Ivancevic_Applied-Diff-Geom

1170 Applied Differential Geometry: A Modern IntroductionWe denoteẋ µ ≡ dxµdσ 0,so that−→ v = ˙−→ xdσ 0dt ,andS = −m∫ √ẋµẋµ dσ 0 .It is the p = 0 case of (6.284) in which we put m (particle’s mass) insteadof γ. The vector of canonical momentum isp µ ≡ ∂L ẋµ= −m√ẋ2 ∂ẋµ . (6.285)The obtained theory is invariant under reparametrization group σ 0 →˜σ 0 = f(σ 0 )), hence its Hamiltonian is zero and by squaring the equation(6.285) one gets a constraint:p 2 − m 2 = 0. (6.286)Note that the information about the sign of p 0 is lost. On the other handone can prove that any constraint has to be linear in un–physical momenta[Golovnev and Prokhorov (2005)]. One can get the solution in the followingway.From the constraint (6.286) one finds√p 0 = ± m 2 + −→ p 2 ,and it is obvious from (6.285) that sgnp 0 = −sgnẋ 0 . Combining these factswe havep 0 + E p sgn(ẋ 0 ) = 0, (6.287)with E p ( −→ √p ) = m 2 + −→ p 2 .The Hamiltonian is zero and the total Dirac Hamiltonian is [Dirac (1982)]H T = v ( p 0 + E p sgn(ẋ 0 ) ) .Here v is the Lagrange multiplier. Strictly speaking (6.287) is not a constraintbecause it contains velocity (and H T is not a Hamiltonian due tothe same reason). But it depends only upon the sign of ẋ, and this factallows us to formulate the quantum theory.We fix the σ 0 ‘time arrow’ by condition∂x 0∂σ 0> 0, (6.288)