Problem
Consider the mechanical system with configuration manifold $Q = R$ and Hamiltonian $H:T^{\star}Q \to R : (q, p) : (q, p) \mapsto p^2$. Verify explicitly that flow along the corresponding Hamiltonian vector field is area-preserving.
Solution
To show the Hamiltonian flow $\Phi_t$ is volume preserving we begin by finding the Hamiltonian vector field $X_H$.
(1)This vector field can be interpreted as the following equations of motion on $T^{\star}Q$,
(2)Solving this linear differential equation yields
(3)Hence, the flow operator $\Phi_t$ is
(4)Finally, we are ready to solve the original problem. Let $A$ be a region in $T^{\star}Q$. The area of of $A$ is
(5)Instead of finding the area of $\Phi_t(A)$ by integrating over $\Phi_t(A)$ (i.e. $\int_{\Phi_t(A)} \mathbf{d}q \mathbf{d}p$), we can compute it by integrating over $A$ with the assistance of $\Phi_t$ and integration_by_substitution.
(6)First we compute $\mathtt{det}(D\Phi_t)$
(7)Thus
(8)Volume is preserved, and Hyrule has been saved.
[kcrane] Here's another solution that does a couple things differently:
- it walks you through the mechanics of finding the Hamiltonian vector field a lot more explicitly
- it does a much simpler check that this vector field is area preserving by simply checking that its divergence is zero everywhere. I.e., if a smooth vector field u is divergence-free then flows through it preserve volume. (You can find a proof of this fact in many places, including Chorin and Marsden, A Mathematical Introduction to Fluid Mechanics, pp. 8-11.)
Given a Hamiltonian vector field $H$, the Hamiltonian vector field is defined as
(9)i.e., the Hamiltonian vector field is the vector field which, when fed to the 2-form $\Omega$, becomes a thing that behaves exactly like the exterior derivative of H. (What is $\Omega?$ it's the canonical symplectic 2-form $\Omega = dq^dp$.) So to figure out what $X_H$ is, we're going to
- compute $i_{X_H}\Omega$
- compute $dH$
- figure out what $X_H$ must be so that the left hand side looks like the right hand side!
Here we go…
We were given a Hamiltonian $H=p^2$. Since $H$ is a 0-form, its exterior derivative is just the differential of $p^2$
(10)We can compute the contraction of $\Omega$ with $X_H$ by imagining we have some additional vector field $v$, doing the usual evaluation of a 2-form, and then figuring out what 1-form would produce the same results if it were fed $v$. First, let's give names to the components of $X_H$ and $v$:
(11)Now evaluate $\Omega$:
(13)So $i_{X_H}$ must be
(14)(Why? Think about what happens when you feed $v$ to this form.)
All we've done at this point is written out the equation that defines our Hamiltonian vector field explicitly in terms of its components $x_q$ and $x_p$. I.e., this equation
(15)became this one
(16)It should be clear that the only way this equality can be satisfied is if $x_q=2p$ and $x_p=0$. Since we've figured out what the components are, we've figured out what the vector field looks like!
(17)How do we check that this vector field is volume-preserving? Easy! As noted at the beginning, we can just check that it's divergence-free, i.e., that
(18)which turns out to be true:
(19)