Conditions for Transformation to be Canonical

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Theorem

Let:

$\ds J_1 \sqbrk {\sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {p_i}_{1 \mathop \le i \mathop \le n} } = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i'-H} \rd x$
$\ds J_2 \sqbrk {\sequence {Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n} } = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^*} \rd x$

be functionals.


Then $\paren {\sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H} \to \paren {\sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$ is a canonical transformation if:

$\ds \sum_{i \mathop = 1}^n p_i y_i' - H = \sum_{i \mathop = 1}^n P_i Y_i' - H^* \pm \dfrac {\d \Phi} {\d x}$

and:

$\ds p_i = \mp \frac {\partial \Phi} {\d y_i}, \quad P_i = \pm \frac {\partial \Phi} {\d Y_i}, \quad H = H^* \mp \frac {\partial \Phi} {\partial x}$


Proof

By Conditions for Integral Functionals to have same Euler's Equations, functionals:

$\ds \int_a^b F_1 \rd x$

and:

$\ds \int_a^b F_2 \rd x = \int_a^b \paren {F_1 \pm \frac {\d \Phi} {\d x} } \rd x$

have same Euler's equations.

Express the first one in canonical variables $\paren {x, \sequence{y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H}$ and the second one in $\paren {x, \sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$:

$\ds \int_a^b F_1 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i' - H} \rd x$
$\ds \int_a^b F_2 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i' - H^*} \rd x$

However:

$\ds \int_a^b \paren {F_2 - F_1} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x} \rd x$

Inserting new expressions for $F_1$ and $F_2$ yields:

$\ds \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^* - \sum_{i \mathop = 1}^n p_i y_i' + H} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x}\rd x$

This is satisfied, if integrands are equal.

Transform the coordinates to $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$ and write out the full derivative of $\Phi$:

$\ds \sum_{i \mathop = 1}^n P_i Y_i' - H^* - \sum_{i \mathop = 1}^n p_i y_i' + H = \pm \frac {\partial \Phi} {\partial x} \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} y_i' \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial Y_i} Y_i'$

Collect terms multiplied by the same the coordinates together:

$\ds \sum_{i \mathop = 1}^n Y_i' \paren {\pm \frac {\partial \Phi} {\partial Y_i} - P_i} + \sum_{i \mathop = 1}^n y_i' \paren {\pm \frac {\partial \Phi} {\partial y_i} + p_i} + \paren {\pm \frac {\partial \Phi} {\partial x} - H + H^*} = 0$

This has to hold for arbitrary values of independent coordinates $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$.

Hence:

$p_i = \mp \dfrac {\partial \Phi} {\d y_i}, \quad P_i = \pm \dfrac {\partial \Phi} {\d Y_i}, \quad H^* = H \mp \dfrac {\partial \Phi} {\partial x}$

$\blacksquare$


This needs considerable tedious hard slog to complete it.
In particular: Eventually cover all four generating functions, most probably in different pages
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Sources

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