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I was reading MG Calkin's text, Lagrangian and Hamiltonian Mechanics. On p147 of his text, he derives the Hamilton-Jacobi equation using the type 2 generating approach. He proceeds to the point that the new Hamiltonian, with its partial derivatives with respect to the new momenta (Alpha) and coordinates (Beta) zero, concluding that that it must be independent of either.

He continues that the new Hamiltonian, $K$, is at most a function of time (which I follow), but then concludes that "we can without a loss take it to be zero", thereby equating the old Hamiltonian $(q,p,t)$ with the partial derivative of the type 2 generating function, $S(q, \alpha, t)$, and the derivation is complete.

Can someone enlighten me with how he justified equating $K = 0$? I've seen derivations where one works backwards, choosing a generating function such that $K = 0$, but not how Calkin proceeds.

https://books.google.co.jp/books?id=uHtIDQAAQBAJ&pg=PA147&lpg=PA147&dq=at+most+a+function+of+time+hamiltonian&source=bl&ots=oegvW4DiSp&sig=ACfU3U3m0rWVW1gXW6ymxdWjfzCHBEHpVw&hl=ja&sa=X&ved=2ahUKEwjA3dLgtProAhXIMd4KHYA2Dg0Q6AEwAHoECAoQAQ#v=onepage&q=at%20most%20a%20function%20of%20time%20hamiltonian&f=false

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Hint: Kamilton's equation are unchanged if we add to the Kamiltonian a term that only depends on time.

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  • $\begingroup$ So the underlying justification is that since the Kamiltonian with or without its explicit time only dependence satisfies the C(K)anonical equations, completely subtracting away said terms produce an equally valid form? I assume it’s an analogous extension to the non-uniqueness of the Lagrangian? $\endgroup$ – Dan Apr 21 at 23:45
  • $\begingroup$ complement to the hint: given the EOMs have derivatives, does adding a constant matter? $\endgroup$ – ZeroTheHero Apr 21 at 23:48
  • $\begingroup$ ${}$ @Dan: Yes. $\endgroup$ – Qmechanic Apr 22 at 7:18

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