Legendre transform

This post has been migrated from my old blog, the math-physics learning seminar. The original post can still be found here.

Yesterday, I gave an introductory talk on Hamiltonian mechanics and symplectic geometry. The starting point is the Legendre transform. First, begin with a configuration space \(Q\). The Lagrangian \(\mathcal{L}\) is a smooth function on \(TQ\). In local coordinates \(q^i\) on \(Q\), we have coordinates \((q_i, v_i)\) on \(TQ\), where the \(v^i\) are the components of the tangent vector

\(v = v_i \partial_i \in T_q Q\). Typically, the Lagrangian will be of the form

\[ \mathcal{L}(q,v) = \frac{1}{2} g(v,v) - V(q), \]

where \(g\) is some metric on \(Q\). Now we introduce new coordinates \(p_i\) defined by

\[ p_i = \frac{\partial \mathcal{L}}{\partial v^i}. \]

If \(\mathcal{L}\) is (strictly?) convex in \(v\) then we can solve for \(v^i\) as a function of \((q^i, p_j)\). It is easy to check that the \(p_i\) transform as covectors, and so this gives a diffeomorphism \(TQ \to T^\ast Q\)(which depends on \(\mathcal{L}\)). For example, in the above Lagrangian,

\[ \frac{\partial \mathcal{L}}{\partial v} = g(v, -), \]

which is just the dual of \(v\) with respect to the metric \(g\). So for Lagrangians of this form, the map \(TQ \to T^\ast Q\) is just the one given by the metric.

Now comes the interesting part. There is a natural way to turn \(\mathcal{L}\), which is a function on \(TQ\), into a function \(H\) on \(T^\ast Q\), in such a way that if we repeat this process, we will get back the original function \(\mathcal{L}\) on \(TQ\). This is the Legendre transform:

\[ \mathcal{H} = pv - L. \]

Now suppose we have a curve \(q(t), \dot{q}(t) \in TQ\) that satisfies the Euler-Lagrange equations. Then by the identification \(TQ = T^\ast Q\), this gives a curve \((q(t), p(t)) \in T^\ast Q\). What equation does it satisfy? We have

\[ \frac{d}{dt} p = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial v} = \frac{\partial \mathcal{L}}{\partial q} = -\frac{\partial H}{\partial q}, \]

and

\[ \frac{d}{dt}q = v = \frac{\partial H}{\partial p}. \]

These are Hamilton’s equations, and they say that the curve \(\gamma = (q(t), p(t)) \in T^\ast Q\) is just an integral curve of the symplectic gradient of \(H\)! So classical mechanics is really just about flows of Hamiltonian vector fields on symplectic manifolds.