Confused about the principle of least action

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SUMMARY

The discussion centers on the principle of least action in Lagrangian mechanics, specifically addressing the correct interpretation of the action integral S and the notation δ. The user initially confuses δ as a time derivative, while it actually represents a change in path. The correct application leads to the Euler-Lagrange equations, which are fundamental to deriving Newton's Laws. The conversation emphasizes the importance of consulting reputable texts on Lagrangian Mechanics or Calculus of Variations for accurate understanding.

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  • Understanding of Lagrangian Mechanics
  • Familiarity with the concept of action integral S
  • Knowledge of Euler-Lagrange equations
  • Basic principles of Calculus of Variations
NEXT STEPS
  • Study the derivation of Euler-Lagrange equations in Lagrangian Mechanics
  • Read "Calculus of Variations" to grasp the concept of path changes
  • Explore resources on the principle of least action in classical mechanics
  • Investigate the relationship between Lagrangian mechanics and Newton's Laws
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Students of physics, particularly those studying classical mechanics, as well as educators and anyone seeking to deepen their understanding of Lagrangian dynamics and the principle of least action.

nomadreid
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To make the confusions both concise and explicit, I will put down some incorrect calculations, and ask for corrections
Take the Lagrangian KE - PE = T - V
Action = S=∫ L dt (with given limits)
Principle of least action: δS= 0: S(t1)-S(t2) =0 if t1-t2 is small (using the (.) as function notation, not multiplication)
I presume δ can be taken as d/dt
Then d(∫ L dt)/dt = L(t1)-L(t2) = T(t1)-T(t2)+V(t2)-V(t1)=0
Next: conservation of energy
KE+PE = k for some k
So d/dt (T+V) = dT/dt + dV/dt = 0
If this is true on all points on the interval (t1,t2), then T(t1)-T(t2)+V(t1)-V(t2) = 0
Putting the two equations together... but obviously I've already gone way too far into the gross errors, and would be grateful for corrections. Thanks.
 
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nomadreid said:
I presume δ can be taken as d/dt

Here's your error. The \delta denotes change in path, not time. Most decent books on Lagrangian Mechanics or Calculus of variations will do a good job of explaining this. The idea, is that you have your action, S. This is a function(al) of the path through space. So if you deform the path a very small amount, that is \delta S. The path for which that is 0, like in normal calculus, is a max or min. I'm not comfortable enough to fully describe the process without going back to a book first, but doing this comes up with the Euler-Lagrange equations which are equivalent to Newton's Laws. I'm sure you can find a derivation online if your book isn't clear.
 
Thank you, DrewD. That helped very much.
 

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