I was reading about the formulation of mechanics in special relativity and found that the action for a massive free point-particle as
$$ S = -mc\int_a^b ds $$
So, I did a few observations, ie.
$$ S = -mc^2 \int_a^b dt $$
and being
$$u_{\mu}u^{\mu} = c^2$$
I wrote the action as
$$ S = -\int_a^b m u_{\mu}u^{\mu} dt $$
which in my opinion resembles more to the classical kinetic energy.
Now, I never saw something like this in a book or text on internet. Everyone seems to work with the classical speed and the old good Euler-Lagrange equations with time as parameter.
So my question is: it's possible to derive the correct equations of relativistic dynamics from this action?
Attempt of solution
The action is
$$ S = \int_a^b L(x,v,t) dt $$
and it should be Lorentz invariant (the action or the lagrangian?). Now, changing the parametrization of the path with $d\tau$ instead of $dt$ shouldn't change the integral (with the condition that everything changes accordingly). So I rewrite the action as
$$ S = \int_{\alpha}^{\beta} L(x_{\mu}, \frac{dx_{\mu}}{d\tau}, \tau)d\tau $$
and do the variation of this and minimize it:
$$ \delta S = \int_{\alpha}^{\beta} \left[ \frac{\partial L}{\partial x_{\mu}}\delta x_{\mu} + \frac{\partial L}{\partial u_{\mu}}\delta u_{\mu}\right] d\tau = 0 $$
Then, using the old manipulations it yields
$$ \frac{d}{d\tau}\left( \frac{\partial L}{\partial u_{\mu}} \right) – \frac{\partial L}{\partial x_{\mu}} = 0 $$
This applied to the lagrangian I wrote
$$ L = m \eta_{\mu\nu}u^{\mu} u^{\nu} $$
seems to yield the correct 4-impulse
$$ p_{\nu} = \frac{\partial L}{\partial u_{\mu}} = m \eta_{\mu\nu}u^{\nu} $$
but the hamiltonian (total energy in the text I'm reading) is zero:
$$ H = p_{\mu}u_{\mu} – L = mu_{\mu}u^{\mu} – mu_{\mu}u^{\mu} = 0 $$
I think I did wrong the derivation of the Euler-Lagrange equations, but not sure.
Best Answer
Comments to the question (v2):
The Minkowski spacetime can be generalizes to a Lorentzian manifold $(M,g)$. We choose the Minkowski signature $(-,+,+,+)$ and put the speed of light $c=1$ equal to one.
OP evidently knows that the action $$S = -E_0~ \Delta \tau\tag{1}$$ for a massive point particle is minus the rest energy $E_0=m_0$ times the change $\Delta \tau$ in proper time, cf. e.g. my Phys.SE answer here. In more details, eq. (1) is the square root action $$S ~=~ \int_{\lambda_i}^{\lambda_f} \!d\lambda~ L, \qquad L~:=~- m_0\sqrt{- \dot{x}^2}, $$ $$ \dot{x}^2~:=~g_{\mu\nu}~\dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\lambda}, \tag{2} $$ where $\lambda$ is a world-line parameter.
The canonical Lagrangian $4$-momentum is precisely the mechanical $4$-momentum $$ p_{\mu}~:=~\frac{\partial L}{\partial \dot{x}^{\mu}} ~=~\frac{m_0\dot{x}_{\mu}}{\sqrt{-\dot{x}^2}}, \qquad \dot{x}_{\mu}~:=~g_{\mu\nu}~\dot{x}^{\nu}. \tag{3}$$
The Lagrangian energy function $$h~:=~p_{\mu}\dot{x}^{\mu}-L~=~0 \tag{4}$$ vanish identically. This is related to the fact that the square root action (2) has world-line reparametrization invariance, which is a gauge symmetry. Note that the $4$-momentum (3) is reparametrization invariant. One often uses the static gauge $$x^0~=~\lambda.\tag{5}$$
OP is essentially pondering if one instead of the square root action (2) could use the non-square root action $$\tilde{S} ~=~ \int_{\lambda_i}^{\lambda_f} \!d\lambda~ \tilde{L}, \qquad \tilde{L}~:=~ \frac{m_0}{2}\dot{x}^2~~? \tag{6} $$ The answer is Yes. The corresponding Euler-Lagrange (EL) equations are in both cases the geodesic equation, cf. e.g. this Phys.SE post.
Note that the non-square action (6) does not have world-line reparametrization invariance. Moreover, for a solution to the EL equations, the world-line parameter $\lambda$ and the proper time $\tau$ are always affinely related, cf. my Phys.SE answer here.
The canonical Lagrangian $4$-momentum $$ \tilde{p}_{\mu}~:=~\frac{\partial \tilde{L}}{\partial \dot{x}^{\mu}} ~=~m_0\dot{x}_{\mu}\tag{7}$$ is the mechanical $4$-momentum if we identify the world-line parameter $\lambda$ and the proper time $\tau$. (We stress that it is not possible to make this identification $\lambda=\tau$ before varying the action. The identification $\lambda=\tau$ is only possible on-shell.)
The Lagrangian energy function $$\tilde{h}~:=~\tilde{p}_{\mu}\dot{x}^{\mu}-\tilde{L}~=~\tilde{L} \tag{8}$$ is just the Lagrangian itself.
The non-square root Lagrangian (6) and its corresponding Hamiltonian is discussed in Refs. 1 and 2.
References:
H. Goldstein, Classical Mechanics, 2nd edition, Sections 7.9 & 8.4.
H. Goldstein, Classical Mechanics, 3rd edition, Sections 7.10 & 8.4.