f11smid3

# f11smid3 - t 1 dt X i ± ∂L ∂q i-d dt ∂L ∂ ˙ 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 ² δq i + " ∂L ∂ ˙ q i δq i # t 2 t 1 = Z t 2 t 1 dt X i ± ∂L ∂q i-d dt ∂L ∂ ˙ q i ² δq i . The last equality holds because of δq i ( t 1 ) = δq i ( t 2 ) = 0. As the variations are inde-pendent, the ﬁnally 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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## This note was uploaded on 12/11/2011 for the course PHY 4936 taught by Professor Berg during the Fall '11 term at FSU.

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