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 $H$.

Related questions:

  • 1
    $\begingroup$ See Statement 1 in my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Feb 25 '18 at 17:44

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.