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Recall how we saw in class that if we add a total time derivative of a function g(g; t) on configuration space to the Lagrangian then the equations of motion were unaffected, i.e. the new Lagrangian gives the same Euler-Lagrange equations as the original Lagrangian L. The reason this worked was that the action only shifted by a boundary term, which doesnt affect the extremization procedure We have also seen that if q is a cyclic coordinate, meaning aL/aq0, then the corresponding generalized momenta, p L/aa, is the constant of the motion associated to the symmetry q-s For L having such a symmetry, consider L as above in the case that g does depend on q. Show that qs is not a symmetry of L, but that the action 6 t1 only shifts by a function of the endpoints. Use the second version of Noethers theorem to find the corresponding constant of the motion, and show that it agrees with the original pi

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