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When using the method of Lagrange undetermined multipliers, it's assumed that the constraint generalized force, $Q_j$, is given by:

$$Q_j=\lambda \cdot \frac{\partial f}{\partial q_{j}}$$

Where $f$ is my constraint equation.

Why is that? I understand that it has to be somehow related to $f$, but why its derivative multiplied by a constant? Is there an intuitive reason for that?

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  • $\begingroup$ Do you know that the gradient of a function is perpendicular to the level sets of the function? Then this just means that the constraint force is parallel to the gradient which is perpendicular to the level sets of the constraint. $\endgroup$
    – Silas
    Commented Mar 29, 2022 at 16:43
  • $\begingroup$ @Silas, I know how to use lagrangian multipliers to optimize a function in a math problem (and the reason behind it), but I don't know how to formally turn this into an optimization problem the way I'd do it in calculus. What would the function I'm optimizing be? I'm not using the gradiant in any other function besides the restriction. $\endgroup$ Commented Mar 29, 2022 at 16:59
  • $\begingroup$ The only way I know to derive this is with D'Alembert's principle but there you conclude that the $Q_j$ live in the space of $\frac{\partial f}{\partial q_j}$ and must therefore be linearly dependent. $\endgroup$
    – Silas
    Commented Mar 29, 2022 at 18:54

2 Answers 2

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lets look at this pendulum example, from the free body diagram you obtain :

enter image description here

the EOM's with Newton second law

$$m\,\ddot x-F_x=0\\ m\,\ddot y+m\,g-F_y=0$$

where $~F_x~,F_y~$ are the constraint forces at the pivot.

additional you have the constraint equation

$$f(x,y)=x^2+y^2-L^2=0$$

if you apply the EL with the kinetic energy

$$T=\frac m2\,(\dot x^2+\dot y^2)$$

and the potential energy

$$U=-m\,g\,y+\lambda\,(x^2+y^2-L^2) $$

you obtain the EOM's

$$m\,\ddot x-\lambda\,2\,x=0\\ m\,\ddot y-\lambda\,2\,y=0$$

hence

$$F_x=\lambda\,2\,x=\lambda\frac{\partial f(x,y)}{\partial x}\\ F_y=\lambda\,2\,y=\lambda\frac{\partial f(x,y)}{\partial y}$$

or with $~Q_1=F_x~,Q_2=F_y~,q_1=x,q_2=y$

$$Q_j=\lambda\,\frac{\partial f}{q_j}$$

notice that unit of $~Q_j~$ is [N], so the unit of $~\lambda~$ in this case is $~[N/m]$

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  1. Given a holonomic constraint $$f(q,t)~\approx ~0,$$ if the Lagrangian is of the form $$L~=~T-U+\lambda f,$$ then the Euler-Lagrange (EL) equations $$0~\approx~\frac{d}{dt}\frac{\partial L}{\partial\dot{q}^j}-\frac{\partial L}{\partial q^j} ~=~\dot{p}_j-\underbrace{Q_j^{\rm cons}}_{\begin{matrix}\text{generalized}\cr\text{conservative}\cr\text{forces}\end{matrix}}- \underbrace{\lambda\frac{\partial f}{\partial q^j}}_{\begin{matrix}\text{generalized}\cr \text{constraint}\cr\text{forces}\end{matrix}} $$ becomes Newton's 2nd law. This essentially answers OP's question.

  2. This can be generalized to semi-holonomic constraints, cf. e.g. this Phys.SE post.

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