# Frictionless bead sliding along a rotating stick

Problem 6.8 in Morin's book on Classical Mechanics has this setup:

A massless stick pivots at its end in a horizontal plane with constant angular velocity $$\omega$$, while a frictionless bead of mass $$m$$ slides along it.

The goal is to compute the Lagrangian $$L$$ and the Hamiltonian $$p\cdot \dot{q}-L=H$$.($$\ast$$)

The solution makes sense to me when, using polar coordinates, the Lagrangian winds up being just the kinetic energy $$L=\frac12 m\dot r^2+\frac12mr^2\dot\theta^2=\frac12 m\dot r^2+\frac12mr^2\omega^2$$

because $$\dot\theta=\omega$$

Then to evaluate $$H=\frac{\partial L}{\partial \dot\theta}\dot\theta+\frac{\partial L}{\partial\dot r}\dot r - L$$, I think that

$$\frac{\partial L}{\partial \dot\theta}=mr\dot\theta$$ and $$\frac{\partial L}{\partial\dot r}=m\dot r$$, so that the terms before $$-L$$ in $$H$$ above become

$$mr\dot\theta^2+m\dot r^2$$

and therefore I thought $$H=\frac12 m\dot r^2 + \frac12mr^2\dot\theta^2=\frac12 m\dot r^2 + \frac12mr^2\omega^2$$.

But according to the solution, this ought to come out to just $$m\dot r^2$$ (as opposed to also having $$mr\dot\theta^2$$) and $$H$$ is supposed to be $$\frac12 m\dot r^2 - \frac12mr^2\omega^2$$.

What it looks like to me, is that rather than continue to use the form of $$L$$ with $$\dot \theta$$ in it to compute the conjugate momenta, $$\dot\theta$$ was immediately replaced with the constant $$\omega$$ so that the term $$\frac12mr^2\omega^2$$ became zero in in $$\frac{\partial L}{\partial \dot\theta}$$ (being a constant with respect to $$\dot\theta$$.)

So that is what I'm asking about: why should I believe we have license to substitute $$\omega$$ for $$\dot\theta$$ into $$L$$ before computing a partial derivative? I do not recall any explicit discussion of that but it may be in there: I only have the sample chapter, not the whole book. Obviously it does not yield the same answer if you substitute at the very end!

($$\ast$$) I guess probably I shouldn't call it the Hamiltonian because the book doesn't do that. The Hamiltonian is supposed to be a function of the coordinates and the conjugate momenta. But as I understand it, $$H$$ and $$L$$ are supposed to be related this way via the Legendre transformation.

• That is because $\theta$ is not really a coordinate here. We are constrained to have $\theta = \theta_0 + \omega t$. This is not just true on shell (i.e. where EL holds), but off shell as well. As such, we should not include it in our variation. Aug 2, 2021 at 2:26
• You're actually dealing with a pretty mathematically subtle system here. If we treat $\theta$ as a coordinate, then the map $\dot{x} \mapsto p_x = \frac{\partial L}{\partial \dot{x}}$ will not be invertible since $p_{\theta} = \frac{\partial L}{\partial \dot{\theta}} = 0$. This means we can't write down the Hamiltonian as $p\dot{q} - L$ because there is no expression for $\dot{q}$ in terms of $p_{\theta} = 0$. We have to pass to the formalism of constrained Hamiltonians, which is actually quite involved. Aug 2, 2021 at 2:32
• @CharlesHudgins thanks for putting your finger on it. I realized that there should be a constraint here, because it cannot simply be the lagrangian of a free mass in polar coordinates. But I was at a loss of how to enforce the mass staying at the correct angle as it slid out. You’ll write an answer I hope? Aug 2, 2021 at 2:44
• @CharlesHudgins I think I should not have mentioned hamiltonians: the take was actually just to compute that conserved quantity. Aug 2, 2021 at 2:45
• To reassure you, I somewhat painstakingly solved this problem in the Lagrangian, Hamiltonian, and Newtonian framework and in each case got the same answer. Suffice to say, not treating $\theta$ as a coordinate is the way to go. It gets you the right answer. Aug 2, 2021 at 3:58

"why should I believe we have license to substitute ω for $$\dot{\theta}$$ into L before computing a partial derivative?"
Good question. The reason is that here, unlike the more common case, $$\theta$$ is completely determined externally, even though it's not constant. The premise of the setup is that no matter what the bead does, its $$\theta$$-value is completely predetermined by the movement of the rod. For that reason it's an external field, not a coordinate.
The fundamental assumption of Lagrangian mechanics is that the system picks the path that minimizes the action (or at least a path that satisfies first-order conditions). For your system, you could surely lower the action still further if you could choose a different value for $$\theta(t)$$, but you're given that you cannot. All you can do is minimize over the coordinates that are given, which in this case is just $$r(t)$$.