7
$\begingroup$

I found this problem in Landau-Lifschitz vol.1 (Mechanics)

A simple pendulum of mass $m$, length $l$ whose point of support moves uniformly on a vertical circle with constant frequency $\gamma$.

$$$$ The problem is, of course, to find the Lagrangian. So I turned to Cartesian coordinates $x$ and $y$ in order to construct the problem in terms of $\phi$ and the rotation terms, i.e. $$ x = a \cos{\gamma t} + l \sin{\phi}$$ $$ y = -a \sin{\gamma t} + l \cos{\phi}$$ I take the time derivative of each, square it, add it together, construct the kinetic and potential term, nothing fancy... (but here's the procedure anyway for the sake of completeness, maybe that's where the mistake is):

$$\dot{x} = -a \gamma \sin(\gamma t) + l \dot{\phi} \cos{\phi}$$ $$\dot{y} = -a \gamma \cos(\gamma t) - l \dot{\phi} \sin{\phi}$$

$$ \dot{x}^2 = a^2 \gamma^2 \sin^2(\gamma t) + l^2 \dot{\phi}^2 \cos^2{\phi} - 2al \gamma \dot{\phi} \sin{(\gamma t)} \cos{\phi}$$ $$ \dot{y}^2 = a^2 \gamma^2 \cos^2(\gamma t) + l^2 \dot{\phi}^2 \sin^2{\phi} + 2al \gamma \dot{\phi} \cos{(\gamma t)} \sin{\phi}$$ $$\dot{x}^2 + \dot{y}^2 = a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2al\gamma\dot{\phi}\sin(\phi -\gamma t) $$ $$\text{also} \hspace{2cm} V=-mgy=-mg(-a \sin{\gamma t} + l \cos{\phi})$$ and I get this Lagrangian: $$\mathcal{L} = \frac{1}{2}m(a^2 \gamma^2 + l^2 \dot{\phi}^2 + 2al\gamma\dot{\phi}\sin(\phi -\gamma t)) + mg(-a \sin(\gamma t) + l\cos{\phi})$$ Then I drop the terms which are total time derivatives, namely these: $$ \frac{1}{2} ma^2 \gamma^2$$ $$-mga \sin(\gamma t)$$ That leaves me with this Lagrangian: $$\mathcal{L}_{Me} = \frac{1}{2} m l^2 \dot{\phi}^2 + mal\gamma\dot{\phi}\sin(\phi -\gamma t) + mgl\cos{\phi}$$

But Landau gives this one: $$\mathcal{L}_{Landau} = \frac{1}{2} m l^2 \dot{\phi}^2 + \color{red}{mal\gamma^2 \sin(\phi -\gamma t) }+ mgl\cos{\phi}$$

Am I missing something? Where did that red term come from?

$\endgroup$
  • $\begingroup$ You'd have to write it more explicitly (the steps inbetween) unless you expect someone to go through the boring algebra (you probably made a small mistake somewhere?) $\endgroup$ – Danu Feb 25 '14 at 13:57
  • 3
    $\begingroup$ I'm pretty sure I didn't make a small mistake somewhere, but one can never be too sure. I'll accept your advice and include the intermediate steps explicitly. $\endgroup$ – Stack Exchange is Cancer Feb 25 '14 at 13:59
  • $\begingroup$ is there a list of errata for L&L? $\endgroup$ – innisfree Feb 25 '14 at 14:13
  • 1
    $\begingroup$ I remember I have struggled with the similar problem, maybe even in the same problem. AFAIR, you can select a bit trickier total time derivative (a derivative of product, IIRC), and then you'll get the L&L's result. $\endgroup$ – Ruslan Feb 25 '14 at 14:16
  • 1
    $\begingroup$ @SchlomoSteinbergerstein I'll be able to tell when I come home, in several hours. I should have this problem solved in my copybook. $\endgroup$ – Ruslan Feb 25 '14 at 14:21
6
$\begingroup$

Let's compare Landau's Lagrangian and the one given by $\mathcal L_{Me}$ in your question:

$$\mathcal{L}_{Landau}-\mathcal{L}_{Me}=mal\gamma^2\sin(\phi-\gamma t)-alm\gamma\dot\phi\sin(\phi-\gamma t)=\\ =mal\gamma\left((\gamma-\dot\phi)\sin(\phi-\gamma t)\right)=\\ =mal\gamma \frac{\text{d}}{\text{d}t}\cos(\phi-\gamma t)$$

Now the difference is obviously a total time derivative, thus you haven't made a mistake, just undersimplified the Lagrangian.

$\endgroup$
0
$\begingroup$

L=T-U is correct. However, to get the equation of motion wrt /phi you need to use: d/dt (partial L/partial phi') - partial L/partial phi =0.

If you follow this, you get the Landau equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.