I recently learnt that one can express all of classical mechanics on a symplectic manifold.
Is it possible to provide an entirely geometric example as to why this is the case? The geometric example should show:
- A choice of manifold $M$, closed 2-form $\omega$, function $H: M \rightarrow \mathbb R$
- How these uniqely determine a vector field $V_H$, whose flow lines "are the dyanmics" --- and some way to see that this choice is unique for a fixed $\omega$ and $H$.
- What mechanical system corresponds to this choice of $(M, H, \omega)?$ Is there a natural classical system that corresponds to the example?
- Why $\omega$ must be a closed form (ie, what fails if it is not). What is the geometry behind this condition?
While algebraically, one can work through the details, I do not see why this entire recipe works. Some sort of picture / visualization / geometry to the whole thing would be greatly appreciated.
I'm not sure if I understand exactly what are you looking for, but before the example, two general answers:
The fact that for given $H\colon M \to \Bbb R$ there is a unique $X_H\in \mathfrak{X}(M)$ such that $\omega(X_H,\cdot) = {\rm d}H$ is due simply to non-degeneracy of $\omega$. Closedness of $\omega$ is not relevant here.
Closedness of $\omega$ comes in the proof of two important facts: (i) the flow of $X_H$ preserves $\omega$ (${\rm d}\omega=0$ simplifies Cartan's magic formula); (ii) the Poisson bracket $\{f,g\} = \omega(X_f,X_g)$ on $\mathcal{C}^\infty(M)$ makes it a Lie algebra (${\rm d}\omega = 0$ is equivalent to the Jacobi identity).
That said, let's look at a harmonic oscillator. That is, in the real line, assume that we have a mass $m>0$ connected to the origin by a string whose elasticity constant is $k>0$, sliding without friction. A Hamiltonian system, in general, is a triple $(M,\omega,H)$, where $M$ is a manifold (to be thought of the space of position and momenta), $\omega$ is a symplectic form, and $H$ is a Hamiltonian function (i.e., a smooth function that describes the time evolution of the mechanical system we're studying). In this case, we have that
(i) $M = T^*\Bbb R \cong \Bbb R^2$ is equipped with coordinates $(x,p)$, $x$ for position, $p$ for momentum;
(ii) $\omega = {\rm d}x \wedge {\rm d}p$ is the canonical symplectic form in $T^*\Bbb R$;
(iii) $\displaystyle{H(x,p) = \frac{p^2}{2m} + \frac{kx^2}{2}}$.
Here we're thinking of the Hamiltonian as "kinetic energy" + "potential energy". Classically we'd think of the kinectic term as $mv^2/2$, but the relation between $p$ and $v$ is given by $p = mv$, so we plug that in to get the first term in $H$ (the actual name for this procedure is "Legendre transform", this might ring a bell). And we know that Hooke's Law says that $F = -kx$, but the relation between the potential $V$ and $F$ is $F = -\nabla V$, which gives us $V(x) = kx^2/2$.
That said, Hamilton's equations become $$x'(t) = \frac{p(t)}{m} \quad \mbox{and}\quad p'(t) = -kx(t),$$or equivalently, the matrix system $$\begin{pmatrix} x'(t) \\ p'(t)\end{pmatrix} = \begin{pmatrix} 0 & 1/m \\ -k & 0\end{pmatrix}\begin{pmatrix}x(t) \\ p(t)\end{pmatrix}$$ Denoting the coefficient matrix by $A$ and fixed initial conditions $x(0) = x_0$ and $p(0) = p_0$, we solve for $$\begin{pmatrix} x(t) \\ p(t)\end{pmatrix} = \exp(tA) \begin{pmatrix} x_0 \\ p_0\end{pmatrix}.$$Since $$\exp(tA) = \cos\left(t\sqrt{\frac{k}{m}}\right){\rm Id}_2 + \sin\left(t\sqrt{\frac{k}{m}}\right) \sqrt{\frac{m}{k}}A,$$we define the angular frequency $\omega = \sqrt{k/m}$ to write $$x(t) = x_0\cos(\omega t) + \frac{p_0}{\sqrt{km}}\sin(\omega t) \quad \mbox{and} \quad p(t) = -x_0 \sqrt{km}\sin(\omega t) + p_0\cos(\omega t).$$The curve $t \mapsto x(t)$ describes the movement of the mass and the curve $t\mapsto (x(t),p(t))$ is the integral curve of $X_H$ starting at $(x_0,p_0)$. So letting $0 \leq t \leq 2\pi/\omega$ range, we're describing the "energy surface" of the Hamiltonian system passing through $(x_0,p_0) \in T^*\Bbb R$. For example, for $(0,0)$ it degenerates to a single point, and for $(1,0)$ and $(0,1)$ we get ellipses. Moreover, you can check using these specific formulas that $t \mapsto H(x(t),p(t))$ is a constant function of $t$. This is a particular case of a more general phenomenon: in any Hamiltonian system $(M,\omega, H)$, $H$ is constant along the integral curves of $X_H$, which is then tangent to the energy surfaces of the system.
Another way to put this is as follows: a function $f$ is a constant of motion for $(M,\omega, H)$ if $f$ is constant along the integral curves of $X_H$, and this is equivalent to having $\{f,H\} = 0$, where $\{\cdot,\cdot\}$ is the Poisson bracket mentioned earlier (which again, has nice properties due to ${\rm d}\omega=0$), and the fact that $H$ itself is a constant of motion is the obvious fact that $\{H,H\} = 0$.