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Confused about the principle of least action

  1. Dec 15, 2012 #1
    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.
     
  2. jcsd
  3. Dec 15, 2012 #2
    Here's your error. The [itex]\delta[/itex] 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, [itex]S[/itex]. This is a function(al) of the path through space. So if you deform the path a very small amount, that is [itex]\delta S[/itex]. 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.
     
  4. Dec 15, 2012 #3
    Thank you, DrewD. That helped very much.
     
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