... (x0) 2 − (x1) 2 − (x 2 ) 2 − (x3) 2 = t 2 − r 2 = t 2 −r 2 1−r 2 2−r 2 3(3.1.46)or, for the in nitesimally small vector dxµ,(dxµ) 2 = dxµgµνdxν= (dt) 2 − (dr) 2 . (3.1.47) In ... coordinates Qr, Prthe Hamiltonian (3.1.6) takes the simpler formH =1 2 K /2 r=−K /2+ 1[PrP†r+ ω 2 rQrQ†r] (3.1. 12) ω 2 r≡ 2 2sin 2 rK+ 2 0. (3.1.13)Thus, in ... substitutions:qk−→q(x)√ak−→1aL0dx −→va(3.1 .21 )and Hamiltonian (3.1.1) takes the following form in the limitH =L0dx1 2 p 2 (x, t) +v 2 ∂q∂x 2 + 2 0q 2 (x, t). (3.1 .22 )Now the degrees of freedom...