Why Lagrangian is negative number for a relativistic massive point particle?

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

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.

Answer 4

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.