Unformatted text preview: 2) = b π 2 = 1 ⇒ b = 2 π so that this path is x ( t ) = 2 π t . For this path the action is given by S [ x ( t )] = 1 2 Z π/ 2 dt " ± 2 π ² 2± 2 π ² 2 t 2 # = 2 π 2 Z π/ 2 dt ³ 1t 2 ´ = 2 π 2 " π 21 3 ± 2 π ² 3 # = 1 π µ 1π 2 12 ! = 1 ππ 12 = 0 . 0565 . . . . 3. Assuming that the previous result is in J · s , we ﬁnd for the large number S/ ¯ h ≈ 5 . 37 × 10 32 ....
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 Fall '11
 berg
 mechanics, Work, Hamiltonian mechanics, Lagrangian mechanics, Noether's theorem, Boundary conditions, Lagrangian, Euler–Lagrange equation

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