Cyclic coordinates in a two body central force problem

[Kashmir]

(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, I've 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 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 is not cyclic.

 

[anuttarasammyak]

$\phi$ is not included in Lagrangian so it is cyclic coordinate. That generates quantity called angular momentum.

 

[Kashmir]

$\phi$ is not included in Lagrangian so cyclic. There is a conservative property called angular momentum.

Goldstein "Hence, an angle coordinate representing rotation about a fixed axis must be **cyclic**" so that includes $\theta$

 

[anuttarasammyak]

There is $\theta$ in L so it is not a cyclic coordinate.
There is not $\phi$ in L so it is a cyclic coordinate.
Do you find anything wrong with it ?

 

[Kashmir]

There is $\theta$ in L so it is not a cyclic coordinate.
There is not $\phi$ in L so it is a cyclic coordinate.
Do you find anything wrong with it ?

No. I agree with this part.

Goldstein said "...spherical symmetry; i.e., any rotation, about any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic"

Suppose my angle coordinate is $\theta$ since it represents rotation then it should be cyclic according to the text. But it's not as the term is in the Lagrangian.

 

[anuttarasammyak]

an angle coordinate representing rotation about a fixed axis must be cyclic"

For angle $\phi$ rotation, the fixed axis is z-axis
For angle $\theta$ rotation you prefer, what is the fixed axis ?

 

[Kashmir]

For angle $\phi$ rotation, the fixed axis is z-axis
For angle $\theta$ rotation you prefer, what is the fixed axis ?

X axis

 

[anuttarasammyak]

X axis

That seems to require $\phi=\pi/2$ a fixed value not a parameter.
Ref. Spehrical coordinate system https://en.wikipedia.org/wiki/Spherical_coordinate_system
Do you think of the system under such a constraint ?

You would see $\theta$ rotation does not have a fixed axis but a non-fixed axis which depends on $\phi$.

 

[Kashmir]

That seems to require $\phi=\pi/2$ a fixed value not a parameter.
Ref. Spehrical coordinate system https://en.wikipedia.org/wiki/Spherical_coordinate_system
Do you think of the system under such a constraint ?

You would see $\theta$ rotation does not have a fixed axis but a non-fixed axis which depends on $\phi$.

Yes i was wrong. I understood it now. Thank you so much. :)

 

[Kashmir]

$\phi$ is not included in Lagrangian so it is cyclic coordinate. That generates quantity called angular momentum.

You mean conservation of angular momentum?

 

[anuttarasammyak]

Yes. A cyclic coordinate assures us a conserved quantity.

 

[Kashmir]

Yes. A cyclic coordinate assures us a conserved quantity.

Since $\phi$ is cyclic so only one component is consreved any idea how to prove that total angular momentum is conserved??

 

[vanhees71]

Goldstein 3rd edition seems to be a nuissance rather than a textbook. It's a distortion of the very good 2nd edition by different authors. I find it unethical to change a well-established and even famous textbook. Why don't they write their own textbook, which then may not be successful, because it's bad?

Of course, $\theta$ is not cyclic here. Why should it be? The introduction of spherical coordinates breaks spherical symmetry, because the polar axis is a distinguished direction of this coordinate system. That's why the symmetry is apparently reduced to rotations around the polar axis, which is represented by a "translation in $\phi$". A variable is cyclic of the action is invariant under a translation of this variable, and that's in this case only $\phi$ but neither $r$ nor $\theta$.

 

[Kashmir]

Goldstein 3rd edition seems to be a nuissance rather than a textbook. It's a distortion of the very good 2nd edition by different authors. I find it unethical to change a well-established and even famous textbook. Why don't they write their own textbook, which then may not be successful, because it's bad?

Of course, $\theta$ is not cyclic here. Why should it be? The introduction of spherical coordinates breaks spherical symmetry, because the polar axis is a distinguished direction of this coordinate system. That's why the symmetry is apparently reduced to rotations around the polar axis, which is represented by a "translation in $\phi$". A variable is cyclic of the action is invariant under a translation of this variable, and that's in this case only $\phi$ but neither $r$ nor $\theta$.

Thank you so much Sir. So $\phi$ is cyclic hence only one component of angular momentum is conserved.
How can we say from the symmetry that the total angular momentum is conserved.

