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I am having some confusion regarding Lagrange multipliers and constraint forces. Consider the simple pendulum with generalized coordinates $r$ and $\phi$. We have the constraint that $r=l$. Orienting our coordinate system such that the pivot of the pendulum is at the origin, the Lagrangian is $$ L = \frac{1}{2} \left(\dot{r}^2 + r^2 \dot{\phi}^2 \right)+mgr \cos \phi $$ If we orient our coordinate system so that the zero of the potential is at the bottom of the pendulum swing, the Lagrangian is $$ L = \frac{1}{2} \left(\dot{r}^2 + r^2 \dot{\phi}^2 \right)-mgr(1- \cos \phi) $$ Usually this difference does not matter in pendulum problems because $r$ is a constant, so the Euler-Lagrange equations are identical. However, introducing the constraint $r = l$ and doing the usual procedure of Lagrange multipliers we obtain two different equations of motion $$ m\left(\ddot{r} - r \dot{\phi}^2 \right) - mg \cos \phi = \lambda $$ and $$ m\left(\ddot{r} - r \dot{\phi}^2 \right) + mg(1- \cos \phi) = \lambda $$ Usually we next associate the tension $\tau$ with $-\lambda$. But in this case that association would give two conflicting values of tension depending on how we defined our coordinate system, which can't be. The difference is not compensated by the value of $\dot{\phi}$, I performed the calculation and obtained the same result regardless of the definition of coordinate system. I am fairly certain that I get the right answer when I carry out the rest of the problem using the first Lagrangian, but I do not understand how to resolve this issue.

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  • $\begingroup$ Without a string to enforce the constraint $r=l$, the correct Lagrangian that governs the motion of the particle is $L = m(\dot{r}^2 + r^2 \dot{\phi}^2)/2 + mgr \cos\phi$. One should introduce the constraint on top of this Lagrangian. $\endgroup$ – higgsss Jun 9 '17 at 14:56
  • $\begingroup$ @higgsss you are right. the potential energy in the second Lagrangian is incorrect because it assumes the constraint. The potential in that coordinate system should be $V = mgl - mgr \cos \phi$. With this definition the two equations of motion are the same. $\endgroup$ – Jackson Burzynski Jun 9 '17 at 15:04
  • $\begingroup$ @higgsss your comment allowed me to see this. If you put it as an answer I'll accept it. $\endgroup$ – Jackson Burzynski Jun 9 '17 at 15:05
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I think that the statement "identify the tension with $-\lambda$" is the problem. When the pendulum is rest you get $\lambda=-mg$ in the first case and $\lambda=0$ in the second. Both are correct, but $-\lambda=$tension is only true in the first case. This is because the extra potential $V_{\rm extra}=mgr$ included in the second Lagrangian is supplying the constraint force already.

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