File:9+10.pdf

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?

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. If this is true, then your equation (3), derivate of P_i w.r.t. p_k, would be equal zero too. maybe it could be useful in solution II and III, that eq. (3) is zero...

Alternative solution:

Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):

Claim: For a generating function of type one (that's the smartman's name for the Phi in our problem) it holds: \( \{ Q, P \} = 1 \).

Proof:

We first fix the \( q \) coordinates in the momentum equation and derive by \( p \), denoted: \( \Phi_p |_q\).

So we get:

Equation 1:

$$ 1 = \Phi_{q Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{q Q}} $$

Equation 2:

Holding \( p \) fixed and derive by \( q \):

$$ 0 = \frac{\partial p}{\partial q} = \Phi_{qq} + \Phi_{qQ}Q_{q}|_p \Rightarrow Q_{q}|_p = - \frac{\Phi_{qq}}{\Phi_{qQ}} $$

Equation 3:

Holding \( q \) fixed and derive by \( p \):

$$ P_p |_q = - \Phi_{QQ}Q_q |_p \Rightarrow P_p |_q = - \frac{\Phi_{QQ}}{\Phi_{qQ}} $$

where we used equation 1 again.

Equation 4:

Holding \( p \) fixed and derive by \( q \):

$$ Italic textP_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$

usinItalic textg equation 2.

We are now prepared to calculate:

$$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}} \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)

( My first idea would have been:

By the script:

$$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$

with \( \nabla = \left( \frac{\partial}{\partial q_1}, ..., \frac{\partial}{\partial p_n} \right) \) and

$$ J = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \\ \end{bmatrix} $$

and this should result in the things we want to show (by some symplectic argument or so...).)

But it might be better to just show this (the property is surely right, found it in a classical mechanics textbook):

\( \vdash : \{ f, g \}_{p, q} = \{ P, Q \} \{ f, g \}_{P, Q} \)

Proof

With simplified notation:

\( \{ f, g \}_{p, q} = \sum_k \left( \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k} \right) = \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial p_k} \right) - \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial p_k} \right) \) \(\left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial q_k} \right) \)

\( = \sum_k \left( \frac{\partial Q}{\partial q_k} \frac{\partial P}{\partial p_k} - \frac{\partial Q}{\partial p_k} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial f}{\partial Q} \frac{\partial g}{\partial P} - \frac{\partial f}{\partial P} \frac{\partial g}{\partial Q} \right) = \{ Q, P \}_{q, p} \cdot \{ f, g \}_{Q, P} \) \( \square \)

Now compute: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, Q_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \{ P_i, P_j \}_{Q, P} = 0 \) which follows from the properties of the Poisson-bracket.

Furthermore: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, P_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \delta_{i j} = \delta_{i j} \)

which solves the problem.

Cheerio, A.