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Hamilton's characteristic function, wave-particle duality and constant-action surfaces.

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?

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?

Hamilton's characteristic function, wave-particle duality and constant-action surfaces.

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:

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

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?

Hamilton's characteristic function, wave-particle duality and constant-action surfaces.

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?

So I'm currently doing some research about the way in which classical physics connects to quantum physics and I came across 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:  $$ \frac{\partial{S(\textbf{q} ,t)}}{\textbf{q}} = \textbf{p} \Rightarrow \nabla_q S(\textbf{q},t) = \textbf{p}$$ 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 conclusion that bothers me is the apparent assumption that the momentum vector is always tangent to the trajectory, can somebody explain under which condditions does such assumption hold or, alternatively, what am I getting wrong about those suggestions?

Hamilton's characteristic function, wave-particle duality and constant-action surfaces.

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: 

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

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?