# Sign of Lagrange multiplier

Hello I have a short question.

Say I would consider a pendulum and define the Lagrangian as usual being \begin{align} L = \frac{1}{2} m(\dot{x}^2 + \dot{y}^2) - mgy \color{red}{-} \lambda (x^2 + y^2 - \ell^2). \end{align} Then I derive the equations of motion. For the $$x$$ component I have \begin{align} m\ddot{x} = \color{red}{-}\lambda x \, . \end{align} So at this stage I noticed that a Lagrangian with a term $$\color{red}{+}\lambda(x^2+y^2-\ell^2)$$ would result the same equation with a reversed sign, but as I understand this would mean a tension force in the reversed direction. Is it true that there is only one sign being correct, the minus sign as usual?

• A redefinition $\tilde{\lambda}:=-\lambda$ does not change the direction of the tension force, if that's your question. May 2, 2019 at 10:10
• $\lambda$ represents the tension force (at least the strength).... May 2, 2019 at 11:32

More generally, if the constraint $$f(q,t)~\approx ~0\tag{1}$$ enters the extended Lagrangian as

$$\widetilde{L}(q,\dot{q},\lambda,t)~=~L(q,\dot{q},t) \color{red}{+} \lambda f(q,t) ,\tag{2}$$

with Lagrange equations

$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}^j}-\frac{\partial L}{\partial q^j}~=~Q^{c}_j, \qquad j~\in \{1,\ldots, n\}, \tag{3}$$

then the $$j$$th generalized constraint force is

$$Q^{c}_j~=~\color{red}{+} \lambda\frac{\partial f}{\partial q^i}.\tag{4}$$

In particular, a linear rescaling of the Lagrange multiplier $$\lambda$$ or the constraint $$f$$ does not change the generalized constraint force $$Q^{c}_i$$.

References:

1. H. Goldstein, Classical Mechanics; Chapter 1 & 2.