Goldstein_2_6_16_31

Goldstein_2_6_16_31 - Homework 10: # 9.2, 9.6, 9.16, 9.31...

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Homework 10: # 9.2, 9.6, 9.16, 9.31 Michael Good Nov 2, 2004 9.2 Show that the transformation for a system of one degree of freedom, Q = q cos α - p sin α P = q sin α + p cos α satisfies the symplectic condition for any value of the parameter α . Find a generating function for the transformation. What is the physical significance of the transformation for α = 0? For α = π/ 2? Does your generating function work for both of these cases? Answer: The symplectic condition is met if MJ ˜ M = J We can find M from ˙ ζ = M ˙ η which is ± ˙ Q ˙ P ² = ± cos α - sin α sin α cos α ²± ˙ q ˙ p ² We know J to be J = ± 0 1 - 1 0 ² Solving MJ ˜ M we get M ( J ˜ M ) = M ± - sin α cos α - cos α - sin α ² MJ ˜ M = ± cos α - sin α sin α cos α ²± - sin α cos α - cos α - sin α ² = ± 0 1 - 1 0 ² 1
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Therefore MJ ˜ M = J and the symplectic condition is met for this transformation. To find the generating function, I will first attempt an F 1 type and proceed to solve, and check at the end if there are problems with it. Rearranging to solve for p ( Q, q ) we have p = - Q sin α + q cos α sin α The related equation for F 1 is p = ∂F 1 ∂q Integrating for F 1 yields F 1 = - Qq sin α + q 2 cos α 2 sin α + g ( Q ) Solve the other one, that is P ( Q, q ), it along with its relevant equation is P = q sin α - Q cos α sin α + q cos 2 α sin α P = - ∂F 1 ∂Q Integrating - F 1 = qQ sin α - Q 2 2 cot α + qQ ( 1 sin α - sin α ) + h ( q ) - F 1 = - Q 2 2 cot α + qQ sin α + h ( q ) F 1 = Q 2 2 cot α - qQ sin α + h ( q ) Using both F 1 ’s we find F 1 = - Qq sin α + 1 2 ( q 2 + Q 2 ) cot α This has a problem. It blows up, sky high, when α = . But otherwise its ok, lets put the condition, α 6 = . If we solve for F 2 we may be able to find out what the generating function is, and have it work for the holes, α = . F 2 ( q, P, t )’s relevant equations are p = ∂F 2 ∂q 2
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p = P cos α - q sin α cos α F 2 = Pq cos α - q 2 2 tan α + f ( P ) and Q = ∂F 2 ∂P Q = q cos α - ( P - q sin α ) tan α F 2 = qP cos α - P 2 2 tan α + qP sin 2 α cos α + g ( q ) F 2 = qP (cos α + 1 cos α - cos α ) - P 2 2 tan α + g ( q ) F 2 = qP cos α - P 2 2 tan α + g ( q ) So therefore F 2 = - 1 2 ( q 2 + P 2 ) tan α + qP cos α This works for α = but blows sky high for α = ( n + 1 2 ) π . So I’ll put a con- dition on F 2 that α 6 = ( n + 1 2 ) π . The physical significance of this transformation
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Goldstein_2_6_16_31 - Homework 10: # 9.2, 9.6, 9.16, 9.31...

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