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?