I saw a proof recently that demonstrated that for(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Why is the time integral of the Lagrangian minimal?

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