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What is the role of time (time interval) in principle of least action?

Physics Asked by user103515 on September 30, 2020

Action is represented by $S[Q(t)]$ where $Q(t)$ is the name of a single complete path in the configuration space of a system. The path starts at the point $q_i$ and ends at the point $q_f$. Suppose that the system is at $q_i$ at the time $t=0$ and then suppose it reaches $q_f$ at $t=T$. Then the action functional for this whole path $Q(t)$is calculated as:
$$S[Q(t)]=int_{0}^{T} L(q,dot{q},t)dt=A [ML^2T^{-1}].$$ Here $L$ is the Lagrangian. Now please suppose that I applied the principle of least action and incidentally found out that $A$ is the least possible action of all the possible paths.

Now suppose that I found another path $Q'(t)$between same end points $q_i$ to $q_f$ but the system takes time $T’$ instead of $T$ to reach at $q_f$ from $q_i$. For this path $$S[Q'(t)]=int_{0}^{T’} L(q,dot{q},t)dt=A'[ML^2T^{-1}].$$

Now if $A’le A$ then what will be the actual path of the system? $Q(t)$ or $Q'(t)$?

So, my question is what is the role of time interval in the principle of least action?

Or should the time taken for each possible paths from $q_i$ to $q_f$ be always same or they can be different?

One Answer

Brief answer: The integration region $[t_i,t_f]$ is kept fixed in the principle of least/extremal/stationary action/Hamilton's principle. To allow variation of the integration region $[t_i,t_f]$, one should use other variational principles, such as e.g. Maupertuis' principle/principle of abbreviated action.

Correct answer by Qmechanic on September 30, 2020

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