A Question about Diffeomorphism Invariant Action I remember that the canonical Hamiltonian of a diffeomorphism-invariant theory, in general, is zero. For example, the geodesic equation is derived from the action of arc length $$S[g(\tau)]=\int_{a}^{b}d\tau\sqrt{g_{\mu\nu}(q)\frac{dq^{\mu}}{d\tau}\frac{dq^{\nu}}{d\tau}}, \tag{0}$$
which is invariant under reparameterization $\tau\rightarrow\tau^{\prime}$. Another example is the Einstein-Hilbert action $$S[g(x)]=\int d^{4}x\sqrt{|g|}R.$$
Yesterday, I was re-visiting the lecture notes from Edmund Bertschinger. On page 3, he studied the Lagrangian whose action suffers from diffeomorphism invariance. The author intended to show that the canonical energy is zero. However, I am very doubtful about his variation of the action.
To begin with, the action is given by $$S[q(t),t]=\int_{a}^{b}dtL(q,\dot{q},t), \tag{1}$$
which is assumed to be invariant under reparameterization of $t$. Here is my first question.
Since the the canonical momentum of the variable $t$ does not show up, does the variable $t$ play the role of a Lagrangian multiplier of Dirac's constrained system?
Next, consider an infinitesimal diffeomorphism $\phi_{\epsilon}:\tau\rightarrow t$ that is connected to the identity (i.e $\phi_{0}=\mathrm{id}$), with $\phi_{\epsilon}(\alpha)=a$, and $\phi_{\epsilon}(\beta)=b$. Denote $q\circ\phi_{\epsilon}=k_{\epsilon}$. To ensure (1) is invariant under the pull-back of the diffeomorphism $t=\phi_{\epsilon}(\tau)$, the Lagrangian $L(q,\dot{q},t)$ must be homogeneous in $\dot{q}$. This is because $\dot{q}$ behaves like a worldline-vector, i.e $$\frac{dq}{dt}=\frac{dq(\phi_{\epsilon}(\tau))}{d\tau}\frac{d\tau}{d\phi_{\epsilon}(\tau)}$$
and $dt$ picks an extra Jacobian factor: $$dt=\frac{d\phi_{\epsilon}(\tau)}{d\tau}d\tau.$$
They cancel out with each other if and only if the Lagrangian is homogeneous in $\dot{q}$, i.e
\begin{align}
& S[q(t),t]=\int_{a}^{b}dtL(q,\dot{q},t)=\int_{\phi_{\epsilon}(\alpha)}^{\phi_{\epsilon}(\beta)}dtL(q(t),\dot{q}(t),t) \\
&=\int_{\alpha}^{\beta}d\tau\frac{d\phi_{\epsilon}(\tau)}{d\tau}L\left(q(\phi_{\epsilon}(\tau)),\frac{dq(\phi_{\epsilon}(\tau))}{d\tau}\frac{d\tau}{d\phi_{\epsilon}(\tau)},\phi_{\epsilon}(\tau)\right) \\
&=\int_{\alpha}^{\beta}d\tau\frac{d\phi_{\epsilon}(\tau)}{d\tau}L\left(k_{\epsilon}(\tau),\frac{d\tau}{d\phi_{\epsilon}(\tau)}\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau)\right) \\
&=\int_{\alpha}^{\beta}d\tau\left(\frac{d\phi_{\epsilon}}{d\tau}\frac{d\tau}{d\phi_{\epsilon}}\right)L\left(k_{\epsilon}(\tau),\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau)\right) \\
&=\int_{\alpha}^{\beta}d\tau L(k_{\epsilon}(\tau),\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau))
\end{align}
This is exactly what happens with the arc length action (0).
However, in Edmund Bertschinger's lecture notes, he seems forgot the extra factors produced from $dt$ and $\dot{q}$ under reparameterization in equation (3) on page 3 in his notes. I also found a similar problem from Kostas Skenderis's lecture notes on string theory (the last line on page 17).
Am I right or did I misunderstand anything in Edmund Bertschinger's lecture notes?
