Question 8.2: Find the canonical transformation generated by F1(q,Q, t) =m......
Find the canonical transformation generated by
F_{1}\left(q,Q, t\right) =\frac{m\left(q − Q\right)^{2}}{2t} (8.26)
and apply it to solving the problem of the motion of a one-dimensional free particle, whose Hamiltonian is H = p²/2m.
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.