Transform velocity vector to spherical coordinates.

[Dinosaur]

I think I authored a similar Thread here or at an other forum, but never got an answer I trusted.

There are well known equations for transforming a Cartesian point (xyz) to Spherical coordinates and vice versa. What are the equations for transforming velocity and/or acceleration vectors?

A while ago, I assumed that taking derivatives of the point conversion formulae would provide correct equations. Then an engineer claimed that a Cartesian point (x, y, z) is a position vector. He further stated that a vector is a vector is a vector. Hence, the equations for transforming a position vector should be correct for transforming velocity and acceleration vectors.

I considered the possibility that both methods were correct and would arrive at the same results. Some work with my MathCad7 software convinced me that this was not true. The two methods do not lead to the same results.

Does anybody have some thoughts on this? Ignoring the formulae for Longitude and Latitude, consider the following equation for the Spherical coordinate radius.

r = SquareRoot( x^2 + y^2 + z^2 )

Derivative analysis results in the following for transforming velocity.

Vr = ( x*Vx + y*Vy + z*Vz ) / r, where Vr, Vx, Vy, Vz are velocities.

Using the position vector formula on the velocity vector results in the following.

Vr = SquareRoot(Vx^2 + Vy^2 + Vz^2 )

Which of the above is correct for Vr? If you have an opinion, please explain.

My vote goes for the first equation, but I am not sure.

BTW: At another forum, it is possible to create subscripts and superscripts/exponents. Is it possible here?

 

[lethe]

http://www.sciforums.com/showthread.php?s=&threadid=12060

here was the old one.

but i guess i will respond in this thread. by the way, stryderunknown answered you about subscripts and superscripts in the last thread.

first of all, in curvilinear coordinates, position is simply not a vector. so throw that idea out. the transformation for position coordinates is not the rule for transforming vectors.

the displacement is the model vector. this is essentially the rule for velocities, with the dt left out.

can i suggest that we consider polar coordinates instead of spherical for the sake of simplicity? i m sure your concerns about transformations will be the same in polar as they are in spherical.

in polar coordinates:

x = r cos q
y = r sin q

so
dx = dr cos q -r dq sin q
dy = dr sin q +r dq cos q

this transformation for the displacement vector works for any vector B:

Bx = Br cos q -r Bq sin q
By = Br sin q +r Bq cos q

this is what your engineer friend means when he says a vector is a vector is a vector. all (contravariant) vectors transform coordinates the same way.

i believe your problem is this here. take the second derivative of the position formula to find acceleration, and you get this:

x'' = r'' cos q - 2r'q' sin q -r q'' sin q -r (q')^2 cos q
y'' = r'' sin q + 2r'q' cos q +r q'' cos q -r (q')^2 sin q

so here it seems like acceleration has a different transformation than the simpler one i gave above, right? i believe this is where you re getting stuck. this two formulae are not the same...

 

[Dinosaur]

Thanx. I feel more comfortable with the equations I worked out by taking derivatives.

I forgot about using < > instead of [ ] for making^{exponents/superscripts} and_subscripts. I included this here in hopes of reinforcing my memory.

 

[lethe]

just so you know, i haven t attempted to answer your question yet. i ve just posed it in a slightly different form, that i think is equivalent. i just want to check that the "paradox" that i have described is also the same question that you are asking, in your opinion.

offhand, i m not immediately sure what the resolution is, although i m pretty comfortable that there is one, and i m just going to tinker with the algebra a bit and get back to you. i think it might have something to do with the basis unit vectors being time dependent in noncartesian coordinates, which adds a bit of complication.

please, read through what i have posted, and perhaps do those calculations yourself, even though you are more comfortable with the calculations that you have done. i think it will be much simpler in this case (polar rather than spherical), and i don t want to get caught on lengthy calculus calculations. i d rather talk about the relationships between coordinate systems and vector representations. with polar coordinates, you won t really need matlab or whatever to do the calculations. also, i noticed you were dealing with derivatives of the inverse transformation (arctans and sqrts and stuff). i don t see the need for that.

i can t offer any more on this tonight, because i have a need to get drunk right now.