Cross Product in Spherical Coordinates - Getting conflicting oppinions

AI Thread Summary
The discussion centers on the challenges of calculating the cross product in spherical coordinates, particularly for angular momentum. While some sources, including Wolfram, suggest that a straightforward determinant approach is not feasible without converting to Cartesian coordinates, others, including professors and textbooks, indicate that it is possible to perform the calculation directly in spherical coordinates. The participant has found that the results align with Cartesian calculations when considering instantaneous values, as the coordinate directions remain mutually perpendicular. There is also a query regarding the correct order of the coordinates in the determinant, with clarification sought on whether it should be (r, theta, phi). Ultimately, the conversation highlights differing opinions on the validity of using spherical coordinates for these calculations.
Damascus Road
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Hey all,
I really need some clarification here.

I've seen problems dealing with the Angular Momentum of a particle, working in spherical coordinates. Wolfram says that there is no simple way to perform this and do the determinant, and you will find many people and other websites claiming this. i.e. you must convert to Cartesian, or I've also seen some kind of operator.

However,
Some of my own professors have said you MAY do this. There are also E-mag textbooks, etc. that do the cross product in spherical coords!

In my own meddling, it seems like it works fine, as long as you look at an instant in time, since the directions are not constant, but they are however, all mutually perpendicular to each other. When I do this, it is identical to what I would get in Cartesian.

Also, if this is allowed, can someone tell me in what order the directions should be in the determinant? (r, theta, phi)
(phi being in the x,y plane)
 
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Maybe W and others were referring to Del X something. The cyclic order r, theta, phi is correct, with anyone of them going first.
 
Damascus Road said:
Hey all,
I really need some clarification here.

I've seen problems dealing with the Angular Momentum of a particle, working in spherical coordinates. Wolfram says that there is no simple way to perform this and do the determinant, and you will find many people and other websites claiming this. i.e. you must convert to Cartesian, or I've also seen some kind of operator.
...

Hmm ... I did a back-of-the-envelope caliculation and got myself

L = m r 2 * (d\theta/dt * phi-direction - sin \theta * d \phi/dt * theta-direction )

Of course, you could convert it into Cartesian, but it's easiest to do it in the coordinate system in which your trajectory is given.
 
What you wrote, xlines, is your result from the cross-product?
 
Damascus Road said:
What you wrote, xlines, is your result from the cross-product?

Yes, that is \vec{L} = m \vec{r} x \vec{v} in spherical coordinates, angular momentum of a pointlike particle.
 
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