# Hamiltonian form of Noether's Theorem

I understand that Noether's Theorem has a Hamiltonian form, whereby {X, H} = 0 iff {H, X} = 0.

The proof of this is trivial, as it follows from the antisymmetry of the Poisson Brackets.

First question:

It remains a question on my mind whether every symmetry, which transforms q(t) into q(t) + k(q(t)) truly has a representation as X(q,p) on the phase space.

In the Butterfield paper, they rely on recitfication to make q a cyclic coordinate, in which case the problem is trivial. Is there an easier way to see it in the general case?

Second question:

What about general non-symmetry transformations? Do they always have a representation X(q,p) on phase space? In other words, suppose I have a know L(q, q'), and p(q, q').

Suppose I define a transformation that changes q(t) into q(t) + k(q(t)), and therefore changes q'(t) into q'(t) + q'(t)dk/dq. Does there exist a corresponding X(q, p) on phase space such that dX/dq = -k(q) and dX/dp = q'(t) dk/dq (after changing variables from q' to p)

• Concerning the Hamiltonian form of Noether's Theorem, see also this Phys.SE post. – Qmechanic Jun 23 '18 at 11:03

## 1 Answer

OP essentially wrote (v1):

What about general non-symmetry transformations? Do they always have a representation [generated by some function $H(q,p)$] on phase space?

Mathematically, OP is asking:

Let $X\in \Gamma(TM)$ be a vector field on a symplectic manifold $(M,\omega)$. Is $X$ a Hamiltonian vector field $X=X_{H}$, where $H\in C^{\infty}(M)$ is a Hamiltonian generator$^1$ of the transformation?

Answer: No, not necessarily. A Hamiltonian vector field $X_H$ always conserves the symplectic structure, which means that the Lie derivative of the symplectic two-form $\omega$ vanishes

$$\tag{1} {\cal L}_{X}\omega ~=~0.$$

Hence condition (1) is a necessary condition for a vector field $X$ to be a Hamiltonian vector field. It turns out that the condition (1) is also sufficient for $X$ to be locally a Hamiltonian vector field, e.g. in a sufficiently small neighborhood $U \subseteq M$. However, condition (1) is in general not sufficient to ensure that $X$ has a globally defined Hamiltonian generator $H\in C^{\infty}(M)$.

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$^1$ For instance, if $X$ is the vector field for time evolution, then $H$ is the actual Hamiltonian of the system.

• Excellent response, that totally cleared it up for me. – Steven Xu Dec 3 '13 at 16:20