I've been studying the Legendre transform and it's been a fun realization to see that the relationship between the Lagrangian and the Hamiltonian is simply a Legendre transform, i.e.,

$$\{H, p\}\rightarrow \{L,v\}.\tag{1}$$

I first saw it in this paper: https://aapt.scitation.org/doi/10.1119/1.3119512

This Legendre transform link between the Hamiltonian/Lagrangian is neat because it implies, $$v \equiv \frac{dH}{dp}.\tag{2}$$

Which we already know from intuition/formulation of the Hamiltonian. Anyways, what I'm wondering is why we necessarilly use the Lagrangian when calculating action. Or more specifically, when demonstrating the principle of least action, i.e.,


Following Feynman's derivation of this for classical mechanics was simplest for me, and I was able to follow it. But I guess my question is more so why the action function uses the Lagrangian and not the Hamiltonian. Now, the result of the above equation for classical mechanics provides,


Which is self evident is showing why the Lagrangian works in the theory. But is there any reason why we wouldn't be prompted to calculate the following?

$$S=\int_{t_1}^{t_2}Hdt. \tag{5}$$

(Other than the fact that the Legendre transform tells us that it should probably be, $S=\int_{t_1}^{t_2}(p\frac{dH}{dp} - H)dt.$) In other words, is it the former simply because the physics is self-consistent? Or am I missing something?

  • 2
    $\begingroup$ Maybe useful. $\endgroup$
    – Charlie
    Commented May 23, 2021 at 17:47
  • $\begingroup$ Could you give an example of an answer you would find satisfying? On some level the answer is "we do it this way because it works." For example, finding a path for which the integral over time of the Lagrangian is stationary reproduces Newton's laws of motion, while the same statement is not true for the integral over time of the Hamiltonian. $\endgroup$
    – Andrew
    Commented May 23, 2021 at 17:47
  • $\begingroup$ @Charlie Actually I think this is exactly what I'm looking for. Just gave it a quick read. $\endgroup$
    – michael b
    Commented May 23, 2021 at 17:59
  • $\begingroup$ @Andrew More so of a theoretical question, I suppose an satisfying answer to the question would be one that follows the mathematics involved $\endgroup$
    – michael b
    Commented May 23, 2021 at 17:59

1 Answer 1


For what its worth: OP's proposed stationary action principle (5) [assuming that the Hamiltonian $H(q,p,t)$ does not depend on $\dot{q}$ and $\dot{p}$] would imply$^1$ $$ \frac{\partial H}{\partial q}~\approx~0~\approx~\frac{\partial H}{\partial p}, $$ which are not Hamilton's equations. For the correct Hamiltonian action principle, see e.g. this related Phys.SE post.


$^1$ The $\approx$ symbol means equality modulo EOM.

  • $\begingroup$ Ah, okay. I haven't heard this term before. Thank you for your response. I did do a cursory search but didn't find this post, so that's on me. So since the Hamiltonian is the total system energy, if integrating all values over time yields a minimum, we accept that the potential field is zero and there's no velocities (at least classically thinking)? $\endgroup$
    – michael b
    Commented May 23, 2021 at 18:12
  • 1
    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Commented May 23, 2021 at 18:14

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