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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.

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This entry was posted on March 5, 2010 at 11:57 pm and is filed under Math and Physics Learning.. Tagged: , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Notes on Goldstein’s Routh’s procedure (continued)”

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

    March 6, 2010 at 3:11 pm

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

    Reply

  2. Asha said

    January 26, 2012 at 4:49 pm

    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.

    Reply

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