Proving that Noether charge generates symmetries in the Lagrangian formalism

Question

I know that similar questions have been asked on this site before, but I haven't been able to find the answer to my specific question.

I want to show that the Noether charge defined in Lagrangian formalism generates corresponding symmetries. More, precisely:

Assume we have a Lagrangian $L(q(t),\dot q(t))$. Let us assume that, under an infinitesimal transformation $$\delta q(t) = \eta(q(t),\dot q(t),t), \ \delta \dot q (t) = \frac{d}{dt}\eta(q(t),\dot q(t),t)\tag{1}$$ the change of Lagrangian is given as: $$\delta L = \frac{d}{dt}K(q(t),\dot q(t),t)\tag{2}$$ Then the Noether charge $Q(q(t),\dot q(t),t)$ is defined by: $$Q(q(t),\dot q(t),t) := \frac{\partial}{\partial \dot q(t)}L(q(t),\dot q(t))\ \eta(q(t),\dot q(t),t) - K(q(t),\dot q(t),t)\tag{3}$$ The claim is that, if the define canonical momentum by $p(q,\dot q)= \frac{\partial}{\partial \dot q}L(q,\dot q)$: $$\delta q(t) = \left(\frac{\partial}{\partial p(t)}Q(q(t),\dot q(t),t)\right)_{q(t)},\qquad \delta p(t) = -\left(\frac{\partial}{\partial q(t)}Q(q(t),\dot q(t),t)\right)_{p(t)}\tag{4}$$ I was able to derive the first of above relations as follows. From, $$\delta L = \frac{\partial}{\partial q}L \delta q(t) + p(t) \delta \dot q(t)\tag{5},$$ by integrating by parts, we obtain: $$ \frac{d}{dt}Q = \delta q \left(\dot p - \frac{\partial}{\partial q}L \right) \\ \quad \quad \quad \quad \quad \ \ =\delta q \left(\frac{\partial p}{\partial q}\dot q + \frac{\partial p}{\partial \dot q}\ddot q - \frac{\partial}{\partial q}L \right) \\ \quad \quad \quad \ = \frac{\partial Q}{\partial t} + \frac{\partial Q}{\partial q}\dot q + \frac{\partial Q}{\partial \dot q}\ddot q\tag{6}$$ By equating terms proportional to $\ddot q$, we obtain: $$ \frac{\partial Q}{\partial \dot q} = \left( \frac{\partial p}{\partial \dot q} \right)_{q} \delta q \tag{7}$$ Then: $$ \left(\frac{\partial Q}{\partial p}\right)_{q} = \frac{\partial Q}{\partial \dot q} \left(\frac{\partial \dot q}{\partial p}\right)_q = \left( \frac{\partial p}{\partial \dot q} \right)_{q} \left(\frac{\partial \dot q}{\partial p}\right)_q \delta q = \delta q\tag{8}$$ My question is how we can show that $$\delta p(t) = -\left(\frac{\partial}{\partial q(t)}Q(q(t),\dot q(t),t)\right)_{p(t)}\tag{9}$$

UPDATE

It seems that it is required to assume equations of motion to obtain above identity. Consider the identity (obtained after removing the terms that include $\ddot q$ in $\frac{dQ}{dt} = \cdots$): $$ \delta q \left(\frac{\partial p}{\partial q}\dot q - \frac{\partial}{\partial q}L \right) = \frac{\partial Q}{\partial t} + \frac{\partial Q}{\partial q}\dot q \tag{10}$$ Now take partial derivative with respect to $\dot q$, and use commutativity of partial derivatives and the identity found above for $\frac{\partial Q}{\partial \dot q}$ to obtain: $$\frac{\partial Q}{\partial q} = \left(\dot q \partial_q p - \partial_q L \right)\partial_{\dot q}\delta q - \partial_t \delta q\partial_{\dot q} p - \dot q\partial_{\dot q}p \partial_q \delta q \tag{11}$$ Now, using $$\left(\frac{\partial}{\partial q}\right)_p = \frac{\partial}{\partial q} + \left(\frac{\partial \dot q}{\partial q}\right)_p\tag{12}$$ and $$\left(\frac{\partial \dot q}{\partial q}\right)_p = -\frac{\partial_{q}p}{\partial_{\dot q} p}\tag{13}$$ we find: $$\left(\frac{\partial Q}{\partial q}\right)_p = - \partial_q p \delta q - \partial_{\dot q}p \left(\dot q\partial_q \delta q + \partial_t \delta q \right) + \left(\dot q \partial_q p - \partial_q L \right) \tag{14}$$ If we assume equations of motion, that is: $$\partial_q L = \frac{d}{dt}p = \dot q \partial_q p + \ddot q \partial_{\dot q} p\tag{15}$$ We obtain: $$\left(\frac{\partial Q}{\partial q}\right)_p = - \partial_q p \delta q - \partial_{\dot q}p \left(\ddot q \partial_{\dot q} \delta q + \dot q\partial_q \delta q + \partial_t \delta q \right) = -\delta p \tag{16}$$ However, I can't see the physical meaning of this assumption.

