Landau, volume 1, chapter 1, problem 4: Lagrangian of a somewhat complex system

[Rick16]

TL;DR: find the Lagrangian of a somewhat complex system

This problem has been addressed before under https://www.physicsforums.com/threads/trouble-understanding-coordinates-for-the-lagrangian.1006528/ I also copied the following problem statement with Landau's very sketchy solution from this old post, because I don't have the English edition of the text.

The old thread only addressed the question of how to determine the coordinates of the masses. My problem is how to get the kinetic energy term from the displacement term. For one mass $m_1$ Landau writes the (square of its) displacement as $dl_1^2=a^2d\theta^2+a^2sin^2\theta d\phi^2$. His kinetic energy term for this mass is $\frac{1}{2}m_1a^2(\dot{\theta}^2+\dot{\phi}^2sin^2\theta)$. I don't know how he gets to this expression from his displacement term.

It seems easier not to lump together the displacement term into one expression right away, but to write the horizontal and vertical coordinates separately. I could then write (for $m_1$): $dy=ad\theta$ and $dx=asin\theta d\phi$. Then I take the derivatives $\dot{y}=a\dot{\theta}$ and $\dot{x}=a\dot{\theta}cos\theta d\phi + a sin\theta \dot{\phi}$. If I now neglect the term $a\dot{\theta}cos\theta d\phi$ because of the infinitesimally small angle $d\phi$ I get Landau's result. Is it legitimate to do it like this? In any case, I don't understand why Landau immediately wrote the square of the displacement and what he did next. Is there a direct and easy way to get from the square of the displacement to the kinetic energy term?

 

[Rick16]

I just thought about something. I am new to Landau and I am not familiar with his style. When he writes the displacement as $dl_1^2 =a^2 d\theta^2 + a^2 sin^2 \theta d\phi^2$, perhaps this is not meant to be an intermediate step towards the final solution, but just a helpful statement?

 

[div_grad]

@Rick16 This (square of the) displacement can be used to obtain the kinetic energy $T = mv^2/2$ because $T$ contains the square of the speed:
$$
v^2 = \left(\frac{\text{d}l}{\text{d}t}\right)^2 = \frac{(\text{d}l)^2}{(\text{d}t)^2} \rm{.}
$$

Then if, as is in your case, $\text{d}l^2$ contains a term like $a^2\text{d}\theta^2$, you use above equation to compute
$$
\frac{a^2\text{d}\theta^2}{(\text{d}t)^2} = a^2 \left(\frac{\text{d}\theta}{\text{d}t}\right)^2 = a^2 \dot{\theta}^2 \rm{,}
$$
and the same goes for the other terms in your displacement $\text{d}l^2$. So you see that finding the displacement can help you quickly write down the kinetic energy, because you may just replace things like $\text{d}\theta^2$ and $\text{d}\phi^2$ by $\dot{\theta}^2$ and $\dot{\phi}^2$, etc., and now you know why.

 

[Rick16]

@Rick16 This (square of the) displacement can be used to obtain the kinetic energy $T = mv^2/2$ because $T$ contains the square of the speed:
$$
v^2 = \left(\frac{\text{d}l}{\text{d}t}\right)^2 = \frac{(\text{d}l)^2}{(\text{d}t)^2} \rm{.}
$$

Then if, as is in your case, $\text{d}l^2$ contains a term like $a^2\text{d}\theta^2$, you use above equation to compute
$$
\frac{a^2\text{d}\theta^2}{(\text{d}t)^2} = a^2 \left(\frac{\text{d}\theta}{\text{d}t}\right)^2 = a^2 \dot{\theta}^2 \rm{,}
$$
and the same goes for the other terms in your displacement $\text{d}l^2$. So you see that finding the displacement can help you quickly write down the kinetic energy, because you may just replace things like $\text{d}\theta^2$ and $\text{d}\phi^2$ by $\dot{\theta}^2$ and $\dot{\phi}^2$, etc., and now you know why.

This is very helpful, thanks a lot.