# Principle of Least Action Question

Let's say we have a particle with no forces on it. The path that this classical particle takes is the one that minimizes the integral $$\frac{1}{2}m\int_{t_i}^{t_f}v^2dt.$$ So if we graph this for the actual path a particle takes it is a straight, horizontal line on the $(t,v^2)$ plane. But couldn't we lessen the integral if we first slow down and then speed up near the end to create a sort of parabolic line that has less area under the $(t,v^2)$ plane? So why doesn't the particle take this path? What am I missing in my thinking?

You have to minimize the integral subject the the constraint that the initial and final positions $x(t_i)$ and $x(t_f)$ are held fixed. In particular, $\Delta x = \int_{t_i}^{t_f} v(t)\, dt$ is held fixed. If the particle slowed down than sped up as you suggested, the action would be less, but it wouldn't have a high enough average speed to cover the full $\Delta x$ in time. You can play around with a few specific trajectories and check for yourself.

• Why not? If you graph both of the possible $(t,v)$ paths you have the right path being a horizontal line and then you could have the other path go up and down and back up with 1 "node" and 2 "antinodes" that has the same area under the curve that satisfies your position. With regard to the average velocity, if I drive 10000 mi/h to right before the finish line and slow down to .000001 mi/h or however slow and wait to have the correct average velocity wouldn't that satisfy the condition? – Shrodinger 2016 Jul 19 '16 at 21:28
• @GavinN. It would, but the time you spent going fast would get weighted heavily (because of the $v^2$) so you'd end with more action than if you'd traveled at the constant speed. If you think I'm wrong, please suggest a specific trajectory that has less action than a constant trajectory with the same average speed. – tparker Jul 19 '16 at 21:43
• @GavinN. This is "arm-wavy" math, but it shows the basic reason why your thinking doesn't work: suppose you do some part of the path with velocity $v - \Delta v$ and another part with $v + \Delta v$. When you square those velocities, you get $v^2 - 2v\Delta v + (\Delta v)^2$ and $v^2 + 2v\Delta v + (\Delta v)^2$. The $\pm 2v\Delta v$ parts of the action might cancel out, but the two $(\Delta v)^2$ parts don't cancel, they add. – alephzero Jul 19 '16 at 22:39
• Ya I calculated some paths and I see what you guys are saying now. that is a good way of seeing it too alephzero. – Shrodinger 2016 Jul 19 '16 at 23:27

You're missing that Dirichlet boundary conditions

$$x(t_i)~=~x_i \quad\text{and} \quad x(t_f)~=~x_i$$

are implicitly implied. The stationary action principle is not well-posed without boundary conditions.

• I'm not sure what you mean, what about my proposal violates this? – Shrodinger 2016 Jul 19 '16 at 21:18
• Your proposal does evidently not observe the boundary conditions. You cannot just reduce the speed in the middle without consequences. Something has to give. – Qmechanic Jul 19 '16 at 21:34

Your proposed path has a VERY LARGE action. As @tparker pointed out, you have to minimize the path subject to the constraint that the average velocity doesn't change. Now, the action is quadratic in velocity. A little fiddling around with the math should convince you that to minimize the integral of $v^2$ subject to the constraint that the average velocity doesn't change, you want the velocity to be constant. Intuitively, going fast for a bit raises the action a LOT (because velocity is squared) while going slow for a bit doesn't do much to lower the action (because velocity is squared).