HW14_prob5_GPS_G39a

HW14_prob5_GPS_G39a - Goldstein 9-39a(3rd ed 9.39a(1 USING...

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Goldstein 9-39a (3 rd ed. 9.39a) (1) USING CARTESIAN COORDINATES Using Cartesian canonical variables, the Hamiltonian for the Kepler problem is H = p 2 2 m k r , (1) where p 2 = p 2 x + p 2 y + p 2 z , (2) and r = p x 2 + y 2 + z 2 . (3) The Laplace-Runge-Lenz vector is A = p × L mk r r . (4) Let x , y ,and z represent the Cartesian basis vectors. We know that L is conserved, and we are free to orient the coordinate system so that the z axis is coincident with L , L = L z z . (5) We also know that r and p lie in the orbital x - y plane, so we may conveniently set z =0 and p z =0 . It follows that p × L is a vector that also lies in the orbital plane. Thus, A has only x and y components, A = A x x + A y y . (6) Since we now have p = p x x + p y y , (7) and r = x x + y y , (8) it follows that A x = p y L z mk r x, (9) and A y = p x L z mk r y. (10) 1
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Now, because A has no explicit time dependence, to prove that A is a constant of the motion, we need to show only that [ A x ,H ]=0 , (11) and [ A y ,H ]=0 . (12) Actually, by taking advantage of the x - y symmetry of this problem, all we have to prove is either Eq.(11) or Eq.(12) because the other will then automatically be true. To see this, just consider what happens to A y ,de f nedinEq
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HW14_prob5_GPS_G39a - Goldstein 9-39a(3rd ed 9.39a(1 USING...

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