Is the following logic correct?:

If you have an Hamiltonian, that has time has a variable explicitly, and you get the Lagrangian $L$ from it, and then you get an equivalent $L'$, since $L$ has the total time derivate of a function, both Lagrangians will lead to the same equations Euler-Lagrange equations right? If so, and if you get you get the Hamiltonian from $L'$, and you find the Hamilton equations from it will they be equal to the original Hamiltonian's Hamilton equations?

If it is of any help, the given Hamiltonian is:

$$H = \frac{p^2}{2m} - bqpe^{-\alpha t} + \frac{ba}{2}e^{-\alpha t}(\alpha + b e^{-\alpha t}) + \frac{kq^2}{2} \, .$$


It seems OP is essentially asking the following two questions:

  1. Is the Legendre transformation an involutive transformation?


  • The answer is Yes in the regular case, i.e. when there is a bijective correspondence $v\leftrightarrow p$ between velocities and momenta. This follows from the symmetric form $$H+L =p_i v^i$$ of the Legendre transformation. See also. e.g. this related Phys.SE post.

  • The singular case (where typically constraints are present) is more subtle, but for physically meaningful situations, it is usually still Yes.

  1. Under a change of the Lagrangian $$\widetilde{L}~=~ L +\frac{dF}{dt},$$ by a total time-derivative, after a Legendre transformation into the Hamiltonian formulation, are the corresponding Hamilton's equations un-changed?

Answer: Well, the function $F$ induces a canonical transformation (CT), so that Hamilton's equations turn into Kamilton's equations. This is the topic of this Phys.SE post.

  • $\begingroup$ but wont the first hamiltonian's hamiltion equation have time explicitly(since the given hamiltonian has time), whereas the second hamiltonian's hamiltion equation not have time(because L' does not have time). how can they be the same? $\endgroup$ – forpointing Dec 14 '17 at 21:23
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Dec 15 '17 at 17:20

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