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Is the Action Integral fundamental

  1. Dec 17, 2003 #1
    What is it about the action integral in physics? Symmetries in the action integral leads to the Euler-lagrange equations. And it seems every physical situation is described by equations that must satisfy the differential equation of the Euler-lagrange.

    This Euler-lagrange equation must be satisfied in 3D and even in the 4D of relativity. Even if the number of parameters increase it must be satisfied over lines and surfaces of classical string theory. So what makes this Action integral so necessary to physics? Is there some underlying geometry that necessitate the Action Integral?

    Can the various functions of physical situations be considered as a type of boundary conditions of the Euler-lagrange differential equation? For example, the momentum is the derivative of the Lagrangian which is a function derived from the particular physical situation, but it is also a specified first order diff eq to the second order diff eq of the Euler-lagrange eq. Isn't this a boundary value problem for a second order diff eq?
  2. jcsd
  3. Dec 17, 2003 #2
    (Force)-(The Same Force)=0 To me, that is all what the Euler-Lagrange equation is all about.
    Sorry, I know this doesn't help, just wanted to check if that statement is kind of true.
  4. Dec 19, 2003 #3
    So the question is: Are there ANY physical situation that can not be described with Lagrangian mechanics, by the variation of the Action integral being zero, which leads to an Euler-lagrange equation?
  5. Dec 20, 2003 #4
    If I remember correctly, the derivation of the EUler Lagrange equations cannot be done when the Hamiltonian has an explicit time dependence in it....
    Not sure though
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