Why Lagrangian is negative number for a relativistic massive point particle? In the special relativistic action for a massive point particle,
$$S=\int_{t_i}^{t_f}\mathcal {L}dt,$$
why is the Lagrangian
$$\mathcal {L}=-E_o\gamma^{-1}$$
a negative number?
 A: At the classical level (meaning $\hbar=0$), to derive the Euler-Lagrange equations (i.e. the special relativistic version of Newton's 2nd law) from the action $S$, an overall non-zero (possibly negative) multiplicative factor is irrelevant. In this case, the normalization is chosen so that the Lagrangian
$$\begin{align} L~=~&-\frac{E_0}{\gamma}~=~-E_0\sqrt{1-\left(\frac{v}{c}\right)^2}\cr 
~\approx~& \frac{1}{2}m_0 v^2 -E_0 \qquad\text{for}\qquad v\ll c\end{align}$$
recovers the well-known expression for the kinetic energy (up to an additive constant) in the non-relativistic limit $v\ll c$. So a bit oversimplified, the negative sign is caused by the huge rest energy $E_0=m_0c^2$. Note that an additive constant in the Lagrangian does not affect the equations of motion.
A: The argument I have seen is that the action is the length of the geodesic i.e.
$$ \text{path length} = \int ds $$
but we know that the trajectory of a free relativistic particle is the one that maximises the path length. So by writing:
$$ S = -m\int ds $$
we get an action that is minimised for the correct path (the $m$ is there to make the dimensions correct).
A: All these notes have important and interesting physical content; however I prefer the solid ground of the proof given in Goldstein's Classical Mechanics. For the hamiltonian to represent the total relativistic energy, the Lagrangian must have a minus sign before the rest energy and in an inhomogeneous way
$L=-\frac{m_0c^2}{\gamma}-V \Longleftrightarrow h=\gamma m_0c^2+V$
Note that this way, both the Lagrangian and the Hamiltonian are unique.
A: The Lagrangian $L = T-V$ describes the energy of motion minus the energy of position. Hence the negative sign of that Lagrangian for a relativistic action for massive point particle describes the deceleration of that massive particle because of the huge potential energy, which will be always greater than its energy of motion.
