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So, I'm currently doing some research about the way in which classical physics connects to quantum physics, and I came across the Hamilton-Jacobi equation, and the implications of Hamilton's characteristic function.

I found out that, since:

$$ \frac{\partial{S(q_i,t)}}{\partial{q_i}} = p_i$$

and, consequently:

$$ \frac{\partial{S(\textbf{q} ,t)}}{\partial\textbf{q}} = \textbf{p} \Rightarrow \nabla_q S(\textbf{q},t) = \textbf{p}$$

the vector composed of generalized momenta is normal to each iso-action surface.

However, I found in multiple articles, a claim that I don't quite understand: this particular relation between iso-action surfaces and the momentum vector, appears to suggest that the trajectory taken by the system in the configuration space is perpendicular to such surfaces.

The thing about such a conclusion that bothers me, is the apparent assumption that the momentum vector is always tangent to the trajectory.

Can somebody explain under which conditions does such an assumption hold or, alternatively, what am I getting wrong about those suggestions?

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    $\begingroup$ "However, I found in multiple articles..." Can you link one or provide the relevant context? $\endgroup$
    – hft
    Commented Aug 5 at 17:38
  • $\begingroup$ @hft I was mainly talking about "From Classical to Quantum Mechanics through Optics" by Jaume Masoliver and Ana Ros, page 5: $ \textit{"On the other hand, the gradient ∇ S= ∇ S is a normal vector to the surface S =constant and, as p has the direction of the trajectory, we can conclude that the trajectories of the particle are perpendicular to the surfaces of constant S"}$ $\endgroup$
    – Matteo Grandinetti
    Commented Aug 6 at 9:12
  • $\begingroup$ Comment to the post (v4): Note that momentum and velocity are not necessarily parallel. $\endgroup$
    – Qmechanic ♦
    Commented Aug 6 at 11:24
  • $\begingroup$ Yes, that's exactly what I was trying to understand, but in that article it seems that the author took such correspondance for granted without much questioning. $\endgroup$
    – Matteo Grandinetti
    Commented Aug 6 at 12:05

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You are right, trajectories of the solutions of EL equations are not normal to the surfaces at constant $S$ in general, if using the Euclidean metric in coordinates which, however, has no physical meaning in general.

However, for Lagrangians whose kinetic energy is quadratic in $\dot{q}^k$ with a non degenerate quadratic form $$L(q,\dot{q})= \sum_{h,k=1}^n \frac{1}{2} g_{hk}(q) \dot{q}^k\dot{q}^h - U(q)$$ we can use this quadratic form as the physically meaningful metric in the configuration space. There, $p$ and $\dot{q}$ are nothing but the contravariant and the covariant form of the normal vector to the $S$ constant surfaces: $$dS_k = p_k\:, \quad p_k = \sum_h g_{kh}\dot{q}^h$$ from HJ equations and the definition of $p_k$ with respect to $L$ respectively.

Referring to that metric the solutions of the EL equations define normal (contravariant) vectors to the $S$-constant surfaces.

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  • $\begingroup$ Thank you so much for your answer! However I would like to extend the question a little bit: since only when a Lagrangian is quadratic in $\dot{q}^k$ with non degenerate quadratic form that correspondance is acceptable, what does that tell us about the relation between geometrical optics and mechanics? I mean, in geometrical optics the rays (which correspond to the "paths" of light in a profane way) are always normal to the wavefronts, whereas the same can't always be said for mechanics, since only in some particular conditions $p$ is tangent to the actual path of the system $\endgroup$
    – Matteo Grandinetti
    Commented Aug 6 at 12:09
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    $\begingroup$ The same is true for mechanics when using the above metric. For a single particle with ($L=T-U$), and the analogy was performed for that case, the analogy works, there $g_{ij}= m\delta_{ij}$ in Cartesian coordinates. The metric used in optics is $\delta_{ij}$ (orthonormal Cartesian coordinates) and everything matches. $\endgroup$
    – Valter Moretti
    Commented Aug 6 at 12:18
  • $\begingroup$ Again, thank you so much for your time, your answer has been extremely helpful. $\endgroup$
    – Matteo Grandinetti
    Commented Aug 6 at 12:19
  • $\begingroup$ You're welcome. $\endgroup$
    – Valter Moretti
    Commented Aug 6 at 12:19
  • $\begingroup$ What does $dS_k$ mean? The same thing as $\frac{\partial S}{\partial q_k}$? $\endgroup$
    – hft
    Commented Aug 6 at 17:18

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