Editing Canonical transformation (section)

=== Generators of dynamical symmetry transformations ===
Consider the transformation where the change of coordinates also depends on the generalized velocities.

<math display="block">\begin{align}
q^r\to q^r+\delta q^r\\
\delta q^r=\epsilon\phi^r(q,\dot{q},t)\\
\end{align}</math>

If the above is a dynamical symmetry, then the lagrangian changes by:

<math display="block">\delta L=\epsilon\frac d {dt}F(q,\dot q,t)</math>

and the new Lagrangian is said to be dynamically equivalent to the old Lagrangian as it ensures the resultant equations of motion being the same. The change in generalized velocity and momentum term can be derived as:

<math display="block">\begin{align}
p=\frac{\partial L}{\partial \dot q}, \quad& \dot q=\frac {dq}{dt}\\
\delta p_r=\frac{\partial^2L}{\partial q^s\partial\dot q^r}\delta q^s+\frac{\partial^2 L}{\partial \dot q^s\partial \dot q^r}\delta \dot q^s,\quad&\delta \dot q^r=\epsilon \frac{\partial \phi^r}{\partial q^s} \dot q^s+\epsilon \frac{\partial \phi^r}{\partial \dot q^s}\ddot q^s+\epsilon\frac{\partial\phi^r}{\partial t}
\\
\end{align}</math>

==== Generator of transformation ====
Using the change in Lagrangian property of a dynamical symmetry:

<math display="block">\frac d{dt}F=\frac{\partial F}{\partial q^r}\dot q^r+\frac{\partial F}{\partial \dot q^r}\ddot q^r+\frac{\partial F}{\partial t}=\frac{\delta L}{\epsilon}=\left(\frac{\partial L}{\partial q^r}\phi^r+\frac{\partial L}{\partial \dot q^r}\frac{\partial \phi^r}{\partial t}\right)+p_s\frac{\partial \phi^s}{\partial q^r}\dot q^r+p_s\frac{\partial \phi^s}{\partial \dot q^r}\ddot q^r</math>

Since the <math>\ddot q</math> terms appear only once in either side, it's coefficients must be equal for this to be true, giving the relation: <math display="inline">p_s\frac{\partial \phi^s}{\partial \dot q^r}=\frac{\partial F}{\partial \dot q^r} </math> using which, it can be shown that 

<math display="block"> \{q^r,\epsilon (p_s\phi^s-F)\}=\delta q^r,\quad \{p_r,\epsilon(p_s\phi^s-F)\}=\delta p_r+\epsilon\left(\frac{\partial L}{\partial q^s}-\frac{d}{dt}\frac{\partial L}{\partial \dot q^s}\right)\frac{\partial \phi^s}{\partial \dot q^r}</math>

Hence, the term <math>p\phi-F</math> generates the canonical dynamical symmetry transformation if either the Euler Lagrange relation gives zero, or if <math>\frac{\partial \phi_s}{\partial \dot q^r}=0\,\forall s,r</math> which is a infinitesimal point transformation. Note that in the point transformation condition, the quantity generates the transformation regardless of if the Euler Lagrange equations are satisfied and since they do not depend on the dynamics of the problem are said to be a purely kinematic relation.<ref>{{Cite journal |last=Mallesh |first=K. S. |last2=Chaturvedi |first2=Subhash |last3=Balakrishnan |first3=V. |last4=Simon |first4=R. |last5=Mukunda |first5=N. |date=2011-02-01 |title=Symmetries and conservation laws in classical and quantum mechanics |url=https://link.springer.com/article/10.1007/s12045-011-0020-5 |journal=Resonance |language=en |volume=16 |issue=2 |pages=129–151 |doi=10.1007/s12045-011-0020-5 |issn=0973-712X|url-access=subscription }}</ref>

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!Proof
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Firstly, the change in momentum can be expressed in a more useful form as follows:<math display="block">\delta p_r=\frac{\partial^2L}{\partial q^s\partial\dot q^r}\delta q^s+\frac{\partial^2 L}{\partial \dot q^s\partial \dot q^r}\delta \dot q^s=\frac{\partial}{\partial \dot q^r}\left(\frac{\partial L}{\partial q^s}\delta q^s+\frac{\partial L}{\partial \dot q^s}\delta \dot q^s\right)-\frac{\partial L}{\partial  q^s} \frac{\partial}{\partial \dot q^r}(\delta q^s)-\frac{\partial L}{\partial \dot q^s} \frac{\partial}{\partial \dot q^r}(\delta\dot  q^s)=\frac{\partial}{\partial \dot q^r}(\delta L)-p_s\frac{\partial}{\partial \dot q^r}(\delta \dot q^s)-\frac{\partial L}{\partial q^s}\frac{\partial}{\partial \dot q^r}(\delta q^s)</math>

