Hamilton's canonical equations

Let
 * $$ g := \begin{pmatrix} q \\ p \end{pmatrix} $$

be a vector of generalized coordinates, ($$q$$ might itself be a vector of positions and $$p$$ a vector of momenta, both of them having the same size)


 * $$ \nabla := \begin{pmatrix} {\partial \over \partial q} \\ {\partial \over \partial p} \end{pmatrix} $$

be a vector differential operator,


 * $$ i := \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} $$

be a Givens rotation matrix for a counterclockwise right-angle rotation. (NB: This is a square root of $$-I$$, where $$I$$ is the identity 2&times;2 matrix; this is a matric version of the imaginary unit.) Let H denote the Hamiltonian (function for total energy).

Then
 * $$ \dot g = \begin{pmatrix} \dot q \\ \dot p \end{pmatrix} $$

where overdot is Newton’s fluxional notation for time derivative.

$$\nabla H = i \dot g $$

This is a compactified form of Hamilton’s canonical equations (of motion). It could also be re-expressed (trivially) as
 * $$ \dot g = -i \nabla H $$

To see that it is the pair of canonical equations, unpack the symbolism:


 * $$ \begin{pmatrix} {\partial H \over \partial q} \\ {\partial H \over \partial p} \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \dot q \\ \dot p \end{pmatrix} = \begin{pmatrix} -\dot p \\ \dot q\end{pmatrix} $$

thus
 * $$ {\partial H \over \partial q} = -\dot p $$
 * $$ {\partial H \over \partial p} = \dot q $$

An immediate consequence of the canonical equation is that
 * $$ \dot g \cdot i \dot g = 0$$,

that is,
 * $$ \begin{pmatrix} \dot q \\ \dot p\end{pmatrix} \cdot \begin{pmatrix} -\dot p \\ \dot q\end{pmatrix} = -\dot q \dot p + \dot p \dot q = 0$$

so $$ \dot g \cdot \nabla H = 0 $$

This means (geometrically) that g moves perpendicularly to H’s gradient, so as to conserve H.

a simple example
Let
 * $$ H = {1\over 2 m} p^2 + {k \over 2} q^2 $$

This corresponds to a simple harmonic oscillator, where m is mass and k is a spring constant. The first term is kinetic energy and the second term is potential energy (of displacement from the stable equilibrium).

Then
 * $${\partial H \over \partial q} = k q = -F = -\dot p = -m \ddot q $$

Note how deriving H by q returns a multiple of q; q is an “eigen-derivator” of H, as it were. ($$k q = -F$$ follows from $${\partial U \over \partial q} = -F$$ being true in general (for conservative forces), where U denotes potential energy; $$F = \dot p$$ is Newton’s second law, and m may be assumed to be constant)

On the other hand,
 * $${\partial H \over \partial p} = {p \over m} = {m \dot q \over m} = \dot q $$

and note how deriving H by p returns a multiple of p, so p is also an “eigen-derivator” of H.