Last time we looked at the volume and surface area of one, two, three, and four dimensional hyperspheres. We found that if we took the integral in spherical coordinates, we could find the volume of a circle, sphere, hypersphere, etc. We saw that to find the surface area of a hypersphere, you just set R constant and integrate the angular portions. It's worth noting that you can also simply take the derivative with respect to r of the volume to get the surface area, for example, the two dimensional case:
Surface Area = ^{d}/_{dr} V(r) = ^{d}/_{dr}(πr^{2}) = 2πr
So how can we extend what we've done for any ndimensional hypersphere? Is there a general expression for the volume of a hypersphere in ndimensions? Turns out it can be calculated (This is a complicated calculation, please see links at bottom for derivations). The result is:
where Γ is the gamma function. The gamma function is a function that looks like:
You can see from the graph above that the gamma function (Γ) is discontinous at 0, 1,2,3,......n). The gamma function is a more general expression of the factorial:
where z can be a negative number, fractional number, or complex number. In fact, the graph above is only the real number graph of the function. The entire function, including the complex parts looks like:
That's a bit more than we need, since we are talking about dimensions which are positive integers (at least in this discussion). As a result, the gamma function in the equation:
the z of the gamma function will be positive integers for even n and positive fractions (3/2, 5/2, etc.) for odd n in the equation above. As a result, we can break out the general solution in terms of even and odd number of dimensions.
and
We break them out this way because of the identities:
and
which allow you to get values for the fractional factorials found in the odd number of dimensions case.
Volume of a Hypersphere with Even Number of Dimensions
We can now calculate the volumes of hyperspheres of different even dimensions (2,4, 6, ... 2n), without integration, since we now have the general solution.
For 2 dimensions we get:
π^{1}(r)^{2}(^{1}/_{1!}) = πr^{2}
For 4 dimensions we get:
π^{2}(r)^{4}(^{1}/_{2!}) = ^{1}/_{2} π^{2}r^{4
}which agrees with the value we got for a four dimensional hypersphere in part one.
Volume of a Hypersphere with Odd Number of Dimensions
We can now calculate the volumes of hyperspheres of different odd dimension (1,3, 5, ... 2n1), without integration, since we now have the general solution.
For 3 dimensions we get:
2^{3}π^{1}(^{1!}/_{3!})r^{3}= (^{8}/_{6})πr^{3} = (^{4}/_{3})πr^{3
}
For 5 dimesions we get:
2^{5}π^{2}(^{2!}/_{5!})r^{5}= (^{64}/_{120})π^{2}r^{5 }= (^{8}/_{15})π^{2}r^{5
}
Volume of a Hypersphere as Dimension Increases
So now we have the volumes of hyperspheres of one, two, three, four, and five dimensions. If we call a hypersphere with radius=1 (r=1), a unit hypersphere, we can see that r^{2}=1, r^{3}=1, r^{4}=1, .... r^{n}=1, so we can get real numerical values for our volumes. Now we can take a look at the volume of the unit hypersphere with increasing dimesion (up to 10 dimensions). We get:
Dimension......Volume
1....................2.0000
2....................3.1416
3....................4.1888
4....................4.9348
5....................5.2638
6....................5.1677
7....................4.7248
8....................4.0587
9....................3.2985
10..................2.5502
Notice that the volume of the unit hypersphere is at a maximum in 5 dimensions and the surface area of the unit hypersphere is at a maximum in 7 dimensions.
If we take the limit as the number of dimensions approaches infinity (Lim n→∞), you would find that the volume approaches zero (V→0). So after 5 dimensions, the volume gradually decreases as more dimensions are added. Also, since surface area is simply the derivative with respect to r of the volume, the surface area also approaches zero (SA→0) as the number of dimensions approaches infinity(Lim n→∞), though it would approach zero slightly slower than the volume does, which explains why surface area peaks in 7 dimensions instead of 5 dimensions.
Of course, that's just the unit hypersphere (radius = 1), what about hyperspheres of larger radius? Does the volume of those hyperspheres go to zero as the number of dimensions goes to infinity? Lets take a look at the volume for an even number of dimensions again:
Lets see if there is any value of r that would produce a nonzero volume as n→∞
If we use Stirlings Approximation:
and make our radius:
r= (^{1}/_{√π})(√πn)^{1/n}(n/2e)
we get:
V = [π^{n/2 }r^{n} (1/(n/2)!)]
V = [π^{n/2 }(1/π^{n/2})(√πn)(n/2)^{n}(1/e)^{n }(1/(n/2)!)]
V = (n/2)! (1/(n/2)!)
V = 1
So the radius of a hypersphere with unit volume (V=1) for a large even number of dimensions (the "large" is a restriction that comes from the stirling's approximation) can be approximated as:
r= (^{1}/_{√π})(√πn)^{1/n}(n/2e)
Notice that if we take the limit as n→∞ that the radius goes to infinity. In other words, as the number of dimensions approaches infinity, the radius must also approach infinity in order for the hypersphere to have unit volume (V=1). Remember, when we kept unit radius earlier (r=1) the volume went to zero as n went to infinity, so it makes sense that to produce a nonzero volume we must increase the radius of the hypersphere as the number of dimensions approaches infinity.
Ok, that's enough for now. I'll see you next week.
Special Thanks to the Following Websites:
http://en.wikipedia.org/wiki/Gamma_function
http://en.wikipedia.org/wiki/Hypersphere
http://mathworld.wolfram.com/Hypersphere.html
http://wwwstaff.lboro.ac.uk/~coael/hypersphere.pdf
