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Let $Q$ be a smooth manifold, let $TQ$ be its tangent bundle and let $T^*Q$ be its cotangent bundle. Lagrangian mechanics takes place on level of the level of velocities $L:TQ\rightarrow \mathbb{R}$. Hamiltonian mechanics takes place on the level of momenta $H:T^*Q\rightarrow \mathbb{R}$. If the configuration manifold, $Q$, is finite dimensional then so is its tangent and cotangent bundle. For simplicity, if the configuration manifold is a Riemannian manifold (as is the case in most mechanical examples), then the Legendre transform is a diffeomorphism between the tangent and cotangent bundles. With a little more work one can show that Hamiltonian and Lagrangian dynamics are equivalent.

Are there cases where the Lagrangian and Hamiltonian are not equivalent. Are there cases there the Lagrangian and Hamiltonian are locally equivalent but not globally equivalent?

Qmechanic
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TheRefrigerator
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1 Answers1

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  • First of all, the definition of the Hamiltonian $H(q,p,t)$ as the Legendre transformation of the Lagrangian $L(q,v,t)$ might not be well-defined, cf. e.g. this Phys.SE.

  • Since the Legendre transformation is defined pointwise wrt. position $q$ and time $t$, the definition is local wrt. $q$ and $t$, so it is not an issue of local versus global in that sense per se.

  • As explained on the Wikipedia page and in my Phys.SE answer here, there are at least two definitions of Legendre transform:

    1. The Legendre-Fenchel transform. It might be singular, but it is always unique. If the Lagrangian is not convex in velocities $v$, then the double Legendre-Fenchel transform does not give the Lagrangian back, i.e. the Legendre-Fenchel transform is then not involutive.

    2. The other definition is not necessarily unique, and may lead to several branches. An example is given in my Phys.SE answer here.

Qmechanic
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