Why Can't an Extremum Path in Lagrangian Mechanics Be a Local Maximum?

[meteorologist1]

Hi, I would like some help in proving the following:

Consider the action for a particle in a potential U. Show that an extremum path is never that of a local maximum for the action.

I think what I have to do is look at the second derivative of the action integral. Then I should somehow argue that this value is always greater or equal zero, so that the extremum is never a local maximum. My problem is how to take the second derivative.

Thanks.

 

[Andrew Mason]

Consider the action for a particle in a potential U. Show that an extremum path is never that of a local maximum for the action.

I think what I have to do is look at the second derivative of the action integral. Then I should somehow argue that this value is always greater or equal zero, so that the extremum is never a local maximum. My problem is how to take the second derivative.

I don't think you have to find the second derivative of the action. (The action is the integral of the difference between the particle's kinetic energy and its potential along the particle's path, over time). I think you have to show that for any deviation from that path the change in the action is > 0. An extremum path is one in which there is no infinitesimal first order change in the action for an infinitesimal change in the path. If the change in action is second order (ie. proportional to dx^2) it would have to be > 0.

This is a very difficult area of physics, conceptually. Feynman's lecture in Vol II, Ch. 19 in his Lectures on Physics is very good as he explains the concepts.

AM

 

[wais]

To prove that an extremum path is never a local maximum for the action, we can use the principle of stationary action in Lagrangian mechanics. The principle states that the path taken by a particle between two points in space and time is the one that minimizes the action integral, which is defined as the integral of the Lagrangian over the path. In other words, the extremum path is the one that makes the action stationary.

Now, to show that this extremum path is never a local maximum, we can consider the second derivative of the action integral with respect to the path. This can be done by using the Euler-Lagrange equation, which is the fundamental equation of Lagrangian mechanics. This equation states that the second derivative of the action integral is equal to the difference between the Lagrangian and its derivative with respect to the path, evaluated at the endpoints of the path.

If we assume that the extremum path is a local maximum, then the second derivative of the action integral should be negative. However, this is not the case. By the principle of stationary action, the extremum path makes the action stationary, which means that the second derivative of the action integral is equal to zero. This implies that the Lagrangian and its derivative at the endpoints must also be equal, which is not possible for a local maximum.

Therefore, we can conclude that an extremum path is never a local maximum for the action. It is either a local minimum or a saddle point. This is a fundamental result in Lagrangian mechanics and is crucial for understanding the behavior of particles in potential energy fields. I hope this helps in your proof. Let me know if you have any further questions.