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 nspace
Let’s start by generalizing a basic definition: in 2space (i.e., the plane) a circle is the locus of all points equidistant from a given point; in 3space, 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 2space the 2sphere (a.k.a. the circle) is the set of all points equidistant from the origin (0, 0). Similarly, the 3sphere 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 4sphere to be the set of points equidistant from the origin (0, 0, 0, 0). In general, then the nsphere is the set of points in nspace equidistant from the origin having n coordinates, all zero.
In Cartesian coordinates, the equation of the 2sphere is x^{2 }+ y^{2} = r^{2}^{ }. The equation for the 3sphere is x^{2 }+ y^{2} + z^{2} = r^{2}^{ }. If we define distance in 4space analogously to the way it’s defined in 2space and 3space (and let’s agree to use w as the name for the fourth independent variable),
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 4sphere). Again, let’s go back to the 2 and 3space cases and try to generalize.
The volume of the 2sphere (otherwise known as the area of the circle) can be found by slicing the figure perpendicular to the xaxis and integrating between the boundaries of defining equation(s) for the 2sphere. I hope it’s clear that the following integral is the correct expression:
_{}
A similar argument in 3space gives an integral that essentially takes circles (2spheres) perpendicular to the xaxis and “adds them all up” to get the volume of the 3sphere.
OK, now we’re ready to move into hyperspace. The 4space sphere (actually, the 4space 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 lowerdimensional analogs). However, at this point, we might as well generalize things to a sphere of radius r . For a fixed value of x, the 3D 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 x^{2 }+ y^{2} + z^{2} + w^{2} = r^{2}. Rearranging terms, we see that y^{2} + z^{2} + w^{2} = r^{2 } x^{2 }. This is the equation of a threedimensional ball whose radius is
_{}
and whose volume is therefore given by the following expression:
Thus, using the integral as the limit of all these “crosssections”, the volume of the 4space 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 ndimensional ball. In general, it won’t surprise you to learn that the volume of an ndimensional ball is proportional to the nth power of the radius of the ball. In fact, the volume is given by the simple expression
where u_{n } is the volume of a ball of unit radius.
The first few values of u_{n } are ones you already know:
u_{1} 
2 
u_{2} 
π 
u_{3} 
_{} 
u_{4} 

A variation on the reasoning above lets you find a recursive formula for u_{n }:
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 u_{n }‘s numerically, you see a very curious pattern:
n 
u_{n} 
u_{n }/ 2^{n} 
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 