A Question about Graph of Function under Diffeomorphism I have a silly question about the graph of a function, which I accidentally encountered in physics.
Consider the graph of the function $q(t)$ in the $q-t$-plane, then it can also be viewed as a curve $\gamma:\mathbb{R}\rightarrow\mathbb{R}^{2}$ in the $q-t$-plane parameterized by time $t$. The curve $\gamma(t)$ then has an local expression $\left\{q(t),t\right\}\in\mathbb{R}^{2}$. In the $q-t$-plane, when treating $q$ and $t$ as independent variables, we consider the following change of coordinates:
Let $\Phi:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ be a diffeomorphism, with the following expression in local coordinates
$$\hat{q}=\Phi^{1}(q,t),\quad\hat{t}=\Phi^{2}(q,t),$$
where $\left\{q,t\right\}\in\mathbb{R}^{2}$, and $\left\{\hat{q},\hat{t}\right\}\in\mathbb{R}^{2}$. Since $\Phi$ is a diffeomorphism, which has an inverse $\Psi$, one can locally solve $q$ and $t$:
$$q=\Psi^{1}(\hat{q},\hat{t}),\quad t=\Psi^{2}(\hat{q},\hat{t}).$$
Under the diffeomorphism $\Phi$, the image of the curve $\gamma(t)$ is mapped to a one dimenional subset of $\mathbb{R}^{2}$ in the $\hat{q}-\hat{t}$-plane. This one dimensional subset in the $\hat{q}-\hat{t}$-plane is the image of some curve $\hat{\gamma}$.
My question is: can $\hat{\gamma}$ in general be locally expressed as a graph $\left\{\hat{q}(\hat{t}),\hat{t}\right\}$?
If the answer is yes, then I am interested in the following case. Consider a continuous one-parameter family of diffeomorphism $\Phi_{\epsilon}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$, such that $\Phi|_{\epsilon=0}=\mathrm{id}$. Then one has
$$q_{\epsilon}=q+\epsilon\eta(q,t)+\mathcal{o}(\epsilon),\quad t_{\epsilon}=t+\epsilon\xi(q,t)+\mathcal{o}(\epsilon),$$
where $\eta$ and $\xi$ are functions $\mathbb{R}^{2}\rightarrow\mathbb{R}$.
In physics, the variation usually means "equal time" variation, i.e
$$\bar{\delta}q(t):=\lim_{\epsilon=0}\frac{1}{\epsilon}\left(q_{\epsilon}(t)-q(t)\right)=\eta(q(t),t).$$
In the most general case when time $t$ is also transformed under diffeomorphism $\Phi_{\epsilon}$, one can define the "total variation",
$$\delta q(t)=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left(q_{\epsilon}(t_{\epsilon})-q(t)\right).$$
Question: how is the equal time variation $\bar{\delta}q(t)$ related with the total variation $\delta q(t)$?
 A: I got some clues of my second question. Here is what I have found so far.
From equations
$$q_{\epsilon}(t)=q(t)+\epsilon\eta(q(t),t)+\mathcal{o}(\epsilon),\quad t_{\epsilon}(t)=t+\epsilon\xi(q(t),t)+\mathcal{o}(\epsilon),$$
one finds
\begin{align}
\delta q(t)&=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left(q_{\epsilon}(t_{\epsilon})-q(t)\right) \\
&=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left(q_{\epsilon}(t_{\epsilon})-q_{\epsilon}(t)+q_{\epsilon}(t)-q(t)\right) \\
&=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}(q_{\epsilon}(t_{\epsilon})-q_{\epsilon}(t))+\bar{\delta}q(t),
\end{align}
where $q_{\epsilon}(t_{\epsilon})=q_{\epsilon}(t+\epsilon\xi(q(t),t)+\mathcal{o}(\epsilon))$ can be Taylor expanded, i.e.
$$q_{\epsilon}(t_{\epsilon})=q_{\epsilon}(t)+\epsilon\xi(q(t),t)\dot{q}_{\epsilon}(t)+\mathcal{o}(\epsilon).$$
Therefore,
$$\delta q(t)=\xi(q(t),t)\dot{q}(t)+\bar{\delta}q(t):=\dot{q}(t)\bar{\delta}t+\bar{\delta}q(t),$$
where $\bar{\delta}t=\xi(q(t),t)$.
Similarly, one finds the difference between the total variation and the equal time variation of the Lagrangian.
\begin{align}
\delta L(q(t),\dot{q}(t),t):&=\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon}\left[L(q_{\epsilon}(t_{\epsilon}),\dot{q}_{\epsilon}(t_{\epsilon}),t_{\epsilon})-L(q(t),\dot{q}(t),t)\right]\\
