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In most of the cases, the Hamiltonian is equal to the total mechanical energy, which is usually conserved. They are equal and conserved when the potential energy is not dependent on velocity and there is no explicit time dependence. So at each point on the path, particle has the same total energy which is $H$ as well.

In a previous question/answer, we derived:

$$H = E = \frac{\partial S}{\partial t_1} = -\frac{\partial S}{\partial t_2}.$$

This is clear, but my book says:

$$H = -\frac{\partial S}{\partial t}\tag{1.4}$$ $$E = -\frac{\partial S}{\partial t}.\tag{1.6}$$ Note here that we're doing with respect to $t$ and not the initial or final point.

Where did we get this from rigorously?

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  • $\begingroup$ Most likely it is just that $t_1=0$ and $t=t_2$ $\endgroup$ Jul 24, 2023 at 4:49

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  1. Briefly speaking, your book is talking about Hamilton's principal function $S(q,P,t)$. Eq. (1.4) is the Hamilton-Jacobi equation. Eq. (1.6) follows by assuming that the Hamiltonian $H=E$ is the energy.

  2. Hamilton's principal function $S(q,P,t)$ should not be conflated with the on-shell action $S(q_f,t_f;q_i,t_t)$ of the previous question, although they are related.

  3. For more details, see this related Phys.SE post.

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