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## Notes on Goldstein’s Routh’s procedure (continued)

Posted by peeterjoot on March 5, 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 started previously.

Now, Goldstein defines the Routhian as

\begin{aligned}R = p_i \dot{q}_i - \mathcal{L},\end{aligned} \hspace{\stretch{1}}(2.16)

where the index $i$ is summed only over the cyclic (ignorable) coordinates. For this spherical pendulum example, this is $q_i = \phi$, and $p_i = m r^2 \sin^2 \theta \dot{\phi}$, for

\begin{aligned}R = \frac{1}{{2}} m r^2 \left( -\dot{\theta}^2 + \dot{\phi}^2 \sin^2 \theta \right) + m g r ( 1 + \cos\theta ).\end{aligned} \hspace{\stretch{1}}(2.17)

Now, we should also have for the non-cyclic coordinates, just like the Euler-Lagrange equations

\begin{aligned}\frac{\partial {R}}{\partial {\theta}} = \frac{d}{dt} \frac{\partial {R}}{\partial {\dot{\theta}}}.\end{aligned} \hspace{\stretch{1}}(2.18)

Evaluating this we have

\begin{aligned}m r^2 \sin\theta \cos\theta \dot{\phi}^2 - m g r \sin\theta = \frac{d}{dt} \left( -m r^2 \dot{\theta} \right).\end{aligned} \hspace{\stretch{1}}(2.19)

It would be reasonable now to compare this the $\theta$ Euler-Lagrange equations, but evaluating those we get

\begin{aligned}m r^2 \sin\theta \cos\theta \dot{\phi}^2 + m g r \sin\theta = \frac{d}{dt} \left( m r^2 \dot{\theta} \right).\end{aligned} \hspace{\stretch{1}}(2.20)

Bugger. We’ve got a sign difference on the $\dot{\phi}^2$ term? But I don’t see any error made.

# Simpler planar example.

Having found an inconsistency with Routhian formalism and the concrete example of the spherical pendulum which has a cyclic coordinate as desired, let’s step back slightly, and try a simpler example, artificially constructed

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m (\dot{x}^2 + \dot{y}^2 ) - V(x).\end{aligned} \hspace{\stretch{1}}(3.21)

Our Hamiltonian and Routhian functions are

\begin{aligned}H &= \frac{1}{{2}} m (\dot{x}^2 + \dot{y}^2 ) + V(x) \\ R &= \frac{1}{{2}} m (-\dot{x}^2 + \dot{y}^2 ) + V(x) \end{aligned} \hspace{\stretch{1}}(3.22)

For the non-cyclic coordinate we should have

\begin{aligned}\frac{\partial {R}}{\partial {x}} = \frac{d}{dt} \frac{\partial {R}}{\partial {\dot{x}}},\end{aligned} \hspace{\stretch{1}}(3.24)

which is

\begin{aligned}V'(x) = \frac{d}{dt}\left( - m \dot{x} \right)\end{aligned} \hspace{\stretch{1}}(3.25)

Okay, good, that’s what is expected, and exactly what we get from the Euler-Lagrange equations. This looks good, so where did things go wrong in the spherical pendulum evaluation.

# References

[1] H. Goldstein. Classical mechanics. Cambridge: Addison-Wesley Press, Inc, 1st edition, 1951.

1. ### Notes on Goldstein’s Routh’s procedure (continued again.) « Peeter Joot's Blog.said

[…] This continues the Routhian procedure notes from last post. […]

2. ### Ashasaid

As you are taking partial derivatives of L, H and R, which are depend on multiple variables, it is helpful (and possibly clear up some confusion early on) to write the respective arguments of the functions for every circumstance.

• ### peeterjootsaid

I ended up figuring out which variables to use in the followup to this, but you are right, that would have helped.