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What is zero action

by Sunfire
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Sunfire
#1
Sep9-13, 05:49 PM
P: 215
Hello All,

I have been reading this thread, trying to understand the concept of action. There is one very intriguing (to me) statement in that thread:
"So your action is the difference between your capability of motion and your actual motion.
The principle of least action says that there is no difference without some sort of constraint - in which case the motion will be such that this difference is as small as possible." /by Simon Bridge/

Do I understand this correctly - the action is always zero for unconstrained motion; and never zero (but minimal) for constrained motion. Is this reasoning true?

Is free fall a constrained motion? Because the action is zero only at y = y_initial / 2, y is the vertical position.

Perhaps someone has an idea...
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UltrafastPED
#2
Sep10-13, 02:45 AM
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See the Feynman lecture on "Least Action" ... it is best to get a complete overview. Otherwise you may never see the forest for the trees!

http://liberzon.csl.illinois.edu/tea...hapter2-19.pdf
Sunfire
#3
Sep10-13, 02:59 PM
P: 215
Thank you, UltrafastPED; I read the chapter you suggested. Quite a few good points there. I am still having trouble to understand something, though. What is "zero action"? Does it have any significance?

In the chapter by Feinman, the free falling body moves downwards and its lagrangian changes from negative to positive. At some point (half the initial height), it becomes zero. Does this zero lagrangian have any physical significance?

DaleSpam
#4
Sep10-13, 05:39 PM
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What is zero action

The actual value of the action is not important, only that it is minimized. The value of the action can be made arbitrarily large or small simply by adding a constant offset to the potential.
Sunfire
#5
Sep12-13, 02:39 PM
P: 215
Quote Quote by DaleSpam View Post
The actual value of the action is not important, only that it is minimized. The value of the action can be made arbitrarily large or small simply by adding a constant offset to the potential.
This makes perfect sense.
What if, in a system,
Ep = f(x) - C, C is arbitrary constant
Ek = f(x)
for all x.

Then L = Ek - Ep = C for all x, thus the action remains constant. Does this ever happen?
DaleSpam
#6
Sep12-13, 05:36 PM
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P: 17,213
In principle, it is possible, but I have never seen such a system. Usually the potential energy is a function of the generalized positions and the kinetic energy is a function of the generalized velocities, so I haven't seen a case where they are the same functions like that.


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