Proving that <PERSON> charge generates symmetries in the Lagrangian formalism
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 <PERSON> 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 $$\delta q(t) = \eta(q(t),\dot q(t),t), \ \delta \dot q (t) = \frac{d}{dt}\eta(q(t),\dot q(t),t)$$ the change of Lagrangian is given as: $$\delta L = \frac{d}{dt}K(q(t),\dot q(t),t)$$Then the Noether charge $Q(q(t),\dot q(t),t)$ is defined by: $$Q(q(t),\dot q(t),t) := \frac{\partial}{\partial \dot q(t)}L(q(t),\dot q(t))\ \eta(q(t),\dot q(t),t) - K(q(t),\dot q(t),t)$$The claim is that, if the define canonical momentum by $p(q,\dot q)= \frac{\partial}{\partial \dot q}L(q,\dot q)$: $$\delta q(t) = \left(\frac{\partial}{\partial p(t)}Q(q(t),\dot q(t),t)\right){q(t)},\ \delta p(t) = -\left(\frac{\partial}{\partial q(t)}Q(q(t),\dot q(t),t)\right){p(t)}$$ I was able to derive the first of above relations as follows. From, $\delta L = \frac{\partial}{\partial q}L \delta q(t) + p(t) \delta \dot q(t)$, by integrating by parts, we obtain: $$ \frac{d}{dt}Q = \delta q \left(\dot p - \frac{\partial}{\partial q}L \right) \ \quad \quad \quad \quad \quad \ \ =\delta q \left(\frac{\partial p}{\partial q}\dot q + \frac{\partial p}{\partial \dot q}\ddot q - \frac{\partial}{\partial q}L \right) \ \quad \quad \quad \ = \frac{\partial Q}{\partial t} + \frac{\partial Q}{\partial q}\dot q + \frac{\partial Q}{\partial \dot q}\ddot q$$ By equating terms proportional to $\ddot q$, we obtain: $$ \frac{\partial Q}{\partial \dot q} = \left( \frac{\partial p}{\partial \dot q} \right){q} \delta q $$ Then: $$ \left(\frac{\partial Q}{\partial p}\right){q} = \frac{\partial Q}{\partial \dot q} \left(\frac{\partial \dot q}{\partial p}\right)q = \left( \frac{\partial p}{\partial \dot q} \right){q} \left(\frac{\partial \dot q}{\partial p}\right)q \delta q = \delta q$$ My question is how we can show that $$\delta p(t) = -\left(\frac{\partial}{\partial q(t)}Q(q(t),\dot q(t),t)\right){p(t)}$$
UPDATE
It seems that it is required to assume equations of motion to obtain above identity. 101