Question 8.2: Find the canonical transformation generated by F1(q,Q, t) =m......

From Eqs. (8.9) we have

$p_{i}= \frac{\partial F_{1}}{\partial q_{i}} , P_{i}=-\frac{\partial F_{1}}{\partial Q_{i}},  i = 1, . . . , n$ (8.9)

$p = \frac{\partial F_{1}}{\partial q}=\frac{m\left(q − Q\right)}{t} ,  P =- \frac{\partial F_{1}}{\partial Q}=\frac{m\left(q − Q\right)}{t},$  (8.27)

or, in direct form,

Q = q −$\frac{pt}{m}$, P = p . (8.28)

On the other hand,

K(Q, P, t) = H(q, p, t) + $\frac{\partial F_{1}}{\partial t}=\frac{p^{2}}{2m}- \frac{m\left(q − Q\right)^{2}}{2t^{2}}= \frac{P^{2}}{2m}-\frac{P^{2}}{2m}=0.$  (8.29)

The new Hamiltonian is identically zero and the transformed Hamilton’s equations are trivially solved:

$\dot{Q}= \frac{\partial K}{\partial P}=0 \Longrightarrow  Q = a , \dot{P}= -\frac{\partial K}{\partial Q}=0 \Longrightarrow  P = b ,$ (8.30)

where a and b are arbitrary constants. Returning to the original variables we finally obtain

$q = a + \frac{b}{m}t,$  (8.31)

which is the general solution to the equation of motion for the free particle.