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I have a very specific difficult on the Problem 3 Item (b)-(c) (page 12) of Landau's book Course of Theoretical Physics: Volume 1 Mechanics Second Edition. The problem is very straightforward:

Find the Lagrangian of a simple pendulum of length $l$ and mass $m$ whose point of support is (b) oscillating horizontally in the plane of motion of the pendulum according to the law $x = a\cos(\gamma t)$ where $a$ and $\gamma$ are constants (c) oscillating vertically in the plane of motion according $y = a\cos(\gamma t)$.

Then what I did was to put an origin in the point where $\cos(\gamma t)=0$ such that $x$ is the direction to the wright and $y$ is the direction of fall. enter image description here

Let work out the case of the image. Then with respect to that origin we get that the point that maps the mass $m$ can be described by

$$(x_m,y_m) = (l\sin(\theta),a\cos(\gamma t)+l\cos(\theta))$$

So

$$T = \frac{m}{2}(\dot{x}_m^2+\dot{y}_m^2) = \frac{m}{2}(l^2\dot{\theta}^2\cos^2(\theta) + a^2\gamma^2\sin^2(\gamma t) + 2al\gamma \dot{\theta}\sin(\gamma t )\sin(\theta) + l^2\dot{\theta}^2\sin^2(\theta))$$

And

$$U = -mg(y_m) = -mg(a\cos(\gamma t) + l\cos(\theta))$$

Such that the Lagrangian would be $\mathcal{L} = T - U$, but, the answer that he presents is

$$\mathcal{L} = \frac{1}{2}ml^2\dot{\theta}^2 + mla\gamma^2\cos(\gamma t)\cos(\theta) + mgl\cos(\theta)$$

My question is how can I get this answer? He says that he's omitting total derivatives but I do not understand what he means and how this could change my answer to his. I also think that he is omitting the terms that only depend on the time and the constants. But the problem is that he gets different trigonometry functions.

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  • $\begingroup$ Why do you add the horizontal displacement ($x=a\cos\gamma t$) along $x$ to the $y$-coordinate of the pendulum? Shouldn't it be $(x_m,y_m)=(l\sin\theta+a\cos\gamma t,l\cos\theta)$? $\endgroup$ – user1583209 Feb 9 '17 at 15:11
  • $\begingroup$ You have to do each case separately so first I did the $y$ case. If you first do the $x$ or the $y$ doesn't matter. What I think you just said is that 'are you doing item (b)?' my answer than is 'no, the example is item (c) when the point just moves vertically' $\endgroup$ – user78217 Feb 9 '17 at 15:14
  • $\begingroup$ Is there any assumption done about $\gamma$ compared to the frequency of the pendulum? $\endgroup$ – user1583209 Feb 9 '17 at 15:40
  • $\begingroup$ Even using $x=a\cos\gamma t$, which would get you the correct gravitational potential term in $\mathcal L$, I don't see how you can eliminate $\dot\theta$ from the kinetic energy term. Could you include more of the text from Landau about this problem? It's possible there's something else in the problem statement/solution that is important. $\endgroup$ – Kyle Kanos Feb 9 '17 at 15:52
  • $\begingroup$ @user1583209 no, gamma is just a constant $\endgroup$ – user78217 Feb 9 '17 at 16:11
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Taking part (a) as an example, where the support moves on a circle. (b) and (c) should have similar reasonings. The coordinates of the mass are:

$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}a\cos(\gamma t)+l\sin\theta\\-a\sin(\gamma t)+l\cos\theta\end{pmatrix}$$

The Lagrangian without any approximations (only collecting terms using trigonometric identities) becomes:

$$\mathcal{L}=\frac{m}{2}l^2\dot{\theta}^2+mla\gamma\dot{\theta}\sin(\theta-\gamma t)+mg\left(l\cos\theta-a\sin(\gamma t)\right)$$

Generally, the equations of motion are invariant on addition of a total time derivative to the Lagrangian (see end of the second section in the book). Specifically in this case:

  1. the last term ($-mga\sin(\gamma t)$) only depends on time and can therefore be ignored (does not contribute to the equations of motion)
  2. the second term can be rewritten using $\dot{\theta}\sin(\theta-\gamma t)=\gamma\sin(\theta-\gamma t) - \frac{d}{dt}\cos(\theta-\gamma t)$, and noting that terms that are total time derivatives (i.e. here $\frac{d}{dt}\cos(\theta-\gamma t)$) can be ignored (because these terms do not contribute to the equations of motion)

This together leads to the Lagrangian from the solution, i.e.: $$\mathcal{L}=\frac{m}{2}l^2\dot{\theta}^2+mla\gamma^2\sin(\theta-\gamma t)+mgl\cos\theta$$

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  • $\begingroup$ Of course! I'll try to use these results on my problems $\endgroup$ – user78217 Feb 9 '17 at 18:11
  • $\begingroup$ This helped me to find the correct result. $\endgroup$ – user78217 Feb 9 '17 at 18:18
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    $\begingroup$ Well, it is as correct as what you had. There are many correct Lagrangians. If you want to compare two Lagrangiangs you could take their difference $\Delta\mathcal{L}=\mathcal{L}'-\mathcal{L}$ and calculate $\frac{d}{dt}\frac{\partial\Delta\mathcal{L}}{\partial\dot{q}}-\frac{\partial\Delta \mathcal{L}}{\partial q}$ which should give 0 if they are both correct. $\endgroup$ – user1583209 Feb 9 '17 at 19:00
  • $\begingroup$ I'l try now to prove that a total derivative (with respect to time) doesn't change the Lagrangian $\endgroup$ – user78217 Feb 9 '17 at 19:03

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