Basic exercise for finding a Lagrangian from the Landau's Mechanics

[Lurco]

Basic exercise for finding a Lagrangian from the Landau's "Mechanics"

Hello everyone!

Homework Statement

I've just started preparing for the classical mechanics course using only Landau & Lifgarbagez, so I'm doing everything according to their formulation.

And so I solved an exercise from the first chapter, but I'm not sure if I made that one correct (the answer is different, but a Lagrangian is ambiguous, isn't it?). The statement is:

Find the Lagrange function of a simple pendulum of length $l$, mass $m$, which pivot rotates along a circle of radius $a$ with a constant frequency $f$ in a homogeneous gravitational field with acceleration $g$.

Homework Equations

Formula for a Lagrangian of an isolated, one-particle system in 2 dimensions:
$$L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+\textbf{F}\cdot \textbf{r}$$

The Attempt at a Solution

I first formulated a more general case:
- the pivot moves around a circle satisfying $\alpha(t)$, ($\alpha$ is the angle between the pivot's radius vector and the y-axis),
- the angle between the pendulum and y-axis is $\phi(t)$,
- I've chosen the $\alpha$ and $\phi$ for the generalized coordinates of this system.

- the Cartesian coordinates of the mass are:
$$x=l \sin{\phi}+a \sin{\alpha}$$
$$y=l \cos{\phi}+a \cos{\alpha}$$

- the velocities:
$$\dot{x}=l \dot{\phi} \cos{\phi}+a \dot\alpha} \cos{\alpha}$$
$$\dot{y}=-l \dot\phi} \sin{\phi}-a \dot\alpha} \sin{\alpha}$$


U=$-\textbf{F}\cdot \textbf{r}=-mgy=-mg(l \cos{\phi}+a \cos{\alpha})$

$$T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)=\frac{1}{2}m(l^2 \dot{\phi}^2 \textrm{cos}^2 \phi+a^2 \dot{\alpha}^2 \textrm{cos}^2\alpha+2 l a \dot{\phi} \dot{\alpha} \cos{\alpha} \cos{\phi}+l^2 \dot{\phi}^2 \textrm{sin}^2 \phi+a^2 \dot{\alpha}^2 \textrm{sin}^2 \alpha+2 l a \dot{\phi} \dot{\alpha} \sin{\alpha} \sin{\phi})$$

$$L=T-U=m \left(\frac{1}{2}(l^2 \dot{\phi}^2+a^2 \dot{\alpha}^2)+l a \dot{\phi} \dot{\alpha} \cos{(\alpha-\phi)}+g(l \cos{\phi}+a \cos{\alpha}) \right)$$

-And now I substitute $\alpha=\alpha(t)=f t$, and my final solution is:
$$L=m \left(\frac{1}{2}(l^2 \dot{\phi}^2+a^2 f^2)+l a \dot{\phi} f \cos{(f t-\phi)}+g(l \cos{\phi}+a \cos{f t}) \right)$$

The answer in Landau & Lifgarbagez is:
$$L=\frac{1}{2}m l^2 \dot{\phi}^2+m l a f^2 \sin{(\phi-f t)}+m g l \cos{\phi}$$

So is my solution correct and if not, what mistakes am I making? I will be grateful for every answer. I will just add that I don't fully understand the idea of the "total derivative" which, as I suspect is crucial for comparing this two solutions.

 

[vela]

A total derivative is where you differentiate everything as opposed to holding some variables constant. For example, say you have a function F(x,t) = x sin t, where x is a function of t. When you take the partial derivative, you hold x constant and get

$$\frac{\partial F}{\partial t} = x \cos t$$

With the total derivative, you allow for x to vary with t, so you'd get

$$\frac{dF}{dt} = \frac{\partial F}{\partial x} \frac{dx}{dt} + \frac{\partial F}{\partial t} = (\sin t)x'(t) + x \cos t$$

So your Lagrangian has two extra terms that you can express as a total derivative with respect to time, namely

$$\frac{1}{2} ma^2f^2 + ag\cos ft = \frac{d}{dt} \left(\frac{1}{2} ma^2f^2t + \frac{ag}{f}\sin ft\right)$$

so you can toss those terms. After doing that, I can almost transform their Lagrangian into yours by considering

$$\frac{d}{dt} [f \cos (\phi-ft)]$$

You can get the two Lagrangians to look almost exactly like each other. You just need to figure out if the remaining difference really matters or if you can somehow get it to go away.