# Valid curves with least action

Greetings,

I'm simulating the principle of least action for simple object motion and reading from Feynman Vol. 2, Chpt 19 -- The Principle of Least Action. He states (with my paraphrasing) that the true path of a trajectory is the one for which the integral over all points of kinetic energy minus potential energy is least.

I created a curve of an object shot straight up at velocity=500m/s from x0=0, g=9.8 and t=80s. It goes up and comes partway down and ends up 8919m above x0 (again, after 80s). I then calculated the integral of KE - PE from 0 to 80 and got a value of -4969173.33. I did the same numerically and came up with a value very close to that. My understanding is that parabolic curve should have the lowest value for the action from (x0,t0) to (x1,t1).

Then I generated a bunch of other curves and sure enough, they all had larger values. Then I drew a straight line from (x0,t0) to (x1,t1) and calculated the integral and came up with a smaller value! Namely, -6490206.667. Hey! The principle of least action didn't hold; a smaller value represents a non-true path. A little more work led me to other curves that "violated" the principle of least action.

The straight line is completely non-physical, of course. It doesn't rise as high a real object, it's not parabolic, etc. However, I think this should be allowable because the principle of least action is used to define the trajectory motion. It should have given a larger value and I could have kept plugging away at curves infinitely and found that the parabola was the smallest.

My guess is that there is more to a valid path that just the starting and ending points being the same as the true path. What are those?

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I suspect the problem is the choice of Lagrangian (this is the name for the function defined by kinetic energy minus potential energy), $\mathcal{L}=T-V$.

The action is the integral you mentioned:

$S=\int^{t_1}_{t_0}\mathcal{L}dt$

Minimising this gives the true path, and as Feynman says, this is carried out using the Calculus of Variations.

The paths you call 'non-physical' would not have the same Lagrangian function as the curve of motion under gravity. For example, looking at the straight line example, it would not have the potential energy function of $V=mgx$.

The principle of least action is a misnomer. It should be called the principle of stationary action. The trajectory of any real motion does not have to be a minimum, local or global, of action. It needs to be its stationary point.

Greetings,

Thanks for your answers. I'm not sure I was clear in my original post. I'm trying to determine the basis or most fundamental principle associated with the principle of least action (or stationary action). The laws of motion can be derived from that principle. Since they can be derived from it, I don't have to rely on them being correct beforehand; that's the whole point of the principle of least action. I should be able to calculate the Lagrangian and the correct equations should just fall out. Feynman seemed to indicate that the most fundamental principle is that the proper laws of motion will follow curves with the smallest action (or stationary action). However, there is one other requirement according to his article -- for the curves to be valid, they must start and end at the same point.

What I found is that his description is too general. I created curves that met the requirements he described and found that the real curve is not minimal (or stationary). There must be some other requirements. What are they?

If I were a physicist of any credibility :-) and he weren't Feynman, I would say his proof was flawed because his assumptions are incorrect. (Of course, I'm sure him, Euler and Lagrange are correct and I'm mistaken, but I don't understand why.)

Regarding the first response, I can still determine the position relative to x0 and therefore calculate the potential energy. I can calculate the kinetic energy because I can determine the velocity by taking the derivative of the position at each point. Ultimately, I can calculate the Lagrangian.

Regarding the second response, yes, I agree with you, but I found curves with actions both larger and smaller than the "real" trajectory, therefore the right answer is not stationary (well, actually, I'm sure it is, but I can't tell why).

The principle of stationary action selects a trajectory that is "special" with respect to trajectories close to it. Thus comparing trajectories that are arbitrarily distant is meaningless. As a simple illustration, consider an ordinary function of one argument: ## y = -x^4 + 2x^2 - 1 ##. Clearly the point x = 0 is special locally - it is its local minimum, yet there is no global minimum.

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WannabeNewton
Your misunderstanding lies in exactly what voko said. You are considering all elements of the function space constrained to connect the two points. The principle of stationary action involves elements belonging to a non-trivial neighborhood of the desired function, with the aforementioned constraint. This is no different from regular single variable calculus. A stationary point is a local property i.e. within a small open ball around that point; it is not a property of the entire curve.

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Thank you. At first I was thinking it doesn't do much good to have to know about the answer beforehand if you're trying to derive it. In thinking about it more, however, I guess you only have to know the start and end points and then you essentially solve for the whole thing at once. At any rate, I think I have enough now that I can explore a little more on my own and come up with the answer or at least a more meaningful question. Thanks again.

The situation is simpler than other posters have suggested. I think you have just made a math error.

I created a curve of an object shot straight up at velocity=500m/s from x0=0, g=9.8 and t=80s. It goes up and comes partway down and ends up 8919m above x0 (again, after 80s).
I get 8640m.

I then calculated the integral of KE - PE from 0 to 80 and got a value of -4969173.33.
I get the same.

Then I drew a straight line from (x0,t0) to (x1,t1) and calculated the integral and came up with a smaller value! Namely, -6490206.667.
I get -2.92032x10^6, which is larger. The principle of least action survives to fight another day!

The straight line is completely non-physical, of course. It doesn't rise as high a real object, it's not parabolic, etc. However, I think this should be allowable because the principle of least action is used to define the trajectory motion. It should have given a larger value and I could have kept plugging away at curves infinitely and found that the parabola was the smallest.
Yes, your expectations are right. I think you just made an arithmetic error.

Others are right that technically the principle of least action should be called the principle of stationary action. However, in this case, and in a large number of realistic cases, the physical path really is the path of least action out of all differentiable paths you can draw between the two endpoints, without exception.

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I don't think I've ever been so happy to have made a math error! Well, actually, I set up my integral wrong, but that's about the same. The weird thing is that I wrote a simple program to double-check, that also had an error, but for a different reason and gave me a similarly wrong answer! I was certain it had to be something else for that reason.

I have it all straightened out now, and indeed the Principle is still surviving and fighting! It's just not fighting me anymore... ;-)

Thanks!!!