Suppose we have a classical Lagrangian $L(q,\dot{q})$. Here $q = q(s,t)$ is a generalized coordinate as a function of time and some parameter $s$ corresponding to a transformation. If this is a symmetry transformation, $L$ by definition changes by a total time derivative: $L' = \dot{X}$ where $'$ means $\frac{\partial}{\partial s}$ and $\cdot$ means $\frac{\partial}{\partial t}$. Noether's theorem then says that the quantity $$ Q = \frac{\partial L}{\partial \dot{q}} q' - X $$ is conserved ($\dot{Q} = 0$), assuming the equations of motion (Euler-Lagrange equations).

Passing to the Hamiltonian formulation of mechanics, we define the canonical momentum $p = \frac{\partial L}{\partial \dot{q}}$ and the Poisson bracket $$\{F,G\} = \frac{\partial F}{\partial q} \frac{\partial G}{\partial p} - \frac{\partial F}{\partial p} \frac{\partial G}{\partial q}$$ for any functions $F(q,p)$ and $G(q,p)$, and we can rewrite $L$ and $Q$ in terms of $q$ and $p$ as $L(q,p)$ and $Q(q,p)$.

I wish to show that the conserved quantity $Q$ (or possibly some multiple of it?) generates the symmetry transformation, meaning $F' = \{F, Q\}$ for any $F$. It suffices to show that $q' = \{q, Q\}$ and $p' = \{p, Q\}$.

I try to show the first part: I expand $$\{q, Q\} = \{q, pq' - X\} = q' + p\{q,q'\} - \{q, X\} = q' + \frac{\partial q'(q,p)}{\partial p} - \frac{\partial X(q,p)}{\partial p}.$$ From here, it would seem I need to show $\frac{\partial}{\partial p}\left(q'(q,p) - X(q,p)\right) = 0$. This isn't straightforward since the functions involved are defined very implicitly. Any pointers on how to show this?

We could look at the special case where $s = t$ (describing time translation): We have $X = L$ and $Q = H$. The Euler-Lagrange equations then give us $$p' = \dot{p} = \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q} = -\frac{\partial H}{\partial q} = \{p, H\}.$$ But this derivation is shady as I have freely confused $L(q,\dot{q})$ with $L(q,p)$ and the same for $H$.

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    $\begingroup$ See Statement 1 in my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Commented Feb 25, 2018 at 17:44

1 Answer 1


Let's consider a transformation $$ q \to q' =q - \delta q$$ $$ p \to p'=p $$ $$ t \to t'=t .$$

The corresponding Noether charge reads

$$Q = \frac{\partial L}{\partial \dot{q}} \delta q - X ,$$ where $X$ is the usual function whose total derivative we are allowed to add to the Lagrangian.

In general, a generator $G$ is related to a finite transformation by $ g = e^{G} = 1 + G + \ldots$. In other words, generators cause infinitesimal transformations:

$$ q \to g_{inf}\circ q = e^{\epsilon G} \circ q = (1 + \epsilon G ) \circ q $$

We therefore say that $Q$ generates the transformation in phase space defined above if $$ (1 + \delta q Q ) \circ q = (1+\delta q) .$$ The natural product in phase space is given by the Poisson bracket and therefore our goal is to check $$ \{ q,Q\} \stackrel{!}{=} \delta q .$$ Using the definition of the Poisson bracket $$\{ A,B \} \equiv \frac{\partial A}{\partial p} \frac{\partial B}{\partial q} - \frac{\partial A}{\partial q} \frac{\partial B}{\partial p} \, .$$ and the formula for the Noether charge from above, this can be shown explicitly: \begin{align} \{ q,Q \} &= \frac{\partial q}{\partial p} \frac{\partial Q}{\partial q} - \frac{\partial q}{\partial q} \frac{\partial Q}{\partial p} \\ &= - \frac{\partial q}{\partial q} \frac{\partial Q}{\partial p} \\ &= - \frac{\partial Q}{\partial p} \\ &= - \frac{\partial ( \frac{\partial L}{\partial \dot{q}} \delta q - X)}{\partial p} \\ &= - \frac{\partial ( \frac{\partial L}{\partial \dot{q}} \delta q )}{\partial p} \\ &= - \frac{\partial ( p \delta q )}{\partial p} \\ &= - \delta q \quad \square \\ \end{align}


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