Goldstein Mechanics chpt 1, Exercise 15

[dracobook]

Homework Statement

A point particle moves in space under the influence of a force derivable from
a generalized potential of the form
[tex] U(r, v) = V (r) + \sigma \cdot L [/tex]
where [tex] r [/tex] is the radius vector from a fixed point, [tex] L [/tex] is the angular momentum about that point, and [tex] \sigma [/tex] is a fixed vector in space.
Find the components of the force on the particle in both Cartesian and
spherical poloar coordinates, on the basis of Lagrangian’s equations with
a generalized potential.

Homework Equations

[tex] Q_j = -\frac{\partial U}{\partial q_j} + \frac{d}{dt} (\frac{\partial U}{\partial \dot{q_j}})[/tex]

The Attempt at a Solution

Using Cartesian, I get :
[tex] F = 2m(\mathbf{\sigma} \times \mathbf{v}) - V' \frac{\mathbf{r}}{r} [/tex] I let [tex] \sigma [/tex] point in the z direction, then it simplifies my cartesian to
[tex] U = V(r) + m\sigma (x\dot{y} - y\dot{x}) [/tex]
If I plug in the x in terms of [tex] r, \theta, \phi [/tex] I get
[tex] Q_r = - \frac{dV}{dr} - 2 m \sigma r \text{sin}^2 \theta \dot{\phi} [/tex].
My Question is, is it possible to go the other way around? That is, what if I plug in the x, y z-> r, theta, phi equivalence of the generalized potential. How can I then restrict it so that sigma is faced in the z-axis? Sorry I know this is perhaps somewhat irrelevant to the question...but it's something that is really bothering me.

In addition if someone could give me an example of situation in which a point particle would undergo such a force that would be great (the angular momentum dot sigma really throws off my intuition).

Any help would be greatly appreciated.

-draco

 

[mark726]

Dear draco,

Thank you for your post. The components of the force on the particle in both Cartesian and spherical polar coordinates can be derived using Lagrangian's equations with a generalized potential. In Cartesian coordinates, the force can be expressed as F = 2m(\mathbf{\sigma} \times \mathbf{v}) - V' \frac{\mathbf{r}}{r}, where \sigma points in the z direction. This expression can be simplified by using the x, y, z-> r, theta, phi equivalence of the generalized potential. However, in order to restrict sigma to be in the z-axis, you would need to set the x and y components of the generalized potential to be zero. This would result in a force that only acts in the z-direction, as desired.

An example of a situation in which a point particle would undergo such a force is in the case of a charged particle moving in a magnetic field. In this case, the force on the particle can be expressed as F = q(\mathbf{v} \times \mathbf{B}), where \mathbf{B} is the magnetic field vector and q is the charge of the particle. This can be rewritten in terms of the generalized potential as F = \frac{\mathbf{r} \times \mathbf{B}}{m} - \frac{q}{m}(\mathbf{r} \times \mathbf{L}), where \mathbf{L} is the angular momentum of the particle.

I hope this helps. Let me know if you have any further questions.