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I've been learning about the Lagrangian formulation recently, and while I'm with the process, I am still struggling somewhat with the theory behind it. As I (rather poorly) understand it, the formulation is based on the principle of least action. The Euler-Lagrange equation is used to find the minimums of the integrals of the Lagrangian (the action), which is then used to solve for whatever value you are interested in.

What I DON'T understand is why the Lagrangian is equal to the difference between the total kinetic and potential energy of the system. The definition seems arbitrary. What is the significance of that particular difference? Every lecture I've found seems to brush past the point, and I've yet to find a satisfying explanation. Can anyone provide a better justification?

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    $\begingroup$ Possible duplicate of Is there a proof from the first principle that the Lagrangian L = T - V? $\endgroup$
    – knzhou
    Commented Jun 8, 2016 at 3:35
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    $\begingroup$ also of physics.stackexchange.com/q/260789 . I like the intuitive argument in the answer by Jahan Claes " Now, what's the intuition? Well, the claim of Lagrangian mechanics is that the path the particle actually takes should minimize the integral of L over time. So in general, when a particle goes along a path, Lagrangian mechanics tells us the particle will spend lots of time at places where L is small, and not a lot of time at places where L is big. $\endgroup$
    – anna v
    Commented Jun 8, 2016 at 3:49
  • $\begingroup$ This makes sense: when L is big, we have LOTS of kinetic energy, so you expect the particle to zoom right by those spots; when L is small, the particle has very little kinetic energy, so it will stay there longer." $\endgroup$
    – anna v
    Commented Jun 8, 2016 at 3:49
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    $\begingroup$ I'll also add that I only mentioned kinetic energy in the post. But minimizing the integral of $T-V$ is the same thing as minimizing $T$ itself, subject to the constraint that total energy is conserved; including the $V$ just makes bookkeeping easier. $\endgroup$
    – Jahan Claes
    Commented Jun 8, 2016 at 4:33

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