Hyperspace

 

If you’re familiar with science fiction stories, you know that hyperspace plays a significant role in many such stories, particularly ones that involve interstellar travel.  For a few minutes, this project will bring hyperspace out of the realm of science fiction and put it in your hands.

 

Introduction to n-space

 

Let’s start by generalizing a basic definition: in 2-space (i.e., the plane) a circle is the locus of all points equidistant from a given point; in 3-space, a sphere is the locus of all points equidistant from a given point.  Because we can translate geometric figures without changing their intrinsic properties, we can take the point in each case to be the origin with (as mathematicians say) no loss of generality.  So, in 2-space the 2-sphere (a.k.a. the circle) is the set of all points equidistant from the origin (0, 0).  Similarly, the 3-sphere is the set of all points equidistant from the origin (0, 0, 0).

 

Under these circumstances, it shouldn’t bother you too much to consider the 4-sphere to be the set of points equidistant from the origin (0, 0, 0, 0).  In general, then the n-sphere is the set of points in n-space equidistant from the origin having n coordinates, all zero. 

 

In Cartesian coordinates, the equation of the 2-sphere is x2 + y2 = r2  .  The equation for the 3-sphere is x2 + y2 + z2 = r2  .  If we define distance in 4-space analogously to the way it’s defined in 2-space and 3-space (and let’s agree to use w  as the name for the fourth independent variable),

 

  1. What’s the distance formula in 4-space?
  2. What’s the Cartesian equation for a sphere in 4-space?

 

Volume

 

We now want to generalize the volume principle so we can look at some hypervolumes of figures like the hypersphere (a common name for the 4-sphere).  Again, let’s go back to the 2- and 3-space cases and try to generalize.

 

The volume of the 2-sphere (otherwise known as the area of the circle) can be found by slicing the figure perpendicular to the x-axis and integrating between the boundaries of defining equation(s) for the 2-sphere.  I hope it’s clear that the following integral is the correct expression:

A similar argument in 3-space gives an integral that essentially takes circles (2-spheres) perpendicular to the x-axis and “adds them all up” to get the volume of the 3-sphere. 

 

  1. What is the requisite integral that produces the volume of the three-sphere of unit radius?

 

OK, now we’re ready to move into hyperspace.  The 4-space sphere (actually, the 4-space ball, since we’re concerned with a solid figure and not just the hollow surface shell) can again be centered at the origin.  If it is, the integration of the unit sphere will go from –1 to 1 (just as in the lower-dimensional analogs).  However, at this point, we might as well generalize things to a sphere of radius r .  For a fixed value of x, the 3-D slice above that x is simply the set of all points (x, y, z, w) that have the given x  value subject to the constraint that x2 + y2 + z2 + w2 = r2.  Rearranging terms, we see that y2 + z2 + w2 = r2 - x2 .  This is the equation of a three-dimensional ball whose radius is

and whose volume is therefore given by the following expression:

Thus, using the integral as the limit of all these “cross-sections”, the volume of the 4-space sphere of radius r  will be given by

.

Because of symmetry considerations, this integral will in turn be equal to

At this point, you can use Mathematica  to come up with a general expression for this integral as a function of r.  However, you should recognize that a trig substitution makes it feasible to do the integral by hand.  If you let x = r sin θ, then this substitution yields

Repeated integration by parts (or use of Mathematica) yields the result that

Knowing the keen intellectual curiosity that motivates you, I know you will want to extend this formula to an n-dimensional ball.  In general, it won’t surprise you to learn that the volume of an n-dimensional ball is proportional to the nth power of the radius of the ball.  In fact, the volume is given by the simple expression

 

where un  is the volume of a ball of unit radius.

 

The first few values of un  are ones you already know:

 

u1

2

u2

π

u3

u4

 

A variation on the reasoning above lets you find a recursive formula for un :

 

 

for any n > 1.  Using integration by parts, you should be able to come up with a general reduction formula for the integral in the above expression as a function of n.  If you do, you can find that

                 

If you tabulate the un ‘s numerically, you see a very curious pattern:

 

n

un

un / 2n

1

2.000

1.000

2

3.142

0.785

3

4.189

0.524

4

4.935

0.308

5

5.264

0.164

6

5.168

0.081

7

4.725

0.037

8

4.059

0.016

9

3.29

0.006

10

2.550

0.002

 

  1. What do you notice about the pattern of volumes of the nth-dimensional unit sphere?
  2. The ratio in the third column is essentially the ratio of the unit sphere’s volume to that of the unit cube.  What happens to this ratio as n increases?
  3. Use integration by parts and the recursive formula for un given above to show that