How Do You Solve the Action Integral for a Free Particle?

[dydxforsn]

I'm trying to solve this simple problem (it's the first problem of Quantum Mechanics and Path Integrals by Feynman, I feel like an idiot not being able to do it...) It's just solving for the action, S, of a free particle (no potential, only kinetic energy..)

So it should just be S = \int_{t_a}^{t_b}{\frac{m}{2} (\frac{dx}{dt})^2 dt}
which according to the book is simply S = \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a}

I've tried a couple of different ways to reason myself into this solution but I can't seem to figure it out.

 

[Mute]

What have you tried so far? What did you plug in for $dx/dt$?

 

[dydxforsn]

What have you tried so far?

Incredibly wrong stuff, heh..

What did you plug in for $dx/dt$?

Yeah I'm an idiot. I was supposed to just plug in v = \left ( \frac{x_{b} - x_{a}}{t_{b} - t_{a}} \right ) because 'v' is constant from the Euler-Lagrange equation..

Thanks for helping me see what should have been obvious >_< I was hell bent on doing things symbolically and didn't seem to care about the appearance of the end point 'x' values.. These should have been very suggestive.

 

[Mute]

Great! You figured it out! Yeah, with a problem like this it helps to remember that the action is a functional of $x(t)$ and $\dot{x}(t)$, so you get different answers depending on which function x(t) you use. Of course, varying the action with respect to x(t) (giving the Euler-Lagrange equations) yields the equation of motion for the classical path. The problem wanted the action of a classical path with boundary values $x(t_a) = x_a$ and $x(t_b) = x_b$.

It can take some practice seeing these sorts of problems a few times before it clicks. =)