Confusion about imposing constraint in the action
I'm totally confused by one thing. I know that I probably shouldn't be confused about that, but at the moment I don't quite know what fails in the following:
Suppose we have a particle of unit mass moving in a potential $V(r)$ on a plane. Then, the action one would consider is given by $$S= \int dt \left[\frac{1}{2}\dot{r}^2 + \frac{1}{2}r^2 \dot{\theta}^2 - V(r)\right].$$ Clearly, $Q=r^2 \dot{\theta}$, the angular momentum, is conserved, so the equation of motion for $r$ reads: $$-\ddot{r} + \frac{Q^2}{r^3}- V^{\prime}(r)=0.$$
So far, everything is fine.
But now I decide to impose angular momentum conservation by hand via a Lagrange multiplier $\lambda$:
$$S^{\prime}= \int dt \left[\frac{1}{2}\dot{r}^2 + \frac{1}{2}r^2 \dot{\theta}^2 - V(r) + \lambda(t)(r^2 \dot{\theta}-Q)\right],$$ where $Q$ is a constant.
To my understanding, this constraint should turn out to be redundant because angular momentum is already constant. But somehow, it doesn't:
First, variation of $S^{\prime}$ w.r.t. $\lambda$ trivially yields $r^2 \dot{\theta}=Q$. 101
Next, I vary w.r.t. $\theta$ and integrate by parts. I get $$\int dt \left[-\partial_t(r^2 \dot{\theta}) - \partial_t(\lambda(t)r^2) \right]\delta \theta=0,$$ i.e. $$\lambda = - \dot{\theta} + \frac{c}{r^2}$$ with constant $c$.
And finally, I vary w.r.t. $r$ and get:
$$0=\int dt \left[-\ddot{r} + r \dot{\theta}^2 - V^{\prime}(r) + 2\lambda(t)r \dot{\theta}\right]\delta r$$
Using the above equations of motion, I can rewrite the EoM for $r$ as:
$$-\ddot{r} - \frac{Q^2}{r^3} - V^{\prime}(r) + \frac{2Qc}{r^3}=0.$$
Unless $c=Q$ (i.e. $\lambda=0$ at any time), this is not the original equation of motion.
But why does the redundant constraint modify the result? Did I change the physics? If so, what is the interpretation behind the constaint and the integration constant $c$?
Thanks in advance!
psm 402