Difference between revisions of "File:9+10.pdf"
Line 45: | Line 45: | ||
We are now prepared to calculate: | We are now prepared to calculate: | ||
− | $$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} | + | $$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} + \frac{1}{\Phi_{p Q}} \left( \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \) |
By the script: | By the script: |
Revision as of 16:46, 16 January 2015
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...
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:
$$ 1 = \Phi_{p Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{p Q}} $$
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 \( q \):
$$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} - \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$
using equation 2.
We are now prepared to calculate:
$$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} + \frac{1}{\Phi_{p Q}} \left( \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)
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 results in showing all the things we wanted to show.
Cheerio, A.
File history
Click on a date/time to view the file as it appeared at that time.
Date/Time | Dimensions | User | Comment | |
---|---|---|---|---|
current | 01:15, 17 June 2015 | (168 KB) | Paddy (Talk | contribs) | Cravens Solution to Problem 9 and 10 |
- You cannot overwrite this file.
File usage
The following page links to this file: