Let $H(z)$ be a Hamiltonian and $\omega_{ij}$ the symplectic form on the phase-space and $\omega^{ij}$ its inverse $\omega_{ij} \omega^{jk} = \delta^k_i$. We know that the Hamilton's equations are then given as $$\dot{z}^i = \omega^{ij} \partial_j H$$ In canonical coordinates $z\to p_i,q^j$ we just have $$\omega^{ij} = \begin{pmatrix} 0 & -\mathbf{1} \\ \mathbf{1} & 0 \end{pmatrix}$$ and thus the usual coordinate form of Hamilton's equations and the Legendre transformation $$L = p_i \frac{\partial H}{\partial p_i} - H(p,q)$$ However, there exist systems (one example would be a Hamiltonian for spinning tops) where $\omega^{ij}$ cannot be globally put into the canonical form given above. How does one then execute a Legendre transform?

In other words: Is there a closed general formula for a Lagrangian $L$ in terms of $H$, general phase-space coordinates $z$, and the symplectic form $\omega_{ij}$?

To add some context: What I want in fact is to write the action on the phase space $$S[z(t)] = \int p_i \dot{q}^i - H(p,q) \mathrm{d}t \,,$$ where $\dot{q}^i$ is not given in terms of phase-space variables. This is then useful in the variational approach to symplectic structure as discussed e.g. by Marsden et al. (1986).

  • $\begingroup$ at first glance, I suspect you cannot do so in general: the Legendre transformation only makes sense once you've distinguished a set of momentum variables, which may not be possible globally if your manifold isn't a cotangent bundle $\endgroup$
    – Christoph
    Aug 29, 2017 at 10:21
  • $\begingroup$ In which direction do you want the Legendre transformation? From Lagrangian to Hamiltonian, or the other way? $\endgroup$
    – Qmechanic
    Aug 29, 2017 at 10:24
  • $\begingroup$ @Qmechanic I have made the statement of the question more clear now, the direction is $H \to L$. $\endgroup$
    – Void
    Aug 29, 2017 at 10:26

1 Answer 1

  1. Given a $2n$-dimensional symplectic manifold $(M,\omega)$, $$\mathrm{d}\omega~=~0 \tag{1}$$ with globally defined Hamiltonian function $H: M \to \mathbb{R}$.

  2. Note that there is no unique notion of position and momentum variables, even locally. So the inverse Legendre transformation from the Hamiltonian to the Lagrangian formalism is not a unique or well-defined notion. But there is no need to perform an inverse Legendre transformation: We can still construct a Hamiltonian action, as shown in my Phys.SE answer here. Here we will just repeat the main action formula (4).

  3. Locally in a contractible open coordinate neighborhood $U\subseteq M$ there exists a symplectic potential 1-form $$\vartheta ~=~\sum_{I=1}^{2n}\vartheta_I~\mathrm{d}z^I ~\in~ \Gamma(T^{\ast}M|_U),\tag{2}$$ such that $$ \omega|_U ~=~\mathrm{d}\vartheta .\tag{3}$$

  4. Given a path $\gamma \subset U$. Define the local Hamiltonian action $$S_U[\gamma]~:=~\int_{\gamma} \left( \vartheta - H ~\mathrm{d}t\right) ~=~\int_{t_i}^{t_f}\! \mathrm{d}t\left( \sum_{I=1}^{2n}\vartheta_I~\dot{z}^I -H\right). \tag{4}$$


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