I am studying the Lagrangian and Hamiltonian description of some dynamical systems. The problem with this description of the particular kind of systems I am studying, is that the Legendre Transformation is degenerate. I could not find the meaning of this on googling but I guessed that there could be multiple Hamiltonians which could be generated from the Legendre transformation of the Lagrangian. I am not sure how is that possible. So in brief, what is the meaning of the 'degenerate Legendre Transformation' and it it possible for only special cases of Lagrangians describing the system (which seems easy to figure out) or in general for a particular kind of system where such a problem usually appears, such as electrical circuits (which seems difficult to figure out). Or could you provide me a reference that describes this ? Thanks. Also, suggestions for better tags is much appreciated.
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1$\begingroup$ If you are interested in the pre-symplectic structure on the space of solutions of your dynamical system (often also called the phase space), then you do not need the Legendre transform. This structure can be defined directly from the Lagrangian, which you already seem to have. That method works equally well for singular and higher order Lagrangians. See my answer to this question for details: mathoverflow.net/q/81800/2622 $\endgroup$– Igor KhavkineCommented Jul 25, 2014 at 18:11
1 Answer
I) We give here a possible definition for a $\mathbb{R}$-vector space $V$, and leave it to the reader to generalize the construction e.g. to a tangent bundle.
Let $C\subseteq V$ be a convex subset. Let there be given a function $L:C\to\mathbb{R}$. Define a new function $L_H: V^{\ast} \times C \to \mathbb{R}$ as
$$ L_H(p,v)~:=~\langle p,v\rangle - L(v).$$
Define the Legendre transform $H: V^{\ast} \to ~]-\infty,\infty]$ as
$$ H(p)~:=~\sup_{v\in C} L_H(p,v).$$
Define the domain $$C^{\ast}~:=~ \{p\in V^{\ast}\mid H(p) <\infty\}.$$
Define for all $p\in V^{\ast}$ the set
$$C(p)~:=~\{ v\in C \mid L_H(p,v)=H(p) \} . $$
Call the Legendre transformation regular or non-degenerate if
$$C^{\ast}=V^{\ast}\qquad \text{and} \qquad \forall p\in V^{\ast}:C(p)\text{ is a singleton}.$$
Call it singular or degenerate otherwise.
II) In the physics literature, one often downplays the role of convexity, cf. this Phys.SE post. In practice, physicists often use another definition of Legendre transform that doesn't rely on convexity, see e.g. this Phys.SE post and Ref. 1. Degenerate Legendre transformations are vital to the topics of gauge theory and constraint dynamics, see e.g. Ref. 2.
References:
Herbert Goldstein, Classical Mechanics.
M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.
