# Why do we insert the Lagrangian with some Lagrange multipliers into the Euler-Lagrange-equation?

Let's assume that we have two coordinates $$q_1$$ and $$q_2$$, a Lagrangian function $$\mathcal{L}(q,\dot{q},t):=\mathcal{L}(q_1, q_2, \dot{q_1}, \dot{q_2},t)$$ and a constraint $$f(q,t) = 0$$. Then I understand the procedure to obtain the equations of motions as constructing a new Lagrangian function $$\mathcal{L}'(q,\dot{q},t) = \mathcal{L}(q,\dot q, t) + \lambda(t) f(q,t)$$ where $$\lambda$$ is the Lagrange multiplier. Then you insert this in the Euler-Lagrange-equations (where you treat $$\lambda$$ like a coordinate $$q_3$$, so you get three equations) and these equations of motions then describe your constrained system.

I don't really understand why this is working. In regular non-physical optimization problems I already constructed Lagrangian functions with the multipliers analogous to the above one, but then I always took the gradient of the Lagrangian function and set it equal to zero. I also understand why this works, because this procedure results in a calculation of points where the gradient of the function of interest and the gradient of the constraint function are parallel, and this is a necessary condition for a maximum which also satisfies the constraint. But in physics we don't do the gradient of the Lagrangian - we just insert it in the Euler-Lagrange-equation (or is this somehow equivalent here? If yes, why?).

• How do you find the extremum of a function when there is a constraint? You must have learned that in you analysis courses. Sep 17 '20 at 11:41
• Oбжорoв: I don't have a problem with the procedure of finding the extremum of a function under constraints like I did it in my analysis course. My problem is that it seems to me that we do it different in physics here. ZeroTheHero: I think it is $\lambda(t)$. I just found the section "Lagrange multipliers and constraints" at this Wikipedia article, and there it also says that it is $\lambda(t)$. I will read it more carefully after this reply, so maybe it will answer my question Sep 17 '20 at 13:06

So we are given a lagrangian of the form $$\mathcal{L}(q_1,q_2,\dot{q_1},\dot{q_2},t)$$ with constraint equation $$f(q_1,q_2,t) = 0$$.

Well, according to Hamilton's principle we have $$\delta\int_{t_1}^{t_2}\mathcal{L}dt=\int_{t_1}^{t_2}\left( \left[ \frac{\partial\mathcal{L}}{\partial q_1}-\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q_1}}\right )\right]\delta q_1 +\left[ \frac{\partial\mathcal{L}}{\partial q_2}-\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q_2}}\right )\right]\delta q_2 \right) dt=0$$ At given instant $$\delta q_1$$ and $$\delta q_2$$ are not independent but $$\delta f= \frac{\partial f}{\partial q_1}\delta q_1+\frac{\partial f}{\partial q_2}\delta q_2=0$$ Thus we have $$\int_{t_1}^{t_2}\left( \left[ \frac{\partial\mathcal{L}}{\partial q_1}-\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q_1}}\right )\right] \frac{1}{\frac{\partial f}{\partial q_1}} -\left[ \frac{\partial\mathcal{L}}{\partial q_2}-\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q_2}}\right )\right]\frac{1}{\frac{\partial f}{\partial q_2}} \right)\delta q_1 dt=0$$ which is true for all choices of $$\delta q_1$$

$$\frac{\frac{\partial\mathcal{L}}{\partial q_1}-\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q_1}}\right )}{\frac{\partial f}{\partial q_1}}=\frac{\frac{\partial\mathcal{L}}{\partial q_2}-\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial \dot{q_2}}\right )}{\frac{\partial f}{\partial q_2}}$$

One obvious way in which we can solve this equation is to separately set both sides equal to the same function of time, which we shall denote $$-\lambda(t)$$. It follows that the Lagrangian equations of motion of the system can be written

$$\displaystyle \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_1}\right)-\frac{\partial L}{\partial q_1} - \lambda(t)\,\frac{\partial f}{\partial q_1}$$ $$\textstyle =$$ $$\displaystyle 0,$$

$$\displaystyle \frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_2}\right)-\frac{\partial L}{\partial q_2} - \lambda(t)\,\frac{\partial f}{\partial q_2}$$ $$\textstyle =$$ $$\displaystyle 0.$$

which is same as to solve $$\mathcal{L}′(q,\dot{q},t)=\mathcal{L}(q,\dot{q},t)+\lambda(t)f(q,t)$$