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.