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

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## Main Question or Discussion Point

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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Orodruin
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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.

Dale
Mentor
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.

anorlunda
Staff Emeritus
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.

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Orodruin
Staff Emeritus
Homework Helper
Gold Member
It sounds like you are misinterpreting "minimum" In this case
Also, this reminds me of my pet peeve:
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.

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?