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In Classical Hamiltonian Mechanics, a canonical transformation of the phase-space coordinates $(p,q,t) \to (P,Q,t)$ is such that the general form of Hamilton's equations is followed and Hamilton's principle is obeyed:

$$\delta \int_{t_0}^{t_1} \left (P_i \dot{Q_i} - K (Q,P,t)\right) \, \mathrm{d}t = 0\tag{1}$$

for some new Hamiltonian $K$ just as before the transformation,

$$\delta \int_{t_0}^{t_1} \left (p_i \dot{q_i} - H (q,p,t)\right) \, \mathrm{d}t = 0\tag{2}$$

for the old Hamiltonian, $H$. Several textbooks then mention the necessary relationship between the integrands in the two equations above to be

$$\lambda (p_i \dot{q_i} - H) = P_i \dot{Q_i} - K + \frac{\mathrm d F}{\mathrm d t}.\tag{3}$$

My question is, how is the relationship above justified?

Qmechanic
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Yejus
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1 Answers1

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That seems to be a misunderstanding. The off-shell relation (3) with $\lambda\neq 0$ is a sufficient (as opposed to a necessary) condition for the stationary action principles (1) & (2) to be equivalent, i.e. have the same stationary paths in phase space. These paths are solutions to Kamilton's and Hamilton's equations, respectively.

For the various definitions of a canonical transformation (CT), see e.g. this Phys.SE post.

Qmechanic
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