Fake time-reparameterization symmetry in the Hamiltonian formulation of a classical system

Question

The following is taken from pages 83-84 of Tom Hartman's lectures on Quantum Gravity and Black Holes.

See related questions here and here.

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Consider a one-dimensional system with

1. a single degree of freedom $q(t)$ and
2. a conjugate momentum $\frac{\partial L}{\partial \dot{q}}$.

The action for the system is \begin{align}I = \int dt\ L = \int dt\ \left[ \frac{\partial L}{\partial \dot{q}} \dot{q}-H(p,q)\right].\end{align}

We can introduce a fake time-reparameterization symmetry by

1. introducing a fake degree of freedom $t(\tau)$ and
2. imposing the constraint $H(p,q) = -\frac{\partial L}{\partial t'},$

where $'$ denotes differentiation with respect to $\tau$.

Then, we find that the action becomes

$$I = \int d\tau \left[ \frac{\partial L}{\partial q'} q' + \frac{\partial L}{\partial t'} t' - N\left(H(p,q)+\frac{\partial L}{\partial t'}\right)\right],$$

where $N$ is a Lagrangian multiplier.

Then, the new Hamiltonian of the system is

$$H' = N \left( H(p,q)+\frac{\partial L}{\partial t'} \right).$$

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The Hamiltonian $H'$, of course, vanishes on-shell, but I do not see why this is because $H'$ generates $\tau$-translations.

Answer 1

1. It is non-standard to call the constraints "the Hamiltonian". Generically, one would rather write a generic constrained Hamiltonian action as $$ S_\text{Ham} = \int (p_i \dot{q}^i - \lambda^i \chi_i(p,q) - H(p,q))\mathrm{d}t,\tag{1}$$ where the $\lambda^i$ are Lagrange multipliers enforcing the constraints $\chi_i \approx 0$ and $H$ is the what we actually call "the Hamiltonian". $H' = \lambda^i \chi_i - H$ is the extended Hamiltonian.

2. Your $H'$ consists entirely of constraints. It's therefore trivial that it vanishes on-shell since the very definition of "on-shell" includes the constraints being satisfied. That one may view the constraints as the generators of gauge transformations, and that the $\tau$-transformations are such gauge transformations generated by $H'$ because it's just a constraint, are secondary concerns that are not really necessary to see why it vanishes on-shell.

3. The non-trivial part here is not why the constrained part of the extended Hamiltonian $H'$ vanishes, it's why there is no $H$ part in it! This is because for the action (1) to be well-defined, $H(p,q)$ has to transform as a scalar density under $t$-transformations. But if $p,q$ transform as scalars (i.e. are invariant) under such transformation, as it is the case here and always in the case where one just artificially made the "actual" time variable a phase space variable, then every function of them is a scalar, not a scalar density, too. Therefore $H=0$ in time-reparametrization invariant systems.

Altogether, the reasoning "$H'\approx 0$ because it generates $\tau$-transformations" is circular - it generates $\tau$-transformations because it consists purely of constraints, which by definition vanish on-shell, but the fact you actually have to explain is why there's only constraints in it.