Adding a constant to potential energy doesn't change action?

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Discussion Overview

The discussion revolves around the implications of adding a constant to potential energy in the context of classical mechanics, specifically regarding its effect on the action of a system. Participants explore the relationship between action, Euler-Lagrange equations, and the concept of stationary action.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that while adding a constant to potential energy does not change the Euler-Lagrange equations or the equations of motion, it raises questions about the action's value and its implications for the true trajectory.
  • Another participant clarifies that adding a constant changes the action's value but does not affect the variation of the action, as the potential is not a dynamical variable.
  • A comparison is made to a parabola, where adding a constant shifts the action's value without changing the location of the minimum, suggesting that the stationary path remains unchanged.
  • Some participants discuss the interpretation of "minimum" in the context of action, arguing that it should be understood as "stationary" rather than strictly minimal, allowing for various stationary points including maxima and saddle points.
  • References to external sources are made to further explore the concept of stationary action and its implications for degenerate points.

Areas of Agreement / Disagreement

Participants express differing views on the implications of adding a constant to potential energy, particularly regarding the interpretation of action and its minimum or stationary nature. There is no consensus on the interpretation of these concepts.

Contextual Notes

Participants highlight the importance of understanding the distinction between action changing in value and the variation of action remaining unchanged. The discussion also touches on the nuances of stationary points versus minima in the context of action.

Who May Find This Useful

This discussion may be of interest to students and enthusiasts of classical mechanics, particularly those exploring the principles of least action and the implications of potential energy in dynamical systems.

blaughli
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Hello. I've been watching Susskind's online Stanford lectures on classical mechanics to review the subject, and I believe he said that adding a constant to the potential energy does not change the action of a system. I see how it doesn't change the Euler-Lagrange equations and therefore doesn't affect the equations of motion (and therefore the trajectories), yet the integral of a constant is non-zero so I don't see how adding a constant to U in A = ∫(T-U)dt wouldn't change the action A. Where have I gone wrong? There seems to be an inconsistency in saying that the action changes (implying it's not at a minimum and therefore doesn't describe the true trajectory) while the E-L equations don't change (implying no change in trajectory). Thank you, sorry if I'm missing something basic and have wasted your time by not thinking about this more on my own :)
 
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It does change the action. However, it does not change the variation of the action. The variation of the action should be taken with respect to the dynamical variables. The potential is not in itself a dynamical variable, but generally a function of them. Adding a constant to the potential does not change the derivatives of the potential.
 
blaughli said:
There seems to be an inconsistency in saying that the action changes (implying it's not at a minimum and therefore doesn't describe the true trajectory) while the E-L equations don't change (implying no change in trajectory).
Think about adding a constant to a parabola. The location of the minimum does not change, but the value does. Similarly, the path that is stationary will not change, even though the value of the action does.
 
blaughli said:
There seems to be an inconsistency in saying that the action changes (implying it's not at a minimum and therefore doesn't describe the true trajectory)

It sounds like you are misinterpreting "minimum" In this case, minimum just means the bottom of a curve, such that any incremental change results in a higher action. For example, see the picture below. The red line in this image could represent the constant. Your comment makes it sound like you think minimum must mean y=0 in the picture.

function-minimum-point.png
 

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anorlunda said:
It sounds like you are misinterpreting "minimum" In this case
Also, this reminds me of my pet peeve:
blaughli said:
implying it's not at a minimum and therefore doesn't describe the true trajectory
The action does not need to be minimal for a path satisfying the equations of motion. The "principle of least action" is a misnomer and it is more accurate to call it the principle of stationary action. Of course, what has been said about minima in this thread also holds for stationary points. A stationary point of an action will continue being a stationary point even if you add a constant to the potential.
 
blaughli said:
Hello. I've been watching Susskind's online Stanford lectures on classical mechanics to review the subject, and I believe he said that adding a constant to the potential energy does not change the action of a system. I see how it doesn't change the Euler-Lagrange equations and therefore doesn't affect the equations of motion (and therefore the trajectories), yet the integral of a constant is non-zero so I don't see how adding a constant to U in A = ∫(T-U)dt wouldn't change the action A. Where have I gone wrong? There seems to be an inconsistency in saying that the action changes (implying it's not at a minimum and therefore doesn't describe the true trajectory) while the E-L equations don't change (implying no change in trajectory). Thank you, sorry if I'm missing something basic and have wasted your time by not thinking about this more on my own :)
Hey! Can you recall which # of Susskind's lectures this was from?
 

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