ball coordinates to cartesian coordinates

[martine]

I am struggeling with the following problem:

give the x,y,z coordinates from the following ball points/vectors

1. (r, theta, phi) = (sqrt3, 3/4pi, 3/4pi)

2. (r, theta, phi) = (1, 1/6pi, 1 1/6pi)

the sollutions I found in my reader are as followed:

1. (x, y, z) = (-1/2 sqrt3, 1/2 sqrt3, -sqrt3/sqrt2)

2. (x, y, z) = 1/4 sqrt3, -1/4, 1/2 sqrt3)

can someone explain to me what was actually done here? I understand the conversion from carthesian coordinates to ball and cylinder coordinates but I can't seem to find the sollution for the other way around. Thanks a lot.

[Muzza]

These (http://mathworld.wolfram.com/SphericalCoordinates.html) equations might be of some use...

[Galileo]

These (http://mathworld.wolfram.com/SphericalCoordinates.html) equations might be of some use...

Yep. It seems the angles \theta and \phi are interchanged though.
It's funny. In my physics books the azimuthal angle is always \phi and in most of my mathematics books it's \theta.
Oh well, guess it doesn`t matter as long as you're aware of it.

[ahrkron]

I would suggest that, instead of plugging this into a set of "conversion equations", you draw the situation (or even build a little model with a box) so that you see how the quantities are related. Once you do this with one problem, the second will be much easier.

[Muzza]

Yep. It seems the angles \theta and \phi are interchanged though.


It brings this up.


Unfortunately, the convention in which the symbols \theta and \phi are reversed is frequently used, especially in physics, leading to unnecessary confusion.


:P

[MiGUi]

That's because notation is not as important as meaning, but we must always specify.

Using astronomy language, I always used \theta for "declination" (angle from vertical axe) and \phi for "Right ascension" (angle from horizontal axe from left to right)