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A regular Lagrangian $L:TQ\to \mathbb{R}$ defines a fibre derivative $\mathbb{F}L:TQ\to T^{\ast }Q$ and the Cartan one-form ${\theta }_L=(\mathbb{F}L{)}^{\ast }\theta$. Its two-form ${\omega }_L=-d{\theta }_L$ is nondegenerate when the velocity Hessian is invertible. The energy is $E_L=v^i{L}_{v^i}-L$. The equation ${\iota }_{{\Gamma }_L}{\omega }_L=d{E}_L$ determines the Euler–Lagrange vector field. A hyperregular Lagrangian makes the fibre derivative a global diffeomorphism and transports this dynamics to Hamiltonian phase space.

The Poisson bracket is antisymmetric, so $\{H,H\}=0$. The Hamiltonian energy balance consequently givesFor a regular Lagrangian the Legendre transform in mechanics gives $H_t=-L_t$, evaluated at the corresponding velocity. In particular, a Lagrangian with no explicit time dependence has time-translation symmetry. Noether theorem associates this continuous symmetry of the action with the conserved quantity ${\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L=H$. For an ordinary natural mechanical Lagrangian $L=T-V$, with $T$ quadratic in the velocities, this quantity is the total energy $T+V$.

Define the canonical momenta byTo obtain an ordinary unconstrained Hamiltonian, assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial {\dot{q}}_i\partial {\dot{q}}_j)$ is invertible. The inverse function theorem then lets us solve locally for $\dot{\mathbf{q}}=\mathbf{v}(\mathbf{q},\mathbf{p},t)$. The phase space is the cotangent bundle $T^{\ast }\mathcal{Q}$, with local canonical coordinates $(\mathbf{q},\mathbf{p})$ and dimension $2n$. The Legendre transform in mechanics definesThe velocities in this formula have been eliminated in favour of phase space variables. A hyperregular Lagrangian makes this construction global; regularity alone guarantees it locally. Hamilton's equations are

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.

The passage to Hamiltonian mechanics uses the Legendre transform in mechanics. Assume a regular Lagrangian: the velocity Hessian matrix $({\partial }^{2}L/\partial v^i\partial v^j)$ is invertible. Thendefines a locally invertible map from the tangent bundle to the cotangent bundle. Write its local inverse as $v=v(t,q,p)$ and define the HamiltonianDifferentiating this expression, all terms containing $dv$ cancel because $p_i=L_{v^i}$. HenceThe Euler-Lagrange equations become Hamilton's equations:Conversely, a solution of Hamilton's equations satisfies $\dot{q}=v(t,q,p)$, hence $p=L_v(t,q,\dot{q})$, and $\dot{p}=L_q$ recovers the Euler-Lagrange equations. This proves local equivalence of the two descriptions.

On the cotangent bundle, choose the canonical symplectic form ${\omega }_{\mathrm{can}}={\sum }_{i}d{q}^i\wedge d{p}_i=-d\lambda$ and the convention ${\iota }_{X_H}{\omega }_{\mathrm{can}}=dH$. Then $X_H={\sum }_i(H_{p_i}{\partial }_{q^i}-H_{q^i}{\partial }_{p_i})$, so its integral curves are exactly the phase-space equations above. This sign convention is used throughout these solutions. For a hyperregular Lagrangian, the Legendre transform in mechanics is globally invertible and gives global equivalence; regularity alone only gives local equivalence. A singular velocity Hessian matrix may instead produce constraints, so the ordinary unconstrained argument does not apply to every Lagrangian.

For ordinary conservative classical mechanics, the Lagrangian is $L(q,\dot{q},t)=T-V$, where $T$ is the kinetic energy and $V$ the potential energy. More generally it is the function whose action $\int Ldt$ satisfies the principle of stationary action. The canonical momenta and Hamiltonian areFor a regular Lagrangian, the first equations can be inverted to express the generalized velocities in terms of $q,p,t$; this defines the Legendre transform on phase space.

Along a solution of the Euler-Lagrange equations, ${\dot{p}}_i=\partial L/\partial q_i$. Taking a total time derivative givesThus an autonomous Lagrangian hasFor the usual time-independent mechanical coordinates and a velocity-independent potential energy, the kinetic energy is quadratic in the generalized velocities, so ${\sum }_i{p}_i{\dot{q}}_i=2T$ and $H=T+V$. The conservation proof itself uses the Euler-Lagrange equations, rather than assuming this special expression for energy.