Get all possible constants of motion given an explicit Hamiltonian

[PeroK]

My bad, what I meant was:

$$\vec p = \vec p_{r} + \vec p_{\phi} + \vec p_{\theta} = \frac{1}{2 \alpha}\Big( \dot r \hat r + r^2 \dot \phi \hat \phi + (r \sin \phi)^2 \dot \theta \hat \theta \Big)$$

I now have an idea on how to get the constant of motion related to angular momentum, thanks.

To get the constant of motion related to conservation of energy I just have to use the Hamiltinian. But could you give me a hint on how should I use it?

The Lagrangian is independent of time, so the total energy if conserved.

 

[strangerep]

[For $f=f(r)$...] we get:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \Big( \frac{x}{\sqrt{ x^2 + y^2 + z^2 }} \Big)$$

Yes.

But what's $\partial f/\partial r$? As you said, $f$ is an arbitrary function of $r$; we have no explicit formula to compute such a derivative.

That's ok. The point of this exercise is to find constants of motion which are applicable to any central potential problem, i.e., where $V = V(r)$.

Besides, I did not use the computation of your hint, so I guess I am missing something.

Not really. You can express $V=V(r)$ or $V=V(r^2)$. Since both $r$ and $r^2$ go from $0$ to $\infty$, the 2 forms are essentially equivalent. Sometimes one form is more convenient than the other. E.g., slightly rewriting what you had above:
$$\frac{\partial f(r)}{\partial x} ~=~ f'(r) \, \Big( \frac{x}{\sqrt{ x^2 + y^2 + z^2 }} \Big) ~,$$ where the prime on $f$ means "derivative wrt to its argument". We could, alternately, have written
$$\frac{\partial f(r^2)}{\partial x} ~=~ \frac{\partial f(r^2)}{\partial r^2} \, \frac{\partial r^2}{\partial x} ~=~ f'(r^2) \, 2x ~.$$

 

[strangerep]

@JD_PM : Since PeroK has mentioned some adhoc techniques for finding constants of the motion, I'll sketch a more systematic approach below.

But first, here's a quick way to simplify any central force problem, and also discover conserved angular momentum...

The equation of motion, written in vectorial notation is
$$m\, {\mathbf a} ~=~ F(r) \, \widehat{\mathbf r} ~=~ - V'(r) \, \widehat{\mathbf r} ~,$$ where bold font denotes a 3-vector and an overhat denotes a unit vector. Therefore, $${\mathbf r} \times \dot{\mathbf v} ~\equiv~ {\mathbf r} \times {\mathbf a} ~=~ \frac{F(r)}{m} \, {\mathbf r} \times \widehat{\mathbf r} ~=~ 0 ~.$$Hence, $$ \frac{d}{dt} ({\mathbf r} \times {\mathbf v} ) ~=~ {\mathbf r} \times \dot{\mathbf v} + \dot{\mathbf r} \times {\mathbf v} ~=~ {\mathbf r} \times {\mathbf a} + {\mathbf v} \times {\mathbf v} ~=~ 0 + 0 ~=~ 0 ~.$$In other words, $${\mathbf L} ~:=~ ({\mathbf r} \times {\mathbf v})$$ is time-independent on solutions of the equations of motion, i.e., a constant of the motion.

If you write out ${\mathbf L}$ explicitly, you'll get something like $${\mathbf L} ~=~ r^2 \dot\theta \hat{\mathbf \phi} ~-~ r^2 \dot\phi \sin\theta \hat{\mathbf \theta} ~.$$(I say "something like" because I'm just copying this stuff across from my personal notes and I'm using the "physicist" convention for the meanings of $\theta,\phi$ whereas you're using the "mathematician" convention which is the other way around. I.e., you'll have to swap the angular variables.)

Then comes the simplifying insight: since ${\mathbf L}$ is constant (i.e., constant magnitude and constant direction) we can rotate our coordinate system so that all motion takes place in a plane corresponding to ${\mathbf L}$. A good choice is $\theta=\pi/2$, hence also $\dot\theta=0$. That makes the whole problem simpler. (Again, remember to swap $\theta,\phi$ compared to your earlier equations.)

