Difference between revisions of "File:9+10.pdf"

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(A sketch on how it might be done. Still to work out a bit.)
(Okay, put everything in order.)
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maybe it could be useful in solution II and III, that eq. (3) is zero...
 
maybe it could be useful in solution II and III, that eq. (3) is zero...
  
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'''Alternative solution:'''
  
 
Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):
 
Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):
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Holding \( p \) fixed and derive by \( q \):  
 
Holding \( p \) fixed and derive by \( q \):  
  
$$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$
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$$ ''Italic text''P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$
  
using equation 2.  
+
usin''Italic text''g equation 2.  
  
 
We are now prepared to calculate:  
 
We are now prepared to calculate:  
  
$$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} d \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}} \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)  
+
$$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}}   \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)  
  
( By the script:
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( ''My first idea would have been:''
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 +
By the script:
  
 
$$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$  
 
$$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$  
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and this should result in the things we want to show (by some symplectic argument or so...).)
 
and this should result in the things we want to show (by some symplectic argument or so...).)
  
Also I read here [http://www.scielo.org.mx/pdf/rmfe/v57n2/v57n2a4.pdf] that \( \{ f, g \}_{p, q} = \{ P, Q \} \{ f, g \}_{P, Q} \) and I think this might lead to the solution by plugging in everything we like for \( f, g \) since the poisson brackets will behave as we wish for \( P, Q \) if we consider them with respect to \( P, Q \). I read that this can be done using the chain rule, so I'm just gonna leave this here...
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''But it might be better to just show this (the property is surely right, found it in a classical mechanics textbook):''
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 +
 
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\( \vdash : \{ f, g \}_{p, q} = \{ P, Q \} \{ f, g \}_{P, Q} \)
 +
 
 +
''Proof''
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 +
With simplified notation:
 +
 
 +
\( \{ f, g \}_{p, q} = \sum_k \left( \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k} \right) = \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial p_k} \right) - \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial p_k} \right) \) \(\left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial q_k} \right) \)
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\( = \sum_k \left(  \frac{\partial Q}{\partial q_k} \frac{\partial P}{\partial p_k} -  \frac{\partial Q}{\partial p_k} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial f}{\partial Q} \frac{\partial g}{\partial P} - \frac{\partial f}{\partial P} \frac{\partial g}{\partial Q} \right) = \{ Q, P \}_{q, p} \cdot \{ f, g \}_{Q, P} \) \( \square \)
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 +
 
 +
 
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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.  
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Furthermore: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, P_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \delta_{i j} = \delta_{i j} \)
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 +
which solves the problem.  
  
 
Cheerio, A.
 
Cheerio, A.

Revision as of 10:28, 17 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...


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.

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