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| − | Could you justify at the end of solution II and solution III the chain rule? Why are the derivative of ''q_i'' with respect to ''Q_k'' zero?
| + | Cravens Solution to Problem 9 and 10 |
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| − | Hallo, why is the derivate of P_s with respect to Q_j equal zero? you do this in the last step of solution III.
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| − | If this is true, then your equation (3), derivate of P_i w.r.t. p_k, would be equal zero too.
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| − | maybe it could be useful in solution II and III, that eq. (3) is zero...
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| − | Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):
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| − | '''Claim:''' For a generating function of type one (that's the smartman's name for the Phi in our problem) it holds: \( \{ Q, P \} = 1 \).
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| − | ''Proof:''
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| − | We first fix the \( q \) coordinates in the momentum equation and derive by \( p \), denoted: \( \Phi_p |_q\).
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| − | So we get:
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| − | '''Equation 1:'''
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| − | $$ 1 = \Phi_{q Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{q Q}} $$
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| − | '''Equation 2:'''
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| − | Holding \( p \) fixed and derive by \( q \):
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| − | $$ 0 = \frac{\partial p}{\partial q} = \Phi_{qq} + \Phi_{qQ}Q_{q}|_p \Rightarrow Q_{q}|_p = - \frac{\Phi_{qq}}{\Phi_{qQ}} $$
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| − | '''Equation 3:'''
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| − | Holding \( q \) fixed and derive by \( p \):
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| − | $$ P_p |_q = - \Phi_{QQ}Q_q |_p \Rightarrow P_p |_q = - \frac{\Phi_{QQ}}{\Phi_{qQ}} $$
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| − | where we used equation 1 again.
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| − | '''Equation 4:'''
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| − | Holding \( p \) fixed and derive by \( q \):
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| − | $$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$
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| − | using equation 2.
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| − | We are now prepared to calculate:
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| − | $$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} d \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}} d \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)
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| − | By the script:
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| − | $$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$
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| − | with \( \nabla = \left( \frac{\partial}{\partial q_1}, ..., \frac{\partial}{\partial p_n} \right) \) and
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| − | $$ J = \begin{bmatrix}
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| − | 0 & I_n \\
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| − | -I_n & 0 \\
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| − | \end{bmatrix} $$
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| − | and this should result in the things we want to show (by some symplectic argument or so...).
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| − | Cheerio, A.
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