Now, moving on to a more systematic approach,...

A dynamical quantity $Z=Z(t,q_i, \dot q_i)$ is a constant of the motion if it has zero total time derivative on trajectories, i.e., if $$0 ~=~ \frac{D Z}{Dt} ~:=~ \frac{\partial Z}{\partial t} + \dot q_k \, \frac{\partial Z}{\partial q_k} + \ddot q_k \, \frac{\partial Z}{\partial \dot q_k} ~$$on solutions of the equation of motion $\ddot q_k = \omega_k(q_i, \dot q_i)$. I.e., if $$\frac{\partial Z}{\partial t} + \dot q_k \, \frac{\partial Z}{\partial q_k} + \omega_k \, \frac{\partial Z}{\partial \dot q_k} ~=~ 0 ~,~~~~~~~~~~~~~~~~~ (1).$$ In principle, one finds constants of the motion (a.k.a. "first integrals") by finding all function $Z$ satisfying eq(1). Usually, this is done using Ansatzen for $Z$ which are up-to-2nd order in the basic variables $q_k, \dot q_k$. Sometimes there are also higher-order solutions (such as the LRL vector in the classical Kepler problem).

Instead, if you have guessed a possible function $Z$ which might be a constant of the motion, you simply check whether it satisfies eq(1). You may wish to try this for the angular momentum and energy functions appropriate for the problem of this thread.

(HTH)

 

[JD_PM]

Thank you PeroK and strangerep for your insight!

I like to finish threads giving answers to the questions I post. However, I would like to do so in a month if you do not mind. This is because I have another question I would like to ask first.

I will also finish the following thread in a month:

https://www.physicsforums.com/threads/using-noethers-theorem-to-get-a-constant-of-motion.982721/

 

[JD_PM]

Back at it! :)

OK After reading the following

@JD_PM : Since PeroK has mentioned some adhoc techniques for finding constants of the motion, I'll sketch a more systematic approach below.

But first, here's a quick way to simplify any central force problem, and also discover conserved angular momentum...

The equation of motion, written in vectorial notation is
$$m\, {\mathbf a} ~=~ F(r) \, \widehat{\mathbf r} ~=~ - V'(r) \, \widehat{\mathbf r} ~,$$ where bold font denotes a 3-vector and an overhat denotes a unit vector. Therefore, $${\mathbf r} \times \dot{\mathbf v} ~\equiv~ {\mathbf r} \times {\mathbf a} ~=~ \frac{F(r)}{m} \, {\mathbf r} \times \widehat{\mathbf r} ~=~ 0 ~.$$Hence, $$ \frac{d}{dt} ({\mathbf r} \times {\mathbf v} ) ~=~ {\mathbf r} \times \dot{\mathbf v} + \dot{\mathbf r} \times {\mathbf v} ~=~ {\mathbf r} \times {\mathbf a} + {\mathbf v} \times {\mathbf v} ~=~ 0 + 0 ~=~ 0 ~.$$In other words, $${\mathbf L} ~:=~ ({\mathbf r} \times {\mathbf v})$$ is time-independent on solutions of the equations of motion, i.e., a constant of the motion.

I now have clear that for any central force problem, angular momentum is conserved.

Alright, I got that the $\theta$ component of angular momentum is conserved:

$$\frac{\partial L}{\partial \dot \theta} = \frac{1}{2 \alpha} (r \sin \theta)^2 \dot \theta = 0$$

But I still have to show that the other component ($\phi$) is conserved to show that angular momentum is indeed conserved.

Based on your insights I know that conserved angular momentum has two components, because as PeroK pointed out

The component of motion in the radial direction makes no contribution to angular momentum.

I have two questions at this point:

1) How can I show that the component related to $\phi$ is indeed conserved? Note that the 'cyclic trick' doesn't work because $\phi$ appears explicitely on the Lagrangian.

2) Why the component of motion in the radial direction makes no contribution to angular momentum? (this question may be naive but I do not see it)

Thank you for your help.

 

[PeroK]

2) Why the component of motion in the radial direction makes no contribution to angular momentum? (this question may be naive but I do not see it)

Thank you for your help.

