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the action is the length of the geodesic

$S=-E_o\int_i^f d\tau$

we get an action that is minimised for the correct path.

if i want action to be positive number then it require that $\tau_i$ be bigger than $\tau_f$

$S=-E_o(\tau_f-\tau_i)$

$S=E_o(\tau_i-\tau_f)$

isn't it true!?

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As has been noted on the site before the "principle of least action" is really a principle of stationary action. Any minimization problem can be converted into a maximization problem by either multiplying by -1 or by inversion. It should be clear that these manipulations don't affect the interpretation of the result. Accordingly we don't care what sign the action has. – dmckee Feb 28 at 21:45

closed as too localized by Qmechanic Apr 30 at 7:40

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1 Answer

Assuming that your action makes sense, $E_0\gt0$ and that the limits are written like

$S=-E_o\int_{\tau_i}^{\tau_f} d\tau,$

the answer is yes.

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protected by Qmechanic Apr 29 at 22:45

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