Canonical transformation in classical mechanics

[dRic2]

Homework Statement

Give the proof of the equation
$$ H' = H + \frac { \partial S } { \partial t }$$
By proceeding as follow. Leave the time t as independent variable and consider a canonical transformation in which the time appears as a parameter. Then obtain the above relation by distinguishing between $dS$ in the canonical integral and $\delta S$ in the definition of canonical transformation. In the first case time is varied, in the second case not.

Relevant Equations

Canonical integral:
$$A = \int ( \sum p_i dq_i - Hdt)$$
Definition of canonical transformation:
$$ \sum p_i \delta q_i = \sum P_i \delta Q_i + \delta S$$

I'm stuck from the beginning. I though I understood the difference between $\delta$ and $d$, but apparently I was wrong, because I don't know how to exploit it here...

Any hint would be greatly appreciated

Thank
Ric

 

[Abhishek11235]

Are you sure you have written correct defination of Canonical Transformation?

$p_i\delta q_i -H= P_i\delta Q_i-H' + dS/dt$

In this case ,the equation is easy to derive by using chain rule to S(q,Q,t) and using linear independence of $\dot q$and $p$

 

[dRic2]

Are you sure you have written correct defination of Canonical

This is the definition found in my book and it makes sense. It seems different from yours though

 

[Abhishek11235]

This is the definition found in my book and it makes sense. It seems different from yours though

Ok. Then use the following:

$$\delta S= dS/dt ~dt + dS/dq ~dq + dS/dQ ~dQ$$

Now subtract $Hdt$ on both sides in your defination of Canonical Transformation

Now use the linear independence of q,Q and t

 

[dRic2]

Your notation is confusion me. $\delta S$ is not $dS$. $\delta$ is used to address variations and, since the problem asks to consider time as a parameter, you don't have to vary wrt to time. $dS$ is obviously the total differential.

Even if you meant $dS$ I don't know how to proceed.

 

[dRic2]

Solved it. I was very confused by the notation of the book.