That's easy:$$\vec L = m(\vec r \times \vec v)$$
and $$\vec r \times \hat r = 0$$

 

[PeroK]

1) How can I show that the component related to $\phi$ is indeed conserved? Note that the 'cyclic trick' doesn't work because $\phi$ appears explicitely on the Lagrangian.

Note that $p_{\theta} = l_z$. The remaining AM is a combination of $l_x$ and $l_y$. It doesn't work out quite as neatly in spherical coordinates as you might hope. One approach is to calculate $l_x, l_y$ and $l^2$ to see what you're aiming for from your Euler-Lagrange equations.

 

[JD_PM]

That's easy:$$\vec L = m(\vec r \times \vec v)$$
and $$\vec r \times \hat r = 0$$

Ahh the cross product of two parallel vectors is zero! Thanks I see it now.

Regarding 1)

I recall you suggested a trick back in #30 to get the $\phi$ component.

One trick, which relies on some physical reasoning, is that you must get motion in a plane. Therefore, you can change your initial coordinates to new (primed) coordinates so that $q'_3 = 0$, which is equivalent to $\phi' = \frac{\pi}{2}$.

If you don't do that, then spherical coordinates don't simplify things as much as you might hope.

Mm but honestly I still do not understand it...

May you please explain further what you meant here?

Thanks.

 

[PeroK]

Ahh the cross product of two parallel vectors is zero! Thanks I see it now.

Regarding 2)

I recall you suggested a trick back in #30 to get the $\phi$ component.
Mm but honestly I still do not understand it...

May you please explain further what you meant here?

Thanks.

There are two separate ideas here:

1) If you want to study central force motion, then a neat trick is first to show that motion stays in a single plane, defined by your initial velocity. Then, you can take it to be the x-y plane, with $\phi = \pi/2$. Most texts on Newtonian gravity do this.

That's not really what we want here. We want to study these equations from a mathematical perspective.

2) A neat trick to show that $l_x, l_y$ are conserved is to note that we chose $\phi$ relative to the z-axis and got $l_z$ as a conserved quantity. And, as we could just as well choose $\phi$ relative to the x or y axis, we must have $l_x, l_y$ as conserved quantities as well.

Note that $l_x$ and $l_y$ are not as simple as you might hope in the usual spherical coordinates. It is simpler, however, to show that $l^2$ is conserved - which is an altenative to showing $l_x$ and $l_y$ are both conserved. That should sound familiar: compatibility of $l_z$ and $l^2$!

 

[strangerep]

[...] But I still have to show that the other component ($\phi$) is conserved to show that angular momentum is indeed conserved.

No, I've already done that for you in my post #38 where I derived $$\frac{d{\mathbf L}}{dt} ~=~ 0 ~.$$ (If you still don't get it, then... what did I say the bold font signified?)

Based on your insights I know that conserved angular momentum has two components, because as PeroK pointed out

The component of motion in the radial direction makes no contribution to angular momentum.

I think you're misinterpreting what PeroK said. Angular momentum is a 3-vector. (Actually, it's a 3-pseudovector, but I'll let you research Wikipedia yourself about that subtlety if you wish).

 

[JD_PM]

2) A neat trick to show that $l_x, l_y$ are conserved is to note that we chose $\phi$ relative to the z-axis and got $l_z$ as a conserved quantity. And, as we could just as well choose $\phi$ relative to the x or y axis, we must have $l_x, l_y$ as conserved quantities as well.

Alright let me show you what is the issue I have at the moment.

We already know how $l_x, l_y$ (I strongly suspect $l_x$ and $l_y$ shown below are not correct; they are simply $p_r$ and $p_{\phi}$)and $l_z$ look like:

$$p_{\theta} = \frac{\partial L}{\partial \dot \theta} = \frac{1}{2 \alpha} (r sin \theta)^2 \dot \theta \hat \theta = l_z$$

$$p_{\phi} = \frac{\partial L}{\partial \dot \phi} = \frac{1}{2 \alpha} r^2 \dot \phi \hat \phi = l_y$$

$$p_{r} = \frac{\partial L}{\partial \dot r} = \frac{1}{2 \alpha} \dot r \hat r = l_x$$

We have shown that $l_z$ is by itself conserved.

