Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Notes on Goldstein’s Routh’s procedure (continued again.)

Posted by peeterjoot on March 6, 2010

[Click here for a PDF of this post with nicer formatting]Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

This continues the Routhian procedure notes from last post.

Polar form example.

The troubles appear to come from when there is a velocity coupling in the Kinetic energy term. Let’s try one more example with a simpler velocity coupling, using polar form coordinates in the plane, and a radial potential. Our Lagrangian, and conjugate momenta, and Hamiltonian, respectively are

\begin{aligned}\mathcal{L} &= \frac{1}{{2}} m \left(\dot{r}^2 + r^2 \dot{\theta}^2 \right) - V(r) \\ p_r &= m \dot{r} \\ p_\theta &= m r^2 \dot{\theta} \\ H &= \frac{1}{{2 m}} \left((p_r)^2 + \frac{1}{{r^2}} (p_\theta)^2 \right) + V(r).\end{aligned} \hspace{\stretch{1}}(4.26)

Evaluation of the Euler-Lagrange equations gives us the equations of motion

\begin{aligned}\frac{d}{dt}\left( m \dot{r} \right) &= m r \dot{\theta}^2 - V'(r) \\ \frac{d}{dt}\left( m r^2 \dot{\theta} \right) &= 0.\end{aligned} \hspace{\stretch{1}}(4.30)

Evaluation of the Hamilitonian equations \partial_p H = \dot{q}, \partial_q H = -\dot{p} should give the same results. First for r this gives

\begin{aligned}\frac{1}{{m}} p_r &= \dot{r} \\ -\frac{1}{{m r^3}} (p_\theta)^2 + V'(r) &= -\dot{p}_r.\end{aligned} \hspace{\stretch{1}}(4.32)

The first just defines the canonical momentum (in this case the linear momentum for the radial aspect of the motion), and the second after some rearrangement is

\begin{aligned}m r (\dot{\theta})^2 - V'(r) &= \frac{d}{dt}\left( m \dot{r} \right),\end{aligned} \hspace{\stretch{1}}(4.34)

which is consistent with the Lagrangian approach. For the \theta evaluation of the Hamiltonian equations we get

\begin{aligned}\frac{p_\theta}{m r^2} &= \dot{\theta} \\ 0 &= -\dot{p}_\theta.\end{aligned} \hspace{\stretch{1}}(4.35)

The first again, is implicitly, the definition of our canonical momentum (angular momentum in this case), while the second is the conservation condition on the angular momentum that we expect associated with this ignorable coordinate. So far so good. Everything is as it should be, and there’s nothing new here. Just Lagrangian and Hamiltonian mechanics as usual. But we have two independently calculated results that are the same and the Routhian procedure should generate the same results.

Now, on to the Routhian. There we have a Hamiltionian like sum of p \dot{q} terms over all cyclic coordinates, minus the Lagrangian. Here the \theta coordinate is observed to be that cyclic coordinate, so this is

\begin{aligned}R &= p_\theta \dot{\theta} - \mathcal{L} \\ &= m r^2 \dot{\theta}^2 - \frac{1}{{2}} m \left(\dot{r}^2 + r^2 \dot{\theta}^2 \right) + V(r) \\ &= \frac{1}{{2}} m r^2 \dot{\theta}^2 - \frac{1}{{2}} m \dot{r}^2 + V(r).\end{aligned}

Now, this Routhian can be written in a few different ways. In particular for the \dot{\theta} dependent term of the kinetic energy we can write

\begin{aligned}\frac{1}{{2}} m r^2 \dot{\theta}^2 = \frac{1}{{2 m r^2 }} (p_\theta)^2 = \frac{1}{{2 }} \dot{\theta} p_\theta\end{aligned} \hspace{\stretch{1}}(4.37)

