f11smid3

f11smid3 - t 1 dt X i L q i-d dt L q i q i + "...

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SOLUTIONS MIDTERM PHY 4936 (Fall 2011) PROBLEM 3 3.1 Principle of least action: Every mechanical system is characterized by a definite function L = L ( q 1 , . . . , q s , ˙ q 1 , . . . , ˙ q s , t ) and the motion of the system is such that the system moves between two fixed positions at different times t 1 and t 2 in a way that for sufficiently short time differences the integral S = Z t 2 t 1 L dt takes the least possible value. 3.2 Deviation of the Euler-Lagrange equations from the least action principle in gen- eral coordinates q i , ˙ q i , i = 1 , . . . s : 0 = δ Z t 2 t 1 dt L ( q i , ˙ q i , t ) = Z t 2 t 1 dt { L ( q i + δq i , ˙ q i + δ ˙ q i , t ) - L ( q i , ˙ q i , t ) } = Z t 2 t 1 dt ± L ( q i , ˙ q i , t ) + X i ∂L ∂q i δq i + X i ∂L ˙ q i δ ˙ q i - L ( q i , ˙ q i , t ) ² = Z t 2 t 1 dt ± X i ∂L ∂q i δq i + X i ∂L ˙ q i d dt δq i ² = Z t 2
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Unformatted text preview: t 1 dt X i L q i-d dt L q i q i + " L q i q i # t 2 t 1 = Z t 2 t 1 dt X i L q i-d dt L q i q i . The last equality holds because of q i ( t 1 ) = q i ( t 2 ) = 0. As the variations are inde-pendent, the nally obtained relation is equivalent to L q i-d dt L q i = 0 for i = 1 , . . . , s . 3.3 L ( q k + k , q k , t ) = L ( q k , q k , t ) implies 0 = L q k = d dt L q k p k = L q k conserved (generalized momentum) ....
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