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(Goldstein 3rd edition pg 72) After reducing two-body problem to one-body problem

We now restrict ourselves to conservative central forces, where the potential is $V(r)$ function of $r$ only, so that the force is always along $\mathbf{r}$. By the results of the preceding section, we need only consider the problem of a single particle of reduced mass $m$ moving about a fixed center of force, which will be taken as the origin of the coordinate system. Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, ahout any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic.

But the 3D kinetic energy has a form $$T = \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\dot\phi^2\right)$$ and hence the Lagrangian $L$ depends on $\theta$ and hence $\theta$ is not cyclic.

Doubt on Mr Joseph's answer :

Goldstein pg 59

It can be shown that if a cyclic coordinate $q_{j}$ is such that $d q_{j}$ corresponds to a rotation of the system of particles around some axis, then the conservation of its conjugate momentum corresponds to conservation of an angular momentum.

In our case $\phi$ is cyclic as is seen from equation of kinetic energy and change in $\phi$ corresponds to a rotation of our system so one component of angular momentum is conserved.

We can't say the same about $\theta$ since its not cyclic to begin with.

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  • $\begingroup$ Your kinetic energy is wrong $\endgroup$
    – Eli
    Oct 15, 2021 at 7:51
  • $\begingroup$ Please see this physics.stackexchange.com/q/183882 $\endgroup$
    – Kashmiri
    Oct 15, 2021 at 9:03
  • $\begingroup$ $\theta$ is not cyclic, neither is it “an angle coordinate representing rotation about a fixed axis”. $\endgroup$ Oct 29, 2021 at 3:42
  • $\begingroup$ @ZeroTheHero, exactly. Took long time to realise it, Thank you :) $\endgroup$
    – Kashmiri
    Oct 29, 2021 at 4:56

1 Answer 1

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The Lagrangian is $$\mathcal L=T-V(r)$$

This problem has spherical symmetry and the potential energy is a function of $r$ only as stated above and the particle is in a conservative central force. This means that the angular momentum is conserved and motion lies in a plane perpendicular to the angular momentum vector.

In such cases, one can chose the motion of the particle to be in a plane at right angles to the polar axis meaning we can reduce the Lagrangian to $$\mathcal{L}=T -V= \frac{1}{2}m\left(\dot r^2 + r^2\dot\theta^2\right)-V(r)$$ by loosing the $\phi$ degree of freedom.

By definition, a coordinate $q_i$ is cyclic if the Lagrangian $\mathcal L$ doesn't explicitly depend on it. That is $$\frac{\mathcal\partial L}{\partial q_i}=0$$

Since the Lagrangian has that $$\frac{\mathcal\partial L}{\partial \theta}=\frac{\mathcal\partial L}{\partial \phi}=0$$ then the coordinates $\theta$ (and $\phi$) are cyclic but note that $\phi$ does not explicitly appear in the original Lagrangian (the kinetic energy expression you quoted in your question).

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  • $\begingroup$ I've a doubt "This problem has spherical symmetry and the potential energy is a function of r only as stated above and the particle is in a conservative central force. This means that the angular momentum is conserved" and I'll add it in the main answer. I'll be thankful if you could help me. :) $\endgroup$
    – Kashmiri
    Oct 15, 2021 at 9:05
  • $\begingroup$ Sure. You mean you need to ask a new question? If you need to ask a new question, rather than add it to this one, please ask it in another question as per our policy. Come back and let me know here, then I'll answer your new question assuming no one beats me to it. Cheers. $\endgroup$
    – joseph h
    Oct 15, 2021 at 9:13
  • $\begingroup$ physics.stackexchange.com/q/671723 $\endgroup$
    – Kashmiri
    Oct 15, 2021 at 9:59

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