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this is relativistic action: $$S=\int_C \mathcal {L}dt$$ where the $\mathcal{L}$ is $-m_oc^2\gamma^{-1}$ what is use of relativistic action!?

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Er...the same use as the non-relativistic one, only it's good over higher relative speeds. –  dmckee Jan 24 '13 at 21:40
    
Related: physics.stackexchange.com/q/44947/2451 –  Qmechanic Jan 25 '13 at 12:30
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The use of an action is do derive all the dynamical equations of a theory from the least-action principle, $\delta S=0$ (action is minimized along the right path). Quantum mechanically, the use of an action is to derive the transition amplitudes from an initial state to the final state by summing over histories weighted by $\exp(iS/\hbar)$ in a quantum theory.

Similarly. The use of a relativistic action is do derive all the dynamical equations of a relativistic theory from the least-action principle, $\delta S=0$ (action is minimized along the right path). Quantum mechanically, the use of a relativistic action is to derive the transition amplitudes from an initial state to the final state by summing over histories weighted by $\exp(iS/\hbar)$ in a relativistic quantum theory.

Yes, I used the copy-and-paste and added the adjective "relativistic" to the second paragraph. ;-)

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