 

[vanhees71]

No, of course all components of angular momentum are conserved. As I said, this is hidden in spherical coordinates, because it distinguishes one direction, the polar axis. The symmetry becomes manifest in the Lagrangian, when using Cartesian coordinates
$$L=\frac{m}{2} \dot{\vec{x}}^2-V(r), \quad r=|\vec{x}|,$$
which is obviously invariant under arbitrary rotations,
$$\vec{x}'=\hat{D} \vec{x}, \quad \hat{D} \in \mathrm{SO}(3).$$
From Noether's theorem it follows that all three components of the angular momentum,
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}$$
are conserved.

Indeed you have, from the Euler-Lagrange equations
$$m \ddot{\vec{x}}=-\vec{\nabla} V(r)=-\frac{\vec{x}}{r} V'(r)$$
and thus
$$\dot{\vec{L}}=m \dot{\vec{x}} \times \dot{\vec{x}} + m \vec{x} \times \ddot{\vec{x}}= \vec{x} \times \vec{x} V'(r)/r=0.$$

 

[Kashmir]

No, of course all components of angular momentum are conserved. As I said, this is hidden in spherical coordinates, because it distinguishes one direction, the polar axis. The symmetry becomes manifest in the Lagrangian, when using Cartesian coordinates
$$L=\frac{m}{2} \dot{\vec{x}}^2-V(r), \quad r=|\vec{x}|,$$
which is obviously invariant under arbitrary rotations,
$$\vec{x}'=\hat{D} \vec{x}, \quad \hat{D} \in \mathrm{SO}(3).$$
From Noether's theorem it follows that all three components of the angular momentum,
$$\vec{L}=m \vec{x} \times \dot{\vec{x}}$$
are conserved.

Indeed you have, from the Euler-Lagrange equations
$$m \ddot{\vec{x}}=-\vec{\nabla} V(r)=-\frac{\vec{x}}{r} V'(r)$$
and thus
$$\dot{\vec{L}}=m \dot{\vec{x}} \times \dot{\vec{x}} + m \vec{x} \times \ddot{\vec{x}}= \vec{x} \times \vec{x} V'(r)/r=0.$$

The author in the previous chapter explained
"It can be shown that if a cyclic coordinate $q_j$ is such that $dq_j$ 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".

I was looking for a proof for angular momentum along these lines.

Using the cyclic nature of $\phi$ i can only say that one component of angular momentum is conserved but can't go beyond that.

 

[vanhees71]

That's right, but of course you can argue that the choice of the polar axis of your spherical coordinate system is arbitrary, because the potential is invariant under arbitray rotations. Thus $\vec{L}$ is conserved and not only one component along one arbitrarily chosen polar axis.

 

[Kashmir]

That's right, but of course you can argue that the choice of the polar axis of your spherical coordinate system is arbitrary, because the potential is invariant under arbitray rotations. Thus $\vec{L}$ is conserved and not only one component along one arbitrarily chosen polar axis.

Exactly. Like I'll choose three mutually perpendicular axis, L is conserved along all three,hence total angular momentum is conserved as well.
Am I correct?

 

[anuttarasammyak]

So we may say
Each cyclic coordinate assures a conserved quantity e.g. $\phi$ makes $L_z$ const.
But not all conserved quantities correspond to cyclic coordinates e.g. constant $L_y, \ L_x$

 

[vanhees71]

Another theorem is that for an integrable system you can always find a complete set of cyclic coordinates. The emphasis is on "integrable" here. You can find these cyclic coordinates in this case by solving the Hamilton-Jacobi partial differential equation, which then is separable.

The claim that you can always find a complete set of cyclic coordinates, made in Landau and Lifshitz, is wrong ;-)). One should note that most systems are in fact not integrable!

 

[ergospherical]

The claim that you can always find a complete set of cyclic coordinates, made in Landau and Lifshitz, is wrong ;-)).

Landau, wrong??
Next you’ll tell me that it’s going to start raining meatballs!

 

[vanhees71]

Well, it's of course not Dau's mistake but Lifshitz's. After all he was writing the books...

 

[ergospherical]

I heard something along those lines, “not a word of Landau and not a thought of Lifshitz”. I also didn’t know that apparently Dau conceived of much of series whilst in a NKDV prison. Those books have an interesting history!

