Editing Canonical transformation (section)

== Canonical transformation conditions ==

=== Canonical transformation relations ===
From: <math>K = H + \frac{\partial G}{\partial t} </math>, calculate <math display="inline">\frac{\partial (K-H)}{\partial P} </math>:

<math display="block">\begin{align}
\left( \frac{\partial (K-H)}{\partial P}\right)_{Q,P,t} &= \frac{\partial K}{\partial P} - \frac{\partial H}{\partial p}\frac{\partial p}{\partial P} - \frac{\partial H}{\partial q}\frac{\partial q}{\partial P} - \frac{\partial H}{\partial t}\left( \frac{\partial t}{\partial P}\right)_{Q,P,t} \\
&= \dot{Q} + \dot{p} \frac{\partial q}{\partial P}  - \dot{q}\frac{\partial p}{\partial P} \\
&= \frac{\partial Q}{\partial t} + \frac{\partial Q}{\partial q} \cdot \dot{q} + \frac{\partial Q}{\partial p} \cdot \dot{p} + \dot{p} \frac{\partial q}{\partial P}  - \dot{q}\frac{\partial p}{\partial P} \\
&=\dot{q}\left(\frac{\partial Q}{\partial q} - \frac{\partial p}{\partial P}\right)+\dot{p}\left(\frac{\partial q}{\partial P} +\frac{\partial Q}{\partial p} \right) + \frac{\partial Q}{\partial t} 
\end{align}</math>
Since the left hand side is <math display="inline">\frac{\partial (K-H)}{\partial P} = \frac \partial {\partial P}\left( \frac{\partial G}{\partial t} \right) \bigg |_{Q,P,t} </math>  which is independent of dynamics of the particles, equating coefficients of <math display="inline">\dot q </math> and <math display="inline">\dot p </math> to zero, canonical transformation rules are obtained. This step is equivalent to equating the left hand side as <math display="inline">\frac{\partial (K-H)}{\partial P} = \frac{\partial Q}{\partial t}   </math>.

Since the left hand side is <math display="inline">\frac{\partial (K-H)}{\partial P} = \frac \partial {\partial P}\left( \frac{\partial G}{\partial t} \right) \bigg |_{Q,P,t} </math>  which is independent of dynamics of the particles, equating coefficients of <math display="inline">\dot q </math> and <math display="inline">\dot p </math> to zero, canonical transformation rules are obtained. This step is equivalent to equating the left hand side as <math display="inline">\frac{\partial (K-H)}{\partial P} = \frac{\partial Q}{\partial t}   </math>.

Similarly:

<math display="block">\begin{align}
\left(\frac{\partial (K-H)}{\partial Q}\right)_{Q,P,t}
&= \frac{\partial K}{\partial Q} - \frac{\partial H}{\partial p}\frac{\partial p}{\partial Q} - \frac{\partial H}{\partial q}\frac{\partial q}{\partial Q} - \frac{\partial H}{\partial t}\left(\frac{\partial t}{\partial Q}\right)_{Q,P,t} \\
&= -\dot{P} + \dot{p} \frac{\partial q}{\partial Q}  - \dot{q}\frac{\partial p}{\partial Q} \\
&= -\frac{\partial P}{\partial t} -\frac{\partial P}{\partial q} \cdot \dot{q} - \frac{\partial P}{\partial p} \cdot \dot{p} + \dot{p} \frac{\partial q}{\partial Q}  - \dot{q}\frac{\partial p}{\partial Q} \\
&=-\left(\dot{q}\left(\frac{\partial P}{\partial q} + \frac{\partial p}{\partial Q}\right)+\dot{p}\left(\frac{\partial P}{\partial p} -\frac{\partial q}{\partial Q} \right) + \frac{\partial P}{\partial t} \right)
\end{align}        </math> 

Similarly the canonical transformation rules are obtained by equating the left hand side as <math display="inline">\frac{\partial (K-H)}{\partial Q} = - \frac{\partial P}{\partial t}   </math>. 

The above two relations can be combined in matrix form as: <math display="inline">J \left(\nabla_\varepsilon \frac{\partial G}{\partial t} \right) = \frac{\partial \varepsilon}{\partial t}  </math> (which will also retain same form for extended canonical transformation) where the result <math display="inline">\frac{\partial G}{\partial t} = K-H </math>, has been used. The canonical transformation relations are hence said to be equivalent to <math display="inline">J \left(\nabla_\varepsilon \frac{\partial G}{\partial t} \right) = \frac{\partial \varepsilon}{\partial t}  </math> in this context.