Answer 1

1. In this answer let us for simplicity restrict to the case of a regular Legendre transformation in a point mechanical setting, cf. this related Phys.SE post. (Generalizations to field theory and gauge theory are in principle possible, with appropriate modifications of conclusions.)

   Within the Lagrangian formalism, let there be given an infinitesimal vertical quasi-symmetry of the form $$\delta q^i ~=~y^i(q,\dot{q},t)\varepsilon, \qquad i~\in~\{1,\ldots,n\}, \qquad\delta t~=~0. \tag{A} $$ Noether's theorem then yields a corresponding Noether charge $Q_L(q,\dot{q},t)$ that is conserved on-shell.

2. OP seems to essentially be asking the following question.

   Does the Noether charge $Q_L(q,\dot{q},t)$ generate the quasi-symmetry (A)?

   The answer is: Yes, the quasi-symmetry $$y^i(q,\dot{q},t)~=~\{q^i(t),Q_L(t)\}, \qquad i~\in~\{1,\ldots,n\},\tag{B} $$ is generated via the Peierls bracket $$\{ F,G \}~:=~\iint\!dt~dt^{\prime}~\sum_{i,k=1}^{n} \frac{\delta F }{\delta q^i(t)}~G^{ik}_{\rm ret}(t,t^{\prime})~\frac{\delta G }{\delta q^k(t^{\prime})} - (F\leftrightarrow G),\tag{C} $$ where $G^{ik}_{\rm ret}(t,t^{\prime})$ is the retarded Green's function, see e.g. various textbooks by Bryce S. DeWitt and this Phys.SE answer by user AccidentalFourierTransform. Unfortunately, within the Lagrangian formalism, we do not know the explicit form of the retarded Green's function $G^{ik}_{\rm ret}(t,t^{\prime})$, except in special cases. This makes OP's purely Lagrangian endeavour challenging & calculations cumbersome.

3. However, we know that there is a bijective correspondence to a corresponding Hamiltonian formulation via a Legendre transformation, cf. e.g. this Phys.SE post. Moreover, there exists a bijective correspondence between conserved quantities $$Q_L(q,\dot{q},t) ~\approx~Q_H(q,p,t)\tag{D}$$ in the Lagrangian and the Hamiltonian formulations, cf. this Phys.SE post. A similar bijective correspondence exists for the corresponding quasi-symmetries. Therefore everything can be translated to the corresponding Hamiltonian formulation. We shall apply this strategy in this answer.

4. In the Hamiltonian setting, OP's question becomes Statement 1 in my Phys.SE answer here, i.e. $$Y^I(z,t)~=~\{z^I(t),Q_H(t)\},\qquad I~\in~\{1,\ldots,2n\} .\tag{E} $$ Within the Hamiltonian formalism, the retarded Green's function $$G^{IK}_{\rm ret}(t,t^{\prime}) ~\simeq~\omega^{IK}~\theta(t\!-\!t^{\prime}),\qquad I,K~\in~\{1,\ldots,2n\},\tag{F} $$ is known explicitly in the adiabatic limit (where the Hamiltonian $H$ can be ignored). Here $\omega^{IK}$ denotes the symplectic unit. Therefore, the equal-time Peierls bracket becomes the canonical Poisson bracket.

Answer 2

I had a similar question and found my answer in the book "Classical Mechanics: Hamiltonian and Lagrangian Formalism" by Alexei Deriglazov. You can find it here. In section 7.12.3, he proves the equivalence between the Lagrangian and Hamiltonian symmetry transformations.

In short: Given a Hamiltonian Noether charge $Q_H(q,p,t)$, the Lagrangian charge $Q_L(q, \dot{q},t)$ is obtained through the transformation:

$\delta q = - \{q,Q_H\}\Big|_{p \to \frac{\partial L}{\partial \dot{q}}} = - \frac{\partial Q_H}{\partial p}\Big|_{p \to \frac{\partial L}{\partial \dot{q}}} = - \frac{\partial Q_H}{\partial \left(\frac{\partial L}{\partial \dot{q}} \right)}$