# Classical Mechanics proof Lagrangian constraint forces

I've got a simple mathematical question. I was studying the Lagrangian approach of classical mechanics and in this part I had the intention of proving that the differential of the Lagrangian is equal to $$-Fr.dr$$, where $$Fr$$ is the restriction force and $$dr$$ is an infinitesimal displacement orthogonal to $$Fr$$, this would result in the differential being 0 and the proof that the Lagrangian method works even with constraint forces, like in the section 7.4 of the book Classical Mechanics from John Taylor. But I think I am missing something, a plus or minus sign from somewhere, and not being able to find the solution. If someone could help.

If you want to proof that the Lagrange method also works for constraint forces I think you're approaching the problem the wrong way. I recommend you use not L but rather L´, such that

$$L' = L + \sum_{j=1}^{m} \lambda_j f_j(x,y, t)$$

, where f(x,y,t) is a holonomic constraint (this wasn't specifically mentioned in your question, but I assume that these are the constraints you are talking about) and $$\lambda$$ is a lagragian multiplier. Lets now say that for the sake of this proof m = 1 (only one constraint force), so we get

$$L' = L + \lambda f(x,y, t)$$

we now need

$$\delta S = \delta \int_{t1}^{t2} (L + \lambda f(x,y, t)) dt = 0$$

in order for the motion to happen. We know that if we had no constraints this equation is true iff

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0$$

, where $$q_i$$ is a coordinate of the system. So what we do in order to account for the constraint force is calculate $$\lambda$$ as a new coordinate to the system. We hence get L' = L'(x,y,$$\lambda$$,t). So in addition to $$q_1 = x$$ and $$q_2 = y$$ we simply let $$q_3 = \lambda$$. We thus get that

$$\delta S = 0$$

$$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot\lambda} \right) - \frac{\partial L}{\partial \lambda} = 0$$.
$$\vec{F} = \lambda \vec{\nabla}f$$
This is true because f has to describe a plane and the Force can not contribute to the acceleration in this plane, which I assume is what you mean by "Fr.dr = 0". So the Force $$\vec{F}$$ has to be proportional to the normal vector of the plane. It is now easy to show that this normal vector is always parallel to $$\vec{\nabla}f$$. So that the constraint force can be always written as stated above and this means that you get equations of motion from the lagragian method even when there are constraint forces.