The canonical transformation relations can now be restated to include time dependance:

<math display="block">\begin{align}
\left( \frac{\partial Q_{m}}{\partial p_{n}}\right)_{\mathbf{q}, \mathbf{p},t} &= - \left( \frac{\partial q_{n}}{\partial P_{m}}\right)_{\mathbf{Q}, \mathbf{P},t} \\
\left( \frac{\partial Q_{m}}{\partial q_{n}}\right)_{\mathbf{q}, \mathbf{p},t} &= \left( \frac{\partial p_{n}}{\partial P_{m}}\right)_{\mathbf{Q}, \mathbf{P},t} 
\end{align} </math>

<math display="block">\begin{align}
\left( \frac{\partial P_{m}}{\partial p_{n}}\right)_{\mathbf{q}, \mathbf{p},t} &=  \left( \frac{\partial q_{n}}{\partial Q_{m}}\right)_{\mathbf{Q}, \mathbf{P},t} \\
\left( \frac{\partial P_{m}}{\partial q_{n}}\right)_{\mathbf{q}, \mathbf{p},t} &= -  \left( \frac{\partial p_{n}}{\partial Q_{m}}\right)_{\mathbf{Q}, \mathbf{P},t}
\end{align}</math>

Since <math display="inline">\frac{\partial (K-H)}{\partial P} = \frac{\partial Q}{\partial t}   </math> and <math display="inline">\frac{\partial (K-H)}{\partial Q} = - \frac{\partial P}{\partial t}   </math>, if {{math|'''Q'''}} and {{math|'''P'''}} do not explicitly depend on time, <math display="inline">K= H + \frac{\partial G}{\partial t}(t)</math> can be taken. The analysis of restricted canonical transformations is hence consistent with this generalization.

=== Symplectic condition ===
Applying transformation of co-ordinates formula for <math> \nabla_\eta H = M^T \nabla_\varepsilon H   
</math>, in Hamiltonian's equations gives:

<math display="block">\dot{\eta}=J\nabla_\eta H =J ( M^T \nabla_\varepsilon H)
</math>

Similarly for <math display="inline">\dot{\varepsilon}
</math>:      

<math display="block">\dot{\varepsilon}=M\dot{\eta} + \frac{\partial \varepsilon}{\partial t} =M J M^T \nabla_\varepsilon H + \frac{\partial \varepsilon}{\partial t}
</math>      

or:      

<math display="block">\dot{\varepsilon}=J \nabla_\varepsilon K  =  J \nabla_\varepsilon H + J \nabla_\varepsilon 
 \left( \frac{\partial G}{\partial t}\right)
</math>      

Where the last terms of each equation cancel due to <math display="inline">J \left(\nabla_\varepsilon \frac{\partial G}{\partial t} \right) = \frac{\partial \varepsilon}{\partial t}  </math> condition from canonical transformations. Hence leaving the symplectic relation: <math display="inline">M J M^T = J
</math> which is also equivalent with the condition <math display="inline">M^T J M = J
</math>. It follows from the above two equations that the symplectic condition implies the equation <math display="inline">J \left(\nabla_\varepsilon \frac{\partial G}{\partial t} \right) = \frac{\partial \varepsilon}{\partial t}  </math>, from which the indirect conditions can be recovered. Thus, symplectic conditions and indirect conditions can be said to be equivalent in the context of using generating functions.      

=== Invariance of the Poisson and Lagrange brackets ===
Since <math display="inline">\mathcal P_{ij}(\varepsilon) = \{ \varepsilon_i,\varepsilon_j\}_\eta =(M J M^T )_{ij} = J_{ij} 
</math> and <math display="inline">\mathcal L_{ij}(\eta) =[\eta_i,\eta_j]_\varepsilon=(M^T J M )_{ij} = J_{ij} 
</math> where the symplectic condition is used in the last equalities. Using <math display="inline">\{\varepsilon_i,\varepsilon_j\}_\varepsilon=[\eta_i,\eta_j]_\eta = J_{ij} 
</math>, the equalities <math display="inline">\{ \varepsilon_i,\varepsilon_j\}_\eta= \{ \varepsilon_i,\varepsilon_j\}_\varepsilon 
</math> and <math display="inline">[\eta_i,\eta_j]_\varepsilon= [\eta_i,\eta_j]_\eta 
</math> are obtained which imply the invariance of Poisson and Lagrange brackets.