Not particularly related to the books, I read an article about his school of theoretical physics which was written by a physicist B. Ioffe who attended there, which had some fun insights. (Can’t imagine how stressful it would be to be asked to present at one of his “seminars”)

https://arxiv.org/pdf/hep-ph/0204295.pdf

 

[vanhees71]

Very interesting! Also Sommerfeld started to write his famous 6-volume textbooks after he was enforced to retire from his chair at the university of Munich. Heisenberg was not allowed to get the chair but some Nazi physicist, a proponent of the "Deutsche Physik" a la Lenard and Stark, was put on this position. That was of course the abrupt end of one of the most productive schools of theoretical physics ever.

A Polish colleague of mine once told me the story about his PhD. It was the time before the breakdown of the communist era, and his thesis advisor was also imprisoned because of taking part in the demonstrations against the regime by the union Solidarnosc. To discuss his thesis he was allowed to visit his advisor in the prison from time to time. Then he almost failed his PhD defense, because the obligatory essay on marxism, leninism, and dialectic materialism was considered not "good" enough along the party line, but finally he got the PhD, because the physics was too good to fail him...

 

[Kashmir]

That's right, but of course you can argue that the choice of the polar axis of your spherical coordinate system is arbitrary, because the potential is invariant under arbitray rotations. Thus $\vec{L}$ is conserved and not only one component along one arbitrarily chosen polar axis.

I understood it but I have one last confusion. Could you please help me.

Even in Goldstein 2 ed he says "Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, about any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic".

Now we just saw $\phi$ was cyclic but $\theta$ was not. So how do I understand "Hence, an angle coordinate representing rotation about a fixed axis must be cyclic".
Isn't it incorrect then?
Please help me. I don't have a teacher. Thank you

 

[Kashmir]

Very interesting! Also Sommerfeld started to write his famous 6-volume textbooks after he was enforced to retire from his chair at the university of Munich. Heisenberg was not allowed to get the chair but some Nazi physicist, a proponent of the "Deutsche Physik" a la Lenard and Stark, was put on this position. That was of course the abrupt end of one of the most productive schools of theoretical physics ever.

A Polish colleague of mine once told me the story about his PhD. It was the time before the breakdown of the communist era, and his thesis advisor was also imprisoned because of taking part in the demonstrations against the regime by the union Solidarnosc. To discuss his thesis he was allowed to visit his advisor in the prison from time to time. Then he almost failed his PhD defense, because the obligatory essay on marxism, leninism, and dialectic materialism was considered not "good" enough along the party line, but finally he got the PhD, because the physics was too good to fail him...

I once read that some people stole Perelmans proof of poincare conjecture.

 

[Kashmir]

I understood it but I have one last confusion. Could you please help me.

Even in Goldstein 2 ed he says "Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, about any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic".

Now we just saw $\phi$ was cyclic but $\theta$ was not. So how do I understand "Hence, an angle coordinate representing rotation about a fixed axis must be cyclic".
Isn't it incorrect then?
Please help me. I don't have a teacher. Thank you

Sir any comments?

 

[vanhees71]

I understood it but I have one last confusion. Could you please help me.

Even in Goldstein 2 ed he says "Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, about any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic".

Now we just saw $\phi$ was cyclic but $\theta$ was not. So how do I understand "Hence, an angle coordinate representing rotation about a fixed axis must be cyclic".
Isn't it incorrect then?
Please help me. I don't have a teacher. Thank you

But $\theta$ is not a rotation around a fixed axis, but around an axis depending on $\phi$. Just draw the standard spherical coordinates:

https://commons.wikimedia.org/wiki/File:3D_Spherical.svg#/media/File:3D_Spherical.svg

 

[Kashmir]

Everything is crystal clear. Thank you so much :)

 

[vanhees71]

By accident I posted a wrong (copyright-violating!) link from another thread first. I corrected my posting with the correct link pointing to the standard drawing of the spherical coordinates in Wikipedia. Please erase the wrong link in the quote and substitute it with the right one in order not to spread the bad link further!

 

[Kashmir]

By accident I posted a wrong (copyright-violating!) link from another thread first. I corrected my posting with the correct link pointing to the standard drawing of the spherical coordinates in Wikipedia. Please erase the wrong link in the quote and substitute it with the right one in order not to spread the bad link further!

Done.