The issue is that $l_x$ and $l_y$ are not conserved by themselves following the cyclic argument.

So I guess that I am wrong stating that $p_{r} = l_x$ and that $p_{\phi} = l_y$. That is why you said that it’s not that easy getting $l_x$ and $l_y$.

OK you propose another approach: Compute $l^2$ and then use the connexion between $l^2$ and $l_z$ to show that indeed $l^2$ is conserved. That fact reminds me of commutation of operators in QM.

But how can I get l^2?

 

[JD_PM]

No, I've already done that for you in my post #38 where I derived $$\frac{d{\mathbf L}}{dt} ~=~ 0 ~.$$

Yes. You showed AM is conserved. However, I want to get to the same conclusion by working out its components. I have shown $l_z$ is conserved. I still have to find an expression for $l_x$ and $l_y$ and then show these are also conserved.

 

[JD_PM]

2) A neat trick to show that $l_x, l_y$ are conserved is to note that we chose $\phi$ relative to the z-axis and got $l_z$ as a conserved quantity. And, as we could just as well choose $\phi$ relative to the x or y axis, we must have $l_x, l_y$ as conserved quantities as well.

Honestly I still do not grasp this argument. May you please explain it through an example?

Thanks.

 

[PeroK]

Honestly I still do not grasp this argument. May you please explain it through an example?

Thanks.

The standard spherical coordinates involve taking the azimuthal angle ($\phi$ in your convention) from the z-axis. But, of course, there are two alternative spherical coordinate systems taking $\phi$ relative to the x and y axes.

In this case you have symmetry of the Lagrangian in x-y-z. You showed that $p_{\theta} = l_z$ was conserved quantity. It follows that by using the alternative spherical coordinate systems you will get $p_{\theta'} = l_x$ and $p_{\theta''} = l_y$ as conserved quantities.

If you need to you can go through the algebra, but every step will be essentially the same.

That's the quickest way to conclude that $l_x$ and $l_y$ must be conserved.

 

[PeroK]

Alright let me show you what is the issue I have at the moment.

We already know how $l_x, l_y$ (I strongly suspect $l_x$ and $l_y$ shown below are not correct; they are simply $p_r$ and $p_{\phi}$)and $l_z$ look like:

$$p_{\theta} = \frac{\partial L}{\partial \dot \theta} = \frac{1}{2 \alpha} (r sin \theta)^2 \dot \theta \hat \theta = l_z$$

$$p_{\phi} = \frac{\partial L}{\partial \dot \phi} = \frac{1}{2 \alpha} r^2 \dot \phi \hat \phi = l_y$$

$$p_{r} = \frac{\partial L}{\partial \dot r} = \frac{1}{2 \alpha} \dot r \hat r = l_x$$

We have shown that $l_z$ is by itself conserved.

The issue is that $l_x$ and $l_y$ are not conserved by themselves following the cyclic argument.

So I guess that I am wrong stating that $p_{r} = l_x$ and that $p_{\phi} = l_y$. That is why you said that it’s not that easy getting $l_x$ and $l_y$.

OK you propose another approach: Compute $l^2$ and then use the connexion between $l^2$ and $l_z$ to show that indeed $l^2$ is conserved. That fact reminds me of commutation of operators in QM.

But how can I get l^2?

I looked online for somewhere that derived something from the second E-L equation, but didn't find anything, so here are my notes :

We can use the definition of AM and the properties of the spherical coordinate system to obtain, from the cross product $m\vec r \times \vec v$, and the first E-L equation $l_z = mr^2 (\sin^2 \phi) \dot \theta$ the following expression: $$l^2 = m^2r^4[\dot \phi^2 + (\sin^2\phi) \dot \theta^2] = m^2r^4\dot \phi^2 + \frac{l_z^2}{\sin^2 \phi} = p_{\phi}^2 + \frac{p_{\theta}^2}{\sin^2 \phi}$$
We can then take the derivative of $l^2$:
$$\frac{d}{dt}(l^2) = 2p_{\phi}\frac{dp_{\phi}}{dt} -2\frac{l_z^2 \cos \phi}{\sin^3 \phi}$$
Now we use the other E-L equation:
$$\frac{dp_{\phi}}{dt} = mr^2(\sin \phi \cos \phi) \dot \theta^2 = \frac{l_z^2 \cos \phi}{mr^2 \sin^3 \phi}$$
To get:
$$2p_{\phi} \frac{dp_{\phi}}{dt} = (2mr^2\dot \phi)(\frac{l_z^2 \cos \phi}{mr^2 \sin^3 \phi}) = 2\frac{l_z^2 \cos \phi}{\sin^3 \phi} $$
And $\frac{d}{dt}(l^2) = 0$, as required.

