Editing Canonical transformation (section)

=== 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.