Once you've identified the conserved quantities, the question becomes whether you can identify the symmetry group they generate. Then you write down the Lagrangian that is invariant under this group-and check that you, indeed, recover the equations you started with as the corresponding Euler-Lagrange equations.

I'd recommend starting with a specific example, speaking in general isn't helpful.

Thanks Nicolis. Of course I tried to find some symmetric groups, but the result turns out the deduced Lagrangian can always be simplified to zero. So I'm sure it is hard to find a proper symmetric group. This raised my question.

Here you have to be careful, because the Lagrangian *can* vanish for fields that satisfy the equations of motion-but that isn't a problem. Take as an example the Dirac Lagrangian, that does have this property.

Could you clarify what is given in your problem? If you have some symmetry group, then indeed, as Stam says, at least in principle it is possible to describe all lagrangians that admit it, but then you also mention the dynamics, which confuses me.

How do you identify your conserved quantities? Do you already have all equations of motions? If not, how can you know that your conserved quantities are conserved??

Here are some examples those can be checked using Noether's

theorem, I think. I did this course many years ago; our teacher

followed Landau Lifshitz.

1) Conservation of 3D linear momentum is a result of translational

symmetry in 3D.

2) Conservation of angular momentum in 3D is a result of rotation

symmetry in 3D.

In four dimensional theories a slightly more complicated example is

1) Conservation of electric charge is a result of U(1) local

gauge symmetries. Here electric charge has to be defined

in terms of conserved electric current

In all these cases symmetries are continuous.

Thanks, everybody. Could you see my edited question?

The Noether theorem assumes that your equations of motion can be derived from a Lagrangian. That is not the case if the Lagrangian is zero, or a total derivative.

A special case of your example is the system^* q' = p, p' = -q (which is in Hamiltonian form). They can be derived from the Lagrangian L = p q' - (p^2 + q^2)/2, and have a conserved quantity H= (p^2 + q^2) due to invariance under time translation.

^* Note that I use ' for time derivatives.

With

A= ((0,1),(-1,0)) one can write H = (p,-q) A (q,p)^T, where ^T denotes transposition. A general antisymmetric A can be transformed to (essentially) a block diagonal repetition of this case, with one conservation law per block.

Whether a given set of equation can be derived from a Lagrange function (via Hamiltons's principle), possibly after change of variables, is a relevant question which I don't know how to answer in general.

Noether theory may be considered as a "functional" Lie theory. One has to follow the 3 Lie theorems with its inverse versions, as in, e.g, Pontryagin 'Continuous Groups'.

There's something not quite clear about the example presented: if dx/dt =A x then

d(dx/dt)/dt=A dx/dt = A^2 x, so the statement that A x is ``a symmetry'' (?) is true only if A is nilpotent. You surely mean something different, as what you write further on indicates. What you *might* mean is that these are the equations of motion in phase space and A is the symplectic form. Then what you should do is ``choose a polarization'', i.e. take half of the equations as definitions of the momenta. From the symplectic form you can find the Hamiltonian and, from that, by a Legendre transform, the Lagrangian.

If the equations are linear, so A is a constant matrix, then things are simple, of course. They're slightly more complicated if A isn't of full rank-which is the case, when symmetries are present. Look in Abraham and Marsden, ``Foundations of Mechanics'' for more details.

Example: the harmonic oscillator: dx/dt =p, dp/dt = -x. The symmetry is time translations.

We readily find that H = (1/2)(p^2+ x^2) is a constant of motion and choosing, for instance, p = dx/dt as the momentum, the Legendre transform gives L = p (dx/dt)-H= (1/2)( (dx/dt)^2-x^2), which is the expected answer.

(The fact that the canonical equations preserve the symplectic form means that the momenta (and positions) at any one time are combinations of momenta and positions at another time.)

Incidentally, your statement, p=x is false, if x is a commuting coordinate; it is true, if x is an anticommuting coordinate.

I completely agree with Stam Nicolis. I am also not sure what Shuchang means by "symmetry is Ax" in his example. Symmetry usually refers to a group acting on the solution manifold which leave it invariant. Is A a representation of the group under consideration?

I have one question for Stam Nicolis. Is the Hamiltonian obtain from the equation of motion (via polarization) always convex (concave)? If it is (globally), the Legendre transform, to obtain the Lagrangian, is well defined. Perhaps local convexity (concavity) is enough?

@Hansen Chen: I don't have the answer at hand, but I'd say that working out an example might be useful. Take V(x)=(x^2 -a^2)^2, for instance, compared to V(x)=(x^2+a^2)^2, with a real (and we're in classical mechanics). Wouldn't we expect, in both cases, to be able to find H = (p^2/2)+ V(x) and L = (dx/dt)^2-V(x) with one choice of polarization for both cases? Look in Sidney Coleman's Erice lectures, ``Aspects of symmetry'' , ``Secret symmetry''-where he discusses the effective potential (not only on field theory).

@Stam Nicoles: Your example does not answer my question, because there H and L are globally convex (in fact quadratic) with respect to p and dx/dt respectively, while my question asks whether H derived from an equation of motion is always convex. To be more specific, the Legendre transform of a convex function is well (uniquely) defined. The Legendre transform of a convex function is also convex. And the Legendre transform on convex function space is an involution, i.e. of the twice Legendre transform on convex functions is the identity transform. It is not well defined on non-convex functions.

So you're asking about non-quadratic dependence in the velocities, or momenta?

It would be useful to talk about a concrete example, talking in general terms isn't that useful. If the kinetic term of your Lagrangian isn't positive definite, your theory has a ``ghost''.

@Stam Nicolis: "talking in general terms isn't that useful" is not true. Any abstraction is a generalization to some degree. Also, surely you know that not having a counterexample at hand does not prove a proposition is correct. Philosophy aside, that is my question, which is whether a Hamiltonian is always convex in p. I would like to think so, but I do not know. If I already have a counterexample where a Hamiltonian is not only non-quadratic but also nonconvex in p, I would not have asked the question but would not have stated as such, would I?

So, to be specific, are you stating that, not only a Hamiltonian have to be convex but stringently quadratic and nonnegative definite?

Well yes, as I said before, if it's not ``convex in p'' (the term that's commonly used is ``positive definite kinetic term'') then it describes unphysical excitations, called ``ghosts'' in the physics literature. Physical excitations are described by Hamiltonians with positive definite kinetic terms.

The statement ``a Hamiltonian have to be convex but stringently quadratic and nonnegative definite?'' requires qualification, namely:

The Hamiltonian of a physical system is bounded from below-if it *isn't*, this means that either the kinetic energy or the potential energy are unbounded from below. In the first case this implies unphysical excitations that are ghosts, indicating that degrees of freedom are not counted correctly. In the second case it implies that the excitations described by this Hamiltonian aren't the whole story, namely the scattering states must be taken into account more carefully.

These considerations imply that a Hamiltonian that describes a *complete* system is convex and non-negative definite. The potential energy need not be quadratic, the kinetic energy must be. The kinetic energy must be ``convex''.

These statements, of course, imply that we're talking about degrees of freedom, that can be idenitified as ``free particles'', that's what the identification of the kinetic term *means*.

If it doesn't satisfy these conditions then the question is what can we deduce about the missing degrees of freedom, that would provide a *complete* description?