6.7: Checking that We Can Eliminate the q˙i's

We should check that we can in fact write

\begin{equation}
H\left(p_{i}, q_{i}\right)=\sum_{i=1}^{n} p_{i} \dot{q}_{i}-L\left(q_{i}, \dot{q}_{i}\right)
\end{equation}

as a function of just the variables \(\begin{equation}
\left(q_{i}, p_{i}\right)
\end{equation}\), with all trace of the \(\begin{equation}
\dot{q}_{i}
\end{equation}\)’s eliminated. Is this always possible? The answer is yes.

Recall the \(\begin{equation}
\dot{q}_{i}
\end{equation}\)’s only appear in the Lagrangian in the kinetic energy term, which has the general form

\begin{equation}
T=\sum_{i, j} a_{i j}\left(q_{k}\right) \dot{q}_{i} \dot{q}_{j}
\end{equation}

where the coefficients \(\begin{equation}
a_{i j}
\end{equation}\) depend in general on some of the \(\begin{equation}
q_{k}
\end{equation}\)'s but are independent of the velocities, the \(\begin{equation}
\dot{q}_{k} \text { 's. }
\end{equation}\) Therefore, from the definition of the generalized momenta,

\begin{equation}
p_{i}=\dfrac{\partial L}{\partial \dot{q}_{i}}=\sum_{j=1}^{n} a_{i j}\left(q_{k}\right) \dot{q}_{j}
\end{equation}

and we can write this as a vector-matrix equation,

\begin{equation}
\mathbf{p}=\mathbf{A} \dot{\mathbf{q}}
\end{equation}

That is, \(\begin{equation}
p_{i}
\end{equation}\) is a linear function of the \(\begin{equation}
\dot{q}_{j} \text { 's. }
\end{equation}\). Hence, the inverse matrix \(\begin{equation}
\mathbf{A}^{-1}
\end{equation}\) will give us \(\begin{equation}
\dot{q}_{i}
\end{equation}\) as a linear function of the pj's, and then putting this expression for the \(\begin{equation}
\dot{q}_{i}
\end{equation}\) into the Lagrangian gives the Hamiltonian as a function only of the qi's and the pi's, that is, the phase space variables.

The matrix A is always invertible because the kinetic energy is positive definite (as is obvious from its Cartesian representation) and a symmetric positive definite matrix has only positive eigenvalues, and therefore is invertible.