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Typo in Landau Mechanics problem? Nope.

Posted by peeterjoot on July 14, 2012

[Click here for a PDF of this post with nicer formatting]

Motivation

Attempting a mechanics problem from Landau I get a different answer. I wrote up my solution to see if I can spot either where I went wrong, or demonstrate the error, and then posted it to physicsforums. I wasn’t wrong, but the text wasn’t either. Here’s the complete result.

Guts

Question: Pendulum with support moving in circle

section 1 problem 3a of [1] is to calculate the Lagrangian of a
pendulum where the point of support is moving in a circle (figure and full text for problem in this google books reference)

Answer

The coordinates of the mass are

\begin{aligned}p = a e^{i \gamma t} + i l e^{i\phi},\end{aligned} \hspace{\stretch{1}}(1.1)

or in coordinates

\begin{aligned}p = (a \cos\gamma t + l \sin\phi, -a \sin\gamma t + l \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.2)

The velocity is

\begin{aligned}\dot{p} = (-a \gamma \sin\gamma t + l \dot{\phi} \cos\phi, -a \gamma \cos\gamma t - l \dot{\phi} \sin\phi),\end{aligned} \hspace{\stretch{1}}(1.3)

and in the square

\begin{aligned}\dot{p}^2 = a^2 \gamma^2 + l^2 \dot{\phi}^2 - 2 a \gamma \dot{\phi} \sin\gamma t \cos\phi + 2 a \gamma l \dot{\phi} \cos \gamma t \sin\phi=a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi).\end{aligned} \hspace{\stretch{1}}(1.4)

For the potential our height above the minimum is

\begin{aligned}h = 2a + l - a (1 -\cos\gamma t) - l \cos\phi = a ( 1 + \cos\gamma t) + l (1 - \cos\phi).\end{aligned} \hspace{\stretch{1}}(1.5)

In the potential the total derivative \cos\gamma t can be dropped, as can all the constant terms, leaving

\begin{aligned}U = - m g l \cos\phi, \end{aligned} \hspace{\stretch{1}}(1.6)

so by the above the Lagrangian should be (after also dropping the constant term m a^2 \gamma^2/2

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma l \dot{\phi} \sin (\gamma t - \phi) \right) + m g l \cos\phi.\end{aligned} \hspace{\stretch{1}}(1.7)

This is almost the stated value in the text

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \sin (\gamma t - \phi) \right) + m g l \cos\phi.\end{aligned} \hspace{\stretch{1}}(1.8)

We have what appears to be an innocent looking typo (text putting in a \gamma instead of a \dot{\phi}), but the subsequent text also didn’t make sense. That referred to the omission of the total derivative m l a \gamma \cos( \phi - \gamma t), which isn’t even a term that I have in my result.

In the physicsforum response it was cleverly pointed out by Dickfore that 1.7 can be recast into a total derivative

\begin{aligned}m a l \gamma \dot{\phi} \sin (\gamma t - \phi) =m a l \gamma ( \dot{\phi} - \gamma ) \sin (\gamma t - \phi) +m a l \gamma^2 \sin (\gamma t - \phi) =\frac{d{{}}}{dt}\left(m a l \gamma \cos (\gamma t - \phi) \right)+m a l \gamma^2 \sin (\gamma t - \phi),\end{aligned} \hspace{\stretch{1}}(1.9)

which resolves the connundrum!

References

[1] LD Landau and EM Lifshitz. Mechanics, vol. 1. 1976.

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This entry was posted on July 14, 2012 at 12:20 am 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.

2 Responses to “Typo in Landau Mechanics problem? Nope.”

  1. peter said

    July 19, 2012 at 7:45 pm

    Hey Peeter — really interesting stuff. I had run into this in the past, where you can end up w. different looking Lagrangian’s (maybe not this complicated it).
    The obvious question I have is: is there a simple procedure to compare 2 Langranian’s and say “yeah, they’re the same” ?

    Reply

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