It seems that the conservation of $l_z$ and $l^2$ come most easily out of these E-L equations. I'm sure you could eventually show the conservation of $l_x, l_y$, but it looks a lot cleaner to appeal to the symmetry of the problem for those.

PS the second E-L equation doesn't appear to produce anything physically meaningful on its own. But, combined with the first it yields the conservation of $l^2$ without too much work.

 

[strangerep]

Yes. You showed AM is conserved. However, I want to get to the same conclusion by working out its components. I have shown $l_z$ is conserved. I still have to find an expression for $l_x$ and $l_y$ and then show these are also conserved.

You're just flagellating yourself unnecessarily. If you want angular momentum in cartesian components, then convert my vectorial notation from post #38 into component notation, e.g., $${\mathbf L} ~:=~ ({\mathbf r} \times {\mathbf v})$$ becomes $$L_i ~:=~ \varepsilon_{ijk} x_j v_k ~,$$(using implicit summation over repeated indices).

A more productive, more educational exercise (though a little tangential to this thread), would be to derive the explicit expressions for generic velocity and acceleration vectors in spherical polar. It's obvious you don't yet understand them properly, and without that understanding you'll just continue flubbing around like you've been doing so far.

There are various educational youtube videos that explain it. E.g., this one, and others in that series by the same guy.

 

[JD_PM]

NOTE: this message has nothing to do with this thread. Skip it if you wish.

I am going to work out some of the basic algebra that is avoided in this helpful video:



Algebra to get to equations 3:59

OK so simplifying the last equation above we get :

$$r \sin \theta \frac{\partial}{\partial r} + \cos \theta \frac{\partial}{\partial \theta} + \frac{\cos \phi}{\sin \theta \sin \phi} \frac{\partial}{\partial r} = \frac{r}{\sin \phi} \frac{\partial}{\partial y}$$

Thus solving for $\frac{\partial}{\partial y}$ we get

$$\frac{\partial}{\partial y} = \sin \theta \sin \phi \frac{\partial}{\partial r} + \frac{\sin \phi \cos \theta}{r} \frac{\partial}{\partial \theta} + \frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi} \ \ \ \ (1)$$

Plugging $\frac{\partial}{\partial y}$ result into equation $(I)$ and rewriting it we get:

$$\cos \theta \frac{\partial}{\partial z} = \frac{\partial}{\partial r} + \frac{\cos \phi}{r \sin \phi} \frac{\partial}{\partial \phi} - \frac{\sin \theta}{\sin \phi}\Big( \sin \theta \sin \phi \frac{\partial}{\partial r} + \frac{\sin \phi \cos \theta}{r} \frac{\partial}{\partial \theta} + \frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi} \Big)$$

$$\cos \theta \frac{\partial}{\partial z} = \cos^2 \theta \frac{\partial}{\partial r} - \frac{\sin \theta \cos \theta}{r} \frac{\partial}{\partial \theta}$$

Solving for $\frac{\partial}{\partial z}$ we get

$$\frac{\partial}{\partial z} = \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta} \ \ \ \ (2)$$

By solving for $\frac{\partial}{\partial x}$ in equation $(III)$ we got:

$$\frac{\partial}{\partial x} = \frac{1}{r \sin \phi \sin \theta} \Big( r \sin \theta \cos \phi \frac{\partial}{\partial y} - \frac{\partial}{\partial \phi} \Big)$$

We already know $\frac{\partial}{\partial y}$ so we just have to plug it into and get

$$\frac{\partial}{\partial x} = \sin \theta \cos \phi \frac{\partial}{\partial r} + \frac{\cos \theta \cos \phi}{r} \frac{\partial}{\partial \theta} - \frac{\sin \phi}{r \sin \theta}\frac{\partial}{\partial \phi} \ \ \ \ (3)$$

NOTE: Andrew got the wrong sign in the third term of equation (3)'s RHS.

