How does $\dot{q}_i p_i - H = \dot{Q}_i P_i - K + \frac{d}{dt}F$ give the same Euler-Lagrange equations and Equations of motion (EoM) for corresponding coordinates and allow us to determine a canonical transformation? Here $K$ is the Kamiltonian.
Here is the logic my professor used in class:
Recall the the Euler-Lagrange equations are invariant when:
- $L(q, \dot{q}, t) = L(Q,\dot{Q},t) $, where $Q = Q(q,t)$ a point transformation.
- $L'(q, \dot{q}, t) = L + \frac{d}{dt}F(q,t).$
Thus $L(q,\dot{q},t) = L(Q,\dot{Q}, t) + \frac{d}{dt}F(q,t) $ will have the same EoM.
Using L = $\dot{q}_i p_i - H$ we can then write that
$\dot{q}_i p_i - H = \dot{Q}_i P_i - K + \frac{d}{dt}F$ will give the same equations of motion.
This logic seems completely errornous to me since
The generating function $F$ can be a function of $p$ and by invertibility a function of $\dot{q}$. But 2) only holds for $F = F(q,t)$. Never are generating functions restricted only to $q$
Combining 1. and 2. together in this way seems erronous. Perhaps it is meant that $L(q,\dot{q},t) = L(Q,\dot{Q}, t) + \frac{d}{dt}F(q,t) $ is also just a point transformation so it does not change the Euler lagrange equations?
