# Peeter Joot's Blog.

• ## Archives

 ivor on Just Energy Canada nasty busin… A final pre-exam upd… on An updated compilation of note… Anon on About peeterjoot on About Anon on About
• ## People not reading this blog: 6,973,738,433 minus:

• 132,347 hits

# Archive for July, 2012

## Another worked Landau pendulum problem

Posted by peeterjoot on July 14, 2012

## Question: Pendulum with support moving in line

This problem like the last, but with the point of suspension moving in a horizontal line $x = a \cos\gamma t$.

Our mass point has coordinates

\begin{aligned}p &= a \cos\gamma t + l i e^{-i\phi} \\ &= a \cos \gamma t + l i ( \cos \phi - i \sin \phi ) \\ &= ( a \cos \gamma t + l \sin \phi, l \cos \phi ),\end{aligned} \hspace{\stretch{1}}(1.10)

so that the velocity is

\begin{aligned}\dot{p} = ( -a \gamma \sin \gamma t + l \dot{\phi} \cos \phi, -l \dot{\phi} \sin \phi ).\end{aligned} \hspace{\stretch{1}}(1.11)

Our squared velocity is

\begin{aligned}\dot{p}^2 &= a^2 \gamma^2 \sin^2 \gamma t + l^2 \dot{\phi}^2 - 2 a \gamma l \dot{\phi} \sin\gamma t \cos \phi \\ &= \frac{1}{{2}} a^2 \gamma^2 \frac{d{{}}}{dt}\left( t - \frac{1}{{2 \gamma}} \sin 2 \gamma t \right) + l^2 \dot{\phi}^2 - a \gamma l \dot{\phi} ( \sin( \gamma t + \phi) + \sin(\gamma t - \phi)).\end{aligned} \hspace{\stretch{1}}(1.12)

In the last term, we can reduce the sum of sines, finding a total derivative term and a remainder as in the previous problem. That is

\begin{aligned}\dot{\phi} (\sin( \gamma t + \phi) + \sin(\gamma t - \phi)) &= (\dot{\phi} + \gamma)\sin(\gamma t + \phi) - \gamma \sin(\gamma t + \phi)+(\dot{\phi} - \gamma)\sin(\gamma t - \phi) + \gamma \sin(\gamma t - \phi) \\ &= \frac{d{{}}}{dt} \left( -\cos(\gamma t + \phi) + \cos(\gamma t - \phi) \right)+ \gamma ( \sin(\gamma t - \phi) - \sin(\gamma t + \phi) ) \\ &= \frac{d{{}}}{dt} \left( -\cos(\gamma t + \phi) + \cos(\gamma t - \phi) \right)- 2 \gamma \cos \gamma t \sin\phi.\end{aligned} \hspace{\stretch{1}}(1.13)

Putting all the pieces together and dropping the total derivatives we have the stated solution

\begin{aligned}\mathcal{L} = \frac{1}{{2}} m \left( l^2 \dot{\phi}^2 + 2 a \gamma^2 l \cos \gamma t \sin\phi \right) + m g l \cos\phi\end{aligned} \hspace{\stretch{1}}(1.14)

# References

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

## Typo in Landau Mechanics problem? Nope.

Posted by peeterjoot on July 14, 2012

# 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)

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.

Posted in Math and Physics Learning. | Tagged: , , | 2 Comments »

## Updated notes compilations for phy356 and phy456 (QM I & II)

Posted by peeterjoot on July 1, 2012

Here’s two updates of class notes compilations for Quantum Mechanics

The QM I notes updates are strictly cosmetic (the book template is updated to that of classicthesis since it was originally posted). The chapters in QM II are reorganized a bit, grouping things by topic instead by lecture dates.