Hamiltonian of the relativistic point particle

Question

It is unclear to me why the Hamiltonian of a relativistic particle is zero. I know that, given a relativistic free particle, I can write the Lagrangian of it in this way: $$\mathcal{L} = mc^2\sqrt{1 - \frac{v^2}{c^2}}$$ from which, using the Legendre transformation, one can write $$\mathcal{H} = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}};$$ the corresponding action is then $$S = -mc^2\int ds,\qquad ds = cdt\sqrt{1 - \frac{v^2}{c^2}}.$$ Nevertheless, if I write the action in this form: $$S = -mc\int_{\tau_1}^{\tau_2}\sqrt{-\eta_{\mu v} \dot{x}^{\mu}\dot{x}^{v}}d\tau$$ then applying the Legendre transformation, one gets $H = 0$. Being an undergrad, I haven't seen this in any class yet, but I thought that maybe it has to do with the parametrization.