Infinitesimally, if one expands the diffeomorphism in terms of $\epsilon$, i.e $$t=\phi_{\epsilon}(\tau)=\tau+\epsilon\xi(\tau)+\mathcal{o}(\epsilon), \tag{2}$$
where $\xi$ is the generator of diffeomorphism $\phi_{\epsilon}$, then one has $$\frac{d\phi_{\epsilon}(\tau)}{d\tau}=1+\epsilon\dot{\xi}(\tau)+\mathcal{o}(\epsilon),\quad\mathrm{and}\quad\frac{d\tau}{d\phi_{\epsilon}(\tau)}=1-\epsilon\dot{\xi}(\tau)+\mathcal{o}(\epsilon). \tag{3}$$
Also, expanding $q(t)=q(\phi_{\epsilon}(\tau))=k_{\epsilon}(\tau)=q(\tau+\epsilon\xi(\tau)+\mathcal{o}(\epsilon))$ in terms of $\epsilon$ yields $$q(t)=k_{\epsilon}(\tau)=q(\tau)+\epsilon\xi(\tau)\dot{q}(\tau)+\mathcal{o}(\epsilon). \tag{4}$$
Plugging (2), (3) and (4) into (1), one obtains
\begin{align}
& S[q(t),t]=\int_{a}^{b}dtL(q,\dot{q},t)=\int_{\alpha}^{\beta}d\tau\frac{d\phi_{\epsilon}(\tau)}{d\tau}L\left(k_{\epsilon}(\tau),\frac{d\tau}{d\phi_{\epsilon}(\tau)}\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau)\right) \\
&=\int_{\alpha}^{\beta}d\tau\left\{L(q(\tau),\dot{q}(\tau),\tau)+\epsilon\left[\frac{\partial L}{\partial q}\xi(\tau)\dot{q}(\tau)+\frac{\partial L}{\partial\dot{q}}\xi(\tau)\ddot{q}(\tau)+\frac{\partial L}{\partial\tau}+L(q(\tau),\dot{q}(\tau),\tau)\dot{\xi}(\tau)\right]+\mathcal{o}(\epsilon)\right\} \\
&=\int_{\alpha}^{\beta}d\tau\left\{L(q(\tau),\dot{q}(\tau),\tau)+\epsilon\frac{d}{d\tau}\left[L(q(\tau),\dot{q}(\tau),\tau)\xi(\tau)\right]+\mathcal{o}(\epsilon)\right\}
\end{align}
Diffeomorphism invariance requires $$\frac{d}{d\epsilon}\bigg|_{\epsilon=0}S[q(t),t]=0,$$
i.e one ends up with a total derivative $$\int_{\alpha}^{\beta}d\tau\frac{d}{d\tau}[L(q(\tau),\dot{q}(\tau),\tau)\xi(\tau)]=L(q(\beta),\dot{q}(\beta),\beta)\xi(\beta)-L(q(\alpha),\dot{q}(\alpha),\alpha)\xi(\alpha)=0.$$
The above equation is true if and only if the diffeomorphism $t=\phi_{\epsilon}(\tau)$ fixes the end points, i.e $\xi(\alpha)=\xi(\beta)=0$.
So, my conclusion is that the variation of action (1) in the lecture notes of Edmund Bertschinger is wrong. The fact that the canonical Hamiltonian vanishes cannot be derived from variation of the action. Did I make any mistakes from the above calculations?
My last question: Since action (0) suffers from reparameterization invariance, can this gauge redundancy be avoided if one starts from the energy functional $$E[q(\tau)]=\int_{a}^{b}d\tau g_{\mu\nu}(q)\frac{dq^{\mu}}{d\tau}\frac{dq^{\nu}}{d\tau}$$
instead?
 A: I suddenly found that I made a mistake yesterday. Since diffeomorphism means
\begin{align}
S[q(t),t]&=\int_{a}^{b}dtL(q(t),\dot{q}(t),t)=\int_{\alpha}^{\beta}d\tau\frac{d\phi_{\epsilon}(\tau)}{d\tau}L\left(k_{\epsilon}(\tau),\frac{d\tau}{d\phi_{\epsilon}(\tau)}\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau)\right) \\
&=\int_{\alpha}^{\beta}d\tau L(k_{\epsilon}(\tau),\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau)),
\end{align}
infinitesimally, this means
\begin{align}
&\int_{\alpha}^{\beta}d\tau\left\{L(q(\tau),\dot{q}(\tau),\tau)+\epsilon\frac{d}{d\tau}\left[L(q(\tau),\dot{q}(\tau),\tau)\xi(\tau)\right]+\mathcal{o}(\epsilon)\right\} \\
&=\int_{\alpha}^{\beta}d\tau\left\{L(q(\tau),\dot{q}(\tau),\tau)+\epsilon\left[\frac{\partial L}{\partial q}\xi\dot{q}+\frac{\partial L}{\partial\dot{q}}\frac{d}{d\tau}(\xi\dot{q})+\frac{\partial L}{\partial\tau}\xi\right]+\mathcal{\epsilon}(\epsilon)\right\},
\end{align}
where in the last line, I expanded the Lagrangian $L(k_{\epsilon}(\tau),\dot{k}_{\epsilon}(\tau),\phi_{\epsilon}(\tau))$. This implies
$$L(q,\dot{q},\tau)\dot{\xi}=\frac{\partial L}{\partial\dot{q}}\dot{\xi}\dot{q}.$$
This above equation must hold for any $\dot{\xi}(\tau)$, therefore one has
$$H_{c}=\frac{\partial L}{\partial\dot{q}}\dot{q}-L=0,$$
where $H_{c}$ is the canonical Hamiltonian. This is a direct result of Euler's theorem.
It is strange that Edmund Bertschinger seems derived the correct result from incorrect ways.
Finally, since the explicit depends of $t$ in $L$ does not make any difference, $t$ has nothing to do with Dirac's constraint. This answers my first question.
