I Does the value of the action S have classical significance?

AI Thread Summary
The discussion focuses on deriving the equations of motion using the Euler-Lagrange equations, emphasizing that the action is an extremum. The significance of the numerical value of the action, which has units of energy-time, is questioned, particularly in the context of a particle under a constant force. The derived action for this scenario is complex, raising concerns about its relevance to the particle's motion. Additionally, while action shares units with angular momentum, it is not directly related to it, similar to the relationship between energy and torque. The complexity of the action in this simple case prompts a deeper examination of its implications in classical mechanics.
Kostik
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We obtain equations of motion by solving the Euler-Lagrange equations; along this path the action ##S## is an extremum. We are usually interested in the path ##x(t)## (in one dimension) connecting ##x(0)=x_0## and ##x(T)=x_1##. But does the numerical (extremal) value of the resulting action have any significance? Its units are plainly energy-time.

In the extremely simple case of a particle with mass ##m## under a constant force ##F##, the Euler-Lagrange equation gives the obvious ##m \ddot x = F##, but the action itself has the complicated form:
$$S = \frac {(x_1+x_0)FT} {2} + \frac {(x_1-x_0)^2 m} {2T} - \frac {F^2 T^3} {24m}.$$

This strikes me as a peculiarly complicated quantity to have some significance to a particle moving under a constant force.
 
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Can you show your steps at getting this equation?
Other than that, action is the same units as angular momentum, but is not directly related to it, like energy and torque.
 
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