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In the special relativistic action for a massive point particle,

$$\int_{t_i}^{t_f}\mathcal {L}dt,$$

why is the Lagrangian

$$\mathcal {L}=-E_o\gamma^{-1}$$

a negative number?

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I'm just guessing this is the relativistic action for a free point particle, but some clarification of your question would help... – Michael Brown Jan 25 '13 at 9:59
Having you noticed that adding a constant the $\mathcal{L}$ doesn't affect the resulting optimal path? – dmckee Jan 25 '13 at 11:09
Because by convention, we want the action to be minimized in normal cases, not maximized. It's a convention: we could redefine $S\to -S$ which would then be maximized. For a massive pointlike particle, the proper time along the straight path is actually maximized, recall the twin paradox (the traveling twin ages less than the stationary one because of time dilation!), so one has to take $S$ to be a negative multiple of the proper time for the action to be minimized along the classical path. – Luboš Motl Jan 25 '13 at 11:25
Luboš agrees with me! :-) – John Rennie Jan 25 '13 at 11:40
I have even upvoted you, @John, after I noticed that you wrote the same thing and before me! ;-) – Luboš Motl Jan 25 '13 at 12:22

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).

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And how about non relativistic $L=mv^2/2$? – Vladimir Kalitvianski Jan 25 '13 at 10:37
Generally what matters is that the action is stationary $\delta S=0$, as is most easily seen from the path integral. Whether the classical path corresponds to a maximum, minimum or saddle point of the action is practically immaterial. – Michael Brown Jan 25 '13 at 11:21

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 (possibly negative) multiplicative factor is irrelevant. In this case, the normalization is chosen so that the Lagrangian

$$ L~=~-\frac{E_0}{\gamma}~=~-E_0\sqrt{1-\left(\frac{v}{c}\right)^2} ~\approx~ \frac{1}{2}m_0 v^2 -E_0 \qquad\text{for}\qquad v\ll c$$

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 additive constants in the Lagrangian do not affect the equations of motion.

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Right, with choosing $L=E_o(1-\gamma^{-1})$ we obtain the same equations. – Vladimir Kalitvianski Jan 25 '13 at 10:54

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.

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