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For a non-relativistic particle of mass $m$ with a conservative force with potential $U$ acting on the particle and a holonomic constraint given by $f(\mathbf{r},t)=0$, the system can be incorporated into the Lagrangian formulation via introducing a variable additionally to the coordinates $\lambda$, called Lagrange multiplier, with Lagrangian given by $$\mathcal{L}=\frac{1}{2}m|\mathbf{v}|^2-U(\mathbf{r})+\lambda f(\mathbf{r},t),$$ and applying the Euler-Lagrange equation for both the coordinates and $\lambda$, the equations of motion are $$m\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=-\nabla U+\lambda\nabla f; f(\mathbf{r},t)=0,$$ which is the standard Newton's second law and the force of constraint is identified as $\mathbf{F}_{\mathrm{c}}=\lambda \nabla f$.

My question is why bother doing this? I don't have any issues understanding how are they used, I simply want to know the point of them. All mechanics textbooks say that one of the reasons for the introduction of the Lagrangian formulation is so that we can eliminate the need for the constraint force to enter the equations of motion, but here we are reintroducing it after having devised a procedure to eliminate them. Why? If one is interested in the constraint force, why not simply revert to the Newtonian formulation?

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  • $\begingroup$ Does this answer your question? Point of Lagrange multipliers $\endgroup$
    – Babu
    Commented Apr 26, 2023 at 3:02
  • $\begingroup$ No because my question instead is as follows. As every mechanics textbook states, one of the reasons for introducing the Lagrangian formulation is to eliminate the forces of constraint from the equations of motion, so if one is interested on them, why introducing Lagrange multipliers and not simply revert to the Newtonian formulation? $\endgroup$
    – Don Al
    Commented Apr 26, 2023 at 3:08
  • $\begingroup$ When the variables of the Lagrangian are dependent, then the basic EL equations don't work anymore. For eg, if we have smthn like $L(x , \dot{x}, y , \dot{y})$ and $x(y)$ or something. The point is, we want to generalize the lagrangian formulation to deal with such interdependent variable cases as well (imagine pulley system. For that, we can either substitutes the $y$s in terms of $x$s) or introduce the lagrange multiplier term to make the lagrangrian term still work $\endgroup$
    – Babu
    Commented Apr 26, 2023 at 3:12
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    $\begingroup$ Newton=Lagrange basically. So, it doesn't matter. You culd maybe ask, why use lagrangian at all? Maybe some problem is easier because of it so $\endgroup$
    – Babu
    Commented Apr 26, 2023 at 3:37
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    $\begingroup$ It is often the case that one wants to use symmetry adapted coördinates, which often eliminates the constraints and it is thus easy to get solutions in the Lagrangian scheme. After getting some simple solutions, one might often want to take a step back and study the constraint forces. That is where the Lagrange multipliers come in. Minimal changes, so you can verify and identify the simple solutions. If you do Newtonian, you often have to work pretty hard just to get an equivalence between the Lagrangian solution and the Newtonian one. People routinely get stuck. $\endgroup$ Commented Apr 26, 2023 at 5:22

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