Some algebra to get to the final result (see 11:21)

Now it just a matter of adding terms up; you can split this task in three and then add the result of these three parts up. I will do just one as an example.

The terms having $\frac{\partial}{\partial r}$ as a common factor are:

$\sin^2 \theta \cos^2 \phi \hat r \frac{\partial}{\partial r} + \cos \theta \cos^2 \phi \sin \theta \hat \theta \frac{\partial}{\partial r} - \sin \phi \sin \theta \cos \phi \hat \phi \frac{\partial}{\partial \phi} + \sin^2 \phi \cos \theta \sin \theta \hat \theta \frac{\partial}{\partial r} + \sin^2 \phi \sin^2 \theta \hat r \frac{\partial}{\partial r} + \sin \phi \sin \theta \cos \phi \hat \phi \frac{\partial}{\partial r} + \cos^2 \theta \hat r \frac{\partial}{\partial r} - \sin \theta \cos \theta \hat \theta \frac{\partial}{\partial r}$

Note how $\hat \phi$ terms add up to zero. The same happens to $\hat \theta$ terms.

By adding $\hat r$ terms up we get

$$\sin^2 \theta \cos^2 \phi \hat r \frac{\partial}{\partial r} + \sin^2 \phi \sin^2 \theta \hat r \frac{\partial}{\partial r} + \cos^2 \theta \hat r \frac{\partial}{\partial r} = \hat r \frac{\partial}{\partial r}$$

By applying the same idea to $\frac{\partial}{\partial \theta}$ and $\frac{\partial}{\partial \phi}$ and adding those results to what I explicitly showed above we get

$$\vec \nabla = \hat r \frac{\partial}{\partial r} + \frac{1}{r} \hat \theta \frac{\partial}{\partial \theta} + \frac{1}{r \sin \theta} \hat \phi \frac{\partial}{\partial \phi}$$

 

[JD_PM]

A more productive, more educational exercise (though a little tangential to this thread), would be to derive the explicit expressions for generic velocity and acceleration vectors in spherical polar. It's obvious you don't yet understand them properly, and without that understanding you'll just continue flubbing around like you've been doing so far.

There are various educational youtube videos that explain it. E.g., this one, and others in that series by the same guy.

Thanks for the advice! I definitely needed that refresh.

I checked and understood these videos:







I think it was a useful work out for me to derive the gradient in spherical coordinates.

It's obvious you don't yet understand them properly, and without that understanding you'll just continue flubbing around like you've been doing so far.

I think I do now :)

 

[JD_PM]

You showed that $p_{\theta} = l_z$ was conserved quantity. It follows that by using the alternative spherical coordinate systems you will get $p_{\theta'} = l_x$ and $p_{\theta''} = l_y$ as conserved quantities.

If you need to you can go through the algebra, but every step will be essentially the same.

That's the quickest way to conclude that $l_x$ and $l_y$ must be conserved.

Honestly I do not know how to proceed to get $p_{\theta'} = l_x$ and $p_{\theta''} = l_y$. I do not even know what $p_{\theta'}$ or $p_{\theta''}$ mean...

We can use the definition of AM and the properties of the spherical coordinate system to obtain, from the cross product $m\vec r \times \vec v$, and the first E-L equation $l_z = mr^2 (\sin^2 \phi) \dot \theta$ the following expression: $$l^2 = m^2r^4[\dot \phi^2 + (\sin^2\phi) \dot \theta^2] = m^2r^4\dot \phi^2 + \frac{l_z^2}{\sin^2 \phi} = p_{\phi}^2 + \frac{p_{\theta}^2}{\sin^2 \phi}$$

I do not see how to get that equation from the cross product $m\vec r \times \vec v$, and the first E-L equation $l_z = mr^2 (\sin^2 \phi) \dot \theta$

I have been training by working out those videos but it is may be not enough.

