Get all possible constants of motion given an explicit Hamiltonian

[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.