Hamilton's canonical equations

Let

$g:=\left(\begin{array}{c}q\\ p\end{array}\right)$

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)

$\mathrm{\nabla }:=\left(\begin{array}{c}\frac{\partial }{\partial q}\\ \frac{\partial }{\partial p}\end{array}\right)$

be a vector differential operator,

$i:=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$

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

Then

$\dot{g}=\left(\begin{array}{c}\dot{q}\\ \dot{p}\end{array}\right)$

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

$\mathrm{\nabla }H=i\dot{g}$

This is a compactified form of Hamilton’s canonical equations (of motion). It could also be re-expressed (trivially) as

$\dot{g}=-i\mathrm{\nabla }H$

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

$\left(\begin{array}{c}\frac{\partial H}{\partial q}\\ \frac{\partial H}{\partial p}\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}\dot{q}\\ \dot{p}\end{array}\right)=\left(\begin{array}{c}-\dot{p}\\ \dot{q}\end{array}\right)$

thus

$\frac{\partial H}{\partial q}=-\dot{p}$

$\frac{\partial H}{\partial p}=\dot{q}$

An immediate consequence of the canonical equation is that

$\dot{g}\cdot i\dot{g}=0$,

that is,

$\left(\begin{array}{c}\dot{q}\\ \dot{p}\end{array}\right)\cdot \left(\begin{array}{c}-\dot{p}\\ \dot{q}\end{array}\right)=-\dot{q}\dot{p}+\dot{p}\dot{q}=0$

so

$\dot{g}\cdot \mathrm{\nabla }H=0$

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

a simple example

Let

$H=\frac{1}{2m}{p}^2+\frac{k}{2}{q}^2$

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

$\frac{\partial H}{\partial q}=kq=-F=-\dot{p}=-m\ddot{q}$

Note how deriving H by q returns a multiple of q; q is an “eigen-derivator” of H, as it were. ($kq=-F$ follows from $\frac{\partial U}{\partial q}=-F$ 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,

$\frac{\partial H}{\partial p}=\frac{p}{m}=\frac{m\dot{q}}{m}=\dot{q}$

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