=\frac{dL}{dt}\bar{\delta}t+\bar{\delta}L,\\
\end{align}
where $\bar{\delta}L=\frac{\partial L}{\partial q}\bar{\delta}q(t)+\frac{\partial L}{\partial\dot{q}}\bar{\delta}\dot{q}(t)$ is the usual equal time variation in classical mechanics.
Since equal time variation commutes with time derivative, one has
$$\delta L=\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\bar{\delta}q(t)+\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\bar{\delta}q\right)+\frac{dL}{dt}\bar{\delta}t \tag{A}$$
Suppose the diffeomorphism $\Phi_{\epsilon}$ is a gauge redundancy of the system, then the action should stay the same under an arbitrary infinitesimal diffeomorphism. i.e.
\begin{align}
&S[q,t]=\int_{t_{1}}^{t_{2}}dt L(q(t),\dot{q}(t),t) \\
&=S[q_{\epsilon},t_{\epsilon}]=\int_{t_{\epsilon}(t_{1})}^{t_{\epsilon}(t_{2})}dt_{\epsilon} L(q_{\epsilon}(t_{\epsilon}),\dot{q}_{\epsilon}(t_{\epsilon}),t_{\epsilon}) \\
&=\int_{t_{1}}^{t_{2}}dt\frac{dt_{\epsilon}}{dt}L(q_{\epsilon}(t_{\epsilon}),\dot{q}_{\epsilon}(t_{\epsilon}),t_{\epsilon}(t)).
\end{align}
Thus, one has
$$\frac{dt_{\epsilon}}{dt}L(q_{\epsilon}(t_{\epsilon}),\dot{q}_{\epsilon}(t_{\epsilon}),t_{\epsilon}(t))=L(q(t),\dot{q}(t),t)+\frac{d}{dt}\Sigma_{\epsilon}(q(t),t),$$
where $\Sigma_{\epsilon}$ is some possible boundary term that, in general, depends on the parameter $\epsilon$.
From previous equations, one has
$$\frac{dt_{\epsilon}}{dt}=1+\epsilon\dot{\xi}(q(t),t)+\mathcal{o}(\epsilon),\quad L(q_{\epsilon}(t_{\epsilon}),\dot{q}_{\epsilon}(t_{\epsilon}),t_{\epsilon})=L(q(t),\dot{q}(t),t)+\epsilon\delta L+\mathcal{o}(\epsilon).$$
Also, assume that the boundary term takes the form
$$\Sigma_{\epsilon}(q,t)=\Sigma_{0}(q,t)+\epsilon\sigma(q,t)+\mathcal{o}(\epsilon).$$
Then one finds the following equation
$$\delta L+\dot{\xi}L=\frac{d\sigma}{dt} \tag{B}.$$
Combining equation (A) and (B), one finds
$$\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\right)\right)\bar{\delta}q(t)+\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}}\bar{\delta}q(t)+L\bar{\delta}t-\sigma(q(t),t)\right)\equiv 0.$$
The first term is the equation of motion, which vanishes on-shell. The last term inside the parentheses is thus a charge which is conserved on-shell.
But this is not what I expected to find in the first place. Noether's (first) theorem says that for each global symmetry, there exists a gauge-invariant charge which is conserved on-shell. Here what I have been studying is a (local) gauge redundancy.
For gauge redundancies on the other hand, one usually finds:

*

*Charge conserved on-shell but is not gauge invariant.


*Gauge invariant charge which is conserved universally.
An example of the first scenario is $U(1)$-gauge in QED. Under a gauge transformation $\delta A_{\mu}=\partial_{\mu}\Lambda$, one finds an on-shell current which takes the form
$$J^{\mu}=\frac{\partial\mathcal L}{\partial(\partial_{\mu}A_{\nu})}\delta A_{\nu}=\partial_{\nu}\left(F^{\mu\nu}\Lambda(x)\right).$$
It is manifestly conserved, but is not gauge invariant.
An example of the second scenario is the Hilbert stress-energy tensor, which is universally conserved due to diffeomorphism invariance, and is also gauge invariant.
The charge from the above equation is
\begin{align}
Q(q,\dot{q},t)&=\frac{\partial L}{\partial\dot{q}}\bar{\delta}q+L\bar{\delta}t-\sigma \\
&=p\bar{\delta}q+L\bar{\delta}t-\sigma \\
&=p\delta q+(L-p\dot{q})\bar{\delta}t-\sigma \\
&=p\delta q+H(p,q)\bar{\delta}t-\sigma.
\end{align}
How does this charge transform under diffeomorphism $\Phi_{\epsilon}$?
