The Holographic Sphere
The Holographic Principle talks about regions of space where the surface area is just as important as the volume. So I decided to try and construct an example.
I did. And the result is both simple and elegant.
Consider objects that have their surface area equal to their volume.
For a circle of radius r in 2D we have surface area (circumference)= 2*pi*r and volume (area)=pi*r^2. Equating these two gives r=2. Which is the dimension of the space!
Doing the same for a sphere in 3D we get r=3. Which is the dimension of the space!
Is this always true for spheres? Yes!
Take a point in n-dimensional Euclidean space (x1,x2,x3,...,xn)
Then the surface of a sphere of radius r is the set of points with:
and the volume consists of all the points with:
x1^2+x2^2+x3^2+....+xn^2 less than or equal to r^2
The volume of the sphere is c(n)*(r^n) and its surface area is n*c(n)*r^(n-1)
So equating them gives r=n.
Note that the function c(n) cancels out so its value is not needed, but c(n)=(pi^(n/2))/gamma(1+n/2). Where gamma is the Euler Gamma Function.
So we can say in general..
"In n-dimensional Euclidean space a sphere with surface area=volume has radius n"
The radius of the sphere is equal to the dimension of the space!
This may be related to the Holographic Principle because it equates surface area and volume.
Content written and posted by Ken Abbott firstname.lastname@example.org