Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 59 / 3 / Solution

For a finite-dimensional mechanical system let $Q$ be its configuration space, a smooth manifold incorporating the holonomic constraints. Lagrangian mechanics starts with $L:TQ\to \mathbb{R}$; a curve $q(t)$ makes the action stationary under fixed-endpoint variations. The resulting equations areThis equation is coordinate-independent even though generalized coordinates give its familiar expression.

The fibre derivative $\mathbb{F}L(q,v)=(q,p)$, with $p_i=L_{v^i}$, defines the Legendre transform in mechanics. The Cartan form of a regular mechanical Lagrangian isFor a regular Lagrangian, the velocity Hessian is invertible, ${\omega }_L$ is symplectic, and ${\iota }_{{\Gamma }_L}{\omega }_L=d{E}_L$ determines a second-order vector field on $TQ$. In coordinates this recovers the displayed Euler–Lagrange equations: the velocity components of the vector field are $v^i$, and its acceleration components are fixed by the invertible Hessian.

Hamiltonian mechanics takes place on $T^{\ast }Q$. Its tautological one-form is $\theta =p_{i}d{q}^i$, and its canonical symplectic form is $\omega =-d\theta =d{q}^i\wedge d{p}_i$. With ${\iota }_{X_H}\omega =dH$,The associated Poisson bracket is $\{f,g\}=f_{q^i}{g}_{p_i}-f_{p_i}{g}_{q^i}$, so $X_{H}f=\{f,H\}$. For a hyperregular Lagrangian, $\mathbb{F}L$ is a global diffeomorphism, $H=E_L\circ (\mathbb{F}L{)}^{-1}$ and $(\mathbb{F}L{)}^{\ast }\omega ={\omega }_L$. The two formulations then describe the same trajectories. Regularity gives only local equivalence; degenerate Lagrangians require constraint analysis instead. These geometric structures distinguish equations of motion from arbitrary coordinate formulae and make symmetry reduction precise.