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Consider a particle executing 1D simple harmonic oscillation. To formulate Lagrangian, we use $V=(1/2)kx^2$ for potential energy, and via Euler-Lagrange's equation we formulate the equation of motion $F=ma=-kx$.

  1. Now, my 1st question is, if we initially did not know anything about the force, then how could we use $V=(1/2)kx^2$ as potential energy for the system?

  2. My second question is: if we knew the force $F=-kx$, then what makes the sense of using Lagrange's equation? I mean what is the implication of using Lagrange's equation to find the equation of motion?

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    $\begingroup$ Your first subquestion (v3) seems to make no sense: If we know nothing about the force, we would know nothing about the potential energy as well, or if it even exists. $\endgroup$
    – Qmechanic
    Commented Aug 14, 2017 at 8:26

2 Answers 2

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If your objective is to simply obtain the equations of motion for the simple harmonic oscillator in 1d, there is little advantage in using a Lagrangian. However, in more complicated problems, this approach is much more powerful that Newton's law (see for instance this problem).

The Lagrangian is a scalar, which means it does not depend on the coordinate system (although of course its exact form will depend on the coordinate used). This makes the Lagrangian well adapted for problems where it would be otherwise difficult to add vectors. In particular, since kinetic and potential energies are additive, it is easy to construct the Lagrangian for a multi-particle system.

The equations of motion can be obtain from a single function: the Lagrangian. The Lagrangian is also efficient, in the sense that you do not need to consider action-reaction pairs; this makes it easy to handle constraints.

The resulting equations of motions are elegant: through calculus of variation, the solutions to the Euler-Lagrange equations minimize the action integral. The extension of the equations of motion from particles to fields is immediate.

Finally, the Lagrangian formulation is a stepping stone to the Hamiltonian formulation and its extensions (including Hamilton-Jacobi).

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  • $\begingroup$ @user163104 did just that. :) $\endgroup$ Commented Aug 14, 2017 at 13:56
  • $\begingroup$ sorry! I did not understand this part "There is a single Lagrangian for the entire system". As far I know, Lagrangian is not unique for a system $\endgroup$
    – sid
    Commented Aug 14, 2017 at 15:06
  • $\begingroup$ @sid that's not what I meant. I meant that you can obtain the equations of motion from a single function. I've edited to accommodate the clarification. $\endgroup$ Commented Aug 14, 2017 at 15:31
  • $\begingroup$ Actually I have several questions in my mind regarding the validity of the equation L=T-V. Still, Thanks for the assistance $\endgroup$
    – sid
    Commented Aug 14, 2017 at 15:46
  • $\begingroup$ @sid you might care to consult Landau&Lifshitz's Mechanics, which has an insightful discussion of this. $\endgroup$ Commented Aug 14, 2017 at 15:48
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As the comment points point out. If we know nothing of the force, we would know nothing of the potential either, and we wouldn't be able to do anything really.

As for 2), why we need the Lagrangian. Unfortunately your example is so trivial that the Lagrangian formulation becomes unnecessarily cumbersome in this case.

The true power of Lagrangian mechanics shines through when the systems become more complicated, such as when there are constraints on the motion or then several particles are interacting.

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  • $\begingroup$ Should we necessarily consider only the potentials due to applied forces while formulating lagrangian? or both the constraint and applied force? Actually the problem is we don't really know the potentials due to constraint forces. $\endgroup$
    – sid
    Commented Aug 14, 2017 at 15:39
  • $\begingroup$ Actaullt the whole point with the lagrangian formalism is that you need not bother with constraint forces at all (at least when constraints limit the positions of the platices). For instance, if you know that particle 1 and 2 have to be a disance $r$ from each other, (while other stuff happen to them), this will be automatically built in using Langrangians, without the need to keep track of constraint forces. $\endgroup$ Commented Aug 14, 2017 at 15:48

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