Angular Momentum Homework: Visualizing Lr, Lθ & Lφ

In summary, the particle has a generalized potential that looks like this:U(r,r&) = V(r) + σ ⋅LDeduce the generalized force Q = (Qr, Qθ, Qφ ) in spherical polar coordinates.Hence derive Lagrange’s equations of motion.
  • #1
Oerg
352
0

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.

--------------------------

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.
 
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  • #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" 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]
 
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  • #3
jdwood983 said:
(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.

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].
(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).

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]
 
  • #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?
 
  • #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?
 

Related to Angular Momentum Homework: Visualizing Lr, Lθ & Lφ

1. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object, and is equal to the product of its moment of inertia and angular velocity.

2. How is angular momentum calculated?

Angular momentum is calculated by multiplying the moment of inertia (I) by the angular velocity (ω), using the formula L = Iω.

3. What is the significance of Lr, Lθ, and Lφ in angular momentum homework?

Lr, Lθ, and Lφ represent the components of angular momentum in the radial, tangential, and perpendicular directions, respectively. They are used to visualize the different aspects of an object's rotational motion.

4. How can I visualize Lr, Lθ, and Lφ?

You can visualize Lr, Lθ, and Lφ by using vector diagrams. Lr is represented by a vector pointing in the direction of the object's rotation axis, Lθ by a vector perpendicular to Lr in the direction of the object's tangential motion, and Lφ by a vector perpendicular to both Lr and Lθ in the direction of the object's perpendicular motion.

5. What are some real-life examples of angular momentum?

Some real-life examples of angular momentum include the rotation of a spinning top, the orbit of planets around the sun, and the movement of a figure skater performing a spin.

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