8.7: How Can p, q Really Be Independent Variables?

It may seem a little odd at first that varying p,q as independent variables leads to the same equations as the Lagrangian minimization, where we only varied q, and that variation “locked in” the variation of \(\begin{equation}
\dot{q}
\end{equation}\). And, isn’t p defined in terms of \(\begin{equation}
q, \dot{q} \text { bу } p=\partial L / \partial \dot{q}
\end{equation}\) which is some function of \(\begin{equation}
q, \dot{q} ?
\end{equation}\)? So wouldn’t varying q automatically determine the variation of p?

The answer is, no, p is not defined as \(\begin{equation}
p=\partial L / \partial \dot{q}
\end{equation}\) from the start in Hamilton’s formulation. In this Hamiltonian approach, p,q really are taken as independent variables, then varying them to find the minimum path gives the equations of motion, including the relation between p and \(\begin{equation}
q, \dot{q}
\end{equation}\)

This comes about as follows: Along the minimum action path, we just established that

\begin{equation}
d H(p, q)=\dot{q} d p-\dot{p} d q
\end{equation}

We also have that \(\begin{equation}
L=p \dot{q}-H
\end{equation}\) so (Legendre transformation!)

\begin{equation}
d L(q, \dot{q})=p d \dot{q}+\dot{p} d q
\end{equation}

from which, along the physical path, \(\begin{equation}
p=(\partial L / \partial \dot{q})_{q \text { constant }}
\end{equation}\). So this identity, previously written as the definition of p, now arises as a consequence of the action minimization in phase space.