We want to find a generating function $S(q_i,P_i,t)$ such that we get the best possible canonical transformations. So it must satisfy the Hamilton-Jacobi equation: $$H(q_i,\frac{\partial S}{\partial q_i},t)+\frac{\partial S}{\partial t}=0.\tag{1}$$ Let's even take the easier case where the hamiltonian is time-independent. Then we can search for the Hamilton's principal function in the form: $$S(q_i,P_i,t)=W(q_i,P_i)-\alpha_0t.\tag{2}$$ However from the theory of canonical transformations we know that in order for a generating function to work it has to satify the following condition: $$F(t_2)-F(t_1)=0.\tag{3}$$ Well in our case we have: $$S(q_i(t_2),P_i,t_2)-S(q_i(t_1),P_i,t_1)=W(q_i(t_2),P_i)-W(q_i(t_1),P_i)-\alpha_0(t_2-t_1).\tag{4}$$ My question is how are we ensuring that this is equal to zero always? And maybe thinking about it I don't understand how do we ensure this in any canonical transformation not just the Hamilton-Jacobi equation. How are we sure what the functions values will be before knowing the trajectory as a function of time?
1 Answers
OP's condition (3) is not there/not necessary. Instead a type-2 generating function $F_2(q,P,t)$ satisfies the following $2n+1$ conditions: $$ p_j~=~\frac{\partial F_2}{\partial q^j},\qquad Q^j~=~\frac{\partial F_2}{\partial P_j},\qquad K-H~=~\frac{\partial F_2}{\partial t}. \tag{9.17}$$
For a Hamiltonian $H(q,p)$ without explicit time dependence, OP seems to ponder how the Hamilton's principal function $S(q,P,t)$ and the Hamilton's characteristic function $W(q,P)$ can both serve the role as an $F_2(q,P,t)$ generating function if we also identify$^1$ $$ S(q,P,t)~=~W(q,P)-P_1 t~?\tag{10.17}$$
The resolution is to pick the Kamiltonian differently in the 2 cases: $$ K_S~:=~0\quad\text{and}\quad K_W~:=~P_1. $$
References:
- H. Goldstein, Classical Mechanics; section 10.3.
$^1$ Note that the constants of integration $\alpha_j$ are identified with the new momenta $P_j$, cf. e.g. this Phys.SE post.
- 220,844