Conditions for Transformation to be Canonical

Theorem
Let


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


 * $ \displaystyle J_2[{ \langle Y_i \rangle_{1 \le i \le n}, \langle P_i \rangle_{1 \le i \le n } }] = \int_a^b \left({ \sum_{i=1}^n P_i Y_i'-H^* } \right) \mathrm d 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=1}^n p_i y_i'-H=\sum_{i=1}^n P_i Y_i'-H^* \pm \frac{ \mathrm d \Phi }{ \mathrm d x}$

and


 * $ \displaystyle p_i= \mp \frac{ \partial \Phi }{ \mathrm d y_i}, \quad P_i= \pm \frac{ \partial \Phi }{ \mathrm 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


 * $ \displaystyle \int_a^b F_1 \mathrm d x$

and
 * $ \displaystyle \int_a^b F_2 \mathrm d x=\int_a^b \left({ F_1 \pm \frac{ \mathrm d \Phi }{ \mathrm d x} } \right) \mathrm d 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 \mathrm d x= \int_a^b \left({ \sum_{i=1}^n p_i y_i'-H } \right) \mathrm d x$
 * $ \displaystyle \int_a^b F_2 \mathrm d x= \int_a^b \left({ \sum_{i=1}^n P_i Y_i'-H^* } \right) \mathrm d x$

However,

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


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