Looking at the troubles obtaining the correct equations of motion from the Routhian, it appears likely that this freedom is where things go wrong. In the Cartesian coordinate description, where there was no coupling between the coordinates in the kinetic energy we had no such freedom. Looking back to Goldstein, I see that he writes the Routhian in terms of a set of explicit variables

\begin{aligned}R = R(q_1, \cdots q_n, p_1, \cdots p_s, \dot{q}_{s+1}, \cdots \dot{q}_n, t) = \sum_{i=1}^{s} \dot{q}_i p_i - \mathcal{L}\end{aligned} \hspace{\stretch{1}}(4.38)

where q_1, \cdots q_s were the cyclic coordinates. Additionally, taking the differential he writes

\begin{aligned}dR &=   \sum_{i=1}^{s} \dot{q}_i dp_i - \sum_{i=1}^s \frac{\partial {\mathcal{L}}}{\partial {q_i}} dq_i- \sum_{i=1}^n \frac{\partial {\mathcal{L}}}{\partial {\dot{q}_i}} d\dot{q}_i- \frac{\partial {\mathcal{L}}}{\partial {t}} dt \\ &=\frac{\partial {R}}{\partial {p_i}} dp_i + \frac{\partial {R}}{\partial {q_i}} dq_i + \frac{\partial {R}}{\partial {\dot{q}_i}} d\dot{q}_i+ \frac{\partial {R}}{\partial {t}} dt,\end{aligned} \hspace{\stretch{1}}(4.39)

with sums implied in the second total differential. It was term by term equivalence of these that led to the Routhian equivalent of the Euler-Lagrange equations for the non-cyclic coordinates, from which we should recover the desired equations of motion. Notable here is that we have no \dot{q}_i for any of the cyclic coordinates q_i.

For this planar radial Lagrangian, it appears that we must write the Routhian, specifically as R = R(r, \theta, p_\theta, \dot{r}), so that we have no explicit dependence on the radial conjugate momentum. That is

\begin{aligned}R &= \frac{1}{{2 m r^2}} (p_\theta)^2 - \frac{1}{{2}} m \dot{r}^2 + V(r).\end{aligned} \hspace{\stretch{1}}(4.41)

As a consequence of 4.39 we should recover the equations of motion by evaluating \delta R/\delta r = 0, and doing so for 4.41 we have

\begin{aligned}\frac{\delta R}{\delta r} = V'(r) - \frac{1}{{m r^3}} (p_\theta)^2 - \frac{d}{dt}\left( -m \dot{r} \right) = 0.\end{aligned} \hspace{\stretch{1}}(4.44)

Good. This agrees with our result from the Lagrangian and Hamiltonian formalisms. On the other hand, if we evaluate this variational derivative for

\begin{aligned}R &= \frac{1}{{2}} m r^2 \dot{\theta}^2 - \frac{1}{{2}} m \dot{r}^2 + V(r),\end{aligned} \hspace{\stretch{1}}(4.43)

something that is formally identical, but written in terms of the “wrong” variables, we get a result that is in fact wrong

\begin{aligned}\frac{\delta R}{\delta r} = m r \dot{\theta}^2 + V'(r) - \frac{d}{dt}\left( -m \dot{r} \right) = 0.\end{aligned} \hspace{\stretch{1}}(4.44)

Here the term that comes from the \dot{\theta} dependent part of the Kinetic energy has an incorrect sign. This was precisely the problem observed in the initial attempt to work the spherical pendulum equations of motion starting from the Routhian.

What variables to use to express the equations is a rather subtle difference, but if we do not get that exactly right the results are garbage. Next step here is go back and revisit the spherical polar pendulum and verify that being more careful with the variables used to express R allows the correct answer to be obtained. That exersize is probably for a different day, and probably a paper only job.

Now, I note that Goldstein includes no problems for this Routhian formalism now that I look, and having worked an example successfully and seeing how we can go wrong, it is not quite clear what his point including this was. Perhaps that will become clearer later. I’d guess that some of the value of this formalism could be once one attempts numerical solutions and finds the cyclic coordinates as a result of a linear approximation of the system equations around the nieghbourhood of some phase space point.

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