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Why is the time integral of the Lagrangian minimal?

  1. May 31, 2012 #1
    I saw a proof recently that demonstrated that for

    F= ∫L(x, y(x), y'(x))dx, if y(x) is such that F is minimal (no other y(x) could produce a smaller F), then dL/dy - d/dx(dL/(dy/dx)) = 0.

    I understood the proof, and I was able to see that with a basic definition of Energy = (m/2)(dx/dt)^2 + V(x), that carrying out this calculation for the lagrangian = (m/2)(dx/dt)^2 - V(x), that doing this always results in m(d^2x/dt^2) + dV(x)/dr = ma - ma = 0.

    Now however, I'd like to know if this means that the time integral of T - V is minimized. Maybe I'm having a brain fart, but it seems like ∫T + Vdt would be minimized and not ∫T - Vdt because integrating the energy would give you ∫Tdt + ∫Vdt = E(t2 - t1), which would seem to indicate that for any given time interval, the value of the integral of energy is irrelevant of the particular equations for T and V, which would mean that no ∫T + Vdt is any larger than any other, which technically satisfies the requirement that it be minimum. I do not know what ∫T - Vdt looks like.
     
  2. jcsd
  3. May 31, 2012 #2
    The Hamilton's Principle states that the the action is NOT MINIMISED but has a stationary value. Think of it like this:

    When the derivative of a single-variable function is zero, that doesn't mean that the function is minimized or maximized in that point. A function which has a zero derivative but has not a minimal or maximal value is f(x) = x^3.

    Energy is not minimized. Try taking the Euler-Lagrange equation but put the energy instead of the Lagrangian. The Euler-Lagrange equation won't give you the correct Newton's 2nd Law.

    To clarify some things: the Euler-Lagrange equation is equivalent to Newton's Laws. That means that:

    With Newton's Laws you can derive Hamilton's Principle (see wikipedia for derivation)
    and from Hamilton's Principle you can derive Newton's Laws (see Mechanics, Landau Lisfhitz 1st chapter for derivation).

    Also, I cannot find any physical meaning for the Lagrangian. If I made any language mistakes, sorry, I'm Greek :P
     
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