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Circles on the Surface of a Sphere


Date: 08/11/99 at 12:32:56
From: Mark Shen
Subject: Circles on the Surface of a Unit Sphere and their 
Transformation from Cartesian to Spherical Coordinates

The Problem:

A circle in the (X,Y,Z) coordinate system has the equation:

     x^2 + z^2 = r^2   and   y = k   ..........................(Eq. 1)

Here, 'r' is the radius and 'k' is a constant. How do I convert this 
equation/relation that is in terms of X, Y, and Z into a function that 
is defined in terms of theta and phi, spherical coordinates? I want to 
be able to make a 2-D graph (with axes theta and phi) of this new 
relation. The spherical coordinate rho is not taken into consideration 
in this 2-D graph.

I believe that when converted into theta and phi coordinates, the 2-D 
graph will be a circle. I can show it is numerically - that is, I can 
convert individual points on the (X,Y,Z) circle into theta and phi 
coordinates. The result was a circle on the theta and phi coordinates. 
 The specific circle equation used was:

     x^2 + z^2 = 0.25   and   y = sqrt(0.75)   ................(Eq. 2)

Obviously, 0.5 is the radius and y = sqrt(0.75) is the 2-D plane in 
which the circle exists. The numbers were picked for convenience (but 
not at first glance) so that rho (in the spherical system) would equal 
one. Using the conversions below, I was able to generate a circle on 
theta and phi axes.

     x = sin(phi)*cos(theta)     rho = 1
     y = sin(phi)*sin(theta)
     z = cos(phi)

How can the equation below (Eq. 3, which is Eq. 2 in theta and phi) be 
converted to a circle equation?

     (sin(phi)*cos(theta))^2 + (cos(phi))^2 = 0.25   ..........(Eq. 3)

=======================================================

This problem is a small part of a larger problem: On a unit sphere in 
the Cartesian coordinate system, you project a cone, with half-angle 
gamma and apex at the origin, onto the surface of the sphere. The 
resulting intersection of these two objects is a circle. If you 
convert this circle, which is in the Cartesian system, into spherical 
coordinates - theta and phi, is the resulting graph a circle as well? 
 How do you prove that this is true, with the exception of a few 
special combinations of theta and phi? At first glance, it seemed easy 
to prove; and the proof almost seems to be in the 'definition' of the 
spherical coordinate system. However, I must be too dense to see it.

Please help. Thanks.

Mark Shen


Date: 08/11/99 at 15:20:59
From: Doctor Rob
Subject: Re: Circles on the Surface of a Unit Sphere and their 
Transformation from Cartesian to Spherical Coordinates

Thanks for writing to Ask Dr. Math!

Probably the simplest way to see that the curve is a circle is to use 
cylindrical coordinates instead of spherical ones:

     x = R*cos(theta)
     y = Z
     z = R*sin(theta)

In your situation, the equations become

     Z = sqrt(0.75)
     R = sqrt(0.25) = 0.5

The equation R = 0.5 is that of a right circular cylinder with radius 
0.5 and axis the Z-axis. The equation Z = constant is the equation of 
a plane perpendicular to the Z-axis. Their intersection is a circle, 
which is the circle you were seeking.

If you must use spherical coordinates, you have to use the additional 
equation:

     y = sqrt(0.75),

or

     sin(phi)*sin(theta) = sqrt(0.75).

Using this to graph the phi-versus-theta relation should reveal the 
graph of a circle.

By the way, if you square this last equation and use the identity 
sin^2(u) + cos^2(u) = 1, you can derive your Eq. 3 above.

You would be even better advised to choose the following coordinate 
transformation instead of the one you used:

     x = sin(phi)*cos(theta)
     y = cos(phi)
     z = sin(phi)*sin(theta)

Then your equations will reduce to:

     rho = 1
     sin^2(phi) = 0.25
     cos(phi) = sqrt(0.75)
     phi = Pi/6

This is a cone with vertex at the origin and gamma = Pi/6, intersected 
with the sphere rho = 1. Here theta can take any value at all.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Higher-Dimensional Geometry
College Trigonometry

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