0
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$ Aug 2, 2021 at 2:26
  • $\begingroup$ 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. $\endgroup$ Aug 2, 2021 at 2:32
  • $\begingroup$ @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? $\endgroup$
    – rschwieb
    Aug 2, 2021 at 2:44
  • $\begingroup$ @CharlesHudgins I think I should not have mentioned hamiltonians: the take was actually just to compute that conserved quantity. $\endgroup$
    – rschwieb
    Aug 2, 2021 at 2:45
  • $\begingroup$ 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. $\endgroup$ Aug 2, 2021 at 3:58

1 Answer 1

2
$\begingroup$

"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 $r(t)$.

$\endgroup$
3
  • $\begingroup$ So in fact, I should have just considered this as a one dimensional problem. Thank you. That makes sense now in retrospect. $\endgroup$
    – rschwieb
    Aug 2, 2021 at 2:47
  • 1
    $\begingroup$ This would be more or less the content of my answer if I wrote one up, so you should just accept this answer. The world of constrained hamiltonians is fascinating though and you should study it if you get the chance. $\endgroup$ Aug 2, 2021 at 2:53
  • $\begingroup$ @rschwieb Yep, physically it is 2D but in terms of the number of coordinates it's definitely a one-dimensional problem. In related work, I'm examining a mechanical structure with have time-varying fields. Once you realize that the shape of the structure is fixed and the only thing that's allowed to vary is its center of mass (that is, it's changing shape but in a "rigid" way) it reduces to a one-dimensional problem. In your case, the 1D problem is basically a mass attached to a spring with negative spring constant, since the mass naturally slides away from the center rather than towards it. $\endgroup$
    – David
    Aug 2, 2021 at 2:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .