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

1. Dec 28, 2017

### Kostik

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.

2. Jan 2, 2018

### DuckAmuck

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.