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Why lagrangian is negative number?

In the special relativistic action for a massive point particle, $$\int_{t_i}^{t_f}\mathcal {L}dt,$$ where the Lagrangian $$\mathcal {L}=-E_o\gamma^{-1}$$ is a negative number then the action is a negative or positive number?

(in my research it was negative, the result was correct but minus sign make me to ask this question)

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marked as duplicate by dmckee Jan 25 '13 at 19:50

This question was marked as an exact duplicate of an existing question.

As was explained at least twice in the comments of your last question on this topic and it's associated answer the correct answer is "Yes, the action can be either negative or positive." Stop worrying about this and pay attention to the way you use the thing: you don't care about the value, you care about paths for which the integral is stationary. – dmckee Jan 25 '13 at 19:48

The value of action itself is of no meaning, you can always add something to it, if you do not like its sign.

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