Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Group of symmetries of Lagrange's equations

+ 4 like - 0 dislike
1502 views

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

+ 5 like - 0 dislike

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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOv$\varnothing$rflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...