Angular Momentum Homework: Visualizing Lr, Lθ & Lφ

[Oerg]

Homework Statement

A point particle moves in space under the influence of a force derivable from a
generalized potential of the form

U(r,r&) = V(r) + σ ⋅L

where r is the radius vector from a fixed point, L is the angular momentum about
that point and σ is a fixed vector in space.
Deduce the generalized force Q = (Qr, Qθ, Qφ ) in spherical polar coordinates.
Hence derive Lagrange’s equations of motion.

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I actually have the solution to this question, but I do not really understand part of the solution. This is the part that I do not understand from the solution:

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Let the polar axis of the polar spherical coordinates (r, θ, ϕ) be in the direction
of σ. Note that $$L =(0, - mrv_\phi,mrv_\theta)$$, where m is the mass of the particle.

$$U(r, v) = V (r) + \sigma \cdot \vec{L}$$
$$= V (r) + \sigma ( L_r \cos \theta - L_\theta \sin \theta )$$
$$= V (r) + \sigma mv_\phi r \sin \theta$$

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Firstly, $$L_r$$ is zero? I have difficulty visualizing $$L_r$$.
Why is there a negative in front of $$mrv_\phi$$?
Which is the polar axis for spherical coordinates?
It would be good if someone could provide a link or explain how to visualize $$L_r , L_\theta$$ and $$L_\phi$$.

Any help would be appreciated, thanks.

 

[jdwood983]

(1) $L_r$ is not zero, but the $\theta$ component of $\sigma$ is zero, that's why that term disappears.

(2) Through geometry and trigonometry you should be able to see that $L_\theta$ is negative.

(3) I've always defined $\phi$ to be the polar angle (the one that sweeps the x-y plane), but mathematicians do it oppositely and call $\theta$ to be this angle. It seems this problem follows what I use (that is, $\phi$ is your polar angle).

(4) http://quantummechanics.ucsd.edu/ph130a/130_notes/node216.html" is pretty good at giving the geometry of angular momentum in spherical coordinates (from Cartesian coordinates). In the figure they give, magenta is $L_\theta$, blue is $L_r$ and green is $L_\phi$

 

[gabbagabbahey]

(1) $L_r$ is not zero, but the $\theta$ component of $\sigma$ is zero, that's why that term disappears.

Really?...Remember, $L_r\equiv \textbf{L}\cdot\mathbf{\hat{r}}=(\textbf{r}\times\textbf{p})\cdot\mathbf{\hat{r}}$, and by definition of the cross product, $\textbf{r}\times\textbf{p}$ must be perpendicular to $\textbf{r}$.

(3) I've always defined $\phi$ to be the polar angle (the one that sweeps the x-y plane), but mathematicians do it oppositely and call $\theta$ to be this angle. It seems this problem follows what I use (that is, $\phi$ is your polar angle).

I disagree. To me, it looks like $\theta$ is the polar angle here.

@Oerg... To more directly answer you question, the polar axis is usually taken to be the z-axis. So, "choosing $\mathbf{\sigma}$ to be directed along the polar axis" is the same as choosing your coordinate system so that the z-axis is aligned with $\mathbf{\sigma}$

 

[Oerg]

I don't really understand.

So $$\sigma$$ is also in spherical coordinates, then it must have a $$\sigma_r$$ component?

Also, the r coordinate in spherical coordinates is scalar and represents the magnitude? If it is then $$L_r$$ should be non-zero if the angular momentum is non zero and the dot product with $$\sigma$$ should also produce a non-zero r coordinate component?

 

[Oerg]

Ahh, I think I might have got it.

If I do the product in cartesian coordinates, then sigma is aligned to the z component of the angular momentum and the z component of the angular momentum is the angular momentum associated with the rotation in the x-y plane which has an angle of $$\phi$$. The answer that I get will be the same as the solution,

but my questions still remain. Also, is the dot product for spherical coordinates different from the dot product in cartesian coordinates?