100 Years of Relativity Space-Time Structure: Einstein and Beyond ...

430 R. Gambini and J. Pullinmotion and the only remaining constraint (Hamiltonian) are˙ A = N(Λ + m 2 φ 2 )sgn(E)E 2 (38)Ė = 2NE 2 A (39)˙φ = 0 (40)˙π = −2N|E| 3 m 2 φ (41)A 2 = (Λ + m 2 φ 2 )|E| (42)It immediately follows from the large mass approximation that φ =constant. To solve for the rest of the variables, we need to distinguish fourcases, depending on the signs of E and A. Let us call ɛ = sgn(E) andχ = sgn(A). Then the solution (with the choice of lapse α = 1) is,A = χ exp(χɛt √ )Λ + m 2 φ 2 (43)exp(2χɛt √ )Λ + m 2 φ 2E = ɛΛ + m 2 φ 2 (44)There are four possibilities according to the signs ɛ, χ. If ɛ = χ = 1 orɛ = χ = −1 we have a universe that expands. If both have different signs,the universe contracts. This just reflects that the Lagrangian is invariantif one changes the sign of either A or E and the sign of time. It is alsoinvariant if one changes simultaneously the sign of both A and E.Let us turn to the observables of the theory (quantities that have vanishingPoisson brackets with the constraint (42) and therefore are constants ofthe motion). The theory has four phase space degrees of freedom with oneconstraint, therefore there should be two independent observables. Immediatelyone can construct an observable O 1 = φ, since the latter is conserveddue to the large mass approximation. To construct the second observablewe write the equation for the trajectory,dπdA = −2Em2 φΛ + m 2 φ 2 = − 2A2 m 2 φ(Λ + m 2 φ 2 2sgnE (45))where in the latter identity we have used the constraint. Integrating, we getthe observable,O 2 = π + 2 m 2 φ3 (Λ + m 2 φ 2 ) 2 A3 sgnE (46)