Conditions for Transformation to be Canonical

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Theorem

Let

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

be functionals.


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

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

and:

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


Proof

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

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

and

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

have same Euler's equations.

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

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

However,

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

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

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

This is satisfied, if integrands are equal.

Transform the coordinates to $\paren{x,\sequence{y_i}_{1\le i\le n},\sequence{Y_i}_{1\le i\le n} }$ 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:

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

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

Hence:

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

$\blacksquare$



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