# What Landau did in there to get the answer of problem 3 of the first book?

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.

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.

• 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 $(x_m,y_m)=(l\sin\theta+a\cos\gamma t,l\cos\theta)$? – user1583209 Feb 9 '17 at 15:11
• 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' – user78217 Feb 9 '17 at 15:14
• Is there any assumption done about $\gamma$ compared to the frequency of the pendulum? – user1583209 Feb 9 '17 at 15:40
• 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. – Kyle Kanos Feb 9 '17 at 15:52
• @user1583209 no, gamma is just a constant – user78217 Feb 9 '17 at 16:11

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.: $$\mathcal{L}=\frac{m}{2}l^2\dot{\theta}^2+mla\gamma^2\sin(\theta-\gamma t)+mgl\cos\theta$$

• Of course! I'll try to use these results on my problems – user78217 Feb 9 '17 at 18:11
• This helped me to find the correct result. – user78217 Feb 9 '17 at 18:18
• 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. – user1583209 Feb 9 '17 at 19:00
• I'l try now to prove that a total derivative (with respect to time) doesn't change the Lagrangian – user78217 Feb 9 '17 at 19:03