Please let me know if it is better if I keep working on similar problems before trying this one again. Are you aware of any book/videos that deal with these kind of problems?

Thanks.

 

[PeroK]

Honestly I do not know how to proceed to get $p_{\theta'} = l_x$ and $p_{\theta''} = l_y$. I do not even know what $p_{\theta'}$ or $p_{\theta''}$ mean...
I do not see how to get that equation from the cross product $m\vec r \times \vec v$, and the first E-L equation $l_z = mr^2 (\sin^2 \phi) \dot \theta$

I have been training by working out those videos but it is may be not enough.

Please let me know if it is better if I keep working on similar problems before trying this one again. Are you aware of any book/videos that deal with these kind of problems?

Thanks.

I assumed it would be obvious that $r, \theta', \phi'$ are spherical coordinates with the polar angle $\phi'$ with respect to the x-axis; and, $r, \theta'', \phi''$ spherical coordinates using the y-axis.

There's nothing special about the z-axis. In any case, the only reason I mentioned alternative coordinates explicitly was because you seemed not to understand the appeal to "symmetry". There should be no need to do any calculations. An understanding of the symmetry of the problem should be sufficient.

Regarding the cross product, $\hat r, \hat \phi , \hat \theta$ form an right-handed orthonormal set of basis vectors at each point. And angular momentum is given by:
$$\vec l = m\vec r \times \vec v = mr \hat r \times (v_r \hat r + v_{\phi} \hat \phi + v_{\theta} \hat \theta)$$
In assume you know how to calculate the velocity components in spherical coordinates. See the Kinematics section.

https://en.wikipedia.org/wiki/Spherical_coordinate_system

Note: you probably know you are using the "mathematical" convention for $\phi, \theta$. The Kinematics section on that page uses the "physics" convenstion, where $\phi, \theta$ are swapped.

 

[JD_PM]

I assumed it would be obvious that $r, \theta', \phi'$ are spherical coordinates with the polar angle $\phi'$ with respect to the x-axis; and, $r, \theta'', \phi''$ spherical coordinates using the y-axis.

There's nothing special about the z-axis. In any case, the only reason I mentioned alternative coordinates explicitly was because you seemed not to understand the appeal to "symmetry". There should be no need to do any calculations. An understanding of the symmetry of the problem should be sufficient.

Ahhh I think I see it now! So, by symmetry, $l_x$ and $l_y$ are conserved as well.

Regarding the cross product, $\hat r, \hat \phi , \hat \theta$ form an right-handed orthonormal set of basis vectors at each point. And angular momentum is given by:
$$\vec l = m\vec r \times \vec v = mr \hat r \times (v_r \hat r + v_{\phi} \hat \phi + v_{\theta} \hat \theta)$$
In assume you know how to calculate the velocity components in spherical coordinates. See the Kinematics section.

https://en.wikipedia.org/wiki/Spherical_coordinate_system

Note: you probably know you are using the "mathematical" convention for $\phi, \theta$. The Kinematics section on that page uses the "physics" convenstion, where $\phi, \theta$ are swapped.

Calculating the velocity components in spherical coordinates (as strangerep also suggested) looks like a nice work out. I will check the link, thanks for sharing.

 

[JD_PM]

Hi @PeroK

I've been thinking about the symmetry argument to justify conservation of angular momentum.

I think I understand it, but I wanted to find another example to simply enforce my understanding.

Say a system has the following Lagrangian:

$$L = \frac 1 2 (M + 2m)(\dot x^2 + \dot y^2 + \dot z^2) + m(\dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2 \theta) + \frac{1}{24} M L^2 (\dot \theta^2 + \dot \phi^2 \sin^2 \theta) - K(r-r_0)^2$$

We notice $\phi$ is a cyclic coordinate and thus:

$$\frac{\partial L}{\partial \dot \phi} = \Big( 2mr^2 + \frac{1}{12} M L^2\Big) \dot \phi \sin^2 \theta = constant$$

We note that the constant term we get is the z-component of the angular momentum.

At this point I would argue that the other two components are also conserved because the axis has been chosen arbitrarily (i.e by symmetry argument). Thus AM of the system is conserved.

Would you be satisfied with such a justification or could we argue it better?

Thank you.