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Angular momentum

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data
    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.


    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:


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

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


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

    Any help would be appreciated, thanks.
    Last edited: Nov 27, 2009
  2. jcsd
  3. Nov 27, 2009 #2
    (1) [itex]L_r[/itex] is not zero, but the [itex]\theta[/itex] component of [itex]\sigma[/itex] is zero, that's why that term disappears.

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

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

    (4) http://quantummechanics.ucsd.edu/ph130a/130_notes/node216.html" [Broken] is pretty good at giving the geometry of angular momentum in spherical coordinates (from Cartesian coordinates). In the figure they give, magenta is [itex]L_\theta[/itex], blue is [itex]L_r[/itex] and green is [itex]L_\phi[/itex]
    Last edited by a moderator: May 4, 2017
  4. Nov 27, 2009 #3


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    Homework Helper
    Gold Member

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

    I disagree. To me, it looks like [itex]\theta[/itex] 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 [itex]\mathbf{\sigma}[/itex] to be directed along the polar axis" is the same as choosing your coordinate system so that the z-axis is aligned with [itex]\mathbf{\sigma}[/itex]
  5. Nov 30, 2009 #4
    I don't really understand.

    So [tex] \sigma [/tex] is also in spherical coordinates, then it must have a [tex] \sigma_r [/tex] component?

    Also, the r coordinate in spherical coordinates is scalar and represents the magnitude? If it is then [tex] L_r [/tex] should be non-zero if the angular momentum is non zero and the dot product with [tex] \sigma[/tex] should also produce a non-zero r coordinate component?
  6. Nov 30, 2009 #5
    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 [tex] \phi [/tex]. 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?
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