lagrangian formalism - Hamiltonian for relativistic free particle is zero

One possible Lagrangian for a point particle moving in (possibly curved) spacetime is

$$L = -m \sqrt{-g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu},$$

where a dot is a derivative with respect to a parameter $\lambda$. This Lagrangian gives an action proportional to proper time, and it is reparametrization invariant ($\lambda$ need not be an affine parameter).

If we try to go to the Hamiltonian picture, we have the momenta

$$p_\mu = \frac{m}{\sqrt{-\dot{x}^2}} g_{\mu\nu} \dot{x}^\nu,$$

which obey the relation $p^2+m^2=0$. We then get that the Hamiltonian $$H = p_\mu \dot{x}^\mu - L$$ is identically zero.

I understand that this is not a problem because, since we have a constraint $\phi(x,p) = p^2 + m^2 = 0$, according to the Dirac method we should really use the Hamiltonian $H' = H + c \phi$, as explained for example in this post. But what I would like to know is, why do we get a zero Hamiltonian? I suspect that this is due to the reparametrization invariance, and the fact that we don't have a preferred notion of time. Will this always happen? Why?

Answer

...what I would like to know is why we get a zero Hamiltonian. I suspect that this is due to the reparametrization invariance... Will this always happen? Why?

Yes, it is due to reparameterization invariance. In other words, the zero-Hamiltonian result holds for any reparameterization-invariant action, not just for the relativistic particle. In this sense, the answer to "Will this always happen" is yes. And one way to answer the "Why?" question is to give a general proof. That's what I'll do here.

I'll denote the parameter as $t$ instead of $\lambda$, because it's easier to type.

Consider any model with an action of the form $$ S=\int dt\ L(t) \hskip2cm L(t) = L\big(\phi(t),\dot\phi(t)\big) \tag{1} $$ where $\phi_1(t),\phi_2(t),...$ is a collection of dynamic variables. If the action is invariant under rigid translations in $t$, then Noether's theorem gives us a corresponding conserved quantity: the Hamiltonian. If the action is invariant under reparameterizations in $t$, then we might expect to get a stronger result because of the more extreme symmetry, and we do: the conservation law still holds, but the conserved quantity is identically zero (and therefore useless). The goal is to prove that the larger symmetry leads to this stronger result.

Suppose that the action is invariant under all transformations of the form $$ \phi_n(t)\rightarrow\phi_n(t+\epsilon) \tag{2} $$ where $\epsilon(t)$ is allowed to be any smooth function for which the map $t\rightarrow t+\epsilon(t)$ is invertible. This is reparameterization invariance. For infinitesimal $\epsilon$, \begin{equation} \delta\phi_n(t) = \dot\phi_n(t)\epsilon. \tag{3} \end{equation} Take the derivative of this with respect to $t$ to get \begin{equation} \delta\dot\phi_n(t) = \frac{d}{dt}\Big(\dot\phi_n(t)\epsilon\Big). \tag{4} \end{equation} Now consider the identity \begin{equation} \delta S = \int dt\ \delta L \tag{5} \end{equation} with \begin{equation} \delta L = \sum_n\left( \frac{\partial L}{\partial \phi_n}\delta\phi_n + \frac{\partial L}{\partial \dot\phi_n}\delta\dot\phi_n \right), \tag{6} \end{equation} which is valid for any transformation of the $\phi$s. For the particular transformation (3)-(4), equations (4)-(5) become \begin{equation} \delta S = \sum_n\int dt\ \left(\frac{\partial L}{\partial \phi_n}\dot\phi_n\epsilon + \frac{\partial L}{\partial \dot\phi_n}\frac{d}{dt}(\dot\phi_n\epsilon) \right). \tag{7} \end{equation} Compare this to the identity $$ \frac{d}{dt}(L\epsilon) = \sum_n\left(\frac{\partial L}{\partial \phi_n}\dot\phi_n + \frac{\partial L}{\partial \dot\phi_n}\frac{d}{dt}\dot\phi_n\right)\epsilon + L\frac{d}{dt}\epsilon \tag{8} $$ to see that (7) may also be written \begin{equation} \delta S = \int dt\ \left(\frac{d}{dt}(L\epsilon) + \left[\sum_n\frac{\partial L}{\partial \dot\phi_n}\dot\phi_n-L\right] \frac{d}{dt}\epsilon \right). \tag{9} \end{equation} For any finite integration interval, the first term is zero if $\epsilon(t)$ is zero at the endpoints of the integration interval. Since $d\epsilon/dt$ is arbitrary within this interval, and since this holds for any interval, the invariance of the action ($\delta S=0$) implies that the quantity in square brackets must be zero. The quantity in square brackets is the Hamiltonian, so this completes the proof that the Hamiltonian is identically zero in this class of models.