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 $r^2 \dot{\theta}=Q$. 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!



The main idea.

It makes sense that the "redundant" constraint modifies the result because when you introduce the Lagrange multiplier, you not only restrict your attention to the paths that satisfy the constraint, but you also restrict your attention to variations that preserve the constraint.

In more detail.

For clarity, let $\mathscr P$ be the set of all admissible paths, and let $\mathscr P'\subseteq \mathscr P$ be the set of admissible paths whose angular momenta are constant.

If $\gamma$ is a critical point of $S$, then it has the following properties:

  • (S1) Its angular momentum is constant.
  • (S2) If you move in any "direction" in $\mathscr P$ away from $\gamma$, then the action $S$ won't change to first order.

On the other hand, if $\gamma$ is a critical point of $S'$, then it has the property (S1) but (S2) is now replaced by the following:

  • (S'2) If you move in any "direction" in $\mathscr P'$ away from $\gamma$, then the action $S$ won't change to first order.

Notice that (S2) is a stronger requirement than (S'2) because (S'2) only allows for perturbations of the path that keep its angular momentum constant, so the set of paths that satisfy (S1) + (S2) is a subset of the set of paths that satisfy (S1) + (S'2). This is precisely what you see in your math.

An analogy.

Consider some two-dimensional surface, like the surface of the Earth in the vicinity of a mountain range where there are a bunch of hills and valleys etc. Now suppose someone tells you to find all points on the surface for which if you walk a small amount in any direction along the surface away from that point, the elevation won't change to first order in how far you walk. You do this, and to your surprise you find that at every such point there is a purple flower. How odd!

Now suppose someone gives you a second task: find every point on the surface at which there is a purple flower such that if you walk in any direction in which there are also purple flowers, the elevation doesn't change to first order. In this case, you might find a bunch more points than in the previous case because there might be points where there is a purple flower such that if you move in any direction in which there are purple flowers, the elevation doesn't change to first order, but if you move in any other direction, the elevation does change to first order.

If points on the surface are like paths in $\mathscr P$, and points at which there is a purple flower are like paths in $\mathscr P'$, then the first task is like finding the critical points of $S$, and the second task is like finding the critical points of $S'$.

A related example.

Consider optimizing the function $$S(x,y) = x^2 + y^2.$$ This function has a single critical point at the origin $(0,0)$, a global minimum.

Consider, instead, optimizing $S$ subject to the constraint $(x-1)^2 + y^2 = 1$. This can be accomplished by introducing a Lagrange multiplier $\lambda$ and optimizing instead the function $$S'(x,y) = x^2 + y^2 + \lambda\big((x-1)^2 = y^2 - 1\big).$$ One finds that in this constrained optimization problem, there is still a global constrained minimum at $(0,0)$, but now there is also a global constrained maximum at $(2,0)$!

TO get intuition for this example, try drawing a picture of the surface defined by $z=f(x,y)$ and then consider the curve lying on this surface defined by only those points satisfying the additional constraint. You'll see immediately what's going on.

Here, as in the variational problem you're considering, the constrained optimization problem yields more critical points than the unconstrained problem.

  • $\begingroup$ Thank you very much, Josh! It's a very nice and clear answer! In fact, it turns out that I missed a crucial point for what I intend to think about next: Looking at a QFT version of this problem under the path integral. $\endgroup$ – psm Jun 22 '15 at 9:51
  • $\begingroup$ @psm Sure thing! By the way, I thought about the physics of this a lot more yesterday, and I think that you are changing the physics when you include the constraint because the "centrifugal" potential changes depending on the value of $c$. By varying the value of $c$, you can find paths that satisfy angular momentum conservation that you couldn't have had before because you've effectively added an additional force of constraint. $\endgroup$ – joshphysics Jun 22 '15 at 14:23

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