Canonical Momentum

## Summation Convention

We use the Einstein Summation convention here, i.e. twice repeated indices will be summed over. So,

(1)
\begin{align} \dot{x_{i}}\dot{x_{i}} \equiv \sum_{i = 1}^{N}\dot{x_{i}}^2 \end{align}

Note that the kinetic energy will be denoted, not by $m_{i}\dot{x_{i}}\dot{x_{i}}$ which would be abuse of notation, but rather by $m_{i}\dot{x_{i}}^2$. The bounds on the dummy variable i will be clear from the context. Here, since $x_{i}$ denotes position of the i-th particle of an N particle system interacting with each other, i ranges from 1 to N. The forces will be conservative.

Consider the Lagrangian of such a system,

(2)
\begin{align} L(x, \dot{x}, t) = \frac{1}{2}m_{i}\dot{x_{i}}^2 - V(x,t) \end{align}

In general, V may depend explicitly on time, but not on $\dot{x}$. Then, we have

(3)
\begin{align} \frac{\partial L}{\partial \dot{x_{i}}} = m_{i}\dot{x_{i}} \end{align}

which is a component of the linear momentum. We define canonical or generalized momentum by

(4)
\begin{align} p_{n} = \frac{\partial L}{\partial \dot{q_{n}}} \end{align}

in terms of the generalized position $q_{n}$. These need not be momenta in cartesian space, for instance in polar coordinates, we would have $p_{\theta} = \partial L/\partial \dot{\theta}}$, which doesn't even have the dimensions of linear momentum.

Next, consider the time rate of change of the Lagrangian:

(5)
\begin{align} \frac{dL}{dt} = \frac{\partial L}{\partial q_{n}}\dot{q_{n}} + \frac{\partial L}{\partial \dot{q_n}}\ddot{q_{n}} + \frac{\partial L}{\partial t} \end{align}

From the Euler-Lagrange equations, we have

(6)
\begin{align} \frac{\partial L}{\partial q_n} = \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q_{n}}}\right) \end{align}

We define a quantity called the Hamiltonian (H), by

(7)
\begin{align} H = p_{n}\dot{q_{n}} - L \end{align}

This gives, after some algebra

(8)
\begin{align} \frac{dH}{dt} + \frac{\partial L}{\partial t} = 0 \end{align}

If L is not an explicit function of time, then $\partial L/\partial t = 0$ and so, the Hamiltonian is constant in time. If the forces are all conservative, we have also

(9)
\begin{align} \frac{\partial L}{\partial \dot{x_{i}}} = m_{i}\dot{x_{i}} \end{align}

and hence,

(10)
\begin{align} H = m_{i}\dot{x_{i}}^2 - L = 2T - (T - V) = T + V \end{align}

So if L and V are both not explicitly dependent on time, then the Hamiltonian is just the total energy of the system and is constant in time (this proves energy conservation in a classical conservative dynamical system).

The definition yields

(11)
\begin{align} dH = \dot{q_n}dp_n + p_{n}dq_{n} - \frac{\partial L}{\partial q_{n}}dq_{n} - \frac{\partial L}{\partial \dot{q_{n}}}d\dot{q_{n}} - \frac{\partial L}{\partial t}dt \end{align}

which simplifies to

(12)
\begin{align} dH = \dot{q_n}dp_n - \dot{p_{n}}dq_{n} - \frac{\partial L}{\partial t}dt \end{align}

This gives

(13)
\begin{align} \dot{q_{n}} = \frac{\partial H}{\partial p_n} \end{align}
(14)
\begin{align} \dot{p_n} = -\frac{\partial H}{\partial q_{n}} \end{align}

and

(15)
\begin{align} \frac{\partial L}{\partial t} = -\frac{\partial H}{\partial t} = -\frac{dH}{dt} \end{align}

which are Hamilton's equations. The set of generalized position and momenta, $(q_{n},p_{n})$ can be replaced by another set $(Q_{n},P_{n})$ where Q and P are in general, functions of q, p and t. If Hamilton's equations remain invariant under such a transformation, then the transformation is said to be a canonical transformation.

page revision: 6, last edited: 27 Jul 2007 08:29