Simplifying the required poisson brackets,

<math> \begin{align}
\{q^r,\epsilon (p_s\phi^s-F)\}=\epsilon \left(\phi_r+\frac{\partial \dot q^m}{\partial p_r}\cancelto{=0}{\left(p_s\frac{\partial \phi^s}{\partial \dot q^m}-\frac{\partial F}{\partial \dot q^m}\right)}\right)&=\delta q^r\\
\{p_r,\epsilon(p_s\phi^s-F)\}=\epsilon\left(-p_s\frac{\partial \phi^s}{\partial q^r}+\frac{\partial F}{\partial q^r}+\cancelto{=0}{\left(\frac{\partial F}{\partial \dot q^m}-p_s\frac{\partial \phi^s}{\partial \dot q^m}\right)}\left(\frac{\partial \dot q^m}{\partial q^r}\right)_{q,p,t}\right) &=\epsilon\left(-p_s\frac{\partial \phi^s}{\partial q^r}+\frac{\partial F}{\partial q^r}\right)\\
\end{align} </math>


As a preliminary result, for any function of <math>(q,\dot q,t)</math>,  

<math> \frac{\partial}{\partial \dot q^r}\frac{d}{dt}-\frac{d}{dt}\frac{\partial}{\partial \dot q^r}=\frac{\partial}{\partial q^r}+\frac{\partial \ddot q^s}{\partial \dot q^r}\frac{\partial}{\partial \dot q^s}</math>

which can be used to calculate the quantity:

<math> \frac{\partial}{\partial \dot q^r}\left(\frac {dF}{dt}\right)-p_s\left(\frac{\partial}{\partial \dot q^r}\left(\frac {d}{dt}\phi^s\right)\right)-\dot p_s\frac{\partial}{\partial \dot q^r}(\phi^s)=\frac{d}{dt}\cancel{\left(\frac{\partial}{\partial \dot q^r}F-p_s\frac{\partial}{\partial \dot q^r}\phi^s\right)}+\frac{\partial \ddot q^s}{\partial \dot q^r}\cancel{\left(\frac{\partial}{\partial \dot q^s}F-p_m\frac{\partial}{\partial \dot q^s}\phi^m\right)}-p_s\frac{\partial \phi^s}{\partial q^r}+\frac{\partial F}{\partial q^r}=\{p_r,(p\phi-F)\}</math>


This relation can be restated and combined with the formula for <math>\delta p_r</math>to give the required relation for momentum.

<math>\{ p_r,\epsilon(p_s\phi^s-F)\}=\frac{\partial}{\partial \dot q^r}(\delta L)-p_s\frac{\partial}{\partial \dot q^r}(\delta \dot q^s) - \dot p_s \frac{\partial}{\partial \dot q^r}(\delta q^s)=\delta p_r+\epsilon\left(\frac{\partial L}{\partial q^s}-\frac{d}{dt}\frac{\partial L}{\partial \dot q^s}\right)\frac{\partial \phi^s}{\partial \dot q^r} </math>

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==== Noether Invariant ====
Using Euler Lagrange relation for the provided Lagrangian, the invariants of motion can be derived as:<math display="block">\delta L-\epsilon\frac d {dt}F(q,\dot q,t)= \epsilon\phi\cancelto{=0}{\left(\frac{\partial}{\partial q}-\frac{d}{dt}\frac{\partial}{\partial \dot q}\right)L}+\epsilon\frac{d}{dt}\left(\phi\frac{\partial}{\partial \dot q}L- F\right)=\epsilon\frac{d}{dt}\left(\phi\frac{\partial}{\partial \dot q}L- F\right)=0</math>

Hence <math>\left(\phi\frac{\partial}{\partial \dot q}L-F\right)=p\phi-F</math> is a constant of motion. Hence, the derived Noether invariant also generates the same symmetry transformation as shown previously.