Conditions for Transformation to be Canonical

Theorem
Let


 * $\displaystyle J_1 \left[{\left\langle{y_i}\right\rangle_{1 \mathop \le i \mathop \le n}, \left\langle{p_i}\right\rangle_{1 \mathop \le i \mathop \le n} }\right] = \int_a^b \left({\sum_{i \mathop = 1}^n p_i y_i'-H}\right) \rd x$


 * $\displaystyle J_2 \left[{\left\langle{Y_i}\right\rangle_{1 \mathop \le i \mathop \le n}, \left\langle{P_i}\right\rangle_{1 \mathop \le i \mathop \le n} }\right] = \int_a^b \left({ \sum_{i \mathop = 1}^n P_i Y_i'-H^*}\right) \rd x$

be functionals.

Then $ \left({ \langle y_i \rangle_{1 \le i \le n}, \langle p_i \rangle_{1 \le i \le n }, H } \right) \to \left({  \langle Y_i \rangle_{1 \le i \le n}, \langle P_i \rangle_{1 \le i \le n }, H^* } \right)$ is a canonical transformation if:


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

and:


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

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


 * $\displaystyle \int_a^b F_1 \rd x$

and
 * $\displaystyle \int_a^b F_2 \rd x = \int_a^b \left({ F_1 \pm \frac {\rd \Phi } { \rd x} } \right) \rd x$

have same Euler's equations.

Express the first one in canonical variables $ \left({ x, \langle y_i \rangle_{1 \le i \le n}, \langle p_i \rangle_{1 \le i \le n }, H } \right) $ and the second one in $ \left({ x, \langle Y_i \rangle_{1 \le i \le n}, \langle P_i \rangle_{1 \le i \le n }, H^* } \right) $:


 * $\displaystyle \int_a^b F_1 \rd x = \int_a^b \left({ \sum_{i \mathop = 1}^n p_i y_i' - H} \right) \rd x$
 * $\displaystyle \int_a^b F_2 \rd x = \int_a^b \left({ \sum_{i \mathop = 1}^n P_i Y_i' - H^*} \right) \rd x$

However,


 * $\displaystyle \int_a^b \left({ F_2-F_1 } \right) \rd x = \int_a^b \pm \frac{ \rd \Phi}{ \rd x} \rd x$

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


 * $\displaystyle \int_a^b \left({ \sum_{i=1}^n P_i Y_i'-H^* - \sum_{i=1}^n p_i y_i' + H} \right) \rd x = \int_a^b \pm \frac{ \rd \Phi }{ \rd x} \rd x$

This is satisfied, if integrands are equal.

Transform the coordinates to $ \left({ x, \langle y_i \rangle_{1 \le i \le n}, \langle Y_i \rangle_{1 \le i \le n} } \right)$ and write out the full derivative of $ \Phi$:


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

Collect terms multiplied by the same the coordinates together:


 * $\displaystyle \sum_{i=1}^n Y_i' \left({ \pm \frac{ \partial \Phi}{ \partial Y_i} - P_i } \right)+ \sum_{i=1}^n y_i' \left({ \pm \frac{ \partial \Phi}{ \partial y_i} +p_i } \right) + \left({ \pm \frac{ \partial \Phi}{ \partial x}-H+H^* } \right)=0$

This has to hold for arbitrary values of independent coordinates $\left({x, \langle y_i \rangle_{1 \le i \le n}, \langle Y_i \rangle_{1 \le i \le n} } \right)$.

Hence:


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