Angle ranges in hyperspherical coordinates

[NanakiXIII]

In spherical coordinates, you have one angle that goes from $0$ to $\pi$ and one that goes from $0$ to $2 \pi$. I'm having a hard time reasoning out what the angle ranges would be in hyperspherical coordinates, i.e. in more than three dimensions. Wikipedia suggests that you have one angle that ranges to $2 \pi$ and the rest ranges to $\pi$. Is this correct and if so, is there an intuitive way to see that it is?

 

[Marin]

Hi, NanakiXIII!

I hope the picture is clear in 3D (I'll omit the r-coordinate, since it's totally superfluous to this discussion):

the first angle (usually denoted as \phi, or azimuth) is given a full 2Pi rotational freedom and the second one (\theta) does not need all of it, since the mapping

polar --> cartesian coordinates

would not be one-to-one*, hence the 0 to Pi domain for \theta.

Now let's just add one more dimension, say, we're now dealing with the 4D-sphere (n dimensions would go similarly). Now I don't know for you but I can't draw a picture of it in my mind no matter how hard I try, and I assume (almost) no one can...

Here comes a clever trick, invented by mathematicians - the projection. But let's define some coordinates for our 4D sphere for better accuracy: \phi will be the azimuthal angle, then we have \theta (same as in 3D), and now, call the last angle \psi. We want to know what the domain of \psi is. Notice that we already know that of \theta, but we're having problems with the fourth dimension (given by \psi). Therefore, project one dimension out by setting \theta = Pi/2 (in the 3D case this projects the spherical coordinates to 2D polar coordinates in the xy-plane). Now the problem is 3D and free coordinates are \phi and \psi. But we already know what the domain of \psi has to be, in order to get one-to-one mapping...

*if you take a closer look you will see that the end points of the domain of the angles in the 3D spherical coordinates are taken out. Therefore there are some sets (curves) on the sphere that cannot be represented by this parametrisation. However, this restriction allows mathematicians to classify the above mapping as a diffeomorphism (check it out!) which has many nice properties.

 

[NanakiXIII]

Thanks, Marin, that seems a good way to look at it. One thing, though. You could also project by setting $\phi$ constant. In 3D, you don't end up with coordinates that map the entire plane, and I suppose in 4D you don't get coordinates that map all of 3-space, but this isn't entirely obvious to me.

 

[Marin]

You could also project by setting LaTeX Code: \\phi constant. In 3D, you don't end up with coordinates that map the entire plane

No, you don't.

Typically the sphere has any symmetries you can imagine, therefore the two angles \theta and \phi are (before you decide for where to put the domains [0, 2Pi] and [0,Pi]) to be seen as having the same rights. Only after you do this (here and above I used a standard convention) can you use this projection argument.

Of course, a natural question arises as, if all additional angles in higher dimension 'have the same rights' and are the \theta or \phi like. To answer it, we have to take a look at how we obtain the n dimensional sphere from the n-1 dimensional:

Let me just mention that I'll be doing this for the unit sphere and not the ball since the radius is (almost) irrelevant but the procedure is literally the same for the ball.

The sphere S^0 is just the set {+1, -1}. In order to obtain the S^1 sphere out of it, we rotate the end points of the line [-1,1] in the dimension above. Notice that this line is not the radius but the diameter of S^0. In order to keep the map one-to-one we only need to rotate from 0 to Pi. Now we have obtained the unit circle. Proceeding further, take the unit circle and rotate it along its diameter in the dimension above, to obtain the sphere S^2. The rotation is again only 0 to Pi. Now, by induction, we obtain the sphere S^n. So any new angle that enters the game in higher dimension take values from 0 to Pi

But where does the 0 to 2Pi domain of \phi then come from?

Well, this is just a reparametrisation of the two dimensional disc B^2. First, we don't like the diameter, so we switch to a new variable - the radius. But by doing so, having \phi only from 0 to Pi, we have to allow r to take negative values... ugly! (or just unconventional). The easiest way of surrounding this obstacle is to take away the negative r's and let \phi make a full rotation 0 to 2Pi. Hence, the oddness of the \phi variable.

 

[NanakiXIII]

That makes complete sense, a very clear explanation. Thanks!