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  Group of symmetries of Lagrange's equations

+ 4 like - 0 dislike

Consider the following statements, for a classical system whose configuration space has dimension $d$:

  1. Lagrange equations admit a smaller group of "symmetries" (coordinate change under which equations are formally unchanged) than Hamilton's;

  2. The 'symplectic diffeomorfism' (=coordinate changes whose jacobian is a symplectic $d$-parametric matrix) Lie group has dimension greater than $\dim G$, $G$ being the (Lie?) group of symmetries of point one.

The first is well known to be true. What about the second? There exists such a $G$ (at a first sight it seemed to me to be the whole $Diff(M)$; but if it is so, then 2 is false)? If it is true, can point 2 explain point 1?

This post has been migrated from (A51.SE)
asked Dec 13, 2011 in Theoretical Physics by tetrapharmakon (65 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
(I can't create the "lagrangian-mechanics" tag)

This post has been migrated from (A51.SE)
Fixed the tag...

This post has been migrated from (A51.SE)

2 Answers

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Here Hamiltonian systems on cotangent bundles $(T^{*}M, \omega_M, H)$ of a manifold $M$ will be considered.

A symmetry of the Hamiltonian system is a diffeomorphism which preserves 1) The cotangent bundle structure, 2) the canonical symplectic form $\omega_M$ and 3) the Hamiltonian $H$.

A point (or Noether) symmetry of the Lagrangian is required in addition to be generated by a vector field on $M$ whose canonical lift to $T^{*}M$ generates a Hamiltonian symmetry.

The group of symplectomorphism is only required to preserve the symplectic form and not associated with a specific hamiltonian, thus it is the largset group and in general infinite dimensional.

This post has been migrated from (A51.SE)
answered Dec 14, 2011 by David Bar Moshe (4,355 points) [ no revision ]
+ 2 like - 0 dislike

0) Let us for simplicity assume that the Legendre transformation from Lagrangian to Hamiltonian formulation is regular.

1) The Lagrangian action $S_L[q]:=\int dt~L$ is invariant under the infinite-dimensional group of diffeomorphisms of the $n$-dimensional (generalized) position space $M$.

2) The Hamiltonian action $S_H[q,p]:=\int dt(p_i \dot{q}^i -H)$ is invariant (up to boundary terms) under the infinite-dimensional group of symplectomorphisms of the $2n$-dimensional phase space $T^*M$.

3) The group of diffeomorphisms of position space can be prolonged onto a subgroup inside the group of symplectomorphisms. (But the group of symplectomorphism is much bigger.) The above is phrased in the active picture. We can also rephrase it in the passive picture of coordinate transformations. Then we can prolong a coordinate transformation

$$q^i ~\longrightarrow~ q^{\prime j}~=~q^{\prime j}(q)$$

into the cotangent bundle $T^*M$ in the standard fashion

$$ p_i ~=~ p^{\prime}_j \frac{\partial q^{\prime j} }{\partial q^i} ~.$$

It is not hard to check that the symplectic two-form becomes invariant

$$dp^{\prime}_j \wedge dq^{\prime j}~=~ dp_i \wedge dq^i $$

(which corresponds to a symplectomorphism in the active picture).

This post has been migrated from (A51.SE)
answered Dec 14, 2011 by Qmechanic (3,120 points) [ no revision ]

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