] R. Banerjee et. al. JHEP 05 (2006) 077

17 Free particle motion Assuming the Hamiltonian of the system to be H T = λ ( P µ P µ m 2 c 2) (26) we write Hamilton s equation Ẋ µ = {X µ, H T } DB = 2λP µ ( 1 ( β + β ) P µ ) P µ, (27) Ṗ µ = {P µ, H T } DB = 0. (28) We rewrite equation (27) in the form Ẋ 0 = 2λP 0 ( 1 ( β + β ) P µ ) P µ, (29) Ẋ i = 2λP i ( 1 ( β + β ) P µ ) P µ. (30) Here index i enumerates the spatial variables.

18 Free particle motion Division (30) by (29) yields dx i dx 0 = Pi P 0. (31) After integration of the latter equation we obtain an unspoiled equation of motion: X i X i 0 = Pi P 0 (X 0 X 0 0 ). (32) The reason of "virginity" of the equation of motion is hidden in the deformation itself, since space and time are deformed simultaneously leaving the velocity undeformed.

19 Thank you for your attention! Дякую за увагу!